The curve is shown in the graph, for .
\nThe curve passes through the following points.
\nIt is required to find the area bounded by the curve, the -axis, the -axis and the line .
\nThe function is given by , where , , , are integers.
\nThe graph of passes through the point (0, 0).
\nThe graph of also passes through the point (3, 18).
\nThe graph of also passes through the points (1, 0) and (–1, –10).
\nWrite down the value of .
\nShow that .
\nWrite down the other two linear equations in , and .
\nWrite down these three equations as a matrix equation.
\nSolve this matrix equation.
\nThe function can also be written where and are integers. Find and .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n= 0 A1 N1
\n[1 mark]
\nAttempting to substitute (3, 18) (M1)
\nA1
\nAG N0
\n[2 marks]
\n+ + = 0 A1 N1
\n− + − = −10 A1 N1
\n[2 marks]
\nEvidence of attempting to set up a matrix equation (M1)
\nCorrect matrix equation representing the given equations A2 N3
\neg
\n[3 marks]
\nA1A1A1 N3
\n[3 marks]
\nFactorizing (M1)
\neg ,
\n(accept ) A1A1 N3
\n[3 marks]
\nWrite down the inverse of the matrix A = .
\nHence or otherwise solve
\n\n
\n
\n
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nA−1 = A2 N2
\n[2 marks]
\nFor recognizing that the equations may be written as A (M1)
\nFor attempting to calculate (M1)
\n= 4, = 1, = −6 A2 N4
\n[4 marks]
\nLet A = and B = .
\nFind AB.
\nThe matrix C = and 2AB = C. Find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nAttempting to multiply matrices (M1)
\nA1A1 N3
\n[3 marks]
\nSetting up equation M1
\neg , ,
\n(A1)
\nA1 N2
\n[3 marks]
\nLet A = and B = .
\nFind A + B.
\nFind −3A.
\nFind AB.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nevidence of addition (M1)
\neg at least two correct elements
\nA + B = A1 N2
\n[2 marks]
\nevidence of multiplication (M1)
\neg at least two correct elements
\n−3A = A1 N2
\n[2 marks]
\nevidence of matrix multiplication (in correct order) (M1)
\neg AB =
\nAB = A2 N3
\n[3 marks]
\nLet A = .
\nLet B = .
\nFind A−1.
\nFind A2.
\nGiven that 2A + B = , find the value of and of .
\nHence find A−1B.
\nLet X be a 2 × 2 matrix such that AX = B. Find X.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nA−1 = A2 N2
\n[2 marks]
\nA2 = A2 N2
\n[2 marks]
\n(M1)
\n= 2, = 3 A1A1 N3
\nEvidence of attempt to multiply (M1)
\neg A−1B =
\nA−1B = A1 N2
\n[2 marks]
\nEvidence of correct approach (M1)
\neg X = A−1B, setting up a system of equations
\nX = A1 N2
\n[2 marks]
\nLet M = .
\nWrite down the determinant of M.
\nWrite down M−1.
\nHence solve M.
\n\n
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ndet M = −4 A1 N1
\n[1 mark]
\nM−1 = A1A1 N2
\nNote: Award A1 for and A1 for the correct matrix.
\n[2 marks]
\nX = M−1 M1
\nX = (, ) A1A1 N0
\n\n
Note: Award no marks for an algebraic solution of the system , .
\n[3 marks]
\nThe curve is shown in the graph, for .
\nThe curve passes through the following points.
\nIt is required to find the area bounded by the curve, the -axis, the -axis and the line .
\nOne possible model for the curve is a cubic function.
\nUse the trapezoidal rule to find an estimate for the area.
\nUse all the coordinates in the table to find the equation of the least squares cubic regression curve.
\nWrite down the coefficient of determination.
\nWrite down an expression for the area enclosed by the cubic regression curve, the -axis, the -axis and the line .
\nFind the value of this area.
\nArea = M1A1
\nArea = 43 A1
\n[3 marks]
\nM1A2
\n[3 marks]
\nA1
\n[1 mark]
\nArea = A1
\n[1 mark]
\n42.5 A2
\n[2 marks]
\nPaul wants to buy a car. He needs to take out a loan for $7000. The car salesman offers him a loan with an interest rate of 8%, compounded annually. Paul considers two options to repay the loan.
\nOption 1: Pay $200 each month, until the loan is fully repaid
\nOption 2: Make 24 equal monthly payments.
\nUse option 1 to calculate
\nUse option 2 to calculate
\nGive a reason why Paul might choose
\nthe number of months it will take for Paul to repay the loan.
\nthe total amount that Paul has to pay.
\nthe amount Paul pays each month.
\nthe total amount that Paul has to pay.
\noption 1.
\noption 2.
\nevidence of using Finance solver on GDC M1
\nA1
\nIt will take 40 months A1
\n[3 marks]
\nM1A1
\n[2 marks]
\nMonthly payment = $316 ($315.70) M1A1
\n[2 marks]
\nM1A1
\n[2 marks]
\nThe monthly repayment is lower, he might not be able to afford $316 per month. R1
\n[1 mark]
\nthe total amount to repay is lower. R1
\n[1 mark]
\nSophie is planning to buy a house. She needs to take out a mortgage for $120000. She is considering two possible options.
\nOption 1: Repay the mortgage over 20 years, at an annual interest rate of 5%, compounded annually.
\nOption 2: Pay $1000 every month, at an annual interest rate of 6%, compounded annually, until the loan is fully repaid.
\nGive a reason why Sophie might choose
\nSophie decides to choose option 1. At the end of 10 years, the interest rate is changed to 7%, compounded annually.
\nCalculate the monthly repayment using option 1.
\nCalculate the total amount Sophie would pay, using option 1.
\nCalculate the number of months it will take to repay the mortgage using option 2.
\nCalculate the total amount Sophie would pay, using option 2.
\noption 1.
\noption 2.
\nUse your answer to part (a)(i) to calculate the amount remaining on her mortgage after the first 10 years.
\nHence calculate her monthly repayment for the final 10 years.
\nevidence of using Finance solver on GDC M1
\nMonthly payment = $785 ($784.60) A1
\n[2 marks]
\nM1A1
\n[2 marks]
\nM1A1
\nIt will take 181 months A1
\n[3 marks]
\nM1A1
\n[2 marks]
\nThe monthly repayment is lower, she might not be able to afford $1000 per month. R1
\n[1 mark]
\nthe total amount to repay is lower. R1
\n[1 mark]
\n$74400 (accept $74300) M1A1
\n[2 marks]
\nUse of finance solver with N =120, PV = $74400, I = 7% A1
\n$855 (accept $854 − $856) A1
\n[2 marks]
\nGive your answers to this question correct to two decimal places.
\nGen invests $2400 in a savings account that pays interest at a rate of 4% per year, compounded annually. She leaves the money in her account for 10 years, and she does not invest or withdraw any money during this time.
\nCalculate the value of her savings after 10 years.
\nThe rate of inflation during this 10 year period is 1.5% per year.
\nCalculate the real value of her savings after 10 years.
\nM1A1
\n[2 marks]
\nreal interest rate = A1
\nM1A1
\n[3 marks]
\nUrvashi wants to model the height of a moving object. She collects the following data showing the height, metres, of the object at time seconds.
\nShe believes the height can be modeled by a quadratic function, , where .
\nHence find
\nShow that .
\nWrite down two more equations for , and .
\nSolve this system of three equations to find the value of , and .
\nwhen the height of the object is zero.
\nthe maximum height of the object.
\nM1
\nAG
\n[1 mark]
\nattempt to substitute either (5, 38) or (7, 24) M1
\nA1
\nA1
\n[3 marks]
\nM1A1A1A1
\n[3 marks]
\nM1
\nseconds M1A1
\n[3 marks]
\nattempt to find maximum height, e.g. sketch of graph M1
\nmetres A1
\n[2 marks]
\nBeth goes for a run. She uses a fitness app to record her distance, km, and time, minutes. A graph of her distance against time is shown.
\nBeth runs at a constant speed of 2.3 ms–1 for the first 8 minutes.
\nBetween 8 and 20 minutes, her distance can be modeled by a cubic function, . She reads the following data from her app.
\nHence find
\nCalculate her distance after 8 minutes. Give your answer in km, correct to 3 decimal places.
\nFind the value of , , and .
\nthe distance she runs in 20 minutes.
\nher maximum speed, in ms–1.
\nM1A1
\n[2 marks]
\neither using a cubic regression or solving a system of 4 equations M1
\nA1A1A1A1
\n[5 marks]
\nkm (Note: Condone km obtained from using rounded values.) M1A1
\n[2 marks]
\nEITHER finding maximum of OR solving M1
\nmaximum speed = 0.390… km per minute A1
\nmaximum speed = 6.51 ms–1 M1A1
\n[4 marks]
\nCharles wants to measure the strength of the relationship between the price of a house and its distance from the city centre where he lives. He chooses houses of a similar size and plots a graph of price, (in thousands of dollars) against distance from the city centre, (km).
\nThe data from the graph is shown in the table.
\nExplain why it is not appropriate to use Pearson’s product moment correlation coefficient to measure the strength of the relationship between and .
\nExplain why it is appropriate to use Spearman’s rank correlation coefficient to measure the strength of the relationship between and .
\nCalculate Spearman’s rank correlation coefficient for this data.
\nState what conclusion Charles can make from the answer in part (c).
\nthe data is not linear R1
\n[1 mark]
\nthe data is (montonically) decreasing. R1
\n[1 mark]
\nassign ranks M1
\naverage equal prices M1
\n A1A1
(Note: condone ) A2
\n[6 marks]
\nThere is a strong, negative relationship between the price of a house and its distance from the city centre. R1
\n[1 mark]
\nThis question will investigate the solution to a coupled system of differential equations when there is only one eigenvalue.
\nIt is desired to solve the coupled system of differential equations
\n\n
\n
The general solution to the coupled system of differential equations is hence given by
\n\n
As the trajectory approaches an asymptote.
\nShow that the matrix has (sadly) only one eigenvalue. Find this eigenvalue and an associated eigenvector.
\nHence, verify that is a solution to the above system.
\nVerify that is also a solution.
\nIf initially at find the particular solution.
\nFind the values of and when .
\nFind the equation of this asymptote.
\nState the direction of the trajectory, including the quadrant it is in as it approaches this asymptote.
\nM1A1
\nA1A1
\nSo only one solution AGA1
\nM1
\nSo an eigenvector is A1
\n[7 marks]
\n\n
So M1A1A1
\nand M1A1
\nshowing that is a solution AG
\n[5 marks]
\nM1A1
\nM1A1A1
\nVerifying that is also a solution AG
\n[5 marks]
\nRequire M1A1
\nA1
\n[3 marks]
\nA1A1
\n[2 marks]
\nAs M1A1
\nso asymptote is A1
\n[3 marks]
\nWill approach the asymptote in the 4th quadrant, moving away from the origin. R1
\n[1 mark]
\nIt is believed that two variables, and are related. Experimental values of and are obtained. A graph of against shows a straight line passing through (2.1, 7.3) and (5.6, 2.4).
\nHence, find
\nFind the equation of the straight line, giving your answer in the form , where .
\na formula for in terms of .
\nthe value of when .
\ngradient = −1.4 A1
\nM1
\nA1
\n[3 marks]
\nA1
\n[1 mark]
\n28000 A1
\n[1 mark]
\nIt is believed that two variables, and are related by the equation , where . Experimental values of and are obtained. A graph of against shows a straight line passing through (−1.7, 4.3) and (7.1, 17.5).
\nFind the value of and of .
\nM1A1
\ngradient M1
\nA1
\n-intercept M1
\nM1A1
\n[7 marks]
\nAdesh wants to model the cooling of a metal rod. He heats the rod and records its temperature as it cools.
\nHe believes the temperature can be modeled by , where .
\nHence
\nShow that .
\nFind the equation of the regression line of on .
\nfind the value of and of .
\npredict the temperature of the metal rod after 3 minutes.
\nM1
\nA1
\nAG
\n[2 marks]
\nM1A1A1
\n[3 marks]
\nA1
\nM1A1
\n[3 marks]
\nM1A1
\n[2 marks]
\nThis question will investigate the solution to a coupled system of differential equations and how to transform it to a system that can be solved by the eigenvector method.
\nIt is desired to solve the coupled system of differential equations
\n\n
\n
where and represent the population of two types of symbiotic coral and is time measured in decades.
\nFind the equilibrium point for this system.
\nIf initially and use Euler’s method with an time increment of 0.1 to find an approximation for the values of and when .
\nExtend this method to conjecture the limit of the ratio as .
\n\n
Show how using the substitution transforms the system of differential equations into .
\nSolve this system of equations by the eigenvalue method and hence find the general solution for of the original system.
\nFind the particular solution to the original system, given the initial conditions of part (b).
\nHence find the exact values of and when , giving the answers to 4 significant figures.
\nUse part (f) to find limit of the ratio as .
\nWith the initial conditions as given in part (b) state if the equilibrium point is stable or unstable.
\nIf instead the initial conditions were given as and , find the particular solution for of the original system, in this case.
\nWith the initial conditions as given in part (j), determine if the equilibrium point is stable or unstable.
\nM1A1
\n[2 marks]
\nUsing
\nGives M1A1A1
\n[3 marks]
\nBy extending the table, conjecture that M1A1
\n[2 marks]
\nR1
\nM1A1AG
\n[3 marks]
\nM1A1A1
\nan eigenvector is
\nan eigenvector is M1A1A1
\nA1A1
\n\n
[8 marks]
\nM1
\nA1
\n[2 marks]
\nA1A1
\n[2 marks]
\nDominant term is so M1A1
\n[2 marks]
\nThe equilibrium point is unstable. R1
\n[1 mark]
\nM1
\nA1
\n[2 marks]
\nAs as the equilibrium point is stable. R1A1
\n[2 marks]
\nKayla wants to measure the extent to which two judges in a gymnastics competition are in agreement. Each judge has ranked the seven competitors, as shown in the table, where 1 is the highest ranking and 7 is the lowest.
\nCalculate Spearman’s rank correlation coefficient for this data.
\nState what conclusion Kayla can make from the answer in part (a).
\naverage equal ranks M1
\n A1A1
A2
\n[5 marks]
\nThere is strong agreement between the two judges. R1
\n[1 mark]
\nA canal system divides a city into six land masses connected by fifteen bridges, as shown in the diagram below.
\nState with reasons whether or not this graph has
\nDraw a graph to represent this map.
\nWrite down the adjacency matrix of the graph.
\nList the degrees of each of the vertices.
\nan Eulerian circuit.
\nan Eulerian trail.
\nFind the number of walks of length 4 from E to F.
\n A2
[2 marks]
\nM = A2
\nNote: Award A1 for one error or omission, A0 for more than one error or omission. Two symmetrical errors count as one error.
\n[2 marks]
\nA B C D E F
\n(8, 4 4, 3 5, 6) A2
\nNote: Award no more than A1 for one error, A0 for more than one error.
\n[2 marks]
\nno, because there are odd vertices M1A1
\n[2 marks]
\nyes, because there are exactly two odd vertices M1A1
\n[2 marks]
\nM4 = (M1)A1
\nnumber of walks of length 4 is 170
\nNote: The complete matrix need not be shown. Only one of the FE has to be shown.
\n[2 marks]
\nThe diagram below is part of a Voronoi diagram.
\nDiagram not to scale
A and B are sites with B having the co-ordinates of (4, 6). L is an edge; the equation of this perpendicular bisector of the line segment from A to B is
\nFind the co-ordinates of the point A.
\nLine from A to B will have the form M1A1
\nThrough so line is M1A1
\nIntersection of and is (2, 5) M1A1
\nLet then M1A1A1
\n\n
[9 marks]
\nThe adjacency matrix of the graph G, with vertices P, Q, R, S, T is given by:
\n\n
Draw the graph of G.
\nDefine an Eulerian circuit.
\nWrite down an Eulerian circuit in G starting at P.
\nDefine a Hamiltonian cycle.
\nExplain why it is not possible to have a Hamiltonian cycle in G.
\nFind the number of walks of length 5 from P to Q.
\nWhich pairs of distinct vertices have more than 15 walks of length 3 between them?
\n A3
\n
Note: Award A2 for one missing or misplaced edge,
\nA1 for two missing or misplaced edges.
\n[3 marks]
\nan Eulerian circuit is one that contains every edge of the graph exactly once A1
\n[1 mark]
\na possible Eulerian circuit is
\nP → Q → S → P → Q → Q → R → T → R → R → P A2
\n[2 marks]
\na Hamiltonian cycle passes through each vertex of the graph A1
\nexactly once A1
\n[2 marks]
\nto pass through T, you must have come from R and must return to R. R3
\nhence there is no Hamiltonian cycle
\n[3 marks]
\nusing the adjacency matrix A = , (M1)
\nwe need the entry in the first row second column of the matrix A5 (M1)
\nA5 = (A1)
\nhence there are 309 ways A1
\n[4 marks]
\nA3 = (M1)
\nhence the pairs of vertices are PQ, PR and QR A1A1A1
\n[4 marks]
\nA king rules a small mountain kingdom which is in the form of a square of length 4 kilometres. The square is described by the co-ordinate system .
\nThe king has four adult children, each of which has a luxury palace located at the points . Each child owns all the land that is nearer their palace than any other palace.
\nThe king has a fifth (youngest) child who is now just growing up. He installs her in a new palace situated at point (2, 2). The rule about ownership of land is then reapplied.
\nSketch a Voronoi diagram to represent this information.
\nSketch a new Voronoi diagram to represent this new situation.
\nState what the shape of the land, owned by the youngest child, is.
\nFind the area of the youngest child’s land.
\nFind how much land an older child has lost.
\nState, with a reason, if all five children now own an equal amount of land.
\n A2
[2 marks]
\n A2
[2 marks]
\nBy symmetry a square A1
\n[1 mark]
\nDistance from (2, 2) to (1, 3) is M1A1
\nSo length of youngest child’s square is and thus area is 2. M1A1
\n[4 marks]
\nBy symmetry each older child must lose A1
\n[1 mark]
\nNo, youngest child has less as each older child has A1R1
\n[2 marks]
\nLet G be the graph below.
\nFind the total number of Hamiltonian cycles in G, starting at vertex A. Explain your answer.
\nFind a minimum spanning tree for the subgraph obtained by deleting A from G.
\nHence, find a lower bound for the travelling salesman problem for G.
\nGive an upper bound for the travelling salesman problem for the graph above.
\nShow that the lower bound you have obtained is not the best possible for the solution to the travelling salesman problem for G.
\nStarting from vertex A there are 4 choices. From the next vertex there are three choices, etc… M1R1
\nSo the number of Hamiltonian cycles is 4! = 24. A1 N1
\n[3 marks]
\nStart (for instance) at B, using Prim′s algorithm Then D is the nearest vertex M1
\nNext E is the nearest vertex A1
\nFinally C is the nearest vertex So a minimum spanning tree is B → D → E → C A1 N1
\n[3 marks]
\nA lower bound for the travelling salesman problem is then obtained by adding the weights of AB and AE to the weight of the minimum M1
\nspanning tree (ie 20) A1
\nA lower bound is then 20 + 7 + 6 = 33 A1 N1
\n[3 marks]
\nABCDE is an Hamiltonian cycle A1
\nThus an upper bound is given by 7 + 9 + 9 + 8 + 6 = 39 A1
\n[2 marks]
\nEliminating C from G a minimum spanning tree is E → A → B → D M1
\nof weight 18 A1
\nAdding BC to CE(18 + 9 + 7) gives a lower bound of 34 > 33 A1
\nSo 33 not the best lower bound. AG N0
\n[3 marks]
\nFind a relationship between and if the matrices and commute under matrix multiplication.
\nFind the value of if the determinant of matrix is −1.
\n\n
Write down for this value of .
\n\n
M1A1
\nSo require M1A1
\n[4 marks]
\nM1A1
\n[2 marks]
\nA1
\n[1 mark]
\nSue sometimes goes out for lunch. If she goes out for lunch on a particular day then the probability that she will go out for lunch on the following day is 0.4. If she does not go out for lunch on a particular day then the probability she will go out for lunch on the following day is 0.3.
\nWrite down the transition matrix for this Markov chain.
\nWe know that she went out for lunch on a particular Sunday, find the probability that she went out for lunch on the following Tuesday.
\nFind the steady state probability vector for this Markov chain.
\nM1A1
\n[2 marks]
\nM1
\nSo probability is 0.34 A1
\n[2 marks]
\nM1A1
\nSo vector is A1
\n[or by investigating high powers of the transition matrix]
\n[3 marks]
\nLet .
\nThe matrix A is defined by A = .
\nDeduce that
\nShow that .
\nHence find the value of .
\nA3 = –I.
\nA–1 = I – A.
\nMETHOD 1
\nas is a root of then M1R1
\nAG
\nNote: Award M1 for the use of in any way.
\nAward R1 for a correct reasoned approach.
\n\n
METHOD 2
\nM1
\nA1
\n\n
[2 marks]
\nMETHOD 1
\n(M1)
\nA1
\n(M1)
\n\n
A1
\n\n
METHOD 2
\n\n
M1A1
\nNote: Award M1 for attempt at binomial expansion.
\nuse of any previous result e.g. M1
\nA1
\nNote: As the question uses the word ‘hence’, other methods that do not use previous results are awarded no marks.
\n\n
[4 marks]
\nA2 = A – I
\n⇒ A3 = A2 – A M1A1
\n= A – I – A A1
\n= –I AG
\nNote: Allow other valid methods.
\n[3 marks]
\nI = A – A2
\nA–1 = A–1A – A–1A2 M1A1
\n⇒ A–1 = I – A AG
\nNote: Allow other valid methods.
\n[2 marks]
\nA transition matrix for a Markov chain will have the form .
\nShow that is always an eigenvalue for M and find the other eigenvalue in terms of and .
\nFind the steady state probability vector for M in terms of and .
\nM1A1
\nA1
\nAGA1
\n[4 marks]
\nM1A1
\nM1
\nSo vector is A1A1
\n[5 marks]
\nThis question will connect Markov chains and directed graphs.
\nAbi is playing a game that involves a fair coin with heads on one side and tails on the other, together with two tokens, one with a fish’s head on it and one with a fish’s tail on it. She starts off with no tokens and wishes to win them both. On each turn she tosses the coin, if she gets a head she can claim the fish’s head token, provided that she does not have it already and if she gets a tail she can claim the fish’s tail token, provided she does not have it already. There are 4 states to describe the tokens in her possession; A: no tokens, B: only a fish’s head token, C: only a fish’s tail token, D: both tokens. So for example if she is in state B and tosses a tail she moves to state D, whereas if she tosses a head she remains in state B.
\nAfter throws the probability vector, for the 4 states, is given by where the numbers represent the probability of being in that particular state, e.g. is the probability of being in state B after throws. Initially .
\nDraw a transition state diagram for this Markov chain problem.
\nExplain why for any transition state diagram the sum of the out degrees of the directed edges from a vertex (state) must add up to +1.
\nWrite down the transition matrix M, for this Markov chain problem.
\nFind the steady state probability vector for this Markov chain problem.
\nExplain which part of the transition state diagram confirms this.
\nExplain why having a steady state probability vector means that the matrix M must have an eigenvalue of .
\nFind .
\nHence, deduce the form of .
\nExplain how your answer to part (f) fits with your answer to part (c).
\nFind the minimum number of tosses of the coin that Abi will have to make to be at least 95% certain of having finished the game by reaching state C.
\n M1A2
[3 marks]
\nYou must leave the state along one of the edges directed out of the vertex. R1
\n[1 mark]
\nM1A2
\n[3 marks]
\nM1
\nsince so steady state vector is . A1R1A1
\n[4 marks]
\nThere is a loop with probability of 1 from state D to itself. A1
\n[1 mark]
\nLet the steady state probability vector be s then Ms = 1s showing that (\\lambda = 1\\) is an eigenvalue with associated eigenvector of s. A1R1
\n[2 marks]
\nA1A1A1A1
\n[4 marks]
\nA2
\n[2 marks]
\nthe steady state probability vector M1R1
\n[2 marks]
\nRequire (e.g. by use of table) R1M1A2
\n[4 marks]
\nLet A = and B = .
\nGiven that X = B – A–1 and Y = B–1 – A,
\nYou are told that , for .
\nGiven that , for ,
\nfind X and Y.
\ndoes X–1 + Y–1 have an inverse? Justify your conclusion.
\nfind and in terms of .
\nand hence find an expression for .
\nX = B – A–1 = A1
\nY = B–1 – A = A1
\n\n
[2 marks]
\nX–1 + Y–1 = (A1)
\nX–1 + Y–1 has no inverse A1
\nas det(X–1 + Y–1) = 0 R1
\n[3 marks]
\nM1
\nA1
\nsolve simultaneous equations to obtain
\nand M1
\nand A1A1N2
\n[5 marks]
\nA1
\n\n
[1 mark]
\nLet M2 = M where M = .
\nShow that .
\nFind an expression for in terms of .
\nHence show that M is a singular matrix.
\nIf all of the elements of M are positive, find the range of possible values for .
\nShow that (I − M)2 = I − M where I is the identity matrix.
\nAttempting to find M2 M1
\nM2 = A1
\nor A1
\nHence (as or ) AG N0
\n[3 marks]
\nM1
\nA1 N1
\n[2 marks]
\nMETHOD 1
\nUsing det M = M1
\ndet M = or det M =
\n(or equivalent) A1
\nusing or to simplify their expression R1
\nHence M is a singular matrix AG N0
\n\n
METHOD 2
\nUsing and to obtain M1A1
\ndet M = and as R1
\nHence M is a singular matrix AG N0
\n\n
[3 marks]
\n(M1)
\n0 < < 1 A1A1 N3
\nNote: Award A1 for correct endpoints and A1 for correct inequality signs.
\n[3 marks]
\nMETHOD 1
\nAttempting to expand (I − M)2 M1
\n(I − M)2 = I − 2M + M2 A1
\n= I − 2M + M A1
\n= I − M AG N0
\n\n
METHOD 2
\nAttempting to expand (I − M)2 = (or equivalent) M1
\n(I − M)2 =
\n(or equivalent) A1
\nUse of and to show desired result. M1
\nHence (I − M)2 = AG N0
\n\n
[3 marks]
\nSameer is trying to design a road system to connect six towns, A, B, C, D, E and F.
\nThe possible roads and the costs of building them are shown in the graph below. Each vertex represents a town, each edge represents a road and the weight of each edge is the cost of building that road. He needs to design the lowest cost road system that will connect the six towns.
\nName an algorithm that will allow Sameer to find the lowest cost road system.
\nFind the lowest cost road system and state the cost of building it. Show clearly the steps of the algorithm.
\nEITHER
\nPrim’s algorithm A1
\nOR
\nKruskal’s algorithm A1
\n[1 mark]
\nEITHER
\nusing Prim’s algorithm, starting at A
\n A1A1A1A1A1
lowest cost road system contains roads AC, CD, CF, FE and AB A1
\ncost is 20 A1
\nOR
\nusing Kruskal’s algorithm
\n A1A1A1A1A1
lowest cost road system contains roads CD, CF, FE, AC and AB A1
\ncost is 20 A1
\nNote: Accept alternative correct solutions.
\n[7 marks]
\nThis question will diagonalize a matrix and apply this to the transformation of a curve.
\nLet the matrix .
\nLet .
\nLet .
\nLet .
\nHence state the geometrical shape represented by
\nFind the eigenvalues for . For each eigenvalue find the set of associated eigenvectors.
\nShow that the matrix equation is equivalent to the Cartesian equation .
\nShow that and are unit eigenvectors and that they correspond to different eigenvalues.
\nHence, show that .
\nFind matrix R.
\nShow that .
\nVerify that .
\nHence, find the Cartesian equation satisfied by and .
\nFind the Cartesian equation satisfied by and and state the geometric shape that this curve represents.
\nState geometrically what transformation the matrix represents.
\nthe curve in and in part (e) (ii), giving a reason.
\nthe curve in and in part (b).
\nWrite down the equations of two lines of symmetry for the curve in and in part (b).
\nM1M1A1A1
\neigenvalues are of the form M1A1
\neigenvalues are of the form M1A1
\n[8 marks]
\nM1A1
\nAG
\n[2 marks]
\ncorresponding to , corresponding to R1R1
\n[2 marks]
\nA1AG
\n[1 mark]
\nDeterminant is 1. M1A1
\n[2 marks]
\nso post multiplying by gives M1AG
\n[1 mark]
\nM1A1
\nand completing the proof A1AG
\n[3 marks]
\n\n
M1A1
\n[2 marks]
\n, a circle (centre at the origin radius of 1) A1A1
\n[2 marks]
\nA rotation about the origin through an angle of 45° anticlockwise. A1A1
\n[2 marks]
\nan ellipse, since the matrix represents a vertical and a horizontal stretch R1A1
\n[2 marks]
\nan ellipse A1
\n[1 mark]
\n, A1A1
\n[2 marks]
\nThe above diagram shows the weighted graph G.
\nWrite down the adjacency matrix for G.
\nFind the number of distinct walks of length 4 beginning and ending at A.
\nStarting at A, use Prim’s algorithm to find and draw the minimum spanning tree for G.
\nYour solution should indicate clearly the way in which the tree is constructed.
\nM = A1
\n[1 mark]
\nWe require the (A, A) element of M4 which is 13. M1A2
\n[3 marks]
\n A1A1A1A1A1
[5 marks]
\nThe following graph shows the two parts of the curve defined by the equation , and the normal to the curve at the point P(2 , 1).
\n\n
Show that there are exactly two points on the curve where the gradient is zero.
\nFind the equation of the normal to the curve at the point P.
\nThe normal at P cuts the curve again at the point Q. Find the -coordinate of Q.
\nThe shaded region is rotated by 2 about the -axis. Find the volume of the solid formed.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
differentiating implicitly: M1
\nA1A1
\nNote: Award A1 for each side.
\nif then either or M1A1
\ntwo solutions for R1
\nnot possible (as 0 ≠ 5) R1
\nhence exactly two points AG
\nNote: For a solution that only refers to the graph giving two solutions at and no solutions for award R1 only.
\n[7 marks]
\nat (2, 1) M1
\n(A1)
\ngradient of normal is 2 M1
\n1 = 4 + c (M1)
\nequation of normal is A1
\n[5 marks]
\nsubstituting (M1)
\nor (A1)
\nA1
\n[3 marks]
\nrecognition of two volumes (M1)
\nvolume M1A1A1
\nNote: Award M1 for attempt to use , A1 for limits, A1 for Condone omission of at this stage.
\nvolume 2
\nEITHER
\n(M1)(A1)
\nOR
\n(M1)(A1)
\nTHEN
\ntotal volume = 19.9 A1
\n[7 marks]
\nConsider the matrix A = .
\nFind the matrix A2.
\nIf det A2 = 16, determine the possible values of .
\nA2 = (M1)A1
\n[2 marks]
\nMETHOD 1
\ndet A2 = M1
\n= ±2 A1A1 N2
\n\n
METHOD 2
\ndet A = M1
\ndet A = ±4
\n= ±2 A1A1 N2
\n[3 marks]
\nConsider the matrices
\nA = , B = .
\nFind BA.
\nCalculate det (BA).
\nFind A(A–1B + 2A–1)A.
\nBA = A2
\nNote: Award A1 for one error, A0 for two or more errors.
\n[2 marks]
\ndet(BA) = (72 – 56) = 16 (M1)A1
\n[2 marks]
\nEITHER
\nA(A–1B + 2A–1)A = BA + 2A (M1)A1
\nA1
\nOR
\nA–1 (A1)
\nan attempt to evaluate (M1)
\nA–1B + 2A–1
\nA(A–1B + 2A–1)A =
\nA1
\n[3 marks]
\nThe function is defined by , 0 < < 3.
\nDraw a set of axes showing and values between −3 and 3. On these axes
\nFind .
\nHence, or otherwise, find the coordinates of the point of inflexion on the graph of .
\nsketch the graph of , showing clearly any axis intercepts and giving the equations of any asymptotes.
\nsketch the graph of , showing clearly any axis intercepts and giving the equations of any asymptotes.
\nHence, or otherwise, solve the inequality .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1
\n(M1)A1A1A1
\nNote: Award M1 for attempt at quotient rule, A1A1 for numerator and A1 for denominator.
\n\n
METHOD 2
\n(A1)
\n(M1)A1A1
\nNote: Award M1 for attempt at product rule, A1 for first term, A1 for second term.
\n\n
[4 marks]
\nfinding turning point of or finding root of (M1)
\nA1
\n(M1)A1
\n(0.899, −0.375)
\nNote: Do not accept . Accept y-coordinates rounding to −0.37 or −0.375 but not −0.38.
[4 marks]
\nsmooth curve over the correct domain which does not cross the y-axis
\nand is concave down for > 1 A1
\n-intercept at 0.607 A1
\nequations of asymptotes given as = 0 and = 3 (the latter must be drawn) A1A1
[4 marks]
\nattempt to reflect graph of in = (M1)
\nsmooth curve over the correct domain which does not cross the -axis and is concave down for > 1 A1
\n-intercept at 0.607 A1
\nequations of asymptotes given as = 0 and = 3 (the latter must be drawn) A1
\nNote: For FT from (i) to (ii) award max M1A0A1A0.
\n
[4 marks]
solve or to get = 0.372 (M1)A1
\n0 < < 0.372 A1
\nNote: Do not award FT marks.
\n
[3 marks]
Let A, B and C be non-singular 2×2 matrices, I the 2×2 identity matrix and k a scalar. The following statements are incorrect. For each statement, write down the correct version of the right hand side.
\n(A + B)2 = A2 + 2AB + B2
\n(A – kI)3 = A3 – 3kA2 + 3k2A – k3
\nCA = B C =
\n(A + B)2 = A2 + AB + BA + B2 A2
\nNote: Award A1 in parts (a) to (c) if error is correctly identified, but not corrected.
\n[2 marks]
\n(A – kI)3 = A3 – 3kA2 + 3k2A – k3I A2
\nNote: Award A1 in parts (a) to (c) if error is correctly identified, but not corrected.
\n[2 marks]
\nCA = B ⇒ C = BA–1 A2
\nNote: Award A1 in parts (a) to (c) if error is correctly identified, but not corrected.
\n[2 marks]
\nLet .
\nConsider the function defined by .
\nThe curvature at any point on a graph is defined as .
\nFind an expression for .
\nShow that .
\nShow that the function has a local maximum value when .
\nFind the -coordinate of the point of inflexion of the graph of .
\nSketch the graph of , clearly indicating the position of the local maximum point, the point of inflexion and the axes intercepts.
\nFind the area of the region enclosed by the graph of and the -axis.
\n\n
Find the value of the curvature of the graph of at the local maximum point.
\nFind the value for and comment on its meaning with respect to the shape of the graph.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nM1A1
\n[2 marks]
\nM1A1
\nAG
\n[2 marks]
\nR1
\nR1
\nhence maximum at AG
\n[2 marks]
\nM1
\nA1
\n\n
Note: Award M1A0 if extra zeros are seen.
\n\n
[2 marks]
\n![\"N16/5/MATHL/HP1/ENG/TZ0/11.e/M\"](\"../assets/uploads/tinymce_asset/asset/3293/Schermafbeelding_2017-02-28_om_14.29.02.png\"/)
correct shape and correct domain A1
\nmax at , point of inflexion at A1
\nzeros at and A1
\n\n
Note: Penalize incorrect domain with first A mark; allow FT from (d) on extra points of inflexion.
\n\n
[3 marks]
\nEITHER
\nM1A1
\nA1
\nOR
\nM1A1
\nA1
\nTHEN
\nM1A1
\nA1
\n[6 marks]
\n(A1)
\n(A1)
\nA1
\n[3 marks]
\nA1
\nthe graph is approximated by a straight line R1
\n[2 marks]
\nConsider the matrix A , where .
\nFind the value of for which A is singular.
\nfinding det A = or equivalent A1
\nA is singular ⇒ det A = 0 (R1)
\n\n
A1
\nsolving for (M1)
\n> 0 (or equivalent explanation) (R1)
\n\n
ln 2 (only) A1 N0
\n[6 marks]
\nLet M = where and are non-zero real numbers.
\nShow that M is non-singular.
\nCalculate M2.
\nShow that det(M2) is positive.
\nfinding det M A1
\n, therefore M is non-singular or equivalent statement R1
\n[2 marks]
\nM2 = M1A1
\n[2 marks]
\nEITHER
\ndet(M2) A1
\ndet(M2)
\nsince the first term is non-negative and the second is positive R1
\ntherefore det(M2) > 0
\nNote: Do not penalise first term stated as positive.
\nOR
\ndet(M2) = (det M)2 A1
\nsince det M is positive so too is det (M2) R1
\n[2 marks]
\nMatrices A, B and C are defined as
\nA = , B = , C = .
\nGiven that AB = , find .
\nHence, or otherwise, find A–1.
\nFind the matrix X, such that AX = C.
\nA1
\n[1 mark]
\nA–1 = (M1)A1
\n[2 marks]
\nAX = C ⇒ X = A–1C (M1)
\n\n
A1
\n[2 marks]
\nFind the determinant of A, where A = .
\ndet A = −2 A2
\n[2 marks]
\nIf A = and A2 is a matrix whose entries are all 0, find .
\nA2 = M1
\nA2
\nNote: Award A2 for 4 correct, A1 for 2 or 3 correct.
\nM1
\nA1
\n[5 marks]
\nGiven that M = and that M2 – 6M + kI = 0 find k.
\nM2 = M1A1
\n+ kI = 0 (M1)
\n+ kI = 0 (A1)
\n⇒ k = 5 A1
\n[5 marks]
\nThe graph of , 0 ≤ ≤ 5 is shown in the following diagram. The curve intercepts the -axis at (1, 0) and (4, 0) and has a local minimum at (3, −1).
\nThe shaded area enclosed by the curve , the -axis and the -axis is 0.5. Given that ,
\nThe area enclosed by the curve and the -axis between and is 2.5 .
\nWrite down the -coordinate of the point of inflexion on the graph of .
\nfind the value of .
\nfind the value of .
\nSketch the curve , 0 ≤ ≤ 5 indicating clearly the coordinates of the maximum and minimum points and any intercepts with the coordinate axes.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
3 A1
\n[1 mark]
\nattempt to use definite integral of (M1)
\n\n
(A1)
\n\n
= 3.5 A1
\n[3 marks]
\n(A1)
\nNote: (A1) is for −2.5.
\n\n
\n
= 1 A1
\n[2 marks]
\n A1A1A1
A1 for correct shape over approximately the correct domain
A1 for maximum and minimum (coordinates or horizontal lines from 3.5 and 1 are required),
A1 for -intercept at 3
[3 marks]
\nThe square matrix X is such that X3 = 0. Show that the inverse of the matrix (I – X) is I + X + X2.
\nFor multiplying (I – X)(I + X + X2) M1
\n= I2 + IX + IX2 – XI – X2 – X3 = I + X + X2 – X – X2 – X3 (A1)(A1)
\n= I – X3 A1
\n= I A1
\nAB = I ⇒ A–1 = B (R1)
\n(I – X) (I + X + X2) = I ⇒ (I – X)–1 = I + X + X2 AG N0
\n[5 marks]
\nWrite down the inverse of the matrix
\nA =
\nHence, find the point of intersection of the three planes.
\n\n
A fourth plane with equation passes through the point of intersection. Find the value of .
\nA–1 = A2 N2
\n[2 marks]
\nFor attempting to calculate = A−1 M1
\n(so the point is (1.2, 0.6, 1.6)) A2 N2
\n[3 marks]
\n(1.2, 0.6, 1.6) lies on
\nA1 N1
\n[1 mark]
\nApply Prim’s algorithm to the weighted graph given below to obtain the minimal spanning tree starting with the vertex A.
\nFind the weight of the minimal spanning tree.
\nWe start with point A and write S as the set of vertices and T as the set of edges.
The weights on each edge will be used in applying Prim’s algorithm.
Initially, S = {A}, T = Φ. In each subsequent stage, we shall update S and T.
Step 1: Add edge h: So S = {A, D}, T = {h}
Step 2: Add edge e: So S = {A, D, E} T = {h, e}
Step 3: Add edge d: Then S = {A, D, E, F} T = {h, e, d}
Step 4: Add edge a: Then S = {A, D, E, F, B} T = {h, e, d, a}
Step 5: Add edge i: Then S = {A, D, E, F, B, G} T = {h, e, d, a, i}
Step 6: Add edge g: Then S = (A, D, E, F, B, G, C} T = {h, e, d, a, I, g} (M4)(A3)
Notes: Award (M4)(A3) for all 6 correct,
(M4)(A2) for 5 correct;
(M3)(A2) for 4 correct,
(M3)(A1) for 3 correct;
(M1)(A1) for 2 correct,
(M1)(AO) for 1 correct
OR
(M2) for the correct definition of Prim’s algorithm,
(M2) for the correct application of Prim’s algorithm,
(A3) for the correct answers at the last three stages.
Now S has all the vertices and the minimal spanning tree is obtained.
\nThe weight of the edges in T is 5 + 3 + 5 + 7 + 5 + 6
\n= 31 (A1)
\n[8 marks]
\nJohn rings a church bell 120 times. The time interval, , between two successive rings is a random variable with mean of 2 seconds and variance of .
\nEach time interval, , is independent of the other time intervals. Let be the total time between the first ring and the last ring.
\nThe church vicar subsequently becomes suspicious that John has stopped coming to ring the bell and that he is letting his friend Ray do it. When Ray rings the bell the time interval, has a mean of 2 seconds and variance of .
\nThe church vicar makes the following hypotheses:
\n: Ray is ringing the bell; : John is ringing the bell.
\nHe records four values of . He decides on the following decision rule:
\nIf for all four values of he accepts , otherwise he accepts .
\nFind
\n(i) ;
\n(ii) .
\nExplain why a normal distribution can be used to give an approximate model for .
\nUse this model to find the values of and such that , where and are symmetrical about the mean of .
\nCalculate the probability that he makes a Type II error.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(i) A1
\n(ii) (M1)A1
\n\n
Note: If 120 is used instead of 119 award A0(M1)A0 for part (a) and apply follow through for parts (b)-(d). (b) is unaffected and in (c) the interval becomes . In (d) the first 2 A1 marks are for and so the final answer will round to 0.017.
\n\n
[3 marks]
\njustified by the Central Limit Theorem R1
\nsince is large A1
\n\n
Note: Accept .
\n\n
[2 marks]
\n\n
(M1)(A1)
\n(A1)
\nso (R1)
\n(M1)
\ninterval is A1A1
\n\n
Notes: Accept the use of inverse normal applied to the distribution of .
\nAlternative is to use the GDC to find a pretend confidence interval for a mean and then convert by multiplying by 119.
\nEither or correct implies the five implied marks.
\nAccept any numbers that round to these 3sf numbers.
\n\n
[7 marks]
\nunder (M1)
\n(A1)
\nprobability that all 4 values of lie in this interval is
\n(M1)(A1)
\nso probability of a Type II error is 0.0304 (3sf) A1
\n\n
Note: Accept any answer that rounds to 0.030.
\n\n
[5 marks]
\nIn this part, marks will only be awarded if you show the correct application of the required algorithms, and show all your working.
\nIn an offshore drilling site for a large oil company, the distances between the planned wells are given below in metres.
\nIt is intended to construct a network of paths to connect the different wells in a way that minimises the sum of the distances between them.
\nUse Prim’s algorithm, starting at vertex 3, to find a network of paths of minimum total length that can span the whole site.
\n (R2)(A4)(M1)
(A1)
\nNote: Award (R2) for correct algorithms, (R1) for 1 error, (R0) for 2 or more errors.
Award (A4) for correct calculations, (A3) for 1 error, (A2) for 2 errors, (A1) for 3 errors, (A0) for 4 or more errors.
Award (M1) for tree/table/method.
Award (A1) for minimum weight.
[8 marks]
\nThe diagram below shows a weighted graph.
\nUse Prim’s algorithms to find a minimal spanning tree, starting at J. Draw the tree, and find its total weight.
\n (C4)
OR
\n (C4)
Total weight = 17 (A2)
\nNote: There are other possible spanning trees.
\n[6 marks]
\nLet G be a weighted graph with 6 vertices L, M, N, P, Q, and R. The weight of the edges joining the vertices is given in the table below:
\nFor example the weight of the edge joining the vertices L and N is 3.
\nUse Prim’s algorithm to draw a minimum spanning tree starting at M.
\nWhat is the total weight of the tree?
\nM → Q (M1)
\nQ → L (A1)
\nM → P (A1)
\nP → N → R (A1)
\n (A1)
Note: There are other correct answers.
\n[5 marks]
\nThe total weight is 2 + 1 + 3 + 2 + 3 = 11. (A1)
\n[1 mark]
\nIn a large population of hens, the weight of a hen is normally distributed with mean kg and standard deviation kg. A random sample of 100 hens is taken from the population.
\nThe mean weight for the sample is denoted by .
\nThe sample values are summarized by and where kg is the weight of a hen.
\nIt is found that = 0.27 . It is decided to test, at the 1 % level of significance, the null hypothesis = 1.95 against the alternative hypothesis > 1.95.
\nState the distribution of giving its mean and variance.
\nFind an unbiased estimate for .
\nFind an unbiased estimate for .
\nFind a 90 % confidence interval for .
\nFind the -value for the test.
\nWrite down the conclusion reached.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
\nNote: Accept in place of 100.
\n[1 mark]
\nA1
\nNote: Accept 2.00, 2.0 and 2.
\n[1 mark]
\n(M1)
\n= 0.086864
\nunbiased estimate for is 0.0869 A1
\nNote: Accept any answer which rounds to 0.087.
\n[2 marks]
\n90 % confidence interval is (M1)
\n= (1.95, 2.05) A1A1
\nNote: FT their from (c).
\nNote: Condone the use of the -value 1.645 since is large.
\nNote: Accept any values that round to 1.95 and 2.05.
\n[3 marks]
\n-value is 0.0377 A2
\nNote: Award A1 for the 2-tail value 0.0754.
\nNote: Award A2 for 0.0377 and A1 for any other value that rounds to 0.038.
\nNote: FT their estimated mean from (b), note that 2 gives = 0.032(0).
\n[2 marks]
\naccept the null hypothesis A1
\nNote: FT their -value.
\n[1 mark]
\nIn this question, give all answers to two decimal places.
\nVelina travels from New York to Copenhagen with 1200 US dollars (USD). She exchanges her money to Danish kroner (DKK). The exchange rate is 1 USD = 7.0208 DKK.
\nAt the end of her trip Velina has 3450 DKK left that she exchanges to USD. The bank charges a 5 % commission. The exchange rate is still 1 USD = 7.0208 DKK .
\nCalculate the amount that Velina receives in DKK.
\nCalculate the amount, in DKK, that will be left to exchange after commission.
\nHence, calculate the amount of USD she receives.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
1200 × 7.0208 (M1)
\nNote: Award (M1) for multiplying by 7.0208.
\n8424.96 (DKK) (A1) (C2)
\n[2 marks]
\n0.95 × 3450 (M1)
\nNote: Award (M1) for multiplying 3450 by 0.95 (or equivalent).
\n3277.50 (DKK) (A1) (C2)
\nNote: The answer must be given to two decimal places unless already penalized in part (a).
\n[2 marks]
\n(M1)
\nNote: Follow through from part (b)(i). Award (M1) for dividing their part (b)(i) by 7.0208.
\n466.83 (USD) (A1)(ft) (C2)
\nNote: The answer must be given to two decimal places unless already penalized in parts (a) or (b)(i).
\n[2 marks]
\nPhil takes out a bank loan of $150 000 to buy a house, at an annual interest rate of 3.5%. The interest is calculated at the end of each year and added to the amount outstanding.
\nTo pay off the loan, Phil makes annual deposits of $P at the end of every year in a savings account, paying an annual interest rate of 2% . He makes his first deposit at the end of the first year after taking out the loan.
\nDavid visits a different bank and makes a single deposit of $Q , the annual interest rate being 2.8%.
\nFind the amount Phil would owe the bank after 20 years. Give your answer to the nearest dollar.
\nShow that the total value of Phil’s savings after 20 years is .
\nGiven that Phil’s aim is to own the house after 20 years, find the value for to the nearest dollar.
\nDavid wishes to withdraw $5000 at the end of each year for a period of years. Show that an expression for the minimum value of is
\n.
\nHence or otherwise, find the minimum value of that would permit David to withdraw annual amounts of $5000 indefinitely. Give your answer to the nearest dollar.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(M1)(A1)
\nA1
\n\n
Note: Only accept answers to the nearest dollar. Accept $298469.
\n\n
[3 marks]
\nattempt to look for a pattern by considering 1 year, 2 years etc (M1)
\nrecognising a geometric series with first term and common ratio 1.02 (M1)
\nEITHER
\nA1
\nOR
\nexplicitly identify and (may be seen as ). A1
\nTHEN
\nAG
\n[3 marks]
\n(M1)(A1)
\nA1
\n\n
Note: Accept answers which round to 12284.
\n\n
[3 marks]
\nMETHOD 1
\nM1A1
\nA1
\nAG
\n\n
METHOD 2
\nthe initial value of the first withdrawal is A1
\nthe initial value of the second withdrawal is R1
\nthe investment required for these two withdrawals is R1
\nAG
\n\n
[3 Marks]
\nsum to infinity is (M1)(A1)
\n\n
so minimum amount is $178572 A1
\n\n
Note: Accept answers which round to $178571 or $178572.
\n\n
[3 Marks]
\nThe following table shows four different sets of numbers: , , and .
\nComplete the second column of the table by giving one example of a number from each set.
\nJosh states: “Every integer is a natural number”.
\nWrite down whether Josh’s statement is correct. Justify your answer.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(A1)(A1)(C4)
[4 marks]
\n\n
Incorrect (A1)
\nNatural numbers are positive integers. Integers can also be negative. (or equivalent) (R1) (C2)
\nNote: Accept a correct justification. Do not award (R0)(A1).
Accept: a statement with an example of an integer which is not natural.
[2 marks]
\nAn arithmetic sequence has and common difference . Given that and are the first three terms of a geometric sequence
\nGiven that
\nfind the value of .
\ndetermine the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
use of M1
\n(or equivalent) M1A1
\nA1
\n[4 marks]
\n\n
(A1)
\n(M1)
\nA1
\n[3 marks]
\nPlace the numbers and in the correct position on the Venn diagram.
\nIn the table indicate which two of the given statements are true by placing a tick (✔) in the right hand column.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(A1)(A1) (C4)
Note: Award (A1) for each number in the correct position.
\n[4 marks]
\n (A1)(A1) (C2)
Note: Award (A1) for each correctly placed tick.
\n[2 marks]
\nConsider the following sets:
\nThe universal set consists of all positive integers less than 15;
is the set of all numbers which are multiples of 3;
is the set of all even numbers.
Write down the elements that belong to .
\nWrite down the elements that belong to .
\nWrite down .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
= {3, 6, 9, 12} AND = {2, 4, 6, 8, 10, 12, 14} (M1)
\nNote: Award (M1) for listing all elements of sets and . May be seen in part (b). Condone the inclusion of 15 in set when awarding the (M1).
\n6, 12 (A1)(A1) (C3)
\nNote: Award (A1) for each correct element. Award (A1)(A0) if one additional value seen. Award (A0)(A0) if two or more additional values are seen.
\n[3 marks]
\n3, 9 (A1)(ft)(A1)(ft) (C2)
\nNote: Follow through from part (a) but only if their and are explicitly listed.
Award (A1)(ft) for each correct element. Award (A1)(A0) if one additional value seen. Award (A0)(A0) if two or more additional values are seen.
[2 marks]
\n2 (A1)(ft) (C1)
\nNote: Follow through from part (b)(i).
\n[1 mark]
\nThe 1st, 4th and 8th terms of an arithmetic sequence, with common difference , , are the first three terms of a geometric sequence, with common ratio . Given that the 1st term of both sequences is 9 find
\nthe value of ;
\nthe value of ;
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
EITHER
\nthe first three terms of the geometric sequence are , and (M1)
\nand (A1)
\nattempt to solve simultaneously (M1)
\n\n
OR
\nthe , and terms of the arithmetic sequence are
\n(M1)
\n(A1)
\nattempt to solve (M1)
\nTHEN
\nA1
\n[4 marks]
\nA1
\n\n
Note: Accept answers where a candidate obtains by finding first. The first two marks in either method for part (a) are awarded for the same ideas and the third mark is awarded for attempting to solve an equation in .
\n\n
[1 mark]
\nThe 3rd term of an arithmetic sequence is 1407 and the 10th term is 1183.
\nFind the first term and the common difference of the sequence.
\nCalculate the number of positive terms in the sequence.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
u1 + 2d = 1407, u1 + 9d = 1183 (M1)(A1)
\nu1 = 1471, d = −32 A1A1
\n[4 marks]
\n1471 + (n − 1)(−32) > 0 (M1)
\n⇒ n <
\nn < 46.96… (A1)
\nso 46 positive terms A1
\n[3 marks]
\nThe Tower of Pisa is well known worldwide for how it leans.
\nGiovanni visits the Tower and wants to investigate how much it is leaning. He draws a diagram showing a non-right triangle, ABC.
\nOn Giovanni’s diagram the length of AB is 56 m, the length of BC is 37 m, and angle ACB is 60°. AX is the perpendicular height from A to BC.
\nGiovanni’s tourist guidebook says that the actual horizontal displacement of the Tower, BX, is 3.9 metres.
\nUse Giovanni’s diagram to show that angle ABC, the angle at which the Tower is leaning relative to the
horizontal, is 85° to the nearest degree.
Use Giovanni's diagram to calculate the length of AX.
\nUse Giovanni's diagram to find the length of BX, the horizontal displacement of the Tower.
\nFind the percentage error on Giovanni’s diagram.
\nGiovanni adds a point D to his diagram, such that BD = 45 m, and another triangle is formed.
\nFind the angle of elevation of A from D.
\n(M1)(A1)
\nNote: Award (M1) for substituting the sine rule formula, (A1) for correct substitution.
\nangle = 34.9034…° (A1)
\nNote: Award (A0) if unrounded answer does not round to 35. Award (G2) if 34.9034… seen without working.
\nangle = 180 − (34.9034… + 60) (M1)
\nNote: Award (M1) for subtracting their angle BAC + 60 from 180.
\n85.0965…° (A1)
\n85° (AG)
\nNote: Both the unrounded and rounded value must be seen for the final (A1) to be awarded. If the candidate rounds 34.9034...° to 35° while substituting to find angle , the final (A1) can be awarded but only if both 34.9034...° and 35° are seen.
If 85 is used as part of the workings, award at most (M1)(A0)(A0)(M0)(A0)(AG). This is the reverse process and not accepted.
sin 85… × 56 (M1)
\n= 55.8 (55.7869…) (m) (A1)(G2)
\nNote: Award (M1) for correct substitution in trigonometric ratio.
\n(M1)
\nNote: Award (M1) for correct substitution in the Pythagoras theorem formula. Follow through from part (a)(ii).
\nOR
\ncos(85) × 56 (M1)
\nNote: Award (M1) for correct substitution in trigonometric ratio.
\n= 4.88 (4.88072…) (m) (A1)(ft)(G2)
\nNote: Accept 4.73 (4.72863…) (m) from using their 3 s.f answer. Accept equivalent methods.
\n[2 marks]
\n(M1)
\nNote: Award (M1) for correct substitution into the percentage error formula.
\n= 25.1 (25.1282) (%) (A1)(ft)(G2)
\nNote: Follow through from part (a)(iii).
\n[2 marks]
\n(A1)(ft)(M1)
\nNote: Award (A1)(ft) for their 40.11927… seen. Award (M1) for correct substitution into trigonometric ratio.
\nOR
\n(37 − 4.88072…)2 + 55.7869…2
\n(AC =) 64.3725…
\n64.3726…2 + 82 − 2 × 8 × 64.3726… × cos120
\n(AD =) 68.7226…
\n(A1)(ft)(M1)
\nNote: Award (A1)(ft) for their correct values seen, (M1) for correct substitution into the sine formula.
\n= 54.3° (54.2781…°) (A1)(ft)(G2)
\nNote: Follow through from part (a). Accept equivalent methods.
\n[3 marks]
\nIt is known that the number of fish in a given lake will decrease by 7% each year unless some new fish are added. At the end of each year, 250 new fish are added to the lake.
\nAt the start of 2018, there are 2500 fish in the lake.
\nShow that there will be approximately 2645 fish in the lake at the start of 2020.
\nFind the approximate number of fish in the lake at the start of 2042.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
EITHER
\n2019: 2500 × 0.93 + 250 = 2575 (M1)A1
\n2020: 2575 × 0.93 + 250 M1
\nOR
\n2020: 2500 × 0.932 + 250(0.93 + 1) M1M1A1
\nNote: Award M1 for starting with 2500, M1 for multiplying by 0.93 and adding 250 twice. A1 for correct expression. Can be shown in recursive form.
\nTHEN
\n(= 2644.75) = 2645 AG
\n[3 marks]
\n2020: 2500 × 0.932 + 250(0.93 + 1)
2042: 2500 × 0.9324 + 250(0.9323 + 0.9322 + … + 1) (M1)(A1)
(M1)(A1)
\n=3384 A1
\nNote: If recursive formula used, award M1 for un = 0.93 un−1 and u0 or u1 seen (can be awarded if seen in part (a)). Then award M1A1 for attempt to find u24 or u25 respectively (different term if other than 2500 used) (M1A0 if incorrect term is being found) and A2 for correct answer.
\nNote: Accept all answers that round to 3380.
\n[5 marks]
\nThe geometric sequence u1, u2, u3, … has common ratio r.
\nConsider the sequence .
\nShow that A is an arithmetic sequence, stating its common difference d in terms of r.
\nA particular geometric sequence has u1 = 3 and a sum to infinity of 4.
\nFind the value of d.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nstate that (or equivalent) A1
\nattempt to consider and use of at least one log rule M1
\nA1
\n(which is an AP) with (and 1st term ) A1
\nso A is an arithmetic sequence AG
\nNote: Condone absence of modulus signs.
\nNote: The final A mark may be awarded independently.
\nNote: Consideration of the first two or three terms only will score M0.
\n[4 marks]
\n\n
METHOD 2
\nconsideration of M1
\n\n
M1
\nA1
\nwhich is constant R1
\nNote: Condone absence of modulus signs.
\nNote: The final A mark may be awarded independently.
\nNote: Consideration of the first two or three terms only will score M0.
\nattempting to solve M1
\nA1
\nA1
\n[3 marks]
\nAbdallah owns a plot of land, near the river Nile, in the form of a quadrilateral ABCD.
\nThe lengths of the sides are and angle .
\nThis information is shown on the diagram.
\n![\"N17/5/MATSD/SP2/ENG/TZ0/03\"](\"../assets/uploads/tinymce_asset/asset/4186/Schermafbeelding_2018-02-13_om_14.24.18.png\"/)
The formula that the ancient Egyptians used to estimate the area of a quadrilateral ABCD is
\n.
\nAbdallah uses this formula to estimate the area of his plot of land.
\nShow that correct to the nearest metre.
\nCalculate angle .
\nFind the area of ABCD.
\nCalculate Abdallah’s estimate for the area.
\nFind the percentage error in Abdallah’s estimate.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\n\n
Note: Award (M1) for correct substitution into Pythagoras.
\nAccept correct substitution into cosine rule.
\n(A1)
\n(AG)
\n\n
Note: Both the rounded and unrounded value must be seen for the (A1) to be awarded.
\n\n
[2 marks]
\n(M1)(A1)
\n\n
Note: Award (M1) for substitution into cosine formula, (A1) for correct substitutions.
\n\n
(A1)(G2)
\n[3 marks]
\n(M1)(M1)(A1)(ft)
\n\n
Note: Award (M1) for correct substitution into right-angle triangle area. Award (M1) for substitution into area of triangle formula and (A1)(ft) for correct substitution.
\n\n
(A1)(ft)(G3)
\n\n
Notes: Follow through from part (b).
\n\n
[4 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution in the area formula used by ‘Ancient Egyptians’.
\n\n
(A1)(G2)
\n\n
\n
[2 marks]
\n(M1)
\n\n
Notes: Award (M1) for correct substitution into percentage error formula.
\n\n
(A1)(ft)(G2)
\n\n
Notes: Follow through from parts (c) and (d)(i).
\n\n
[2 marks]
\nSuppose that is the first term of a geometric series with common ratio .
\nProve, by mathematical induction, that the sum of the first terms, is given by
\n, where .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
, so true for R1
\nassume true for , ie. M1
\nNote: Award M0 for statements such as “let ”.
\nNote: Subsequent marks after the first M1 are independent of this mark and can be awarded.
\nM1
\nA1
\n\n
A1
\nA1
\ntrue for and if true for then true for , the statement is true for any positive integer (or equivalent). R1
\nNote: Award the final R1 mark provided at least four of the previous marks are gained.
\n[7 marks]
\nConsider a geometric sequence with a first term of 4 and a fourth term of −2.916.
\nFind the common ratio of this sequence.
\nFind the sum to infinity of this sequence.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(A1)
\nsolving, (M1)A1
\n\n
[3 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nThe sum of the first terms of a sequence is given by , where .
\nWrite down the value of .
\nFind the value of .
\nProve that is an arithmetic sequence, stating clearly its common difference.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
\n[1 mark]
\nM1A1
\n[2 marks]
\nM1
\n\n
\n
A1
\nR1
\n\n
\n
(constant) A1
\n\n
Notes: Award R1 only if candidate provides a clear argument that proves that the difference between ANY two consecutive terms of the sequence is constant. Do not accept examples involving particular terms of the sequence nor circular reasoning arguments (eg use of formulas of APs to prove that it is an AP). Last A1 is independent of R1.
\n\n
[4 marks]
\nThe following table shows the average body weight, , and the average weight of the brain, , of seven species of mammal. Both measured in kilograms (kg).
\n![\"M17/5/MATSD/SP2/ENG/TZ1/01\"](\"../assets/uploads/tinymce_asset/asset/3598/Schermafbeelding_2017-08-16_om_10.57.27.png\"/)
The average body weight of grey wolves is 36 kg.
\nIn fact, the average weight of the brain of grey wolves is 0.120 kg.
\nFind the range of the average body weights for these seven species of mammal.
\nFor the data from these seven species calculate , the Pearson’s product–moment correlation coefficient;
\nFor the data from these seven species describe the correlation between the average body weight and the average weight of the brain.
\nWrite down the equation of the regression line on , in the form .
\nUse your regression line to estimate the average weight of the brain of grey wolves.
\nFind the percentage error in your estimate in part (d).
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\n(A1)(G2)
\n[2 marks]
\n(G2)
\n[2 marks]
\n(very) strong, positive (A1)(ft)(A1)(ft)
\n\n
Note: Follow through from part (b)(i).
\n\n
[2 marks]
\n(A1)(A1)
\n\n
Note: Award (A1) for , (A1) for 0.0923.
\nAward a maximum of (A1)(A0) if the answer is not an equation in the form .
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for substituting 36 into their equation.
\n\n
(A1)(ft)(G2)
\n\n
Note: Follow through from part (c). The final (A1) is awarded only if their answer is positive.
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for their correct substitution into percentage error formula.
\n\n
(A1)(ft)(G2)
\n\n
Note: Follow through from part (d). Do not accept a negative answer.
\n\n
[2 marks]
\nOn the day of her birth, 1st January 1998, Mary’s grandparents invested in a savings account. They continued to deposit on the first day of each month thereafter.
\nThe account paid a fixed rate of 0.4% interest per month. The interest was calculated on the last day of each month and added to the account.
\nLet be the amount in Mary’s account on the last day of the month, immediately after the interest had been added.
\nFind an expression for and show that .
\n(i) Write down a similar expression for and .
\n(ii) Hence show that the amount in Mary’s account the day before she turned 10 years old is given by .
\nWrite down an expression for in terms of on the day before Mary turned 18 years old showing clearly the value of .
\nMary’s grandparents wished for the amount in her account to be at least the day before she was 18. Determine the minimum value of the monthly deposit required to achieve this. Give your answer correct to the nearest dollar.
\nAs soon as Mary was 18 she decided to invest of this money in an account of the same type earning 0.4% interest per month. She withdraws every year on her birthday to buy herself a present. Determine how long it will take until there is no money in the account.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
\nA1
\nAG
\n\n
Note: Accept an argument in words for example, first deposit has been in for two months and second deposit has been in for one month.
\n\n
[2 marks]
\n(i) (M1)A1
\nA1
\n(ii) (A1)
\nM1A1
\nAG
\n[6 marks]
\nA1
\n[1 mark]
\n(A1)(M1)(A1)
\n\n
Note: Award (A1) for , (M1) for attempting to solve and (A1) for .
\n\n
A1
\n\n
Note: Accept . Accept .
\n\n
[4 marks]
\n(M1)
\n(A1)(M1)(A1)
\n\n
Note: Award (A1) for the equation (with their value of ), (M1) for attempting to solve for and (A1) for
\n\n
A1
\n\n
Note: Accept .
\n\n
[5 marks]
\nFor a study, a researcher collected 200 leaves from oak trees. After measuring the lengths of the leaves, in cm, she produced the following cumulative frequency graph.
\n![\"M17/5/MATSD/SP1/ENG/TZ2/06\"](\"../assets/uploads/tinymce_asset/asset/3583/Schermafbeelding_2017-08-15_om_17.29.13.png\"/)
Write down the median length of these leaves.
\nWrite down the number of leaves with a length less than or equal to 8 cm.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
9 (cm) (A1) (C1)
\n[1 mark]
\n40 (leaves) (A1) (C1)
\n[1 mark]
\nConsider the equation , where , , , .
\nThe equation has three distinct real roots which can be written as , and .
\nThe equation also has two imaginary roots, one of which is where .
\nThe values , , and are consecutive terms in a geometric sequence.
\nShow that .
\nShow that one of the real roots is equal to 1.
\nGiven that , find the other two real roots.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nrecognition of the other root (A1)
\nM1A1
\nNote: Award M1 for sum of the roots, A1 for 3. Award A0M1A0 for just .
\n(M1)
\nA1
\nAG
\n[5 marks]
\nMETHOD 1
\nlet the geometric series be , ,
\nM1
\nA1
\nhence one of the roots is R1
\n\n
METHOD 2
\n\n
M1
\nA1
\nhence one of the roots is R1
\n\n
[3 marks]
\nMETHOD 1
\nproduct of the roots is (M1)(A1)
\nA1
\nsum of the roots is (M1)(A1)
\nA1
\nsolving simultaneously (M1)
\n, A1A1
\n\n
METHOD 2
\nproduct of the roots M1A1
\nA1
\nEITHER
\n, , can be written as , , M1
\n\n
attempt to solve M1
\n\n
\n
A1A1
\nOR
\n, , can be written as , , M1
\n\n
attempt to solve M1
\nA1A1
\nTHEN
\nand are (A1)
\nroots are −2, 4 A1
\n\n
[9 marks]
\nLet ,
\nwhere and .
\nWrite your answer to part (b)(ii) in the form , where .
\n(A1)(ft)(A1)(ft) (C2)
\n\n
Note: Answer should be consistent with their answer to part (b)(ii). Award (A1)(ft) for 3.91, and (A1)(ft) for . Follow through from part (b)(ii).
\n\n
[2 marks]
\nThe marks obtained by nine Mathematical Studies SL students in their projects (x) and their final IB examination scores (y) were recorded. These data were used to determine whether the project mark is a good predictor of the examination score. The results are shown in the table.
\nThe equation of the regression line y on x is y = mx + c.
\nA tenth student, Jerome, obtained a project mark of 17.
\nUse your graphic display calculator to write down , the mean examination score.
\nUse your graphic display calculator to write down r , Pearson’s product–moment correlation coefficient.
\nFind the exact value of m and of c for these data.
\nUse the regression line y on x to estimate Jerome’s examination score.
\nJustify whether it is valid to use the regression line y on x to estimate Jerome’s examination score.
\n54 (G1)
\n\n
[1 mark]
\n0.5 (G2)
\n\n
[2 marks]
\nm = 0.875, c = 41.75 (A1)(A1)
\nNote: Award (A1) for 0.875 seen. Award (A1) for 41.75 seen. If 41.75 is rounded to 41.8 do not award (A1).
\n\n
[2 marks]
\ny = 0.875(17) + 41.75 (M1)
\nNote: Award (M1) for correct substitution into their regression line.
\n\n
= 56.6 (56.625) (A1)(ft)(G2)
\nNote: Follow through from part (b)(i).
\n\n
[2 marks]
\nthe estimate is valid (A1)
\nsince this is interpolation and the correlation coefficient is large enough (R1)
\nOR
\nthe estimate is not valid (A1)
\nsince the correlation coefficient is not large enough (R1)
\nNote: Do not award (A1)(R0). The (R1) may be awarded for reasoning based on strength of correlation, but do not accept “correlation coefficient is not strong enough” or “correlation is not large enough”.
\nAward (A0)(R0) for this method if no numerical answer to part (a)(iii) is seen.
\n\n
[2 marks]
\nIn an arithmetic sequence, the sum of the 3rd and 8th terms is 1.
\nGiven that the sum of the first seven terms is 35, determine the first term and the common difference.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempting to form two equations involving and M1
\nand
\n\n
A1
\nNote: Award A1 for any two correct equations
\nattempting to solve their equations: M1
\n, A1
\n[4 marks]
\nConsider the graph G represented in the following diagram.
\nThe graph G is a plan of a holiday resort where each vertex represents a villa and the edges represent the roads between villas. The weights of the edges are the times, in minutes, Mr José, the security guard, takes to walk along each of the roads. Mr José is based at villa A.
\nState, with a reason, whether or not G has an Eulerian circuit.
\nUse Kruskal’s algorithm to find a minimum spanning tree for G, stating its total weight. Indicate clearly the order in which the edges are added.
\nUse a suitable algorithm to show that the minimum time in which Mr José can get from A to E is 13 minutes.
\nFind the minimum time it takes Mr José to patrol the resort if he has to walk along every road at least once, starting and ending at A. State clearly which roads need to be repeated.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
no because the graph has vertices (A, B, D, F) of odd degree R1
\n\n
[1 mark]
\nthe edges are added in the order
\nBI 5
\nDH 5 A1
\nAB 6
\nAF 6
\nCI 6 A1
\nCD 7
\nEF 7 A1
\ntotal weight = 42 A1
\nNote: The orders of the edges with the same weight are interchangeable.
Accept indication of correct edge order on a diagram.
\n
[4 marks]
\nclear indication of using Dijkstra for example M1
\n\n
[5 marks]
\nthere are 4 vertices of odd degree (A, F, B and D) (A1)
\nattempting to list at least 2 possible pairings of odd vertices M1
\nA → F and B → D has minimum weight 6 + 17 = 23
\nA → B and F → D has minimum weight 6 + 18 = 24
\nA → D and F → B has minimum weight 20 + 12 = 32 A1A1
\nNote: Award A1A0 for 2 pairs.
\n\n
minimum time is (116 + 23 =) 139 (mins) (M1)A1
\nroads repeated are AF, BC and CD A1
\n\n
[7 marks]
\nThe random variable has the Poisson distribution . Given that , find the value of in the form where is an integer.
\nThe random variable has the Poisson distribution . Find in the form where and are integers.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(M1)
\nor equivalent A1
\nA1
\n[3 marks]
\n(M1)
\nA1
\nrecognition that (A1)
\nA1
\n[4 marks]
\nThe simple, complete graph has vertices . The weight of the edge from to is given by the number .
\nConsider the general graph .
\n(i) Draw the graph including the weights of all the edges.
\n(ii) Use the nearest-neighbour algorithm, starting at vertex , to find a Hamiltonian cycle.
\n(iii) Hence, find an upper bound to the travelling salesman problem for this weighted graph.
\nConsider the graph . Use the deleted vertex algorithm, with as the deleted vertex, to find a lower bound to the travelling salesman problem for this weighted graph.
\n(i) Use the nearest-neighbour algorithm, starting at vertex , to find a Hamiltonian cycle.
\n(ii) Hence find and simplify an expression in , for an upper bound to the travelling salesman problem for this weighted graph.
\nBy splitting the weight of the edge into two parts or otherwise, show that all Hamiltonian cycles of have the same total weight, equal to the answer found in (c)(ii).
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(i) ![\"N16/5/MATHL/HP3/ENG/TZ/DM/M/04.a\"](\"../assets/uploads/tinymce_asset/asset/3301/Schermafbeelding_2017-03-02_om_08.49.36.png\"/)
A1A1
\n
Note: A1 for the graph, A1 for the weights.
\n\n
(ii) cycle is A1
\n(iii) upper bound is A1
\n[4 marks]
\nwith deleted, (applying Kruskal’s Algorithm) the minimum spanning tree will consist of the edges , of weights 3, 4, 5 (M1)A1
\nthe two edges of smallest weight from are and of weights 6 and 7 (M1)A1
\nso lower bound is A1
\n[5 marks]
\n(i) starting at we go
\nwe now have to take
\nthus the cycle is A1A1
\n\n
Note: Final A1 is for .
\n(ii) smallest edge from is of weight 3, smallest edge from (to a new vertex) is of weight 5, smallest edge from (to a new vertex) is of weight (M1)
\nweight of is
\nweight is A1
\nM1A1
\n(which is an upper bound) A1
\n\n
Note: Follow through is not applicable.
\n\n
[7 marks]
\nput a marker on each edge so that of the weight belongs to vertex and of the weight belongs to vertex M1
\nthe Hamiltonian cycle visits each vertex once and only once and for vertex there will be weight (belonging to vertex ) both going in and coming out R1
\nso the total weight will be A1AG
\n\n
Note: Accept other methods for example induction.
\n\n
[3 marks]
\nIt is given that .
\nShow that where .
\nExpress in terms of . Give your answer in the form , where p , q are constants.
\nThe region R, is bounded by the graph of the function found in part (b), the x-axis, and the lines and where . The area of R is .
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nM1A1
\nAG
\n[2 marks]
\n\n
METHOD 2
\nM1
\nA1
\nAG
\n[2 marks]
\n\n
METHOD 1
\n\n
M1
\nM1
\n\n
M1A1
\nA1
\nNote: For the final A mark, must be expressed in the form .
\n[5 marks]
\n\n
METHOD 2
\n\n
M1
\nM1
\nM1
\nA1
\nA1
\nNote: For the final A mark, must be expressed in the form .
\n[5 marks]
\n\n
the area of R is M1
\nA1
\nA1
\nM1
\nA1
\nNote: Only follow through from part (b) if is in the form
\n[5 marks]
\nSolve .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
EITHER
\nM1
\nA1
\nOR
\nM1A1
\nTHEN
\nor A1
\nor (M1)A1
\nNote: (M1) is for an appropriate use of a log law in either case, dependent on the previous M1 being awarded, A1 for both correct answers.
\nsolution is A1
\n[6 marks]
\nMathilde delivers books to five libraries, A, B, C, D and E. She starts her deliveries at library D and travels to each of the other libraries once, before returning to library D. Mathilde wishes to keep her travelling distance to a minimum.
\nThe weighted graph , representing the distances, measured in kilometres, between the five libraries, has the following table.
\n![\"N17/5/MATHL/HP3/ENG/TZ0/DM/01\"](\"../assets/uploads/tinymce_asset/asset/4126/Schermafbeelding_2018-02-09_om_05.58.38.png\"/)
Draw the weighted graph .
\nStarting at library D use the nearest-neighbour algorithm, to find an upper bound for Mathilde’s minimum travelling distance. Indicate clearly the order in which the edges are selected.
\nBy first removing library C, use the deleted vertex algorithm, to find a lower bound for Mathilde’s minimum travelling distance.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"N17/5/MATHL/HP3/ENG/TZ0/DM/M/01.a\"](\"../assets/uploads/tinymce_asset/asset/4127/Schermafbeelding_2018-02-09_om_06.16.03.png\"/)
complete graph on 5 vertices A1
\nweights correctly marked on graph A1
\n[2 marks]
\nclear indication that the nearest-neighbour algorithm has been applied M1
\nDA (or 16) A1
\nAB (or 18) then BC (or 15) A1
\nCE (or 17) then ED (or 19) A1
\nA1
\n[5 marks]
\nan attempt to find the minimum spanning tree (M1)
\nDA (16) then BE (17) then AB (18) (total 51) A1
\nreconnect C with the two edges of least weight, namely CB (15) and CE (17) M1
\nA1
\n[4 marks]
\nSolve the equation .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to form a quadratic in M1
\nA1
\nM1
\nR1
\nA1
\n\n
Note: Award R0 A1 if final answer is .
\n\n
[5 marks]
\nSolve the equation .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
(M1)
\nM1A1
\n\n
(A1)
\nA1
\n[5 marks]
\nFind the solution of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
collecting at least two log terms (M1)
\neg
\nobtaining a correct equation without logs (M1)
\negOR (A1)
\nA1
\n[4 marks]
\nThe weights of the edges in the complete graph are given in the following table.
\n![\"M17/5/MATHL/HP3/ENG/TZ0/DM/02\"](\"../assets/uploads/tinymce_asset/asset/3521/Schermafbeelding_2017-08-10_om_09.18.27.png\"/)
Starting at A , use the nearest neighbour algorithm to find an upper bound for the travelling salesman problem for .
\nBy first deleting vertex A , use the deleted vertex algorithm together with Kruskal’s algorithm to find a lower bound for the travelling salesman problem for .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
the edges are traversed in the following order
\nAB A1
\nBC
\nCF A1
\nFE
\nED A1
\nDA A1
\nA1
\n[5 marks]
\nhaving deleted A, the order in which the edges are added is
\nBC A1
\nCF A1
\nCD A1
\nEF A1
\n\n
Note: Accept indication of the correct order on a diagram.
\n\n
to find the lower bound, we now reconnect A using the two edges with the lowest weights, that is AB and AF (M1)(A1)
\nA1
\n\n
Note: Award (M1)(A1)A1 for obtained either from an incorrect order of correct edges or where order is not indicated.
\n\n
[7 marks]
\nIn an arithmetic sequence, the first term is 8 and the second term is 5.
\nFind the common difference.
\nFind the tenth term.
\nFind the sum of the first ten terms.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
subtracting terms (M1)
\neg
\nA1 N2
\n[2 marks]
\ncorrect substitution into formula (A1)
\neg
\nA1 N2
\n[2 marks]
\ncorrect substitution into formula for sum (A1)
\neg
\nA1 N2
\n[2 marks]
\nConsider
\nThese four points form the vertices of a quadrilateral, Q.
\nExpress w2 and w3 in modulus-argument form.
\nSketch on an Argand diagram the points represented by w0 , w1 , w2 and w3.
\nShow that the area of the quadrilateral Q is .
\nLet . The points represented on an Argand diagram by form the vertices of a polygon .
\nShow that the area of the polygon can be expressed in the form , where .
\n(M1)A1A1
\nNote: Accept Euler form.
\nNote: M1 can be awarded for either both correct moduli or both correct arguments.
\nNote: Allow multiplication of correct Cartesian form for M1, final answers must be in modulus-argument form.
\n[3 marks]
\n A1A1
[2 marks]
\nuse of area = M1
\nA1A1
\nNote: Award A1 for , A1 for correct moduli.
\nAG
\nNote: Other methods of splitting the area may receive full marks.
\n[3 marks]
\nM1A1
\nNote: Award M1 for powers of 2, A1 for any correct expression including both the first and last term.
\n\n
identifying a geometric series with common ratio 22(= 4) (M1)A1
\nM1
\nNote: Award M1 for use of formula for sum of geometric series.
\nA1
\n[6 marks]
\nThe first three terms of a geometric sequence are , , , for .
\nFind the common ratio.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ncorrect use A1
\neg
\nvalid approach to find (M1)
\neg
\nA1 N2
\n[3 marks]
\nConsider the complex numbers and .
\nBy expressing and in modulus-argument form write down the modulus of ;
\nBy expressing and in modulus-argument form write down the argument of .
\nFind the smallest positive integer value of , such that is a real number.
\nand A1A1
\n\n
Note: Award A1A0 for correct moduli and arguments found, but not written in mod-arg form.
\n\n
A1
\n[3 marks]
\nand A1A1
\n\n
Note: Award A1A0 for correct moduli and arguments found, but not written in mod-arg form.
\n\n
A1
\n\n
Notes: Allow FT from incorrect answers for and in modulus-argument form.
\n\n
[1 mark]
\nEITHER
\n(M1)
\nOR
\n(M1)
\n\n
THEN
\nA1
\n[2 marks]
\nLet , for x > 0.
\nThe k th maximum point on the graph of f has x-coordinate xk where .
\nGiven that xk + 1 = xk + a, find a.
\nHence find the value of n such that .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach to find maxima (M1)
\neg one correct value of xk, sketch of f
\nany two correct consecutive values of xk (A1)(A1)
\neg x1 = 1, x2 = 5
\na = 4 A1 N3
\n[4 marks]
\nrecognizing the sequence x1, x2, x3, …, xn is arithmetic (M1)
\neg d = 4
\ncorrect expression for sum (A1)
eg
\nvalid attempt to solve for n (M1)
\neg graph, 2n2 − n − 861 = 0
\nn = 21 A1 N2
\n[4 marks]
\nAn arithmetic sequence has and , where and .
\nLet and . Find the value of .
\nMETHOD 1 (finding and d)
\nrecognizing (seen anywhere) (A1)
\nattempt to find or d using (M1)
eg , , correct value of or d
= 2, d = 3 (seen anywhere) (A1)(A1)
\ncorrect working (A1)
eg
= 610 A1 N2
\n\n
METHOD 2 (expressing S in terms of c)
\nrecognizing (seen anywhere) (A1)
\ncorrect expression for S in terms of c (A1)
eg
(seen anywhere) (A1)(A1)
\ncorrect working (A1)
\neg
\n= 610 A1 N2
\n\n
METHOD 3 (expressing S in terms of c)
\nrecognizing (seen anywhere) (A1)
\ncorrect expression for S in terms of c (A1)
eg
correct application of log law (A1)
eg
correct application of definition of log (A1)
eg
correct working (A1)
eg
= 610 A1 N2
\n[6 marks]
\n\n
Let , , and let .
\nShow the points represented by and on the following Argand diagram.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
Note: Award A1 for in first quadrant and its reflection in the -axis.
\n[1 mark]
\nLet .
\nSolve .
\nShow that .
\nFind the modulus and argument of in terms of . Express each answer in its simplest form.
\nHence find the cube roots of in modulus-argument form.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
(M1)(A1)
\nA1
\n\n
M1
\nA1
\n[5 marks]
\nEITHER
\nchoosing two appropriate angles, for example 60° and 45° M1
\nand
\n(A1)
\nA1
\nAG
\nOR
\nattempt to square the expression M1
\n\n
A1
\nA1
\nAG
\n\n
[3 marks]
\nEITHER
\n\n
M1
\nA1
\nA1
\n\n
A1
\nlet
\nM1
\n(A1)
\nA1
\nA1
\nA1
\nOR
\n\n
M1A1
\n(A1)
\nM1A1
\nM1A1
\nA1
\nA1
\n[9 marks]
\nattempt to apply De Moivre’s theorem M1
\nA1A1A1
\n\n
Note: A1 for modulus, A1 for dividing argument of by 3 and A1 for .
\n\n
Hence cube roots are the above expression when . Equivalent forms are acceptable. A1
\n[5 marks]
\nThe first term of an infinite geometric sequence is 4. The sum of the infinite sequence is 200.
\nFind the common ratio.
\nFind the sum of the first 8 terms.
\nFind the least value of n for which Sn > 163.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ncorrect substitution into infinite sum (A1)
eg
r = 0.98 (exact) A1 N2
\n[2 marks]
\ncorrect substitution (A1)
\n\n
29.8473
\n29.8 A1 N2
\n[2 marks]
\nattempt to set up inequality (accept equation) (M1)
eg
correct inequality for n (accept equation) or crossover values (A1)
eg n > 83.5234, n = 83.5234, S83 = 162.606 and S84 = 163.354
n = 84 A1 N1
\n[3 marks]
\nThe following diagram shows [AB], with length 2 cm. The line is divided into an infinite number of line segments. The diagram shows the first three segments.
\n![\"N17/5/MATME/SP1/ENG/TZ0/10.a\"](\"../assets/uploads/tinymce_asset/asset/4156/Schermafbeelding_2018-02-12_om_09.33.46.png\"/)
The length of the line segments are , where .
\nShow that .
\nThe following diagram shows [CD], with length , where . Squares with side lengths , where , are drawn along [CD]. This process is carried on indefinitely. The diagram shows the first three squares.
\n![\"N17/5/MATME/SP1/ENG/TZ0/10.b\"](\"../assets/uploads/tinymce_asset/asset/4157/Schermafbeelding_2018-02-12_om_09.48.48.png\"/)
The total sum of the areas of all the squares is . Find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ninfinite sum of segments is 2 (seen anywhere) (A1)
\neg
\nrecognizing GP (M1)
\negratio is
\ncorrect substitution into formula (may be seen in equation) A1
\neg
\ncorrect equation (A1)
\neg
\ncorrect working leading to answer A1
\neg
\nAG N0
\n[5 marks]
\nrecognizing infinite geometric series with squares (M1)
\neg
\ncorrect substitution into (must substitute into formula) (A2)
\neg
\ncorrect working (A1)
\neg
\n(seen anywhere) A1
\nvalid approach with segments and CD (may be seen earlier) (M1)
\neg
\ncorrect expression for in terms of (may be seen earlier) (A1)
\neg
\nsubstituting their value of into their formula for (M1)
\neg
\nA1 N3
\n[9 marks]
\nIn an arithmetic sequence, , and .
\nConsider the terms, , of this sequence such that ≤ .
\nLet be the sum of the terms for which is not a multiple of 3.
\nFind the value of .
\nFind the exact value of .
\nShow that .
\nAn infinite geometric series is given as , .
\nFind the largest value of such that .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to find (M1)
\neg 1.4 − 1.3 , ,
\n(may be seen in expression for ) (A1)
\ncorrect equation (A1)
\neg ,
\nA1 N3
\n[4 marks]
\ncorrect substitution (A1)
\neg , ,
\nA1 N2
\n[2 marks]
\nrecognizing need to find the sequence of multiples of 3 (seen anywhere) (M1)
\neg first term is (= 1.5) (accept notation ) ,
\n(= 0.3) , 100 terms (accept ), last term is 31.2
\n(accept notation ) , (accept )
\ncorrect working for sum of sequence where n is a multiple of 3 A2
\n, , 1635
\nvalid approach (seen anywhere) (M1)
\neg , , (their sum for )
\ncorrect working (seen anywhere) A1
\neg , 4875 − 1635
\nAG N0
\n[5 marks]
\nattempt to find (M1)
eg dividing consecutive terms
correct value of (seen anywhere, including in formula)
eg , 0.707106… ,
\ncorrect working (accept equation) (A1)
eg
\ncorrect working A1
\n
METHOD 1 (analytical)
\neg , , 948.974
\nMETHOD 2 (using table, must find both values)
eg when , AND when ,
\nA1 N2
\n[5 marks]
\nIn the following Argand diagram the point A represents the complex number and the point B represents the complex number . The shape of ABCD is a square. Determine the complex numbers represented by the points C and D.
\n![\"M17/5/MATHL/HP1/ENG/TZ2/05\"](\"../assets/uploads/tinymce_asset/asset/3510/Schermafbeelding_2017-08-09_om_06.11.20.png\"/)
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
C represents the complex number A2
\nD represents the complex number A2
\n[4 marks]
\nBoxes of mixed fruit are on sale at a local supermarket.
\nBox A contains 2 bananas, 3 kiwifruit and 4 melons, and costs $6.58.
\nBox B contains 5 bananas, 2 kiwifruit and 8 melons and costs $12.32.
\nBox C contains 5 bananas and 4 kiwifruit and costs $3.00.
\nFind the cost of each type of fruit.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
let be the cost of one banana, the cost of one kiwifruit, and the cost of one melon
\nattempt to set up three linear equations (M1)
\n\n
\n
(A1)
\nattempt to solve three simultaneous equations (M1)
\n\n
banana costs ($)0.36, kiwifruit costs ($)0.30, melon costs ($)1.24 A1
\n[4 marks]
\nConsider the following system of equations where .
\n\n
\n
.
\nFind the value of for which the system of equations does not have a unique solution.
\nFind the solution of the system of equations when .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
an attempt at a valid method eg by inspection or row reduction (M1)
\n\n
A1
\n\n
[2 marks]
\nusing elimination or row reduction to eliminate one variable (M1)
\ncorrect pair of equations in 2 variables, such as
\nA1
\nNote: Award A1 for = 0 and one other equation in two variables.
\n\n
attempting to solve for these two variables (M1)
\n, , A1A1
\nNote: Award A1A0 for only two correct values, and A0A0 for only one.
\nNote: Award marks in part (b) for equivalent steps seen in part (a).
\n\n
[5 marks]
\nConsider a geometric sequence where the first term is 768 and the second term is 576.
\nFind the least value of such that the th term of the sequence is less than 7.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nattempt to find (M1)
\neg
\ncorrect expression for (A1)
\neg
\nEITHER (solving inequality)
\nvalid approach (accept equation) (M1)
\neg
\nvalid approach to find M1
\neg, sketch
\ncorrect value
\neg (A1)
\n(must be an integer) A1 N2
\nOR (table of values)
\nvalid approach (M1)
\neg, one correct crossover value
\nboth crossover values, and A2
\n(must be an integer) A1 N2
\nOR (sketch of functions)
\nvalid approach M1
\negsketch of appropriate functions
\nvalid approach (M1)
\negfinding intersections or roots (depending on function sketched)
\ncorrect value
\neg (A1)
\n(must be an integer) A1 N2
\n[6 marks]
\nConsider the functions and defined by , \\ , and , \\ , where , .
\nThe graphs of and intersect at the point P .
\nDescribe the transformation by which is transformed to .
\nState the range of .
\nSketch the graphs of and on the same axes, clearly stating the points of intersection with any axes.
\nFind the coordinates of P.
\nThe tangent to at P passes through the origin (0, 0).
\nDetermine the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ntranslation units to the left (or equivalent) A1
\n[1 mark]
\nrange is A1
\n[1 mark]
\ncorrect shape of A1
\ntheir translated units to left (possibly shown by marked on -axis) A1
\nasymptote included and marked as A1
\nintersects -axis at , A1
\nintersects -axis at , A1
\nintersects -axis at A1
\nNote: Do not penalise candidates if their graphs “cross” as .
\nNote: Do not award FT marks from the candidate’s part (a) to part (c).
\n[6 marks]
\nat P
\nattempt to solve (or equivalent) (M1)
\n(or ) A1
\nP (or P)
\n[2 marks]
\nattempt to differentiate or (M1)
\nA1
\nat P, A1
\nrecognition that tangent passes through origin (M1)
\nA1
\n(A1)
\nA1
\nNote: For candidates who explicitly differentiate (rather than or , award M0A0A1M1A1A1A1.
\n[7 marks]
\nIn an arithmetic sequence, u1 = −5 and d = 3.
\nFind u8.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct working (A1)
\neg −5 + (8 − 1)(3)
\nu8 = 16 A1 N2
\n\n
[2 marks]
\nThe sum of an infinite geometric sequence is 33.25. The second term of the sequence is 7.98. Find the possible values of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct substitution into formula for infinite geometric series (A1)
\neg
\ncorrect substitution into formula for (seen anywhere) (A1)
\neg
\nattempt to express in terms of (or vice-versa) (M1)
\neg , , ,
\ncorrect working (A1)
\neg , , (0.4, 19.95), (0.6, 13.3),
\n, A1A1 N3
\n\n
[6 marks]
\nThe first two terms of an infinite geometric sequence, in order, are
\n, where .
\nThe first three terms of an arithmetic sequence, in order, are
\n, where .
\nLet be the sum of the first 12 terms of the arithmetic sequence.
\nFind .
\nShow that the sum of the infinite sequence is .
\nFind , giving your answer as an integer.
\nShow that .
\nGiven that is equal to half the sum of the infinite geometric sequence, find , giving your answer in the form , where .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of dividing terms (in any order) (M1)
\neg
\nA1 N2
\n[2 marks]
\ncorrect substitution (A1)
\neg
\ncorrect working A1
\neg
\nAG N0
\n[2 marks]
\nevidence of subtracting two terms (in any order) (M1)
\neg
\ncorrect application of the properties of logs (A1)
\neg
\ncorrect working (A1)
\neg
\nA1 N3
\n[4 marks]
\ncorrect substitution into the formula for the sum of an arithmetic sequence (A1)
\neg
\ncorrect working A1
\neg
\nAG N0
\n[2 marks]
\ncorrect equation (A1)
\neg
\ncorrect working (A1)
\neg
\n(accept ) A1 N2
\n[3 marks]
\nConsider
\nThe function is defined by
\nThe function is defined by .
\nFind the largest possible domain for to be a function.
\nSketch the graph of showing clearly the equations of asymptotes and the coordinates of any intercepts with the axes.
\nExplain why is an even function.
\nExplain why the inverse function does not exist.
\nFind the inverse function and state its domain.
\nFind .
\nHence, show that there are no solutions to ;
\nHence, show that there are no solutions to .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\nor A1
\n[2 marks]
\n![\"M17/5/MATHL/HP2/ENG/TZ1/12.b/M\"](\"../assets/uploads/tinymce_asset/asset/3516/Schermafbeelding_2017-08-09_om_15.40.09.png\"/)
shape A1
\nand A1
\n-intercepts A1
\n[3 marks]
\nEITHER
\nis symmetrical about the -axis R1
\nOR
\nR1
\n[1 mark]
\nEITHER
\nis not one-to-one function R1
\nOR
\nhorizontal line cuts twice R1
\n\n
Note: Accept any equivalent correct statement.
\n\n
[1 mark]
\nM1
\nM1
\nA1A1
\n[4 marks]
\nM1A1
\nA1
\n[3 marks]
\nM1
\nwhich is not in the domain of (hence no solutions to ) R1
\n\n
[2 marks]
\nM1
\nas so no solutions to R1
\n\n
Note: Accept: equation has no solutions.
\n\n
[2 marks]
\nTen students were surveyed about the number of hours, , they spent browsing the Internet during week 1 of the school year. The results of the survey are given below.
\n\n
During week 4, the survey was extended to all 200 students in the school. The results are shown in the cumulative frequency graph:
\n![\"N16/5/MATME/SP2/ENG/TZ0/08.d\"](\"../assets/uploads/tinymce_asset/asset/3311/Schermafbeelding_2017-03-03_om_16.35.16.png\"/)
Find the mean number of hours spent browsing the Internet.
\nDuring week 2, the students worked on a major project and they each spent an additional five hours browsing the Internet. For week 2, write down
\n(i) the mean;
\n(ii) the standard deviation.
\nDuring week 3 each student spent 5% less time browsing the Internet than during week 1. For week 3, find
\n(i) the median;
\n(ii) the variance.
\n(i) Find the number of students who spent between 25 and 30 hours browsing the Internet.
\n(ii) Given that 10% of the students spent more than k hours browsing the Internet, find the maximum value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nattempt to substitute into formula for mean (M1)
\neg
\nmean A1 N2
\n[2 marks]
\n(i) mean A1 N1
\n(ii) A1 N1
\n[2 marks]
\n(i) valid approach (M1)
\neg95%, 5% of 27
\ncorrect working (A1)
\neg
\nmedian A1 N2
\n(ii) METHOD 1
\nvariance (seen anywhere) (A1)
\nvalid attempt to find new standard deviation (M1)
\neg
\nvariance A1 N2
\nMETHOD 2
\nvariance (seen anywhere) (A1)
\nvalid attempt to find new variance (M1)
\neg
\nnew variance A1 N2
\n[6 marks]
\n(i) both correct frequencies (A1)
\neg80, 150
\nsubtracting their frequencies in either order (M1)
\neg
\n70 (students) A1 N2
\n(ii) evidence of a valid approach (M1)
\neg10% of 200, 90%
\ncorrect working (A1)
\neg, 180 students
\nA1 N3
\n[6 marks]
\nThe first terms of an infinite geometric sequence, , are 2, 6, 18, 54, …
\nThe first terms of a second infinite geometric sequence, , are 2, −6, 18, −54, …
\nThe terms of a third sequence, , are defined as .
\nThe finite series, , can also be written in the form .
\nWrite down the first three non-zero terms of .
\nFind the value of .
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to add corresponding terms (M1)
\neg
\ncorrect value for (A1)
\neg 324
\n4, 36, 324 (accept 4 + 36 + 324) A1 N3
\n[3 marks]
\nvalid approach (M1)
\neg ,
\n(accept ; may be incorrect) A1 N2
\n[2 marks]
\nrecognition that 225 terms of consists of 113 non-zero terms (M1)
\neg , , 113
\n(accept ; may be incorrect) A1 N2
\n[2 marks]
\nThe function is defined by , for .
\nThe function is defined by
\nFind the inverse function , stating its domain.
\nExpress in the form where A, B are constants.
\nSketch the graph of . State the equations of any asymptotes and the coordinates of any intercepts with the axes.
\nThe function is defined by , for ≥ 0.
\nState the domain and range of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to make the subject of M1
\nA1
\nA1
\nA1
\nNote: Do not allow in place of .
\nA1
\nNote: The final A mark is independent.
\n[5 marks]
\nA1A1
\n[2 marks]
\nhyperbola shape, with single curves in second and fourth quadrants and third quadrant blank, including vertical asymptote A1
\nhorizontal asymptote A1
\nintercepts A1
\n[3 marks]
\nthe domain of is A1A1
\nthe range of is A1A1
\n[4 marks]
\nThe following table shows values of ln x and ln y.
\nThe relationship between ln x and ln y can be modelled by the regression equation ln y = a ln x + b.
\nFind the value of a and of b.
\nUse the regression equation to estimate the value of y when x = 3.57.
\nThe relationship between x and y can be modelled using the formula y = kxn, where k ≠ 0 , n ≠ 0 , n ≠ 1.
\nBy expressing ln y in terms of ln x, find the value of n and of k.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg one correct value
\n−0.453620, 6.14210
\na = −0.454, b = 6.14 A1A1 N3
\n[3 marks]
\ncorrect substitution (A1)
\neg −0.454 ln 3.57 + 6.14
\ncorrect working (A1)
\neg ln y = 5.56484
\n261.083 (260.409 from 3 sf)
\ny = 261, (y = 260 from 3sf) A1 N3
\nNote: If no working shown, award N1 for 5.56484.
If no working shown, award N2 for ln y = 5.56484.
[3 marks]
\nMETHOD 1
\nvalid approach for expressing ln y in terms of ln x (M1)
\neg
\ncorrect application of addition rule for logs (A1)
\neg
\ncorrect application of exponent rule for logs A1
\neg
\ncomparing one term with regression equation (check FT) (M1)
\neg
\ncorrect working for k (A1)
\neg
\n465.030
\n(464 from 3sf) A1A1 N2N2
\n\n
METHOD 2
\nvalid approach (M1)
\neg
\ncorrect use of exponent laws for (A1)
\neg
\ncorrect application of exponent rule for (A1)
\neg
\ncorrect equation in y A1
\neg
\ncomparing one term with equation of model (check FT) (M1)
\neg
\n465.030
\n(464 from 3sf) A1A1 N2N2
\n\n
METHOD 3
\nvalid approach for expressing ln y in terms of ln x (seen anywhere) (M1)
\neg
\ncorrect application of exponent rule for logs (seen anywhere) (A1)
\neg
\ncorrect working for b (seen anywhere) (A1)
\neg
\ncorrect application of addition rule for logs A1
\neg
\ncomparing one term with equation of model (check FT) (M1)
\neg
\n465.030
\n(464 from 3sf) A1A1 N2N2
\n[7 marks]
\nConsider the expression .
\nThe expression can be written as where .
\nLet , β be the roots of , where 0 < < 1.
\nSketch the graph of for .
\nWith reference to your graph, explain why is a function on the given domain.
\nExplain why has no inverse on the given domain.
\nExplain why is not a function for .
\nShow that .
\nSketch the graph of for t ≤ 0. Give the coordinates of any intercepts and the equations of any asymptotes.
\nFind and β in terms of .
\nShow that + β < −2.
\n A1A1
A1 for correct concavity, many to one graph, symmetrical about the midpoint of the domain and with two axes intercepts.
\nNote: Axes intercepts and scales not required.
\nA1 for correct domain
\n[2 marks]
\nfor each value of there is a unique value of A1
\nNote: Accept “passes the vertical line test” or equivalent.
\n[1 mark]
\nno inverse because the function fails the horizontal line test or equivalent R1
\nNote: No FT if the graph is in degrees (one-to-one).
\n[1 mark]
\nthe expression is not valid at either of R1
\n[1 mark]
\nMETHOD 1
\nM1
\nM1A1
\nAG
\n\n
METHOD 2
\n(M1)
\nA1
\nA1
\nAG
\n[3 marks]
\n\n
for t ≤ 0, correct concavity with two axes intercepts and with asymptote = 1 A1
\nt intercept at (−1, 0) A1
\nintercept at (0, 1) A1
\n[3 marks]
\nMETHOD 1
\n, β satisfy M1
\nA1
\nA1
\nattempt at using quadratic formula M1
\n, β or equivalent A1
\n\n
METHOD 2
\n, β satisfy M1
\nM1
\n(or equivalent) A1
\nM1
\n(or equivalent) A1
\nso for eg, , β
\n[5 marks]
\n+ β A1
\nsince R1
\n+ β < −2 AG
\nNote: Accept a valid graphical reasoning.
\n[2 marks]
\nLet and , for , where is a constant.
\nFind .
\nGiven that , find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to form composite (M1)
\neg
\ncorrect function A1 N2
\neg
\n[2 marks]
\nevidence of (M1)
\neg, graph with horizontal asymptote when
\n\n
Note: Award M0 if candidate clearly has incorrect limit, such as .
\n\n
evidence that (seen anywhere) (A1)
\neg, graph of or
\nwith asymptote , graph of composite function with asymptote
\ncorrect working (A1)
\neg
\nA1 N2
\n[4 marks]
\nConsider a function f (x) , for −2 ≤ x ≤ 2 . The following diagram shows the graph of f.
\nOn the grid above, sketch the graph of f −1.
\nA1A1A1A1 N4
Note: Award A1 for evidence of approximately correct reflection in y = x with correct curvature.
\n(y = x does not need to be explicitly seen)
\nOnly if this mark is awarded, award marks as follows:
\nA1 for both correct invariant points in circles,
\nA1 for the three other points in circles,
\nA1 for correct domain.
\n[4 marks]
\nConsider a function . The line L1 with equation is a tangent to the graph of when
\nLet and P be the point on the graph of where .
\nWrite down .
\nFind .
\nShow that the graph of g has a gradient of 6 at P.
\nLet L2 be the tangent to the graph of g at P. L1 intersects L2 at the point Q.
\nFind the y-coordinate of Q.
\nrecognize that is the gradient of the tangent at (M1)
\neg
\n(accept m = 3) A1 N2
\n[2 marks]
\nrecognize that (M1)
\neg
\nA1 N2
\n[2 marks]
\nrecognize that the gradient of the graph of g is (M1)
\nchoosing chain rule to find (M1)
\neg
\nA2
\nA1
\nAG N0
\n[5 marks]
\n\n
\n
at Q, L1 = L2 (seen anywhere) (M1)
\nrecognize that the gradient of L2 is g'(1) (seen anywhere) (M1)
eg m = 6
finding g (1) (seen anywhere) (A1)
eg
attempt to substitute gradient and/or coordinates into equation of a straight line M1
eg
correct equation for L2
\neg A1
\ncorrect working to find Q (A1)
eg same y-intercept,
A1 N2
\n[7 marks]
\n\n
Let and , for .
\nShow that .
\nOn the following grid, sketch the graph of , for .
\n![\"M17/5/MATME/SP2/ENG/TZ2/06.b\"](\"../assets/uploads/tinymce_asset/asset/3560/Schermafbeelding_2017-08-15_om_08.00.33.png\"/)
The equation has exactly two solutions, for . Find the possible values of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to form composite in either order (M1)
\neg
\nA1
\nAG N0
\n[2 marks]
\n![\"M17/5/MATME/SP2/ENG/TZ2/06.b/M\"](\"../assets/uploads/tinymce_asset/asset/3562/Schermafbeelding_2017-08-15_om_08.05.50.png\"/)
A1
A1A1 N3
\n\n
Note: Award A1 for approximately correct shape which changes from concave down to concave up. Only if this A1 is awarded, award the following:
\nA1 for left hand endpoint in circle and right hand endpoint in oval,
\nA1 for minimum in oval.
\n\n
[3 marks]
\nevidence of identifying max/min as relevant points (M1)
\neg
\ncorrect interval (inclusion/exclusion of endpoints must be correct) A2 N3
\neg
\n[3 marks]
\nConsider the function , where .
\nFor , sketch the graph of . Indicate clearly the maximum and minimum values of the function.
\nWrite down the least value of such that has an inverse.
\nFor the value of found in part (b), write down the domain of .
\nFor the value of found in part (b), find an expression for .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nconcave down and symmetrical over correct domain A1
\nindication of maximum and minimum values of the function (correct range) A1A1
\n\n
[3 marks]
\n= 0 A1
\nNote: Award A1 for = 0 only if consistent with their graph.
\n\n
[1 mark]
\nA1
\nNote: Allow FT from their graph.
\n\n
[1 mark]
\n\n
\n
(M1)
\n\n
A1
\n\n
[2 marks]
\nThe following diagram shows the graph of a function , for .
\nThe points and lie on the graph of . There is a minimum point at .
\nWrite down the range of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ncorrect interval A2 N2
\neg, from 0 to 6
\n[2 marks]
\nLet . The line is tangent to the graph of at .
\ncan be expressed in the form r u.
\nThe direction vector of is .
\nFind the gradient of .
\nFind u.
\nFind the acute angle between and .
\nFind .
\nHence, write down .
\nHence or otherwise, find the obtuse angle formed by the tangent line to at and the tangent line to at .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nattempt to find (M1)
\neg , ,
\n−0.25 (exact) A1 N2
\n[2 marks]
\nu or any scalar multiple A2 N2
\n[2 marks]
\ncorrect scalar product and magnitudes (A1)(A1)(A1)
\nscalar product
\nmagnitudes ,
\nsubstitution of their values into correct formula (M1)
\neg , , 2.1112, 120.96°
\n1.03037 , 59.0362°
\nangle = 1.03 , 59.0° A1 N4
\n[5 marks]
\nattempt to form composite (M1)
\neg , ,
\ncorrect working (A1)
\neg ,
\nA1 N2
\n[3 marks]
\n(accept , ) A1 N1
\nNote: Award A0 in part (ii) if part (i) is incorrect.
Award A0 in part (ii) if the candidate has found by interchanging and .
[1 mark]
\nMETHOD 1
\nrecognition of symmetry about (M1)
\neg (2, 8) ⇔ (8, 2)
evidence of doubling their angle (M1)
eg ,
\n2.06075, 118.072°
\n2.06 (radians) (118 degrees) A1 N2
\n\n
METHOD 2
finding direction vector for tangent line at (A1)
eg ,
\nsubstitution of their values into correct formula (must be from vectors) (M1)
eg ,
\n2.06075, 118.072°
\n2.06 (radians) (118 degrees) A1 N2
\n\n
METHOD 3
using trigonometry to find an angle with the horizontal (M1)
eg ,
\nfinding both angles of rotation (A1)
eg ,
\n2.06075, 118.072°
\n2.06 (radians) (118 degrees) A1 N2
\n[3 marks]
\nThe function is defined by .
\nWrite down the range of .
\nFind an expression for .
\nWrite down the domain and range of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1A1
\n\n
Note: A1 for correct end points, A1 for correct inequalities.
\n\n
[2 marks]
\n(M1)A1
\n[2 marks]
\nA1A1
\n[2 marks]
\nSketch the graph of , showing clearly any asymptotes and stating the coordinates of any points of intersection with the axes.
\n![\"N17/5/MATHL/HP1/ENG/TZ0/06.a\"](\"../assets/uploads/tinymce_asset/asset/4115/Schermafbeelding_2018-02-07_om_17.42.06.png\"/)
Hence or otherwise, solve the inequality .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"N17/5/MATHL/HP1/ENG/TZ0/06.a/M\"](\"../assets/uploads/tinymce_asset/asset/4116/Schermafbeelding_2018-02-07_om_17.44.18.png\"/)
correct vertical asymptote A1
\nshape including correct horizontal asymptote A1
\nA1
\nA1
\n\n
Note: Accept and marked on the axes.
\n\n
[4 marks]
\nMETHOD 1
\n![\"N17/5/MATHL/HP1/ENG/TZ0/06.b/M\"](\"../assets/uploads/tinymce_asset/asset/4117/Schermafbeelding_2018-02-07_om_18.03.17.png\"/)
(M1)
\nA1
\n(M1)
\n\n
Note: Award this M1 for the line above or a correct sketch identifying a second critical value.
\n\n
A1
\nsolution is A1
\n\n
METHOD 2
\n\n
(M1)A1
\n\n
\n
A1
\n(M1)
\nsolution is A1
\n\n
METHOD 3
\n\n
consider (M1)
\n\n
Note: Also allow consideration of “>” or “=” for the awarding of the M mark.
\n\n
recognition of critical value at A1
\nconsider (M1)
\n\n
Note: Also allow consideration of “>” or “=” for the awarding of the M mark.
\n\n
recognition of critical value at A1
\nsolution is A1
\n[5 marks]
\n\n
Consider the function .
\nConsider the region bounded by the curve , the -axis and the lines .
\nShow that the -coordinate of the minimum point on the curve satisfies the equation .
\nDetermine the values of for which is a decreasing function.
\nSketch the graph of showing clearly the minimum point and any asymptotic behaviour.
\nFind the coordinates of the point on the graph of where the normal to the graph is parallel to the line .
\nThis region is now rotated through radians about the -axis. Find the volume of revolution.
\nattempt to use quotient rule or product rule M1
\nA1A1
\n\n
Note: Award A1 for or equivalent and A1 for or equivalent.
\n\n
setting M1
\n\n
or equivalent A1
\nAG
\n[5 marks]
\n\n
A1A1
\n\n
Note: Award A1 for and A1 for . Accept .
\n\n
[2 marks]
\n![\"N17/5/MATHL/HP2/ENG/TZ0/10.b/M\"](\"../assets/uploads/tinymce_asset/asset/4122/Schermafbeelding_2018-02-08_om_16.19.25.png\"/)
concave up curve over correct domain with one minimum point above the -axis. A1
\napproaches asymptotically A1
\napproaches asymptotically A1
\n\n
Note: For the final A1 an asymptote must be seen, and must be seen on the -axis or in an equation.
\n\n
[3 marks]
\n(A1)
\nattempt to solve for (M1)
\nA1
\n\n
A1
\n[4 marks]
\n(M1)(A1)
\n\n
Note: M1 is for an integral of the correct squared function (with or without limits and/or ).
\n\n
A1
\n[3 marks]
\nLet and , for .
\nFind .
\nFind .
\nSolve .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to substitute (M1)
\neg
\nA1 N2
\n[2 marks]
\nattempt to form composition (in any order) (M1)
\neg
\nA1 N2
\n[2 marks]
\nvalid approach (M1)
\neg , ![\"N16/5/MATME/SP2/ENG/TZ0/01.c/M\"](\"../assets/uploads/tinymce_asset/asset/3314/Schermafbeelding_2017-03-03_om_17.12.57.png\"/)
\n
A1A1 N3
\n[3 marks]
\nThe following diagram shows the graph of a function , for −4 ≤ x ≤ 2.
\nOn the same axes, sketch the graph of .
\nAnother function, , can be written in the form . The following diagram shows the graph of .
\nWrite down the value of a and of b.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A2 N2
[2 marks]
recognizing horizontal shift/translation of 1 unit (M1)
\neg b = 1, moved 1 right
\nrecognizing vertical stretch/dilation with scale factor 2 (M1)
\neg a = 2, y ×(−2)
\na = −2, b = −1 A1A1 N2N2
\n[4 marks]
\nLet . The following diagram shows part of the graph of .
\nThe function can be written in the form .
\nFind the equation of the axis of symmetry of the graph of .
\nWrite down the value of .
\nFind the value of .
\nThe graph of a second function, , is obtained by a reflection of the graph of in the -axis, followed by a translation of .
\n\n
Find the coordinates of the vertex of the graph of .
\ncorrect working (A1)
\neg ,
\n\n
(must be an equation with ) A1 N2
\n\n
[2 marks]
\n= 2 A1 N1
\n\n
[1 mark]
\nMETHOD 1
\nvalid approach (M1)
\neg (2)
\ncorrect substitution (A1)
\neg (2)2 − 4(2) − 5
\n= −9 A1 N2
\n\n
METHOD 2
\nvalid attempt to complete the square (M1)
\neg 2 − 4 + 4
\ncorrect working (A1)
\neg (2 − 4 + 4) − 4 − 5, ( − 2)2 − 9
\n= −9 A1 N2
\n\n
[3 marks]
\n\n
METHOD 1 (working with vertex)
\nvertex of is at (2, −9) (A1)
\ncorrect horizontal reflection (A1)
\neg = −2, (−2, −9)
\nvalid approach for translation of their or value (M1)
\neg − 3, + 6, , one correct coordinate for vertex
\nvertex of is (−5, −3) (accept = −5, = −3) A1A1 N1N1
\n\n
METHOD 2 (working with function)
\ncorrect approach for horizontal reflection (A1)
\neg (−)
\ncorrect horizontal reflection (A1)
\neg (−)2 −4(−) − 5, 2 + 4 − 5, (− − 2)2 − 9
\nvalid approach for translation of their or value (M1)
\neg ( + 3)2 + 4( + 3) − 5 + 6, 2 + 10 + 22, ( + 5)2 − 3, one correct coordinate for vertex
\nvertex of is (−5, −3) (accept = −5, = −3) A1A1 N1N1
\n\n
[5 marks]
\nThe following diagram shows part of the graph of with -intercept (5, 0) and -intercept (0, 8).
\nFind the -intercept of the graph of .
\nFind the -intercept of the graph of .
\nFind the -intercept of the graph of .
\nDescribe the transformation .
\n-intercept is 11 (accept (0, 11) ) A1 N1
\n[1 mark]
\nvalid approach (M1)
\neg , recognizing stretch of in -direction
\n-intercept is 8 (accept (0, 8) ) A1 N2
\n[2 marks]
\n-intercept is (accept or (2.5, 0) ) A2 N2
\n[2 marks]
\ncorrect name, correct magnitude and direction A1A1 N2
\neg name: translation, (horizontal) shift (do not accept move)
\neg magnitude and direction: 1 unit to the left, , horizontal by –1
\n[2 marks]
\nLet f(x) = ax2 − 4x − c. A horizontal line, L , intersects the graph of f at x = −1 and x = 3.
\nThe equation of the axis of symmetry is x = p. Find p.
\nHence, show that a = 2.
\nMETHOD 1 (using symmetry to find p)
\nvalid approach (M1)
\neg ,
p = 1 A1 N2
\nNote: Award no marks if they work backwards by substituting a = 2 into to find p.
\nDo not accept .
\n\n
METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate a M1
eg a + 4 − c = a(32) − 4(3) − c, f(−1) = f(3)
\ncorrect working A1
\neg 8a = 16
\na = 2 AG N0
\nvalid approach to find p (M1)
\neg
\np = 1 A1 N2
\n[2 marks]
\nMETHOD 1
\nvalid approach M1
\neg (might be seen in (i)), f' (1) = 0
\ncorrect equation A1
\neg = 1, 2a(1) − 4 = 0
\na = 2 AG N0
\n\n
METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate a M1
eg a + 4 − c = a(32) − 4(3) − c, f(−1) = f(3)
\ncorrect working A1
\neg 8a = 16
\na = 2 AG N0
\n[2 marks]
\nLet , where p ≠ 0. Find Find the number of roots for the equation .
\nJustify your answer.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nevidence of discriminant (M1)
eg
correct substitution into discriminant (A1)
eg
correct discriminant A1
eg
A1
\nA1
\nA1
\nhas 2 roots A1 N0
\n\n
METHOD 2
\ny-intercept = −4p (seen anywhere) A1
\nif p is positive, then the y-intercept will be negative A1
\nan upward-opening parabola with a negative y-intercept R1
eg sketch that must indicate p > 0.
if p is negative, then the y-intercept will be positive A1
\na downward-opening parabola with a positive y-intercept R1
eg sketch that must indicate p > 0.
has 2 roots A2 N0
\n[7 marks]
\n\n
All lengths in this question are in metres.
\nLet , for . Mark uses as a model to create a barrel. The region enclosed by the graph of , the -axis, the line and the line is rotated 360° about the -axis. This is shown in the following diagram.
\n![\"N16/5/MATME/SP2/ENG/TZ0/06\"](\"../assets/uploads/tinymce_asset/asset/3310/Schermafbeelding_2017-03-03_om_15.49.19.png\"/)
Use the model to find the volume of the barrel.
\nThe empty barrel is being filled with water. The volume of water in the barrel after minutes is given by . How long will it take for the barrel to be half-full?
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to substitute correct limits or the function into the formula involving
\n\n
eg
\n0.601091
\nvolume A2 N3
\n[3 marks]
\nattempt to equate half their volume to (M1)
\neg, graph
\n4.71104
\n4.71 (minutes) A2 N3
\n[3 marks]
\nLet .
\nThe function can also be expressed in the form .
\nFind the equation of the axis of symmetry of the graph of .
\n(i) Write down the value of .
\n(ii) Find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct approach (A1)
\neg
\n(must be an equation) A1 N2
\n[2 marks]
\n(i) A1 N1
\n(ii) METHOD 1
\nvalid attempt to find (M1)
\neg
\ncorrect substitution into their function (A1)
\neg
\nA1 N2
\nMETHOD 2
\nvalid attempt to complete the square (M1)
\neg
\ncorrect working (A1)
\neg
\nA1 N2
\n[4 marks]
\nThe following diagram shows triangle ABC, with , , and .
\n![\"N17/5/MATME/SP1/ENG/TZ0/04\"](\"../assets/uploads/tinymce_asset/asset/4144/Schermafbeelding_2018-02-11_om_09.17.57.png\"/)
Show that .
\nThe shape in the following diagram is formed by adding a semicircle with diameter [AC] to the triangle.
\n![\"N17/5/MATME/SP1/ENG/TZ0/04.b\"](\"../assets/uploads/tinymce_asset/asset/4151/Schermafbeelding_2018-02-11_om_10.50.00.png\"/)
Find the exact perimeter of this shape.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nevidence of choosing the cosine rule (M1)
\neg
\ncorrect substitution into RHS of cosine rule (A1)
\neg
\nevidence of correct value for (may be seen anywhere, including in cosine rule) A1
\neg
\ncorrect working clearly leading to answer A1
\neg
\nAG N0
\n\n
Note: Award no marks if the only working seen is or (or similar).
\n\n
[4 marks]
\ncorrect substitution for semicircle (A1)
\neg
\nvalid approach (seen anywhere) (M1)
\neg
\nA1 N2
\n[3 marks]
\nThe following diagram shows the chord [AB] in a circle of radius 8 cm, where .
\n![\"M17/5/MATME/SP2/ENG/TZ1/05\"](\"../assets/uploads/tinymce_asset/asset/3549/Schermafbeelding_2017-08-14_om_13.20.17.png\"/)
Find the area of the shaded segment.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nattempt to find the central angle or half central angle (M1)
\neg![\"M17/5/MATME/SP2/ENG/TZ1/05/M\"](\"../assets/uploads/tinymce_asset/asset/3551/Schermafbeelding_2017-08-14_om_13.40.22.png\"/)
, cosine rule, right triangle
correct working (A1)
\neg
\ncorrect angle (seen anywhere)
\neg (A1)
\ncorrect sector area
\neg (A1)
\narea of triangle (seen anywhere) (A1)
\neg
\nappropriate approach (seen anywhere) (M1)
\neg, their sector-their triangle
\n22.5269
\narea of shaded region A1 N4
\n\n
Note: Award M0A0A0A0A1 then M1A0 (if appropriate) for correct triangle area without any attempt to find an angle in triangle OAB.
\n\n
[7 marks]
\nUniversity students were surveyed and asked how many hours, , they worked each month. The results are shown in the following table.
\nUse the table to find the following values.
\nThe first five class intervals, indicated in the table, have been used to draw part of a cumulative frequency curve as shown.
\n.
\n.
\nOn the same grid, complete the cumulative frequency curve for these data.
\nUse the cumulative frequency curve to find an estimate for the number of students who worked at most 35 hours per month.
\n(A1) (C1)
\nNote: Award (A1) for each correct value.
\n[1 mark]
\n(A1) (C1)
\nNote: Award (A1) for each correct value.
\n[1 mark]
\n (A1)(A1) (C2)
Note: Award (A1)(ft) for their 3 correctly plotted points; award (A1)(ft) for completing diagram with a smooth curve through their points. The second (A1)(ft) can follow through from incorrect points, provided the gradient of the curve is never negative. Award (C2) for a completely correct smooth curve that goes through the correct points.
\n[2 marks]
\na straight vertical line drawn at 35 (accept 35 ± 1) (M1)
\n26 (students) (A1) (C2)
\nNote: Accept values between 25 and 27 inclusive.
\n[2 marks]
\nA group of 20 students travelled to a gymnastics tournament together. Their ages, in years, are given in the following table.
\n![\"N17/5/MATSD/SP1/ENG/TZ0/01\"](\"../assets/uploads/tinymce_asset/asset/4169/Schermafbeelding_2018-02-12_om_17.17.22.png\"/)
The lower quartile of the ages is 16 and the upper quartile is 18.5.
\nFor the students in this group write down the median age.
\nDraw a box-and-whisker diagram, for these students’ ages, on the following grid.
\n![\"N17/5/MATSD/SP1/ENG/TZ0/01.b\"](\"../assets/uploads/tinymce_asset/asset/4174/Schermafbeelding_2018-02-12_om_18.53.31.png\"/)
16.5 (A1) (C1)
\n[1 mark]
\n![\"N17/5/MATSD/SP1/ENG/TZ0/01.b/M\"](\"../assets/uploads/tinymce_asset/asset/4175/Schermafbeelding_2018-02-12_om_18.54.55.png\"/)
(A1)(A1)(A1)(ft) (C3)
\n
Note: Award (A1) for correct endpoints, (A1) for correct quartiles, (A1)(ft) for their median. Follow through from part (a)(ii), but only if median is between 16 and 18.5. If a horizontal line goes through the box, award at most (A1)(A1)(A0). Award at most (A0)(A1)(A1) if a ruler has not been used.
\n\n
[3 marks]
\nA transportation company owns 30 buses. The distance that each bus has travelled since being purchased by the company is recorded. The cumulative frequency curve for these data is shown.
\nIt is known that 8 buses travelled more than m kilometres.
\nFind the number of buses that travelled a distance between 15000 and 20000 kilometres.
\nUse the cumulative frequency curve to find the median distance.
\nUse the cumulative frequency curve to find the lower quartile.
\nUse the cumulative frequency curve to find the upper quartile.
\nHence write down the interquartile range.
\nWrite down the percentage of buses that travelled a distance greater than the upper quartile.
\nFind the number of buses that travelled a distance less than or equal to 12 000 km.
\nFind the value of m.
\nThe smallest distance travelled by one of the buses was 2500 km.
The longest distance travelled by one of the buses was 23 000 km.
On graph paper, draw a box-and-whisker diagram for these data. Use a scale of 2 cm to represent 5000 km.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
28 − 20 (A1)
\nNote: Award (A1) for 28 and 20 seen.
\n8 (A1)(G2)
\n[2 marks]
\n13500 (G2)
\nNote: Accept an answer in the range 13500 to 13750.
\n[2 marks]
\n10000 (G1)
\nNote: Accept an answer in the range 10000 to 10250.
\n[1 mark]
\n16000 (G1)
\nNote: Accept an answer in the range 16000 to 16250.
\n[1 mark]
\n6000 (A1)(ft)
\nNote: Follow through from their part (b)(ii) and (iii).
\n[1 mark]
\n25% (A1)
\n[1 mark]
\n11 (G1)
\n[1 mark]
\n30 − 8 OR 22 (M1)
\nNote: Award (M1) for subtracting 30 − 8 or 22 seen.
\n15750 (A1)(G2)
\nNote: Accept 15750 ± 250.
\n[2 marks]
\n(A1)(A1)(A1)(A1)
Note: Award (A1) for correct label and scale; accept “distance” or “km” for label.
\n(A1)(ft) for correct median,
(A1)(ft) for correct quartiles and box,
(A1) for endpoints at 2500 and 23 000 joined to box by straight lines.
Accept ±250 for the median, quartiles and endpoints.
Follow through from their part (b).
The final (A1) is not awarded if the line goes through the box.
[4 marks]
\nA health inspector analysed the amount of sugar in 500 different snacks prepared in various school cafeterias. The collected data are shown in the following box-and-whisker diagram.
\n
Amount of sugar per snack in grams
The health inspector visits two school cafeterias. She inspects the same number of meals at each cafeteria. The data is shown in the following box-and-whisker diagrams.
\nMeals prepared in the school cafeterias are required to have less than 10 grams of sugar.
\nState, giving a reason, which school cafeteria has more meals that do not meet the requirement.
\nCafeteria 2 (A1) (C1)
\n75 % > 50 % (do not meet the requirement) (R1) (C1)
\nOR
\n25 % < 50 % (meet the requirement) (R1) (C1)
\nNote: Do not award (A1)(R0). Award the (R1) for a correct comparison of percentages for both cafeterias, which may be in words. The percentage values or fractions must be seen. It is possible to award (A0)(R1).
\n[2 marks]
\nA sample of 120 oranges was tested for Vitamin C content. The cumulative frequency curve below represents the Vitamin C content, in milligrams, of these oranges.
\n![\"N16/5/MATSD/SP1/ENG/TZ0/02\"](\"../assets/uploads/tinymce_asset/asset/3318/Schermafbeelding_2017-03-06_om_09.11.24.png\"/)
The minimum level of Vitamin C content of an orange in the sample was 30.1 milligrams. The maximum level of Vitamin C content of an orange in the sample was 35.0 milligrams.
\nDraw a box-and-whisker diagram on the grid below to represent the Vitamin C content, in milligrams, for this sample.
\n![\"N16/5/MATSD/SP1/ENG/TZ0/02.b\"](\"../assets/uploads/tinymce_asset/asset/3323/Schermafbeelding_2017-03-06_om_12.47.06.png\"/)
![\"N16/5/MATSD/SP1/ENG/TZ0/02.b/M\"](\"../assets/uploads/tinymce_asset/asset/3324/Schermafbeelding_2017-03-06_om_12.50.10.png\"/)
Note: Award (A1)(ft) for correct median, (A1)(ft) for correct quartiles and box, (A1) for correct end points of whiskers and straight whiskers.
\nAward at most (A1)(A1)(A0) if a horizontal line goes right through the box or if the whiskers are not well aligned with the midpoint of the box.
\nFollow through from part (a).
\n\n
[3 marks]
\nEach month the number of days of rain in Cardiff is recorded.
The following data was collected over a period of 10 months.
11 13 8 11 8 7 8 14 x 15
\nFor these data the median number of days of rain per month is 10.
\nFind the value of x.
\nFind the standard deviation
\nFind the interquartile range.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\nNote: Award (M1) for correct substitution into median formula or for arranging all 9 values into ascending/descending order.
\n(A1) (C2)
\n[2 marks]
\n2.69 (2.69072…) (A2)(ft)
\nNote: Follow through from part (a).
\n\n
[2 marks]
\n13 − 8 (M1)
Note: Award (M1) for 13 and 8 seen.
= 5 (A1)(ft) (C4)
Note: Follow through from part (a).
[2 marks]
\nIn the month before their IB Diploma examinations, eight male students recorded the number of hours they spent on social media.
\nFor each student, the number of hours spent on social media () and the number of IB Diploma points obtained () are shown in the following table.
\n![\"N16/5/MATSD/SP2/ENG/TZ0/01\"](\"../assets/uploads/tinymce_asset/asset/3339/Schermafbeelding_2017-03-07_om_07.43.52.png\"/)
Use your graphic display calculator to find
\nTen female students also recorded the number of hours they spent on social media in the month before their IB Diploma examinations. Each of these female students spent between 3 and 30 hours on social media.
\nThe equation of the regression line y on x for these ten female students is
\n\n
An eleventh girl spent 34 hours on social media in the month before her IB Diploma examinations.
\nOn graph paper, draw a scatter diagram for these data. Use a scale of 2 cm to represent 5 hours on the -axis and 2 cm to represent 10 points on the -axis.
\n(i) , the mean number of hours spent on social media;
\n(ii) , the mean number of IB Diploma points.
\nPlot the point on your scatter diagram and label this point M.
\nWrite down the equation of the regression line on for these eight male students.
\nDraw the regression line, from part (e), on your scatter diagram.
\nUse the given equation of the regression line to estimate the number of IB Diploma points that this girl obtained.
\nWrite down a reason why this estimate is not reliable.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"N16/5/MATSD/SP2/ENG/TZ0/01.a/M\"](\"../assets/uploads/tinymce_asset/asset/3343/Schermafbeelding_2017-03-07_om_08.41.53.png\"/)
(A4)
\n
Notes: Award (A1) for correct scale and labelled axes.
\nAward (A3) for 7 or 8 points correctly plotted,
\n(A2) for 5 or 6 points correctly plotted,
\n(A1) for 3 or 4 points correctly plotted.
\nAward at most (A0)(A3) if axes reversed.
\nAccept and sufficient for labelling.
\nIf graph paper is not used, award (A0).
\nIf an inconsistent scale is used, award (A0). Candidates’ points should be read from this scale where possible and awarded accordingly.
\nA scale which is too small to be meaningful (ie mm instead of cm) earns (A0) for plotted points.
\n\n
[4 marks]
\n(i) (A1)
\n(ii) (A1)
\n[2 marks]
\ncorrectly plotted on graph (A1)(ft)
\nthis point labelled M (A1)
\n\n
Note: Follow through from parts (b)(i) and (b)(ii).
\nOnly accept M for labelling.
\n\n
[2 marks]
\n(A1)(A1)(G2)
\n\n
Notes: Award (A1) for and (A1) . Award a maximum of (A1)(A0) if answer is not an equation.
\n\n
[2 marks]
\nline on graph (A1)(ft)(A1)(ft)
\n\n
Notes: Award (A1)(ft) for straight line that passes through their M, (A1)(ft) for line (extrapolated if necessary) that passes through .
\nIf M is not plotted or labelled, follow through from part (e).
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution.
\n\n
19 (points) (A1)(G2)
\n[2 marks]
\nextrapolation (R1)
\nOR
\n34 hours is outside the given range of data (R1)
\n\n
Note: Do not accept ‘outlier’.
\n\n
[1 mark]
\nThe table below shows the distribution of test grades for 50 IB students at Greendale School.
\n![\"M17/5/MATSD/SP2/ENG/TZ1/05\"](\"../assets/uploads/tinymce_asset/asset/3601/Schermafbeelding_2017-08-16_om_11.25.22.png\"/)
A student is chosen at random from these 50 students.
\nA second student is chosen at random from these 50 students.
\nThe number of minutes that the 50 students spent preparing for the test was normally distributed with a mean of 105 minutes and a standard deviation of 20 minutes.
\nCalculate the mean test grade of the students;
\nCalculate the standard deviation.
\nFind the median test grade of the students.
\nFind the interquartile range.
\nFind the probability that this student scored a grade 5 or higher.
\nGiven that the first student chosen at random scored a grade 5 or higher, find the probability that both students scored a grade 6.
\nCalculate the probability that a student chosen at random spent at least 90 minutes preparing for the test.
\nCalculate the expected number of students that spent at least 90 minutes preparing for the test.
\n(M1)
\n\n
Note: Award (M1) for correct substitution into mean formula.
\n\n
(A1) (G2)
\n[2 marks]
\n(G1)
\n[1 mark]
\n5 (A1)
\n[1 mark]
\n(M1)
\n\n
Note: Award (M1) for 6 and 4 seen.
\n\n
(A1) (G2)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for seen.
\n\n
(A1) (G2)
\n[2 marks]
\n(M1)(M1)
\n\n
Note: Award (M1) for seen, (M1) for multiplying their first probability by .
\n\n
OR
\n\n
\n
Note: Award (M1) for seen, (M1) for dividing their first probability by .
\n\n
(A1)(ft) (G3)
\n\n
Note: Follow through from part (d).
\n\n
[3 marks]
\n(M1)
\nOR
\n![\"M17/5/MATSD/SP2/ENG/TZ1/05.f.i/M\"](\"../assets/uploads/tinymce_asset/asset/3604/Schermafbeelding_2017-08-16_om_15.40.38.png\"/)
(M1)
\n
Note: Award (M1) for a diagram showing the correct shaded region .
\n\n
(A1) (G2)
\n[2 marks]
\n(M1)
\n(A1)(ft) (G2)
\n\n
Note: Follow through from part (f)(i).
\n\n
[2 marks]
\nA survey was conducted on a group of people. The first question asked how many pets they each own. The results are summarized in the following table.
\nThe second question asked each member of the group to state their age and preferred pet. The data obtained is organized in the following table.
\nA test is carried out at the 10 % significance level.
\nWrite down the total number of people, from this group, who are pet owners.
\nWrite down the modal number of pets.
\nFor these data, write down the median number of pets.
\nFor these data, write down the lower quartile.
\nFor these data, write down the upper quartile.
\nWrite down the ratio of teenagers to non-teenagers in its simplest form.
\nState the null hypothesis.
\nState the alternative hypothesis.
\nWrite down the number of degrees of freedom for this test.
\nCalculate the expected number of teenagers that prefer cats.
\nState the conclusion for this test. Give a reason for your answer.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
140 (A1)
\n[1 mark]
\n1 (A1)
\n[1 mark]
\n2 (A1)
\n[1 mark]
\n1 (A1)
\n[1 mark]
\n3 (A1)
\n[1 mark]
\n17:15 OR (A1)
\nNote: Award (A0) for 85:75 or 1.13:1.
\n[1 mark]
\npreferred pet is independent of “whether or not the respondent was a teenager\" or \"age category” (A1)
\nNote: Accept there is no association between pet and age. Do not accept “not related” or “not correlated” or “influenced”.
\n[1 mark]
\npreferred pet is not independent of age (A1)(ft)
\nNote: Follow through from part (e)(i) i.e. award (A1)(ft) if their alternative hypothesis is the negation of their null hypothesis. Accept “associated” or “dependent”.
\n[1 mark]
\n3 (A1)
\n[1 mark]
\nOR (M1)
\n29.2 (29.2187…) (A1)(G2)
\n[2 marks]
\n0.208 > 0.1 (R1)
\naccept null hypothesis OR fail to reject null hypothesis (A1)(ft)
\nNote: Award (R1) for a correct comparison of their -value to the significance level, award (A1)(ft) for the correct result from that comparison. Accept “-value > 0.1” as part of the comparison but only if their -value is explicitly seen in part (h). Follow through from their answer to part (h). Do not award (R0)(A1).
\n[2 marks]
\nThe price per kilogram of tomatoes, in euro, sold in various markets in a city is found to be normally distributed with a mean of 3.22 and a standard deviation of 0.84.
\nOn the following diagram, shade the region representing the probability that the price of a kilogram of tomatoes, chosen at random, will be higher than 3.22 euro.
\nFind the price that is two standard deviations above the mean price.
\nFind the probability that the price of a kilogram of tomatoes, chosen at random, will be between 2.00 and 3.00 euro.
\nTo stimulate reasonable pricing, the city offers a free permit to the sellers whose price of a kilogram of tomatoes is in the lowest 20 %.
\nFind the highest price that a seller can charge and still receive a free permit.
\n (A1) (C1)
Note: Award (A1) for vertical line drawn at the mean (3.22 does not have to be seen) and correct region shaded.
\n[1 mark]
\n4.90 (A1) (C1)
\n[1 mark]
\n0.323 (0.323499…; 32.3 %) (A2) (C2)
\nNote: If final answer is incorrect, (M1)(A0) may be awarded for correct shaded area shown on a sketch, below, or for a correct probability statement “P(2 ≤ ≤ 3)” (accept other variables for or “price” and strict inequalities).
\n[2 marks]
\n2.51 (2.51303…) (A2) (C2)
\nNote: If final answer is incorrect, (M1)(A0) may be awarded for correct shaded area shown on a sketch, below, or for a correct probability statement “P( ≤ ) = 0.2” (accept other variables and strict inequalities).
\n[2 marks]
\nRosewood College has 120 students. The students can join the sports club () and the music club ().
\nFor a student chosen at random from these 120, the probability that they joined both clubs is and the probability that they joined the music club is.
\nThere are 20 students that did not join either club.
\nComplete the Venn diagram for these students.
\n![\"N17/5/MATSD/SP1/ENG/TZ0/07.a\"](\"../assets/uploads/tinymce_asset/asset/4181/Schermafbeelding_2018-02-13_om_08.15.35.png\"/)
One of the students who joined the sports club is chosen at random. Find the probability that this student joined both clubs.
\nDetermine whether the events and are independent.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n![\"N17/5/MATSD/SP1/ENG/TZ0/07.a/M\"](\"../assets/uploads/tinymce_asset/asset/4182/Schermafbeelding_2018-02-13_om_08.19.04.png\"/)
(A1)(A1) (C2)
\n
Note: Award (A1) for 30 in correct area, (A1) for 60 and 10 in the correct areas.
\n\n
[2 marks]
\n(A1)(ft)(A1)(ft) (C2)
\n\n
Note: Award (A1)(ft) for correct numerator of 30, (A1)(ft) for correct denominator of 90. Follow through from their Venn diagram.
\n\n
[2 marks]
\n(R1)
\n\n
Note: Award (R1) for multiplying their by .
\n\n
therefore the events are independent (A1)(ft) (C2)
\n\n
Note: Award (R1)(A1)(ft) for an answer which is consistent with their Venn diagram.
\nDo not award (R0)(A1)(ft).
\nDo not award final (A1) if is not calculated. Follow through from part (a).
\n\n
[2 marks]
\nThe Home Shine factory produces light bulbs, 7% of which are found to be defective.
\nFrancesco buys two light bulbs produced by Home Shine.
\nThe Bright Light factory also produces light bulbs. The probability that a light bulb produced by Bright Light is not defective is .
\nDeborah buys three light bulbs produced by Bright Light.
\nWrite down the probability that a light bulb produced by Home Shine is not defective.
\nFind the probability that both light bulbs are not defective.
\nFind the probability that at least one of Francesco’s light bulbs is defective.
\nWrite down an expression, in terms of , for the probability that at least one of Deborah’s three light bulbs is defective.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
0.93 (93%) (A1) (C1)
\n[1 mark]
\n(M1)
\n\n
Note: Award (M1) for squaring their answer to part (a).
\n\n
0.865 (0.8649; 86.5%) (A1)(ft) (C2)
\n\n
Notes: Follow through from part (a).
\nAccept .
\n\n
[2 marks]
\n(M1)
\n\n
Note: Follow through from their answer to part (b)(i).
\n\n
OR
\n(M1)
\n\n
Note: Follow through from part (a).
\n\n
0.135 (0.1351; 13.5%) (A1)(ft) (C2)
\n[2 marks]
\n(A1) (C1)
\n\n
Note: Accept or equivalent.
\n\n
[1 mark]
\nSara regularly flies from Geneva to London. She takes either a direct flight or a non-directflight that goes via Amsterdam.
\nIf she takes a direct flight, the probability that her baggage does not arrive in London is 0.01.
If she takes a non-direct flight the probability that her baggage arrives in London is 0.95.
The probability that she takes a non-direct flight is 0.2.
\n![\"M17/5/MATSD/SP1/ENG/TZ1/07\"](\"../assets/uploads/tinymce_asset/asset/3568/Schermafbeelding_2017-08-15_om_11.08.43.png\"/)
Complete the tree diagram.
\nFind the probability that Sara’s baggage arrives in London.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"M17/5/MATSD/SP1/ENG/TZ1/07.a/M\"](\"../assets/uploads/tinymce_asset/asset/3578/Schermafbeelding_2017-08-15_om_14.22.22.png\"/)
(A1)(A1)(A1) (C3)
\n
\n
Note: Award (A1) for each correct pair of probabilities.
\n\n
[3 marks]
\n(A1)(ft)(M1)
\n\n
Note: Award (A1)(ft) for two correct products of probabilities taken from their diagram, (M1) for the addition of their products.
\n\n
(A1)(ft) (C3)
\n\n
Note: Follow through from part (a).
\n\n
[3 marks]
\nA bag contains 5 red and 3 blue discs, all identical except for the colour. First, Priyanka takes a disc at random from the bag and then Jorgé takes a disc at random from the bag.
\nComplete the tree diagram.
\nFind the probability that Jorgé chooses a red disc.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(A1) (C3)
Note: Award (A1) for each correct pair of branches.
\n\n
[3 marks]
\n(A1)(ft)(M1)
\nNote: Award (A1)(ft) for their two correct products from their tree diagram. Follow through from part (a), award (M1) for adding their two products. Award (M0) if additional products or terms are added.
\n\n
= (A1)(ft) (C3)
\nNote: Follow through from their tree diagram, only if probabilities are [0,1].
\n\n
[3 marks]
\nThe marks achieved by students taking a college entrance test follow a normal distribution with mean 300 and standard deviation 100.
\nIn this test, 10 % of the students achieved a mark greater than k.
\nMarron College accepts only those students who achieve a mark of at least 450 on the test.
\nFind the value of k.
\nFind the probability that a randomly chosen student will be accepted by Marron College.
\nGiven that Naomi attends Marron College, find the probability that she achieved a mark of at least 500 on the test.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
Note: Award (M1) for diagram that shows the correct shaded area and percentage, k has to be greater than the mean.
\nOR
\nAward (M1) for P(mark > k) = 0.1 or P(mark ≤ k) = 0.9 seen.
\n\n
428 (428.155…) (A1) (C2)
\n\n
[2 marks]
\n (M1)
Note: Award (M1) for diagram that shows the correct shaded area and the value 450 labelled to the right of the mean.
\nOR
\nAward (M1) for P(mark ≥ 450) seen.
\n\n
0.0668 (0.0668072…, 6.68 %, 6.68072… %) (A1) (C2)
\n\n
[2 marks]
\n(M1)
\nNote: Award (M1) for 0.0228 (0.0227500…) seen. Accept 1 − 0.97725.
\n\n
= 0.341 (0.340532…, 34.1 %, 34.0532…%) (A1)(ft) (C2)
\nNote: Follow through from part (b), provided answer is between zero and 1.
\n\n
[2 marks]
\nOn a work day, the probability that Mr Van Winkel wakes up early is .
\nIf he wakes up early, the probability that he is on time for work is .
\nIf he wakes up late, the probability that he is on time for work is .
\nThe probability that Mr Van Winkel arrives on time for work is .
\nComplete the tree diagram below.
\n![\"N16/5/MATSD/SP1/ENG/TZ0/12.a\"](\"../assets/uploads/tinymce_asset/asset/3336/Schermafbeelding_2017-03-07_om_06.20.32.png\"/)
Find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"N16/5/MATSD/SP1/ENG/TZ0/12.a/M\"](\"../assets/uploads/tinymce_asset/asset/3337/Schermafbeelding_2017-03-07_om_06.24.40.png\"/)
(A1)(A1) (C2)
\n
Note: Award (A1) for each correct pair of probabilities.
\n\n
[2 marks]
\n(A1)(ft)(M1)(M1)
\n\n
Note: Award (A1)(ft) for two correct products from part (a), (M1) for adding their products, (M1) for equating the sum of any two probabilities to .
\n\n
(A1)(ft) (C4)
\n\n
Note: Award the final (A1)(ft) only if . Follow through from part (a).
\n\n
[4 marks]
\nMalthouse school opens at 08:00 every morning.
\nThe daily arrival times of the 500 students at Malthouse school follow a normal distribution. The mean arrival time is 52 minutes after the school opens and the standard deviation is 5 minutes.
\nFind the probability that a student, chosen at random arrives at least 60 minutes after the school opens.
\nFind the probability that a student, chosen at random arrives between 45 minutes and 55 minutes after the school opens.
\nA second school, Mulberry Park, also opens at 08:00 every morning. The arrival times of the students at this school follows exactly the same distribution as Malthouse school.
\nGiven that, on one morning, 15 students arrive at least 60 minutes after the school opens, estimate the number of students at Mulberry Park school.
\n0.0548 (0.054799…, 5.48%) (A2) (C2)
\n[2 marks]
\n0.645 (0.6449900…, 64.5%) (A2) (C2)
\n[2 marks]
\n(M1)
\nNote: Award (M1) for dividing 15 by their part (a)(i).
\nAccept an equation of the form 15 = x × 0.0548 for (M1).
\n274 (273.722…) (A1)(ft) (C2)
\nNote: Follow through from part (a)(i). Accept 273.
\n[2 marks]
\nApplicants for a job had to complete a mathematics test. The time they took to complete the test is normally distributed with a mean of 53 minutes and a standard deviation of 16.3. One of the applicants is chosen at random.
\nFor 11% of the applicants it took longer than minutes to complete the test.
\nThere were 400 applicants for the job.
\nFind the probability that this applicant took at least 40 minutes to complete the test.
\nFind the value of .
\nEstimate the number of applicants who completed the test in less than 25 minutes.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
0.787 (0.787433…, 78.7%) (M1)(A1) (C2)
\n\n
Note: Award (M1) for a correct probability statement, , or a correctly shaded normal distribution graph.
\n\n
![\"N17/5/MATSD/SP1/ENG/TZ0/13.a/M\"](\"../assets/uploads/tinymce_asset/asset/4183/Schermafbeelding_2018-02-13_om_13.12.51.png\"/)
[2 marks]
\n73.0 (minutes) (72.9924…) (M1)(A1) (C2)
\n\n
Note: Award (M1) for a correct probability statement, , or a correctly shaded normal distribution graph.
\n\n
![\"N17/5/MATSD/SP1/ENG/TZ0/13.b/M\"](\"../assets/uploads/tinymce_asset/asset/4184/Schermafbeelding_2018-02-13_om_13.16.45.png\"/)
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for multiplying a probability by 400. Do not award (M1) for .
\nUse of a lower bound less than zero gives a probability of 0.0429172….
\n(A1) (C2)
\n\n
Notes: Accept a final answer of 17. Do not accept a final answer of 18. Accept a non-integer final answer either 16.9 (16.9373…) from use of lower bound zero or 17.2 (17.1669…) from use of the default lower bound of .
\n\n
[2 marks]
\nThe mass of a certain type of Chilean corncob follows a normal distribution with a mean of 400 grams and a standard deviation of 50 grams.
\nA farmer labels one of these corncobs as premium if its mass is greater than grams. 25% of these corncobs are labelled as premium.
\nWrite down the probability that the mass of one of these corncobs is greater than 400 grams.
\nFind the value of .
\nEstimate the interquartile range of the distribution.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1) (C1)
\n[1 mark]
\nOR (M1)
\n\n
Note: Award (M1) for a sketch of approximate normal curve with a vertical line drawn to the right of the mean with the area to the right of this line shaded.
\n\n
(A1) (C2)
\n[2 marks]
\n(A1)(ft)(M1)
\n\n
Note: Award (A1)(ft) for seen, award (M1) for multiplying their 33.7244… by 2. Follow through from their answer to part (b).
\n\n
OR
\n(A1)(ft)(M1)
\n\n
Note: Award (A1)(ft) for their seen, (M1) for difference between their answer to (b) and their 366.
\n\n
OR
\n![\"M17/5/MATSD/SP1/ENG/TZ2/11.c/M\"](\"../assets/uploads/tinymce_asset/asset/3593/Schermafbeelding_2017-08-16_om_07.56.55.png\"/)
(A1)(ft)(M1)
\n
Note: Award (A1)(ft) for their seen. Award (M1) for correct symmetrical region indicated on labelled normal curve.
\n\n
67.4 (g) (A1)(ft) (C3)
\n\n
Note: Accept an answer of 68 from use of rounded values. Follow through from part (b).
\n\n
[3 marks]
\nA manufacturer produces 1500 boxes of breakfast cereal every day.
\nThe weights of these boxes are normally distributed with a mean of 502 grams and a standard deviation of 2 grams.
\nAll boxes of cereal with a weight between 497.5 grams and 505 grams are sold. The manufacturer’s income from the sale of each box of cereal is $2.00.
\nThe manufacturer recycles any box of cereal with a weight not between 497.5 grams and 505 grams. The manufacturer’s recycling cost is $0.16 per box.
\nA different manufacturer produces boxes of cereal with weights that are normally distributed with a mean of 350 grams and a standard deviation of 1.8 grams.
\nThis manufacturer sells all boxes of cereal that are above a minimum weight, .
\nThey sell 97% of the cereal boxes produced.
\nDraw a diagram that shows this information.
\n(i) Find the probability that a box of cereal, chosen at random, is sold.
\n(ii) Calculate the manufacturer’s expected daily income from these sales.
\nCalculate the manufacturer’s expected daily recycling cost.
\nCalculate the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"N16/5/MATSD/SP2/ENG/TZ0/04.a/M\"](\"../assets/uploads/tinymce_asset/asset/3346/Schermafbeelding_2017-03-08_om_11.51.39.png\"/)
(A1)(A1)
\n\n
Notes: Award (A1) for bell shape with mean of 502.
\nAward (A1) for an indication of standard deviation eg 500 and 504.
\n\n
[2 marks]
\n(i) (G2)
\n\n
Note: Award (M1) for a diagram showing the correct shaded region.
\n\n
(ii) (M1)
\n(A1)(ft)(G2)
\n\n
Note: Follow through from their answer to part (b)(i).
\n\n
[4 marks]
\n(M1)
\n\n
Notes: Award (A1) for .
\n\n
OR
\n(M1)
\n\n
Notes: Award (M1) for .
\n\n
(A1)(ft)(G2)
\n[2 marks]
\n(G3)
\n\n
Notes: Award (G2) for an answer that rounds to 346.
\nAward (G1) for seen without working (for finding the top 3%).
\n\n
[3 marks]
\nThe weight, W, of basketball players in a tournament is found to be normally distributed with a mean of 65 kg and a standard deviation of 5 kg.
\nThe probability that a basketball player has a weight that is within 1.5 standard deviations of the mean is q.
\nA basketball coach observed 60 of her players to determine whether their performance and their weight were independent of each other. Her observations were recorded as shown in the table.
\nShe decided to conduct a χ 2 test for independence at the 5% significance level.
\nFind the probability that a basketball player has a weight that is less than 61 kg.
\nIn a training session there are 40 basketball players.
\nFind the expected number of players with a weight less than 61 kg in this training session.
\nSketch a normal curve to represent this probability.
\nFind the value of q.
\nGiven that P(W > k) = 0.225 , find the value of k.
\nFor this test state the null hypothesis.
\nFor this test find the p-value.
\nState a conclusion for this test. Justify your answer.
\nP(W < 61) (M1)
\nNote: Award (M1) for correct probability statement.
\nOR
\n (M1)
Note: Award (M1) for correct region labelled and shaded on diagram.
\n= 0.212 (0.21185…, 21.2%) (A1)(G2)
\n[2 marks]
\n40 × 0.21185… (M1)
\nNote: Award (M1) for product of 40 and their 0.212.
\n= 8.47 (8.47421...) (A1)(ft)(G2)
\nNote: Follow through from their part (a)(i) provided their answer to part (a)(i) is less than 1.
\n[2 marks]
\n\n
(A1)(M1)
Note: Award (A1) for two correctly labelled vertical lines in approximately correct positions. The values 57.5 and 72.5, or μ − 1.5σ and μ + 1.5σ are acceptable labels. Award (M1) for correctly shaded region marked by their two vertical lines.
\n[2 marks]
\n0.866 (0.86638…, 86.6%) (A1)(ft)
\nNote: Follow through from their part (b)(i) shaded region if their values are clear.
\n[1 mark]
\nP(W < k) = 0.775 (M1)
\nOR
\n (M1)
Note: Award (A1) for correct region labelled and shaded on diagram.
\n(k =) 68.8 (68.7770…) (A1)(G2)
\n[2 marks]
\n(H0:) performance (of players) and (their) weight are independent. (A1)
\nNote: Accept “there is no association between performance (of players) and (their) weight”. Do not accept \"not related\" or \"not correlated\" or \"not influenced\".
\n[1 mark]
\n0.287 (0.287436…) (G2)
\n[2 marks]
\naccept/ do not reject null hypothesis/H0 (A1)(ft)
\nOR
\nperformance (of players) and (their) weight are independent. (A1)(ft)
\n0.287 > 0.05 (R1)(ft)
\nNote: Accept p-value>significance level provided their p-value is seen in b(ii). Accept 28.7% > 5%. Do not award (A1)(R0). Follow through from part (d).
\n[2 marks]
\nConsider the following graphs of normal distributions.
\nAt an airport, the weights of suitcases (in kg) were measured. The weights are normally distributed with a mean of 20 kg and standard deviation of 3.5 kg.
\nIn the following table, write down the letter of the corresponding graph next to the given mean and standard deviation.
\nFind the probability that a suitcase weighs less than 15 kg.
\nAny suitcase that weighs more than kg is identified as excess baggage.
19.6 % of the suitcases at this airport are identified as excess baggage.
Find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1) (C2)
Note: Award (A1) for each correct entry.
\n[2 marks]
\n (M1)
Note: Award (M1) for sketch with 15 labelled and left tail shaded OR for a correct probability statement, P(X < 15).
\n0.0766 (0.0765637…, 7.66%) (A1) (C2)
\n[2 marks]
\n (M1)
Note: Award (M1) for a sketch showing correctly shaded region to the right of the mean with 19.6% labelled (accept shading of the complement with 80.4% labelled) OR for a correct probability statement, P(X > ) = 0.196 or P(X ≤ ) = 0.804.
\n23.0 (kg) (22.9959… (kg)) (A1) (C2)
\n[2 marks]
\nConsider the quadratic function .
\nThe equation of the line of symmetry of the graph .
\nThe graph intersects the x-axis at the point (−2 , 0).
\nUsing only this information, write down an equation in terms of a and b.
\nUsing this information, write down a second equation in terms of a and b.
\nHence find the value of a and of b.
\nThe graph intersects the x-axis at a second point, P.
\nFind the x-coordinate of P.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(or equivalent) (A1) (C1)
\nNote: Award (A1) for or for or for .
\n[1 mark]
\n(or equivalent) (A1) (C1)
\nNote: Award (A1) for seen.
\nAward (A0) for .
\n[1 mark]
\na = −2, b = 7 (A1)(ft)(A1)(ft) (C2)
\nNote: Follow through from parts (a) and (b).
Accept answers(s) embedded as a coordinate pair.
[2 marks]
\n−2x2 + 7x + 22 = 0 (M1)
\nNote: Award (M1) for correct substitution of a and b into equation and setting to zero. Follow through from part (c).
\n(x =) 5.5 (A1)(ft) (C2)
\nNote: Follow through from parts (a) and (b).
\nOR
\nx-coordinate = 1.75 + (1.75 − (−2)) (M1)
\nNote: Award (M1) for correct use of axis of symmetry and given intercept.
\n(x =) 5.5 (A1) (C2)
\n[2 marks]
\nConsider the function .
\nThe tangent to the graph of at is parallel to the line .
\nFind .
\nShow that .
\nFind the equation of the tangent to the graph of at . Give your answer in the form .
\nUse your answer to part (a) and the value of , to find the -coordinates of the stationary points of the graph of .
\nFind .
\nHence justify that is decreasing at .
\nFind the -coordinate of the local minimum.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(A1)
\n\n
Note: Award (A1) for , (A1) for and (A1) for . Award at most (A1)(A1)(A0) if additional terms are seen.
\n\n
[3 marks]
\n(M1)(M1)
\n\n
Note: Award (M1) for equating their derivative to 21. Award (M1) for substituting 2 into their derivative. The second (M1) should only be awarded if correct working leads to the final answer of .
\nSubstituting in the known value, , invalidates the process; award (M0)(M0).
\n\n
(AG)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for substituting 2 into .
\n\n
(M1)
\n\n
Note: Award (M1) for correct substitution of 21, 2 and their 7 into gradient intercept form.
\n\n
OR
\n\n
(M1)
\n\n
Note: Award (M1) for correct substitution of 21, 2 and their 7 into gradient point form.
\n\n
(A1) (G2)
\n[3 marks]
\n(or equivalent) (M1)
\n\n
Note: Award (M1) for equating their part (a) (with substituted) to zero.
\n\n
(A1)(ft)(A1)(ft)
\n\n
Note: Follow through from part (a).
\n\n
[3 marks]
\n(M1)
\n\n
Note: Award (M1) for substituting into their derivative, with substituted. Follow through from part (a).
\n\n
(A1)(ft) (G2)
\n[2 marks]
\n(therefore is decreasing when ) (R1)
\n[1 marks]
\n(M1)
\n\n
Note: Award (M1) for correctly substituting 6 and their 1 into .
\n\n
(A1)(ft) (G2)
\n\n
Note: Award, at most, (M1)(A0) or (G1) if answer is given as a coordinate pair. Follow through from part (c).
\n\n
[2 marks]
\nA scientist measures the concentration of dissolved oxygen, in milligrams per litre (y) , in a river. She takes 10 readings at different temperatures, measured in degrees Celsius (x).
\nThe results are shown in the table.
\nIt is believed that the concentration of dissolved oxygen in the river varies linearly with the temperature.
\nFor these data, find Pearson’s product-moment correlation coefficient, r.
\nFor these data, find the equation of the regression line y on x.
\nUsing the equation of the regression line, estimate the concentration of dissolved oxygen in the river when the temperature is 18 °C.
\n−0.974 (−0.973745…) (A2)
\nNote: Award (A1) for an answer of 0.974 (minus sign omitted). Award (A1) for an answer of −0.973 (incorrect rounding).
\n[2 marks]
\ny = −0.365x + 17.9 (y = −0.365032…x + 17.9418…) (A1)(A1) (C4)
\nNote: Award (A1) for −0.365x, (A1) for 17.9. Award at most (A1)(A0) if not an equation or if the values are reversed (eg y = 17.9x −0.365).
\n[2 marks]
\ny = −0.365032… × 18 + 17.9418… (M1)
\nNote: Award (M1) for correctly substituting 18 into their part (a)(ii).
\n= 11.4 (11.3712…) (A1)(ft) (C2)
\nNote: Follow through from part (a)(ii).
\n[2 marks]
\nA quadratic function is given by . The points and lie on the graph of .
\nThe -coordinate of the minimum of the graph is 3.
\nFind the value of and of .
\n\n
or equivalent
\nor equivalent
\nor equivalent (M1)
\n\n
Note: Award (M1) for two of the above equations.
\n\n
(A1)(ft)
\n(A1)(ft) (C3)
\n\n
Note: Award at most (M1)(A1)(ft)(A0) if the answers are reversed.
\nFollow through from parts (a) and (b).
\n\n
[3 marks]
\nThe graph of a quadratic function has -intercept 10 and one of its -intercepts is 1.
\nThe -coordinate of the vertex of the graph is 3.
\nThe equation of the quadratic function is in the form .
\nWrite down the value of .
\nFind the value of and of .
\nWrite down the second -intercept of the function.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
10 (A1) (C1)
\n\n
Note: Accept .
\n\n
[1 mark]
\n\n
\n
\n
(M1)(M1)
\n\n
Note: Award (M1) for each of the above equations, provided they are not equivalent, up to a maximum of (M1)(M1). Accept equations that substitute their 10 for .
\n\n
OR
\nsketch graph showing given information: intercepts and and line (M1)
\n(M1)
\n\n
Note: Award (M1) for seen.
\n\n
(A1)(ft)
\n(A1)(ft) (C4)
\n\n
Note: Follow through from part (a).
\nIf it is not clear which is and which is award at most (A0)(A1)(ft).
\n\n
[4 marks]
\n5 (A1) (C1)
\n[1 mark]
\nA factory produces shirts. The cost, C, in Fijian dollars (FJD), of producing x shirts can be modelled by
\nC(x) = (x − 75)2 + 100.
\nThe cost of production should not exceed 500 FJD. To do this the factory needs to produce at least 55 shirts and at most s shirts.
\nFind the cost of producing 70 shirts.
\nFind the value of s.
\nFind the number of shirts produced when the cost of production is lowest.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(70 − 75)2 + 100 (M1)
\nNote: Award (M1) for substituting in x = 70.
\n125 (A1) (C2)
\n[2 marks]
\n(s − 75)2 + 100 = 500 (M1)
\nNote: Award (M1) for equating C(x) to 500. Accept an inequality instead of =.
\nOR
\n (M1)
\n
Note: Award (M1) for sketching correct graph(s).
\n(s =) 95 (A1) (C2)
\n[2 marks]
\n (M1)
Note: Award (M1) for an attempt at finding the minimum point using graph.
\nOR
\n(M1)
\nNote: Award (M1) for attempting to find the mid-point between their part (b) and 55.
\nOR
\n(C'(x) =) 2x − 150 = 0 (M1)
\nNote: Award (M1) for an attempt at differentiation that is correctly equated to zero.
\n75 (A1) (C2)
\n[2 marks]
\nConsider the following graphs of quadratic functions.
\n![\"M17/5/MATSD/SP1/ENG/TZ2/15\"](\"../assets/uploads/tinymce_asset/asset/3591/Schermafbeelding_2017-08-16_om_06.31.16.png\"/)
The equation of each of the quadratic functions can be written in the form , where .
\nEach of the sets of conditions for the constants , and , in the table below, corresponds to one of the graphs above.
\nWrite down the number of the corresponding graph next to each set of conditions.
\n ![\"M17/5/MATSD/SP1/ENG/TZ2/15_02\"](\"../assets/uploads/tinymce_asset/asset/3592/Schermafbeelding_2017-08-16_om_06.39.22.png\"/)
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"M17/5/MATSD/SP1/ENG/TZ2/15/M\"](\"../assets/uploads/tinymce_asset/asset/3595/Schermafbeelding_2017-08-16_om_08.45.18.png\"/)
(A1)(A1)(A1)(A1)(A1)(A1) (C6)
\n
Note: Award (A1) for each correct entry.
\n\n
[6 marks]
\nConsider the function .
\nFind f'(x)
\nFind the gradient of the graph of f at .
\nFind the x-coordinate of the point at which the normal to the graph of f has gradient .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
x3 (A1) (C1)
\nNote: Award (A0) for and not simplified to x3.
\n[1 mark]
\n(M1)
\nNote: Award (M1) for correct substitution of into their derivative.
\n(−0.125) (A1)(ft) (C2)
\nNote: Follow through from their part (a).
\n[2 marks]
\nx3 = 8 (A1)(M1)
\nNote: Award (A1) for 8 seen maybe seen as part of an equation y = 8x + c, (M1) for equating their derivative to 8.
\n(x =) 2 (A1) (C3)
\nNote: Do not accept (2, 4).
\n[3 marks]
\nMaria owns a cheese factory. The amount of cheese, in kilograms, Maria sells in one week, , is given by
\n,
\nwhere is the price of a kilogram of cheese in euros (EUR).
\nMaria earns for each kilogram of cheese sold.
\nTo calculate her weekly profit , in EUR, Maria multiplies the amount of cheese she sells by the amount she earns per kilogram.
\nWrite down how many kilograms of cheese Maria sells in one week if the price of a kilogram of cheese is 8 EUR.
\nFind how much Maria earns in one week, from selling cheese, if the price of a kilogram of cheese is 8 EUR.
\nWrite down an expression for in terms of .
\nFind the price, , that will give Maria the highest weekly profit.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
522 (kg) (A1) (C1)
\n[1 mark]
\nor equivalent (M1)
\n\n
Note: Award (M1) for multiplying their answer to part (a) by .
\n\n
626 (EUR) (626.40) (A1)(ft) (C2)
\n\n
Note: Follow through from part (a).
\n\n
[2 marks]
\n(A1)
\nOR
\n(A1) (C1)
\n[1 mark]
\nsketch of with some indication of the maximum (M1)
\nOR
\n(M1)
\n\n
Note: Award (M1) for equating the correct derivative of their part (c) to zero.
\n\n
OR
\n(M1)
\n\n
Note: Award (M1) for correct substitution into the formula for axis of symmetry.
\n\n
(A1)(ft) (C2)
\n\n
Note: Follow through from their part (c), if the value of is such that .
\n\n
[2 marks]
\nIn an experiment, a number of fruit flies are placed in a container. The population of fruit flies, P , increases and can be modelled by the function
\n\n
where t is the number of days since the fruit flies were placed in the container.
\nFind the number of fruit flies which were placed in the container.
\nFind the number of fruit flies that are in the container after 6 days.
\nThe maximum capacity of the container is 8000 fruit flies.
\nFind the number of days until the container reaches its maximum capacity.
\n(M1)
\nNote: Award (M1) for substituting zero into the equation.
\n= 12 (A1) (C2)
\n[2 marks]
\n(M1)
\nNote: Award (M1) for substituting 6 into the equation.
\n320 (A1) (C2)
\nNote: Accept an answer of 319.756… or 319.
\n[2 marks]
\n(M1)
\nNote: Award (M1) for equating equation to 8000.
Award (M1) for a sketch of P(t) intersecting with the straight line y = 8000.
= 11.9 (11.8848…) (A1) (C2)
\nNote: Accept an answer of 11 or 12.
\n[2 marks]
\n\n
A café serves sandwiches and cakes. Each customer will choose one of the following three options; buy only a sandwich, buy only a cake or buy both a sandwich and a cake.
\nThe probability that a customer buys a sandwich is 0.72 and the probability that a customer buys a cake is 0.45.
\nFind the probability that a customer chosen at random will buy
\nOn a typical day 200 customers come to the café.
\nIt is known that 46 % of the customers who come to the café are male, and that 80 % of these buy a sandwich.
\nboth a sandwich and a cake.
\nonly a sandwich.
\nFind the expected number of cakes sold on a typical day.
\nFind the probability that more than 100 cakes will be sold on a typical day.
\nA customer is selected at random. Find the probability that the customer is male and buys a sandwich.
\nA female customer is selected at random. Find the probability that she buys a sandwich.
\nuse of formula or Venn diagram (M1)
\n0.72 + 0.45 − 1 (A1)
\n= 0.17 A1
\n[3 marks]
\n0.72 − 0.17 = 0.55 A1
\n[1 mark]
\n200 × 0.45 = 90 A1
\n[1 mark]
\nlet X be the number of customers who order cake
\nX ~ B(200,0.45) (M1)
\nP(X > 100) = P(X ≥ 101)(= 1 − P(X ≤ 100)) (M1)
\n= 0.0681 A1
\n\n
[3 marks]
\n0.46 × 0.8 = 0.368 A1
\n[1 mark]
\nMETHOD 1
\nM1A1A1
\nNote: Award M1 for an appropriate tree diagram. Award M1 for LHS, M1 for RHS.
\nA1
\n\n
METHOD 2
\n(M1)
\nA1A1
\nNote: Award A1 for numerator, A1 for denominator.
\nA1
\n\n
[4 marks]
\nThe following function models the growth of a bacteria population in an experiment,
\nP(t) = A × 2t, t ≥ 0
\nwhere A is a constant and t is the time, in hours, since the experiment began.
\nFour hours after the experiment began, the bacteria population is 6400.
\nFind the value of A.
\nInterpret what A represents in this context.
\nFind the time since the experiment began for the bacteria population to be equal to 40A.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
6400 = A × 24 (M1)
\nNote: Award (M1) for correct substitution of 4 and 6400 in equation.
\n(A =) 400 (A1) (C2)
\n[2 marks]
\nthe initial population OR the population at the start of experiment (A1) (C1)
\n[1 mark]
\n40A = A × 2t OR 40 × 400 = 400 × 2t (M1)
\nNote: Award (M1) for correct substitution into equation. Follow through with their A from part (a).
\n40 = 2t (M1)
\nNote: Award (M1) for simplifying.
\n5.32 (5.32192…) (hours) OR 5 hours 19.3 (19.3156…) minutes (A1) (C3)
\n[3 marks]
\n\n
Sejah placed a baking tin, that contained cake mix, in a preheated oven in order to bake a cake. The temperature in the centre of the cake mix, , in degrees Celsius (°C) is given by
\n\n
where is the time, in minutes, since the baking tin was placed in the oven. The graph of is shown in the following diagram.
\n![\"N17/5/MATSD/SP1/ENG/TZ0/12\"](\"../assets/uploads/tinymce_asset/asset/4173/Schermafbeelding_2018-02-12_om_18.27.39.png\"/)
The temperature in the centre of the cake mix was 18 °C when placed in the oven.
\nThe baking tin is removed from the oven 15 minutes after the temperature in the centre of the cake mix has reached 130 °C.
\nWrite down what the value of 150 represents in the context of the question.
\nFind the value of .
\nFind the total time that the baking tin is in the oven.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
the temperature in the oven (A1)
\nOR
\nthe maximum possible temperature of the cake mix (A1) (C1)
\n\n
Note: Award (A0) for “the maximum temperature”.
\n\n
[1 mark]
\n(M1)
\n\n
Note: Award (M1) for correct substitution of 18 and 0. Substitution of 0 can be implied.
\n\n
(A1) (C2)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for substituting their a and equating to 130. Accept an inequality.
\nAward (M1) for a sketch of the horizontal line on the graph.
\n\n
(A1)(ft)
\n\n
Note: Follow through from part (b).
\n\n
34.8 (minutes) (34.7992…, 34 minutes 48 seconds) (A1)(ft) (C3)
\n\n
Note: Award the final (A1) for adding 15 minutes to their value.
\nIn part (c), award (C2) for a final answer of 19.8 with no working.
\n\n
[3 marks]
\nThe mean number of squirrels in a certain area is known to be 3.2 squirrels per hectare of woodland. Within this area, there is a 56 hectare woodland nature reserve. It is known that there are currently at least 168 squirrels in this reserve.
\nAssuming the population of squirrels follow a Poisson distribution, calculate the probability that there are more than 190 squirrels in the reserve.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
X is number of squirrels in reserve
X ∼ Po(179.2) A1
Note: Award A1 if 179.2 or 56 × 3.2 seen or implicit in future calculations.
\nrecognising conditional probability M1
\nP(X > 190 | X ≥ 168)
\n(A1)(A1)
\n= 0.245 A1
\n[5 marks]
\nThe continuous random variable X has a probability density function given by
\n.
\nFind the value of .
\nBy considering the graph of f write down the mean of ;
\nBy considering the graph of f write down the median of ;
\nBy considering the graph of f write down the mode of .
\nShow that .
\nHence state the interquartile range of .
\nCalculate .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to equate integral to 1 (may appear later) M1
\n\n
correct integral A1
\n\n
substituting limits M1
\n\n
A1
\n[4 marks]
\nmean A1
\n\n
Note: Award A1A0A0 for three equal answers in .
\n\n
[1 mark]
\nmedian A1
\n\n
Note: Award A1A0A0 for three equal answers in .
\n\n
[1 mark]
\nmode A1
\n\n
Note: Award A1A0A0 for three equal answers in .
\n\n
[1 mark]
\nM1
\nA1
\n\n
Note: Accept without the at this stage if it is added later.
\n\n
M1
\nAG
\n[4 marks]
\nfrom (c)(i) (A1)
\nas the graph is symmetrical about the middle value (A1)
\nso interquartile range is
\n\n
A1
\n[3 marks]
\n\n
(M1)
\nA1
\n[2 marks]
\nTwo unbiased tetrahedral (four-sided) dice with faces labelled 1, 2, 3, 4 are thrown and the scores recorded. Let the random variable T be the maximum of these two scores.
\nThe probability distribution of T is given in the following table.
\nFind the value of a and the value of b.
\nFind the expected value of T.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
and (M1)A1A1
\n[3 marks]
\nNote: Award M1 for consideration of the possible outcomes when rolling the two dice.
\n(M1)A1
\nNote: Allow follow through from part (a) even if probabilities do not add up to 1.
\n[2 marks]
\nThe continuous random variable X has probability density function given by
\n\n
\n
Show that .
\nFind .
\nGiven that , and that 0.25 < s < 0.4 , find the value of s.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
M1
\nNote: Award the M1 for the total integral equalling 1, or equivalent.
\n(M1)A1
\nAG
\n[3 marks]
\nEITHER
\n(M1)(A1)
\nA1
\nOR
\n(M1)
\nso (M1)A1
\n[3 marks]
\nM1A1
\n\n
(A1)
\nA1
\n\n
\n
equating
\n(A1)
\nattempt to solve for s (M1)
\ns = 0.274 A1
\n[7 marks]
\nA continuous random variable has probability density function given by
\n\n
It is given that .
\nEight independent observations of are now taken and the random variable is the number of observations such that .
\nShow that and .
\nFind .
\nFind .
\nFind the median of .
\nFind .
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1
\nM1A1
\n\n
Note: could be seen/used in place of either of the above equations.
\n\n
evidence of an attempt to solve simultaneously (or check given a,b values are consistent) M1
\nAG
\n[5 marks]
\n(M1)
\nA1
\n[2 marks]
\n(M1)
\nA1
\n[2 marks]
\n(M1)
\n(A1)
\nA1
\n[3 marks]
\n(M1)
\nA1
\n[2 marks]
\nA1
\n[1 mark]
\nSteffi the stray cat often visits Will’s house in search of food. Let be the discrete random variable “the number of times per day that Steffi visits Will’s house”.
\nThe random variable can be modelled by a Poisson distribution with mean 2.1.
\nLet Y be the discrete random variable “the number of times per day that Steffi is fed at Will’s house”. Steffi is only fed on the first four occasions that she visits each day.
\nFind the probability that on a randomly selected day, Steffi does not visit Will’s house.
\nCopy and complete the probability distribution table for Y.
\nHence find the expected number of times per day that Steffi is fed at Will’s house.
\nIn any given year of 365 days, the probability that Steffi does not visit Will for at most days in total is 0.5 (to one decimal place). Find the value of .
\nShow that the expected number of occasions per year on which Steffi visits Will’s house and is not fed is at least 30.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
(M1)A1
\n[2 marks]
\n A1A1A1A1
Note: Award A1 for each correct probability for Y = 1, 2, 3, 4. Accept 0.162 for P(Y = 4).
\n[4 marks]
\n(M1)
\n(A1)
\nA1
\n[3 marks]
\nlet be the no of days per year that Steffi does not visit
\n(M1)
\nrequire (M1)
\n\n
A1
\n[3 marks]
\nMETHOD 1
\nlet be the discrete random variable “number of times Steffi is not fed per day”
\nM1
\nA1
\n= 0.083979... A1
\nexpected no of occasions per year > 0.083979... × 365 = 30.7 A1
\nhence Steffi can expect not to be fed on at least 30 occasions AG
\nNote: Candidates may consider summing more than three terms in their calculation for .
\n\n
METHOD 2
\nM1A1
\n0.0903… × 365 M1
\n= 33.0 > 30 A1AG
\n\n
[4 marks]
\nThe number of bananas that Lucca eats during any particular day follows a Poisson distribution with mean 0.2.
\nFind the probability that Lucca eats at least one banana in a particular day.
\nFind the expected number of weeks in the year in which Lucca eats no bananas.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
let be the number of bananas eaten in one day
\n\n
(M1)
\nA1
\n[2 marks]
\nEITHER
\nlet be the number of bananas eaten in one week
\n(A1)
\n(A1)
\nOR
\nlet be the number of days in one week at least one banana is eaten
\n(A1)
\n(A1)
\nTHEN
\n(M1)
\nA1
\n[4 marks]
\nChloe and Selena play a game where each have four cards showing capital letters A, B, C and D.
Chloe lays her cards face up on the table in order A, B, C, D as shown in the following diagram.
![\"N17/5/MATHL/HP1/ENG/TZ0/10\"](\"../assets/uploads/tinymce_asset/asset/4114/Schermafbeelding_2018-02-07_om_14.39.35.png\"/)
Selena shuffles her cards and lays them face down on the table. She then turns them over one by one to see if her card matches with Chloe’s card directly above.
Chloe wins if no matches occur; otherwise Selena wins.
Chloe and Selena repeat their game so that they play a total of 50 times.
Suppose the discrete random variable X represents the number of times Chloe wins.
Show that the probability that Chloe wins the game is .
\nDetermine the mean of X.
\nDetermine the variance of X.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1
\nnumber of possible “deals” A1
\nconsider ways of achieving “no matches” (Chloe winning):
\nSelena could deal B, C, D (ie, 3 possibilities)
\nas her first card R1
\nfor each of these matches, there are only 3 possible combinations for the remaining 3 cards R1
\nso no. ways achieving no matches M1A1
\nso probability Chloe wins A1AG
\n\n
METHOD 2
\nnumber of possible “deals” A1
\nconsider ways of achieving a match (Selena winning)
\nSelena card A can match with Chloe card A, giving 6 possibilities for this happening R1
\nif Selena deals B as her first card, there are only 3 possible combinations for the remaining 3 cards. Similarly for dealing C and dealing D R1
\nso no. ways achieving one match is M1A1
\nso probability Chloe wins A1AG
\n\n
METHOD 3
\nsystematic attempt to find number of outcomes where Chloe wins (no matches)
\n(using tree diag. or otherwise) M1
\n9 found A1
\neach has probability M1
\nA1
\ntheir 9 multiplied by their M1A1
\nAG
\n\n
[6 marks]
\n(M1)
\n(M1)A1
\n[3 marks]
\n(M1)A1
\n[2 marks]
\nThe number of taxis arriving at Cardiff Central railway station can be modelled by a Poisson distribution. During busy periods of the day, taxis arrive at a mean rate of 5.3 taxis every 10 minutes. Let T represent a random 10 minute busy period.
\nFind the probability that exactly 4 taxis arrive during T.
\nFind the most likely number of taxis that would arrive during T.
\nGiven that more than 5 taxis arrive during T, find the probability that exactly 7 taxis arrive during T.
\nDuring quiet periods of the day, taxis arrive at a mean rate of 1.3 taxis every 10 minutes.
\nFind the probability that during a period of 15 minutes, of which the first 10 minutes is busy and the next 5 minutes is quiet, that exactly 2 taxis arrive.
\n\n
(M1)
\n= 0.164 A1
\n[2 marks]
\nMETHOD 1
\nlisting probabilities (table or graph) M1
\nmode X = 5 (with probability 0.174) A1
\nNote: Award M0A0 for 5 (taxis) or mode = 5 with no justification.
\n\n
METHOD 2
\nmode is the integer part of mean R1
\nE(X) = 5.3 ⇒ mode = 5 A1
\nNote: Do not allow R0A1.
\n[2 marks]
\nattempt at conditional probability (M1)
\nor equivalent A1
\n= 0.267 A1
\n[3 marks]
\nMETHOD 1
\nthe possible arrivals are (2,0), (1,1), (0,2) (A1)
\nA1
\nattempt to compute, using sum and product rule, (M1)
\n0.070106… × 0.52204… + 0.026455… × 0.33932… + 0.0049916… × 0.11028… (A1)(A1)
\nNote: Award A1 for one correct product and A1 for two other correct products.
\n= 0.0461 A1
\n[6 marks]
\n\n
METHOD 2
\nrecognising a sum of 2 independent Poisson variables eg Z = X + Y R1
\n\n
P(Z = 2) = 0.0461 (M1)A3
\n[6 marks]
\n\n
Find the coordinates of the point of intersection of the planes defined by the equations and .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nfor eliminating one variable from two equations (M1)
\neg, A1A1
\nfor finding correctly one coordinate
\neg, A1
\nfor finding correctly the other two coordinates A1
\n\n
the intersection point has coordinates
\nMETHOD 2
\nfor eliminating two variables from two equations or using row reduction (M1)
\neg, or A1A1
\nfor finding correctly the other coordinates A1A1
\nor
\nthe intersection point has coordinates
\nMETHOD 3
\n(A1)
\nattempt to use Cramer’s rule M1
\nA1
\nA1
\nA1
\n\n
Note: Award M1 only if candidate attempts to determine at least one of the variables using this method.
\n\n
[5 marks]
\nThe faces of a fair six-sided die are numbered 1, 2, 2, 4, 4, 6. Let be the discrete random variable that models the score obtained when this die is rolled.
\nComplete the probability distribution table for .
\n![\"N16/5/MATHL/HP1/ENG/TZ0/02.a\"](\"../assets/uploads/tinymce_asset/asset/3291/Schermafbeelding_2017-02-28_om_11.16.45.png\"/)
Find the expected value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"N16/5/MATHL/HP1/ENG/TZ0/02.a/M\"](\"../assets/uploads/tinymce_asset/asset/3292/Schermafbeelding_2017-02-28_om_11.18.41.png\"/)
A1A1
\n
Note: Award A1 for each correct row.
\n\n
[2 marks]
\n(M1)
\nA1
\n\n
Note: If the probabilities in (a) are not values between 0 and 1 or lead to award M1A0 to correct method using the incorrect probabilities; otherwise allow FT marks.
\n\n
[2 marks]
\nPackets of biscuits are produced by a machine. The weights , in grams, of packets of biscuits can be modelled by a normal distribution where . A packet of biscuits is considered to be underweight if it weighs less than 250 grams.
\nThe manufacturer makes the decision that the probability that a packet is underweight should be 0.002. To do this is increased and remains unchanged.
\nThe manufacturer is happy with the decision that the probability that a packet is underweight should be 0.002, but is unhappy with the way in which this was achieved. The machine is now adjusted to reduce and return to 253.
\nGiven that and find the probability that a randomly chosen packet of biscuits is underweight.
\nCalculate the new value of giving your answer correct to two decimal places.
\nCalculate the new value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)A1
\n[2 marks]
\n(M1)(A1)
\nA1
\n\n
Notes: Only award A1 here if the correct 2dp answer is seen. Award M0 for use of .
\n\n
[3 marks]
\n(A1)
\nA1
\n[2 marks]
\nA discrete random variable follows a Poisson distribution .
\nShow that .
\nGiven that and , use part (a) to find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nA1
\nM1A1
\nAG
\nMETHOD 2
\nA1
\nM1A1
\nAG
\nMETHOD 3
\n(M1)
\nA1
\nA1
\nand so AG
\n[3 marks]
\nA1
\nattempting to solve for (M1)
\nA1
\n[3 marks]
\nThe age, L, in years, of a wolf can be modelled by the normal distribution L ~ N(8, 5).
\nFind the probability that a wolf selected at random is at least 5 years old.
\nEight wolves are independently selected at random and their ages recorded.
\nFind the probability that more than six of these wolves are at least 5 years old.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
P(L ≥ 5) = 0.910 (M1)A1
\n[2 marks]
\nX is the number of wolves found to be at least 5 years old recognising binomial distribution M1
\nX ~ B(8, 0.910…)
\nP(X > 6) = 1 − P(X ≤ 6) (M1)
\n= 0.843 A1
\nNote: Award M1A0 for finding P(X ≥ 6).
\n[3 marks]
\nConsider the system of paired differential equations
\n\n
.
\nThis represents the populations of two species of symbiotic toadstools in a large wood.
\nTime is measured in decades.
\nUse the eigenvalue method to find the general solution to this system of equations.
\nGiven the initial conditions that when , , , find the particular solution.
\nHence find the solution when .
\nAs , find an asymptote to the trajectory of the particular solution found in (b)(i) and state if this trajectory will be moving towards or away from the origin.
\nThe characteristic equation is given by
\nM1A1A1A1
\nM1A1
\nM1A1
\nGeneral solution is A1A1
\n[10 marks]
\nRequire M1A1
\nParticular solution is A1
\n[3 marks]
\nA1
\n[1 mark]
\nThe dominant term is so as , M1A1
\nGiving the asymptote as A1
\nThe trajectory is moving away from the origin. A1
\n[4 marks]
\nTimmy owns a shop. His daily income from selling his goods can be modelled as a normal distribution, with a mean daily income of $820, and a standard deviation of $230. To make a profit, Timmy’s daily income needs to be greater than $1000.
\nCalculate the probability that, on a randomly selected day, Timmy makes a profit.
\nThe shop is open for 24 days every month.
\nCalculate the probability that, in a randomly selected month, Timmy makes a profit on between 5 and 10 days (inclusive).
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
X ~ N(820, 2302) (M1)
\nNote: Award M1 for an attempt to use normal distribution. Accept labelled normal graph.
\n⇒P(X > 1000) = 0.217 A1
\n[2 marks]
\nY ~ B(24,0.217...) (M1)
\nNote: Award M1 for recognition of binomial distribution with parameters.
\nP(Y ≤ 10) − P(Y ≤ 4) (M1)
\nNote: Award M1 for an attempt to find P(5 ≤ Y ≤ 10) or P(Y ≤ 10) − P(Y ≤ 4).
\n= 0.613 A1
\n[3 marks]
\nThe cost adjacency matrix for the complete graph K6 is given below.
\nIt represents the distances in kilometres along dusty tracks connecting villages on an island. Find the minimum spanning tree for this graph; in all 3 cases state the order in which the edges are added.
\nIt is desired to tarmac some of these tracks so that it is possible to walk from any village to any other village walking entirely on tarmac.
\nBriefly explain the two differences in the application of Prim’s and Kruskal’s algorithms for finding a minimum spanning tree in a weighted connected graph.
\nUsing Kruskal’s algorithm.
\nUsing Prim’s algorithm starting at vertex A.
\nUsing Prim’s algorithm starting at vertex F.
\nState the total minimum length of the tracks that have to be tarmacked.
\nSketch the tracks that are to be tarmacked.
\nIn Prim’s algorithm you start at a particular (given) vertex, whereas in Kruskal’s you start with the smallest edge. A1
\nIn Prim’s as smallest edges are added (never creating a circuit) the created graph always remains connected, whereas in Kruskal’s this requirement to always be connected is not necessary. A1
\n[2 marks]
\nEdges added in the order
\nAB EF AC AD AE A1A1
\n[note A1 for the first 2 edges A1 for other 3]
\n[2 marks]
\nEdges added in the order
\nAB AC AD AE EF A1A1
\n[note A1 for the first 2 edges A1 for other 3]
\n[2 marks]
\nEdges added in the order
\nFE AE AB AC AD A1A1
\n[note A1 for the first 2 edges A1 for other 3]
\n[2 marks]
\nM1A1
\n[2 marks]
\n A2
[2 marks]
\nThe random variable X has a normal distribution with mean μ = 50 and variance σ 2 = 16 .
\nSketch the probability density function for X, and shade the region representing P(μ − 2σ < X < μ + σ).
\nFind the value of P(μ − 2σ < X < μ + σ).
\nFind the value of k for which P(μ − kσ < X < μ + kσ) = 0.5.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
normal curve centred on 50 A1
\nvertical lines at = 42 and = 54, with shading in between A1
\n[2 marks]
\nP(42 < X < 54) (= P(− 2 < Z < 1)) (M1)
\n= 0.819 A1
\n[2 marks]
\nP(μ − kσ < X < μ + kσ) = 0.5 ⇒ P(X < μ + kσ) = 0.75 (M1)
\nk = 0.674 A1
\nNote: Award M1A0 for k = −0.674.
\n[2 marks]
\nIt is known that 56 % of Infiglow batteries have a life of less than 16 hours, and 94 % have a life less than 17 hours. It can be assumed that battery life is modelled by the normal distribution .
\nFind the value of and the value of .
\nFind the probability that a randomly selected Infiglow battery will have a life of at least 15 hours.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
use of inverse normal (implied by ±0.1509… or ±1.554…) (M1)
\nP(X < 16) = 0.56
\n(A1)
\nP(X < 17) = 0.94
\n(A1)
\nattempt to solve a pair of simultaneous equations (M1)
\n= 15.9, = 0.712 A1A1
\n\n
[6 marks]
\ncorrectly shaded diagram or intent to find P(X ≥ 15) (M1)
\n= 0.895 A1
\nNote: Accept answers rounding to 0.89 or 0.90. Award M1A0 for the answer 0.9.
\n\n
[2 marks]
\nA random variable has a probability distribution given in the following table.
\n![\"N16/5/MATHL/HP2/ENG/TZ0/01\"](\"../assets/uploads/tinymce_asset/asset/3294/Schermafbeelding_2017-02-28_om_17.11.47.png\"/)
Determine the value of .
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)A1
\n[2 marks]
\nMETHOD 1
\n(M1)(A1)
\nA1
\nMETHOD 2
\n(A1)
\nuse of (M1)
\n\n
Note: Award (M1) only if is used correctly.
\n\n
\n
A1
\n\n
Note: Accept 2.11.
\n\n
METHOD 3
\n(A1)
\nuse of (M1)
\n\n
A1
\n[3 marks]
\nThe random variable X has a binomial distribution with parameters n and p.
It is given that E(X) = 3.5.
Find the least possible value of n.
\nIt is further given that P(X ≤ 1) = 0.09478 correct to 4 significant figures.
\nDetermine the value of n and the value of p.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
np = 3.5 (A1)
\np ≤ 1 ⇒ least n = 4 A1
\n[2 marks]
\n(1 − p)n + np(1 − p)n−1 = 0.09478 M1A1
\nattempt to solve above equation with np = 3.5 (M1)
\nn = 12, p = (=0.292) A1A1
\nNote: Do not accept n as a decimal.
\n[5 marks]
\nMr Sailor owns a fish farm and he claims that the weights of the fish in one of his lakes have a mean of 550 grams and standard deviation of 8 grams.
\nAssume that the weights of the fish are normally distributed and that Mr Sailor’s claim is true.
\nKathy is suspicious of Mr Sailor’s claim about the mean and standard deviation of the weights of the fish. She collects a random sample of fish from this lake whose weights are shown in the following table.
\nUsing these data, test at the 5% significance level the null hypothesis against the alternative hypothesis , where grams is the population mean weight.
\nKathy decides to use the same fish sample to test at the 5% significance level whether or not there is a positive association between the weights and the lengths of the fish in the lake. The following table shows the lengths of the fish in the sample. The lengths of the fish can be assumed to be normally distributed.
\nFind the probability that a fish from this lake will have a weight of more than 560 grams.
\nThe maximum weight a hand net can hold is 6 kg. Find the probability that a catch of 11 fish can be carried in the hand net.
\nState the distribution of your test statistic, including the parameter.
\nFind the p-value for the test.
\nState the conclusion of the test, justifying your answer.
\nState suitable hypotheses for the test.
\nFind the product-moment correlation coefficient .
\nState the p-value and interpret it in this context.
\nUse an appropriate regression line to estimate the weight of a fish with length 360 mm.
\nNote: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
\n\n
(550, 82) (M1)
\nA1
\n\n
[2 marks]
\nNote: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
\n\n
(550, 82), ,…, 11
\nlet
\nA1
\n(M1)A1
\nA1
\n\n
[4 marks]
\nNote: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
\n\n
distribution with 7 degrees of freedom A1A1
\n\n
[2 marks]
\nNote: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
\n\n
p = 0.25779…= 0.258 A2
\n\n
[2 marks]
\nNote: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
\n\n
p > 0.05 R1
\ntherefore we conclude that there is no evidence to reject A1
\nNote: FT their p-value.
\nNote: Only award A1 if R1 awarded.
\n\n
[2 marks]
\nNote: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
\n\n
, A1
\nNote: Do not accept in place of .
\n\n
[1 mark]
\nNote: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
\n\n
= 0.782 A2
\n\n
[2 marks]
\nNote: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
\n\n
0.01095… = 0.0110 A1
\nsince 0.0110 < 0.05 R1
\nthere is positive association between weight and length A1
\nNote: FT their p-value.
\nNote: Only award A1 if R1 awarded.
\nNote: Conclusion must be in context.
\n\n
[3 marks]
\nNote: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
\n\n
regression line of (weight) on (length) is (M1)
\n= 0.8267… + 255.96… (A1)
\n= 360 gives = 554 A1
\nNote: Award M1A0A0 for the wrong regression line, that is = 0.7393… – 51.62….
[3 marks]
\nThe cost adjacency matrix below represents the distance in kilometres, along routes between bus stations.
\nAll the values in the matrix are positive, distinct integers.
\nIt is decided to electrify some of the routes, so that it will be possible to travel from any station to any other station solely on electrified routes. In order to achieve this with a minimal total length of electrified routes, Prim’s algorithm for a minimal spanning tree is used, starting at vertex A.
\nThe algorithm adds the edges in the following order:
\nAB AC CD DE.
\nThere is only one minimal spanning tree.
\nFind with a reason, the value of .
\nIf the total length of the minimal spanning tree is 14, find the value of .
\nHence, state, with a reason, what can be deduced about the values of , , .
\nAB must be the length of the smallest edge from A so . R1A1
\n[2 marks]
\nM1A1
\n[2 marks]
\nThe last minimal edge chosen must connect to E , so since each of , , must be ≥ 9. R1A1
\n[2 marks]
\nThe times taken for male runners to complete a marathon can be modelled by a normal distribution with a mean 196 minutes and a standard deviation 24 minutes.
\nIt is found that 5% of the male runners complete the marathon in less than minutes.
\nThe times taken for female runners to complete the marathon can be modelled by a normal distribution with a mean 210 minutes. It is found that 58% of female runners complete the marathon between 185 and 235 minutes.
\nFind the probability that a runner selected at random will complete the marathon in less than 3 hours.
\nCalculate .
\nFind the standard deviation of the times taken by female runners.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
(M1)A1
\n[2 marks]
\n(M1)
\nA1
\n[2 marks]
\n\n
(M1)
\nor equivalent (M1)(A1)
\nA1
\n[4 marks]
\nThe graph of is transformed onto the graph of by a translation of units vertically and a stretch parallel to the -axis of scale factor .
\nWrite down the value of .
\nFind the value of .
\nThe outer dome of a large cathedral has the shape of a hemisphere of diameter 32 m, supported by vertical walls of height 17 m. It is also supported by an inner dome which can be modelled by rotating the curve through 360° about the -axis between = 0 and = 33, as indicated in the diagram.
\nFind the volume of the space between the two domes.
\n= 33 A1
\n[1 mark]
\nM1A1
\n[2 marks]
\nvolume within outer dome
\nM1A1
\nvolume within inner dome
\nM1A1
\nvolume between = 22 250.85 − 3446.92 = 18 803.93 m3 A1
\n[5 marks]
\nLet , where .
\nfor = 2,
\nfind the values of , , and .
\ndraw , , , and on the following Argand diagram.
\nLet .
\nFind the value of for which successive powers of lie on a circle.
\n, , (, , ) (M1)A1
\n[2 marks]
\n A3
Note: Award A1 for correct arguments, award A1 for and −16 clearly indicated, award A1 for | | < 4 and 4 < | | < 16.
\n[3 marks]
\nM1
\nA1
\n[2 marks]
\nThe Osaka Tigers basketball team play in a multilevel stadium.
\nThe most expensive tickets are in the first row. The ticket price, in Yen (¥), for each row forms an arithmetic sequence. Prices for the first three rows are shown in the following table.
\nWrite down the value of the common difference,
\nCalculate the price of a ticket in the 16th row.
\nFind the total cost of buying 2 tickets in each of the first 16 rows.
\n( =) − 250 A1
\n[1 mark]
\nM1
\n(¥)3050 A1
\n[2 marks]
\nM1M1
\nNote: Award M1 for correct substitution into arithmetic series formula.
Award M1 for multiplication by 2 seen.
OR
\nM1M1
\nNote: Award M1 for correct substitution into arithmetic series formula.
Award M1 for multiplication by 2 seen.
(¥)158 000 (157 600) A1
\n[3 marks]
\nIn this question, give all answers to two decimal places.
\nBryan decides to purchase a new car with a price of €14 000, but cannot afford the full amount. The car dealership offers two options to finance a loan.
\nFinance option A:
\nA 6 year loan at a nominal annual interest rate of 14 % compounded quarterly. No deposit required and repayments are made each quarter.
\nFinance option B:
\nA 6 year loan at a nominal annual interest rate of % compounded monthly. Terms of the loan require a 10 % deposit and monthly repayments of €250.
\nFind the repayment made each quarter.
\nFind the total amount paid for the car.
\nFind the interest paid on the loan.
\nFind the amount to be borrowed for this option.
\nFind the annual interest rate, .
\nState which option Bryan should choose. Justify your answer.
\nBryan’s car depreciates at an annual rate of 25 % per year.
\nFind the value of Bryan’s car six years after it is purchased.
\nN = 24
I % = 14
PV = −14000
FV = 0
P/Y = 4
C/Y = 4 (M1)(A1)
Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct. Accept PV = 14000.
\n(€)871.82 A1
\n[3 marks]
\n4 × 6 × 871.82 (M1)
\n(€) 20923.68 A1
\n[2 marks]
\n20923.68 − 14000 (M1)
\n(€) 6923.68 A1
\n[2 marks]
\n0.9 × 14000 (= 14000 − 0.10 × 14000) M1
\n(€) 12600.00 A1
\n[2 marks]
\nN = 72
\nPV = 12600
\nPMT = −250
\nFV = 0
\nP/Y = 12
\nC/Y = 12 (M1)(A1)
\nNote: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct. Accept PV = −12600 provided PMT = 250.
\n12.56(%) A1
\n[3 marks]
\nEITHER
\nBryan should choose Option A A1
\nno deposit is required R1
\nNote: Award R1 for stating that no deposit is required. Award A1 for the correct choice from that fact. Do not award R0A1.
\nOR
\nBryan should choose Option B A1
\ncost of Option A (6923.69) > cost of Option B (72 × 250 − 12600 = 5400) R1
\nNote: Award R1 for a correct comparison of costs. Award A1 for the correct choice from that comparison. Do not award R0A1.
\n[2 marks]
\n(M1)(A1)
\nNote: Award M1 for substitution into compound interest formula.
Award A1 for correct substitutions.
= (€)2491.70 A1
\nOR
\nN = 6
\nI% = −25
\nPV = ±14 000
\nP/Y = 1
\nC/Y = 1 (A1)(M1)
\nNote: Award A1 for PV = ±14 000, M1 for other entries correct.
\n(€)2491.70 A1
\n[3 marks]
\nIn this question, give all answers to two decimal places.
\nBryan decides to purchase a new car with a price of €14 000, but cannot afford the full amount. The car dealership offers two options to finance a loan.
\nFinance option A:
\nA 6 year loan at a nominal annual interest rate of 14 % compounded quarterly. No deposit required and repayments are made each quarter.
\nFinance option B:
\nA 6 year loan at a nominal annual interest rate of % compounded monthly. Terms of the loan require a 10 % deposit and monthly repayments of €250.
\nFind the repayment made each quarter.
\nFind the total amount paid for the car.
\nFind the interest paid on the loan.
\nFind the amount to be borrowed for this option.
\nFind the annual interest rate, .
\nState which option Bryan should choose. Justify your answer.
\nBryan chooses option B. The car dealership invests the money Bryan pays as soon as they receive it.
\nIf they invest it in an account paying 0.4 % interest per month and inflation is 0.1 % per month, calculate the real amount of money the car dealership has received by the end of the 6 year period.
\nN = 24
I % = 14
PV = −14000
FV = 0
P/Y = 4
C/Y = 4 (M1)(A1)
Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct. Accept PV = 14000.
\n(€)871.82 A1
\n[3 marks]
\n4 × 6 × 871.82 (M1)
\n(€) 20923.68 A1
\n[2 marks]
\n20923.68 − 14000 (M1)
\n(€) 6923.68 A1
\n[2 marks]
\n0.9 × 14000 (= 14000 − 0.10 × 14000) M1
\n(€) 12600.00 A1
\n[2 marks]
\nN = 72
\nPV = 12600
\nPMT = −250
\nFV = 0
\nP/Y = 12
\nC/Y = 12 (M1)(A1)
\nNote: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct. Accept PV = −12600 provided PMT = 250.
\n12.56(%) A1
\n[3 marks]
\nEITHER
\nBryan should choose Option A A1
\nno deposit is required R1
\nNote: Award R1 for stating that no deposit is required. Award A1 for the correct choice from that fact. Do not award R0A1.
\nOR
\nBryan should choose Option B A1
\ncost of Option A (6923.69) > cost of Option B (72 × 250 − 12600 = 5400) R1
\nNote: Award R1 for a correct comparison of costs. Award A1 for the correct choice from that comparison. Do not award R0A1.
\n[2 marks]
\nreal interest rate is 0.4 − 0.1 = 0.3% (M1)
\nvalue of other payments 250 + 250 × 1.003 + … + 250 × 1.00371
\nuse of sum of geometric sequence formula or financial app on a GDC (M1)
\n= 20 058.43
\nvalue of deposit at the end of 6 years
\n1400 × (1.003)72 = 1736.98 (A1)
\nTotal value is (€) 21 795.41 A1
\nNote: Both M marks can awarded for a correct use of the GDC’s financial app:
\nN = 72 (6 × 12)
I % = 3.6 (0.3 × 12)
PV = 0
PMT = −250
FV =
P/Y = 12
C/Y = 12
OR
\nN = 72 (6 × 12)
I % = 0.3
PV = 0
PMT = −250
FV =
P/Y = 1
C/Y = 1
[4 marks]
\nThe number of fish that can be caught in one hour from a particular lake can be modelled by a Poisson distribution.
\nThe owner of the lake, Emily, states in her advertising that the average number of fish caught in an hour is three.
\nTom, a keen fisherman, is not convinced and thinks it is less than three. He decides to set up the following test. Tom will fish for one hour and if he catches fewer than two fish he will reject Emily’s claim.
\nState a suitable null and alternative hypotheses for Tom’s test.
\nFind the probability of a Type I error.
\nThe average number of fish caught in an hour is actually 2.5.
\nFind the probability of a Type II error.
\n, A1
\nNote: Accept equivalent statements in words.
\n[1 mark]
\n(let be the number of fish caught)
\nM1A1
\n[2 marks]
\nM1A1
\nNote: Award M1 for using = 2.5 to evaluate a probability, award A1 for also having ≥ 2 .
\n= 0.713 A1
\n[3 marks]
\nThe Malvern Aquatic Center hosted a 3 metre spring board diving event. The judges, Stan and Minsun awarded 8 competitors a score out of 10. The raw data is collated in the following table.
\nThe Commissioner for the event would like to find the Spearman’s rank correlation coefficient.
\nWrite down the value of the Pearson’s product–moment correlation coefficient, .
\nUsing the value of , interpret the relationship between Stan’s score and Minsun’s score.
\nWrite down the equation of the regression line on .
\nUse your regression equation from part (b) to estimate Minsun’s score when Stan awards a perfect 10.
\nState whether this estimate is reliable. Justify your answer.
\nCopy and complete the information in the following table.
\nFind the value of the Spearman’s rank correlation coefficient, .
\nComment on the result obtained for .
\nThe Commissioner believes Minsun’s score for competitor G is too high and so decreases the score from 9.5 to 9.1.
\nExplain why the value of the Spearman’s rank correlation coefficient does not change.
\n0.909 (0.909181…) A2
\n[2 marks]
\n(very) strong and positive A1A1
\nNote: Award A1 for (very) strong A1 for positive.
\n[2 marks]
\nA1A1
\nNote: Award A1 for , A1 for . Award a maximum of A1A0 if the answer is not an equation in the form .
\n[2 marks]
\n1.14 × 10 + 0.578 M1
\n12.0 (11.9814…) A1
\n[2 marks]
\nno the estimate is not reliable A1
\noutside the known data range R1
OR
a score greater than 10 is not possible R1
Note: Do not award A1R0.
\n[2 marks]
\n A1A1
Note: Award A1 for correct ranks for Stan. Award A1 for correct ranks for Minsun.
\n[2 marks]
\n0.933 (0.932673…) A2
\n[2 marks]
\nStan and Minsun strongly agree on the ranking of competitors. A1A1
\nNote: Award A1 for “strongly agree”, A1 for reference to a rank order.
\n[2 marks]
\ndecreasing the score to 9.1, does not change the rank of competitor G A1
\n[1 mark]
\nAt the end of a school day, the Headmaster conducted a survey asking students in how many classes they had used the internet.
\nThe data is shown in the following table.
\nThe mean number of classes in which a student used the internet is 2.
\nState whether the data is discrete or continuous.
\nFind the value of .
\nIt was not possible to ask every person in the school, so the Headmaster arranged the student names in alphabetical order and then asked every 10th person on the list.
\nIdentify the sampling technique used in the survey.
\ndiscrete A1
\n[1 mark]
\nM1A1
\nNote: Award M1 for substitution into the formula for the mean, award A1 for a correct equation.
\nattempt to solve their equation (M1)
\n= 31 A1
\n[4 marks]
\nsystematic A1
\n[1 mark]
\nMr Burke teaches a mathematics class with 15 students. In this class there are 6 female students and 9 male students.
\nEach day Mr Burke randomly chooses one student to answer a homework question.
\nIn the first month, Mr Burke will teach his class 20 times.
\nFind the probability he will choose a female student 8 times.
\nThe Head of Year, Mrs Smith, decides to select a student at random from the year group to read the notices in assembly. There are 80 students in total in the year group. Mrs Smith calculates the probability of picking a male student 8 times in the first 20 assemblies is 0.153357 correct to 6 decimal places.
\nFind the number of male students in the year group.
\nP(X = 8) (M1)
\nNote: Award (M1) for evidence of recognizing binomial probability. eg, P(X = 8), X ∼ B.
\n= 0.180 (0.179705…) A1
\n[2 marks]
\nlet be the number of male students
\nrecognize that probability of selecting a male is equal to (A1)
\n(M1)
\nnumber of male students = 37 (M1)A1
\nNote: Award (M1)A0 for 27.
\n[4 marks]
\nLet be a random variable which follows a normal distribution with mean . Given that , find
\n.
\n.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
use of symmetry eg diagram (M1)
\nA1
\n[2 marks]
\nEITHER
\n(M1)
\n(A1)
\nA1A1
\nNote: A1 for denominator is independent of the previous A marks.
\nOR
\nuse of diagram (M1)
\nNote: Only award (M1) if the region is indicated and used.
\n(A1)
\nNote: Probabilities can be shown on the diagram.
\nM1A1
\nTHEN
\nA1
\n[5 marks]
\nThe rate, , of a chemical reaction at a fixed temperature is related to the concentration of two compounds, and , by the equation
\n, where , , .
\nA scientist measures the three variables three times during the reaction and obtains the following values.
\nFind , and .
\n(M1)
\n\n
\n
M1A1
\nNote: Allow any consistent base, allow numerical equivalents.
\nattempting to solve their system of equations (M1)
\n= 1.53, = 0.505 A1
\n= 0.997 A1
\n[6 marks]
\nProfessor Vinculum investigated the migration season of the Bulbul bird from their natural wetlands to a warmer climate.
\nHe found that during the migration season their population, could be modelled by , ≥ 0 , where is the number of days since the start of the migration season.
\nFind the population of the Bulbul birds at the start of the migration season.
\nFind the population of the Bulbul birds after 5 days.
\nCalculate the time taken for the population to decrease below 1400.
\nAccording to this model, find the smallest possible population of Bulbul birds during the migration season.
\n1750 A1
\n[1 mark]
\n(M1)
\n= 1480 A1
\nNote: Accept 1481.
\n[2 marks]
\n(M1)
\n9.32 (days (9.31885…) (days)) A1
\n[2 marks]
\n1350 A1
\nNote: Accept 1351 as a valid interpretation of the model as = 1350 is an asymptote.
\n[1 mark]
\nThe Happy Straw Company manufactures drinking straws.
\nThe straws are packaged in small closed rectangular boxes, each with length 8 cm, width 4 cm and height 3 cm. The information is shown in the diagram.
\nEach week, the Happy Straw Company sells boxes of straws. It is known that , ≥ 0, where is the weekly profit, in dollars, from the sale of thousand boxes.
\nCalculate the surface area of the box in cm2.
\nCalculate the length AG.
\nFind the number of boxes that should be sold each week to maximize the profit.
\nFind .
\nFind the least number of boxes which must be sold each week in order to make a profit.
\n2(8 × 4 + 3 × 4 + 3 × 8) M1
\n= 136 (cm2) A1
\n[2 marks]
\nM1
\n(AG =) 9.43 (cm) (9.4339…, ) A1
\n[2 marks]
\nM1
\nA1
\n110 000 (boxes) A1
\n[3 marks]
\nM1
\nNote: Award M1 for evidence of integration.
\nA1A1
\nNote: Award A1 for either or award A1 for both correct terms and constant of integration.
\nM1
\n\n
A1
\n[5 marks]
\nM1
\nA1
\n11 006 (boxes) A1
\nNote: Award M1 for their , award A1 for their correct solution to .
Award the final A1 for expressing their solution to the minimum number of boxes. Do not accept 11 005, the nearest integer, nor 11 000, the answer expressed to 3 significant figures, as these will not satisfy the demand of the question.
[3 marks]
\nA random variable is normally distributed with mean and standard deviation , such that and .
\nFind and .
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\neither (M1)
\n(A1)
\n(A1)
\nattempting to solve simultaneously (M1)
\nand A1
\n[6 marks]
\n(or equivalent eg. ) (M1)
\nA1
\n\n
Note: Award (M1)A1 for .
\n[2 marks]
\nThe braking distance of a vehicle is defined as the distance travelled from where the brakes are applied to the point where the vehicle comes to a complete stop.
\nThe speed, , and braking distance, , of a truck were recorded. This information is summarized in the following table.
\nThis information was used to create Model A, where is a function of , ≥ 0.
\nModel A: , where ,
\nAt a speed of , Model A can be represented by the equation .
\nAdditional data was used to create Model B, a revised model for the braking distance of a truck.
\nModel B:
\nThe actual braking distance at is .
\nWrite down a second equation to represent Model A, when the speed is .
\nFind the values of and .
\nFind the coordinates of the vertex of the graph of .
\nUsing the values in the table and your answer to part (b), sketch the graph of for 0 ≤ ≤ 10 and −10 ≤ ≤ 60, clearly showing the vertex.
\nHence, identify why Model A may not be appropriate at lower speeds.
\nUse Model B to calculate an estimate for the braking distance at a speed of .
\nCalculate the percentage error in the estimate in part (e).
\nIt is found that once a driver realizes the need to stop their vehicle, 1.6 seconds will elapse, on average, before the brakes are engaged. During this reaction time, the vehicle will continue to travel at its original speed.
\nA truck approaches an intersection with speed . The driver notices the intersection’s traffic lights are red and they must stop the vehicle within a distance of .
\nUsing model B and taking reaction time into account, calculate the maximum possible speed of the truck if it is to stop before the intersection.
\nM1
\nA1
\n[2 marks]
\n, A1A1
\nNote: If and are both incorrect then award M1A0 for an attempt to solve simultaneous equations.
\n[2 marks]
\n(2, −4) A1A1
\nNote: Award A1 for each correct coordinate.
Award A0A1 if parentheses are missing.
[2 marks]
\n A3
Note: Award A1 for smooth quadratic curve on labelled axes and within correct window.
Award A1 for the curve passing through (0, 0) and (10, 60). Award A1 for the curve passing through their vertex. Follow through from part (b).
[3 marks]
\nthe graph indicates there are negative stopping distances (for low speeds) R1
\nNote: Award R1 for identifying that a feature of their graph results in negative stopping distances (vertex, range of stopping distances…).
\n[1 mark]
\n(M1)
\nA1
\n[2 marks]
\nM1
\n(%) A1
\n[2 marks]
\nM1A1
\nNote: Award M1 for an attempt to find an expression including stopping distance (model B) and reaction distance, equated to 330. Award A1 for a completely correct equation.
\nA1
\n[3 marks]
\nFind the coordinates of the points on the curve at which .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt at implicit differentiation M1
\nA1A1
\nNote: Award A1 for the second & third terms, A1 for the first term, fourth term & RHS equal to zero.
\nsubstitution of M1
\n\n
A1
\nsubstitute either variable into original equation M1
\n(or ) A1
\n(or ) A1
\n, (3, −3) A1
\n[9 marks]
\nThe folium of Descartes is a curve defined by the equation , shown in the following diagram.
\n![\"N17/5/MATHL/HP1/ENG/TZ0/07\"](\"../assets/uploads/tinymce_asset/asset/4118/Schermafbeelding_2018-02-07_om_18.23.15.png\"/)
Determine the exact coordinates of the point P on the curve where the tangent line is parallel to the -axis.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
M1A1
\n\n
Note: Differentiation wrt is also acceptable.
\n\n
(A1)
\n\n
Note: All following marks may be awarded if the denominator is correct, but the numerator incorrect.
\n\n
M1
\nEITHER
\n\n
M1A1
\n\n
\n
A1
\nA1
\nOR
\nM1
\n\n
A1
\n\n
\n
\n
A1
\nA1
\n[8 marks]
\nThe region is enclosed by the graph of , the -axis and the line .
\nWrite down a definite integral to represent the area of .
\nCalculate the area of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\n(M1)
\n(A1)
\nM1A1
\n\n
Note: Award M1 for an attempt to find the difference between two functions, A1 for all correct.
\n\n
METHOD 2
\nwhen A1
\nM1A1
\n\n
Note: Award M1 for an attempt to find the inverse function.
\n\n
A1
\nMETHOD 3
\nM1A1A1A1
\n\n
Note: Award M1 for considering the area below the -axis and above the -axis and A1 for each correct integral.
\n\n
[4 marks]
\nA2
\n[2 marks]
\nIt is given that one in five cups of coffee contain more than 120 mg of caffeine.
It is also known that three in five cups contain more than 110 mg of caffeine.
Assume that the caffeine content of coffee is modelled by a normal distribution.
Find the mean and standard deviation of the caffeine content of coffee.
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
let be the random variable “amount of caffeine content in coffee”
\n(M1)
\n\n
\n
Note: Award M1 for at least one correct probability statement.
\n\n
(M1)(A1)(A1)
\n\n
Note: Award M1 for attempt to find at least one appropriate -value.
\n\n
\n
attempt to solve simultaneous equations (M1)
\nA1
\n[6 marks]
\nAs part of a study into healthy lifestyles, Jing visited Surrey Hills University. Jing recorded a person’s position in the university and how frequently they ate a salad. Results are shown in the table.
\nJing conducted a 2 test for independence at a 5 % level of significance.
\nState the null hypothesis.
\nCalculate the -value for this test.
\nState, giving a reason, whether the null hypothesis should be accepted.
\nnumber of salad meals per week is independent of a person’s position in the university A1
\nNote: Accept “not associated” instead of independent.
\n[1 mark]
\n0.0201 (0.0201118…) A2
\n[2 marks]
\n0.0201 < 0.05 R1
\nthe null hypothesis is rejected A1
Note: Award (R1) for a correct comparison of their -value to the test level, award (A1) for the correct interpretation from that comparison.
Do not award (R0)(A1).
\n[2 marks]
\nA function satisfies the conditions , and its second derivative is , ≥ 0.
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)A1A1
\nNote: A1 for first term, A1 for second term. Withhold one A1 if extra terms are seen.
\n\n
A1
\nNote: Allow FT from incorrect if it is of the form .
\nAccept .
\n\n
attempt to use at least one boundary condition in their (M1)
\n,
\n⇒ A1
\n,
\n⇒
\n⇒ A1
\n\n
\n
[7 marks]
\nGiven that and , find
\n.
\n.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(M1)(A1)
\nA1
\nA1
\n[4 marks]
\n(M1)
\n= 12 A1
\n[2 marks]
\nConsider the function .
\nLet .
\nDetermine an expression for in terms of .
\nSketch a graph of for .
\nFind the -coordinate(s) of the point(s) of inflexion of the graph of , labelling these clearly on the graph of .
\nExpress in terms of .
\nExpress in terms of .
\nHence show that can be expressed as .
\nSolve the equation , giving your answers in the form where .
\n(or equivalent) (M1)A1
\n[2 marks]
\n![\"N17/5/MATHL/HP2/ENG/TZ0/11.a.ii/M\"](\"../assets/uploads/tinymce_asset/asset/4123/Schermafbeelding_2018-02-08_om_16.47.49.png\"/)
A1A1A1A1
\n
Note: Award A1 for correct behaviour at , A1 for correct domain and correct behaviour for , A1 for two clear intersections with -axis and minimum point, A1 for clear maximum point.
\n\n
[4 marks]
\nA1
\nA1
\n[2 marks]
\nattempt to write in terms of only (M1)
\nA1
\n[2 marks]
\n(A1)
\nattempt to use (M1)
\nA1
\n[3 marks]
\n\n
M1
\n(or equivalent) A1
\nAG
\n[2 marks]
\nor (M1)
\nA1
\nA1
\n\n
Note: Only accept answers given the required form.
\n\n
[3 marks]
\nThe function is defined by , ≥ 1 and the function is defined by , ≥ 0.
\nThe region is bounded by the curves , and the lines , and as shown on the following diagram.
\nThe shape of a clay vase can be modelled by rotating the region through 360˚ about the -axis.
\nFind the volume of clay used to make the vase.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
volume (M1)(M1)(M1)(A1)(A1)
\nNote: Award (M1) for use of formula for rotating about -axis, (M1) for finding at least one inverse, (M1) for subtracting volumes, (A1)(A1)for each correct expression, including limits.
\n\n
A2
\n[7 marks]
\nA point P moves in a straight line with velocity ms−1 given by at time t seconds, where t ≥ 0.
\nDetermine the first time t1 at which P has zero velocity.
\nFind an expression for the acceleration of P at time t.
\nFind the value of the acceleration of P at time t1.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to solve for t or equivalent (M1)
\nt1 = 0.441(s) A1
\n[2 marks]
\nM1A1
\nNote: Award M1 for attempting to differentiate using the product rule.
\n[2 marks]
\n(ms−2) A1
\n[1 mark]
\nXavier, the parachutist, jumps out of a plane at a height of metres above the ground. After free falling for 10 seconds his parachute opens. His velocity, , seconds after jumping from the plane, can be modelled by the function
\n\n
His velocity when he reaches the ground is .
\nFind his velocity when .
\nCalculate the vertical distance Xavier travelled in the first 10 seconds.
\nDetermine the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\nA1
\n[2 marks]
\n(M1)
\nA1
\n[2 marks]
\n(M1)
\nA1
\n(M1)(A1)
\nA1
\n[5 marks]
\nEmployees answer the telephone in a customer relations department. The time taken for an employee to deal with a customer is a random variable which can be modelled by a normal distribution with mean 150 seconds and standard deviation 45 seconds.
\nFind the probability that the time taken for a randomly chosen customer to be dealt with by an employee is greater than 180 seconds.
\nFind the probability that the time taken by an employee to deal with a queue of three customers is less than nine minutes.
\nAt the start of the day, one employee, Amanda, has a queue of four customers. A second employee, Brian, has a queue of three customers. You may assume they work independently.
\nFind the probability that Amanda’s queue will be dealt with before Brian’s queue.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
Note: In question 2, accept answers that round correctly to 2 significant figures.
\n\n
(M1)A1
\n[2 marks]
\nNote: In question 2, accept answers that round correctly to 2 significant figures.
\nrequired to find
\nlet
\n(A1)
\n(M1)
\n(A1)
\nA1
\nNote: In (b) and (c) condone incorrect notation, eg, for .
\n[4 marks]
\nNote: In question 2, accept answers that round correctly to 2 significant figures.
\nlet (M1)
\n(A1)
\n(M1)
\n= 14175 (A1)
\nrequired to find (M1)
\n= 0.104 A1
\n[6 marks]
\nA smartphone’s battery life is defined as the number of hours a fully charged battery can be used before the smartphone stops working. A company claims that the battery life of a model of smartphone is, on average, 9.5 hours. To test this claim, an experiment is conducted on a random sample of 20 smartphones of this model. For each smartphone, the battery life, hours, is measured and the sample mean, , calculated. It can be assumed the battery lives are normally distributed with standard deviation 0.4 hours.
\nIt is then found that this model of smartphone has an average battery life of 9.8 hours.
\nState suitable hypotheses for a two-tailed test.
\nFind the critical region for testing at the 5 % significance level.
\nFind the probability of making a Type II error.
\nAnother model of smartphone whose battery life may be assumed to be normally distributed with mean μ hours and standard deviation 1.2 hours is tested. A researcher measures the battery life of six of these smartphones and calculates a confidence interval of [10.2, 11.4] for μ.
\nCalculate the confidence level of this interval.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
Note: In question 3, accept answers that round correctly to 2 significant figures.
\nA1
\n[1 mark]
\nNote: In question 3, accept answers that round correctly to 2 significant figures.
\nthe critical values are (M1)(A1)
\ni.e. 9.3247…, 9.6753…
\nthe critical region is < 9.32, > 9.68 A1A1
\nNote: Award A1 for correct inequalities, A1 for correct values.
\nNote: Award M0 if t-distribution used, note that t(19)97.5 = 2.093 …
\n[4 marks]
\nNote: In question 3, accept answers that round correctly to 2 significant figures.
\n(A1)
\n(M1)
\n=0.0816 A1
\nNote: FT the critical values from (b). Note that critical values of 9.32 and 9.68 give 0.0899.
\n[3 marks]
\nNote: In question 3, accept answers that round correctly to 2 significant figures.
\nMETHOD 1
\n(M1)(A1)
\nP(10.2 < X < 11.4) = 0.7793… (A1)
\nconfidence level is 77.9% A1
\nNote: Accept 78%.
\nMETHOD 2
\n(M1)
\n(A1)
\nP(−1.224… < Z < 1.224…) = 0.7793… (A1)
\nconfidence level is 77.9% A1
\nNote: Accept 78%.
\n[4 marks]
\nA particle moves in a straight line such that at time seconds , its velocity , in , is given by . Find the exact distance travelled by the particle in the first half-second.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
attempt at integration by parts M1
\nA1
\n(A1)
\n\n
Note: Condone absence of limits (or incorrect limits) and missing factor of 10 up to this point.
\n\n
(M1)
\nA1
\n[5 marks]
\nA particle moves along a straight line. Its displacement, metres, at time seconds is given by . The first two times when the particle is at rest are denoted by and , where .
\nFind and .
\nFind the displacement of the particle when
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
M1A1
\nM1
\n\n
A1A1
\n\n
Note: Award A0A0 if answers are given in degrees.
\n\n
[5 marks]
\nA1A1
\n[2 marks]
\nTwo independent random variables and follow Poisson distributions.
\nGiven that and , calculate
\n.
\nVar.
\n.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)A1
\n[2 marks]
\nVar and Var (R1)
\n(M1)
\n= 84 A1
\n\n
[3 marks]
\nuse of (M1)
\n; A1
\n(M1)
\n= −8 A1
\n\n
[4 marks]
\nPoints A(3, 1), B(3, 5), C(11, 7), D(9, 1) and E(7, 3) represent snow shelters in the Blackburn National Forest. These snow shelters are illustrated in the following coordinate axes.
\nHorizontal scale: 1 unit represents 1 km.
\nVertical scale: 1 unit represents 1 km.
\nThe Park Ranger draws three straight lines to form an incomplete Voronoi diagram.
\nCalculate the gradient of the line segment AE.
\nFind the equation of the line which would complete the Voronoi cell containing site E.
\nGive your answer in the form where , , .
\nIn the context of the question, explain the significance of the Voronoi cell containing site E.
\n(M1)
\n= 0.5 A1
\n[2 marks]
\n(A1) (M1)
\nNote: Award (A1) for their −2 seen, award (M1) for the correct substitution of (5, 2) and their normal gradient in equation of a line.
\nA1
\n[3 marks]
\nevery point in the cell is closer to E than any other snow shelter A1
\n[1 mark]
\nThe times , in minutes, taken by a random sample of 75 workers of a company to travel to work can be summarized as follows
\n, .
\nLet be the random variable that represents the time taken to travel to work by a worker of this company.
\nFind unbiased estimates of the mean of .
\nFind unbiased estimates of the variance of .
\nAssuming that is normally distributed, find
\n(i) the 90% confidence interval for the mean time taken to travel to work by the workers of this company,
\n(ii) the 95% confidence interval for the mean time taken to travel to work by the workers of this company.
\nBefore seeing these results the managing director believed that the mean time was 26 minutes.
\nExplain whether your answers to part (b) support her belief.
\nA1
\n\n
[1 mark]
\n(M1)A1
\nNote: Accept all answers that round to 28.9 and 189.
\nNote: Award M0 if division by 75.
\n\n
[2 marks]
\nattempting to find a confidence interval. (M1)
\n(i) 90% interval: (26.2, 31.5) A1
\n(ii) 95% interval: (25.7, 32.0) A1
\n\n
Note: Accept any values which round to within 0.1 of the correct value.
\nNote: Award M1A1A0 if only confidence limits are given in the form 28.9 ± 2.6.
\n\n
[3 marks]
\n26 lies within the 95% interval but not within the 90% interval R1
\nNote: Award R1 for considering whether or not one or two of the intervals contain 26.
\n\n
the belief is supported at the 5% level (accept 95%) A1
\nthe belief is not supported at the 10% level (accept 90%) A1
\nNote: FT their intervals but award R1A1A0 if both intervals give the same conclusion.
\n\n
[3 marks]
\nA particle moves along a horizontal line such that at time seconds, ≥ 0, its acceleration is given by = 2 − 1. When = 6 , its displacement from a fixed origin O is 18.25 m. When = 15, its displacement from O is 922.75 m. Find an expression for in terms of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to integrate to find M1
\n\n
A1
\n\n
A1
\nattempt at substitution of given values (M1)
\nat
\nat
\nsolve simultaneously: (M1)
\nA1
\n\n
\n
[6 marks]
\nThe weights, X kg, of the males of a species of bird may be assumed to be normally distributed with mean 4.8 kg and standard deviation 0.2 kg.
\nThe weights, Y kg, of female birds of the same species may be assumed to be normally distributed with mean 2.7 kg and standard deviation 0.15 kg.
\nFind the probability that a randomly chosen male bird weighs between 4.75 kg and 4.85 kg.
\nFind the probability that the weight of a randomly chosen male bird is more than twice the weight of a randomly chosen female bird.
\nTwo randomly chosen male birds and three randomly chosen female birds are placed on a weighing machine that has a weight limit of 18 kg. Find the probability that the total weight of these five birds is greater than the weight limit.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
Note: In question 1, accept answers that round correctly to 2 significant figures.
\nP(4.75 < X < 4.85) = 0.197 A1
\n[1 mark]
\nNote: In question 1, accept answers that round correctly to 2 significant figures.
\nconsider the random variable X − 2Y (M1)
\nE(X − 2Y) = − 0.6 (A1)
\nVar(X − 2Y) = Var(X) + 4Var(Y) (M1)
\n= 0.13 (A1)
\nX − 2Y ∼ N(−0.6, 0.13)
\nP(X − 2Y > 0) (M1)
\n= 0.0480 A1
\n[6 marks]
\nNote: In question 1, accept answers that round correctly to 2 significant figures.
\nlet W = X1 + X2 + Y1 + Y2 + Y3 be the total weight
\nE(W) = 17.7 (A1)
\nVar(W) = 2Var(X) + 3Var(Y) = 0.1475 (M1)(A1)
\nW ∼ N(17.7, 0.1475)
\nP(W > 18) = 0.217 A1
\n[4 marks]
\n\n
The intensity level of sound, measured in decibels (dB), is a function of the sound intensity, watts per square metre (W m−2). The intensity level is given by the following formula.
\n, ≥ 0.
\nAn orchestra has a sound intensity of 6.4 × 10−3 W m−2 . Calculate the intensity level, of the orchestra.
\nA rock concert has an intensity level of 112 dB. Find the sound intensity, .
\n(M1)
\n= 98.1(dB) (98.06179…) A1
\n[2 marks]
\n(M1)
\n0.158 (W m−2) (0.158489… (W m−2)) A1
\n[2 marks]
\nMs Calhoun measures the heights of students in her mathematics class. She is interested to see if the mean height of male students, , is the same as the mean height of female students, . The information is recorded in the table.
\nAt the 10 % level of significance, a -test was used to compare the means of the two groups. The data is assumed to be normally distributed and the standard deviations are equal between the two groups.
\nState the null hypothesis.
\nState the alternative hypothesis.
\nCalculate the -value for this test.
\nState, giving a reason, whether Ms Calhoun should accept the null hypothesis.
\nA1
\nNote: Accept equivalent statements in words.
\n[1 mark]
\nA1
\nNote: Accept equivalent statements in words.
\n[1 mark]
\n0.296 (0.295739…) A2
\n[2 marks]
\n0.296 > 0.1 R1
\nfail to reject the null hypothesis, there is no difference between the mean height of male and female students A1
\nNote: Award (R1) for a correct comparison of their -value to the test level, award (A1) for the correct interpretation from that comparison.
Do not award R0A1.
[2 marks]
\nAnne is a farmer who grows and sells pumpkins. Interested in the weights of pumpkins produced, she records the weights of eight pumpkins and obtains the following results in kilograms.
\n\n
Assume that these weights form a random sample from a distribution.
\n\n
Anne claims that the mean pumpkin weight is 7.5 kilograms. In order to test this claim, she sets up the null hypothesis .
\nDetermine unbiased estimates for and .
\nUse a two-tailed test to determine the -value for the above results.
\nInterpret your -value at the 5% level of significance, justifying your conclusion.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
UE of is A1
\nUE of is 0.404 (M1)A1
\n\n
Note: Accept answers that round correctly to 2 sf.
\n\n
Note: Condone incorrect notation, ie, instead of UE of and instead of UE of .
\n\n
Note: M0 for squaring giving 0.354, M1A0 for failing to square
\n\n
[3 marks]
\nattempting to use the -test (M1)
\n-value is 0.0566 A2
\n\n
Note: Accept any answer that rounds correctly to 2 sf.
\n\n
[3 marks]
\nR1
\nwe accept the null hypothesis (mean pumpkin weight is 7.5 kg) A1
\n\n
Note: Apply follow through on the candidate’s -value.
\n\n
Note: Do not award A1 if R1 is not awarded.
\n\n
[2 marks]
\nThe following diagram shows part of the graph of , . The shaded region R is bounded by the -axis, -axis and the graph of .
\nWrite down an integral for the area of region R.
\nFind the area of region R.
\nThe three points A(0, 0) , B(3, 10) and C(, 0) define the vertices of a triangle.
\nFind the value of , the -coordinate of C, such that the area of the triangle is equal to the area of region R.
\nA = A1A1
\nNote: Award A1 for the limits = 0, = 2. Award A1 for an integral of .
\n[2 marks]
\n28 A1
\n[1 mark]
\nM1
\nA1
\n[2 marks]
\nIt was pleasing to see that, for those candidates who made a reasonable attempt at the paper, many were able to identify the correct values on the tree diagram.
\nA curve C is given by the implicit equation .
\nThe curve intersects C at P and Q.
\nShow that .
\nFind the coordinates of P and Q.
\nGiven that the gradients of the tangents to C at P and Q are m1 and m2 respectively, show that m1 × m2 = 1.
\nFind the coordinates of the three points on C, nearest the origin, where the tangent is parallel to the line .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt at implicit differentiation M1
\nA1M1A1
\nNote: Award A1 for first two terms. Award M1 for an attempt at chain rule A1 for last term.
\nA1
\nAG
\n[5 marks]
\nEITHER
\nwhen M1
\n(A1)
\nOR
\nor equivalent M1
\n(A1)
\nTHEN
\ntherefore A1
\nor A1
\n[4 marks]
\nm1 = M1A1
\nm2 = A1
\nm1 m2 = 1 AG
\nNote: Award M1A0A0 if decimal approximations are used.
Note: No FT applies.
[3 marks]
\nequate derivative to −1 M1
\n(A1)
\nR1
\nin the first case, attempt to solve M1
\n(0.486,0.486) A1
\nin the second case, and (M1)
\n(0,1), (1,0) A1
\n[7 marks]
\nConsider the system of paired differential equations
\n\n
.
\nThis system is going to be solved by using the eigenvalue method.
\n\n
If the system has a pair of purely imaginary eigenvalues
\nShow that if the system has two distinct real eigenvalues then .
\nFind two conditions that must be satisfied by , , , .
\nExplain why and must have opposite signs.
\nIn the case when there is a pair of purely imaginary eigenvalues you are told that the solution will form an ellipse. You are also told that the initial conditions are such that the ellipse is large enough that it will cross both the positive and negative axes and the positive and negative axes.
\nBy considering the differential equations at these four crossing point investigate if the trajectory is in a clockwise or anticlockwise direction round the ellipse. Give your decision in terms of and . Using part (b) (ii) show that your conclusions are consistent.
\nThe characteristic equation is given by
\nM1A1A1
\n\n
For two distinct real roots require R1
\nA1A1
\nAG
\n[6 marks]
\nUsing the working from part (a) (or using the characteristic equation) for a pair of purely imaginary eigenvalues require
\nand R1A1A1
\nand A1A1
\n[5 marks]
\nso and must have opposite signs M1AG
\n[1 mark]
\nWhen crossing the axes, so M1A1
\nWhen crossing the positive axes, has the sign of . A1
\nWhen crossing the negative axes, has the sign of . A1
\nHence if is positive the trajectory is anticlockwise and if is negative the trajectory is clockwise. R1R1
\nWhen crossing the axes, so M1A1
\nWhen crossing the positive axis, has the sign of . A1
\nWhen crossing the negative axes, has the sign of . A1
\nHence if is positive the trajectory is clockwise and if is negative the trajectory is anticlockwise. R1R1
\nSince by (b)(ii), and have opposite signs the above conditions agree with each other. R1
\n[13 marks]
\nJae Hee plays a game involving a biased six-sided die.
\nThe faces of the die are labelled −3, −1, 0, 1, 2 and 5.
\nThe score for the game, X, is the number which lands face up after the die is rolled.
\nThe following table shows the probability distribution for X.
\nJae Hee plays the game once.
\nFind the exact value of .
\nCalculate the expected score.
\nJae Hee plays the game twice and adds the two scores together.
\nFind the probability Jae Hee has a total score of −3.
\nA1
\n[1 mark]
\n(M1)
\nNote: Award (M1) for their correct substitution into the formula for expected value.
\nA1
\n[2 marks]
\n(M1)(M1)
\nNote: Award (M1) for , award (M1) for multiplying their product by 2.
\nA1
\n[3 marks]
\nHelen is building a cabin using cylindrical logs of length 2.4 m and radius 8.4 cm. A wedge is cut from one log and the cross-section of this log is illustrated in the following diagram.
\nFind the volume of this log.
\nvolume M1M1M1
\nNote: Award M1 240 × area, award M1 for correctly substituting area sector formula, award M1 for subtraction of their area of the sector from area of circle.
\n= 45800 (= 45811.96071) A1
\n[4 marks]
\nThe curve is defined by equation .
\nFind in terms of and .
\nDetermine the equation of the tangent to at the point
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1A1
\n\n
Note: Award A1 for the first two terms, A1 for the third term and the 0.
\n\n
A1
\n\n
Note: Accept .
\n\n
Note: Accept .
\n\n
[4 marks]
\n(M1)
\n(A1)
\n\n
or equivalent A1
\n\n
Note: Accept .
\n\n
[3 marks]
\nA camera at point C is 3 m from the edge of a straight section of road as shown in the following diagram. The camera detects a car travelling along the road at = 0. It then rotates, always pointing at the car, until the car passes O, the point on the edge of the road closest to the camera.
\nA car travels along the road at a speed of 24 ms−1. Let the position of the car be X and let OĈX = θ.
\nFind , the rate of rotation of the camera, in radians per second, at the instant the car passes the point O .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
let OX =
\nMETHOD 1
\n(or −24) (A1)
\n(M1)
\nA1
\nEITHER
\nA1
\n\n
attempt to substitute for into their differential equation M1
\nOR
\n\n
A1
\n\n
attempt to substitute for into their differential equation M1
\nTHEN
\n(rad s−1) A1
\nNote: Accept −8 rad s−1.
\n\n
METHOD 2
\n(or −24) (A1)
\nA1
\nattempt to differentiate implicitly with respect to M1
\nA1
\n\n
attempt to substitute for into their differential equation M1
\n(rad s−1) A1
\nNote: Accept −8 rad s−1.
\nNote: Can be done by consideration of CX, use of Pythagoras.
\n\n
METHOD 3
\nlet the position of the car be at time be from O (A1)
\nM1
\nNote: For award A0M1 and follow through.
\nEITHER
\nattempt to differentiate implicitly with respect to M1
\nA1
\nattempt to substitute for into their differential equation M1
\nOR
\nM1
\nA1
\nat O, A1
\nTHEN
\nA1
\n\n
[6 marks]
\nLet be the tangent to the curve at the point (1, ).
\nFind the coordinates of the point where meets the -axis.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1
\nequation of tangent is OR (M1)(A1)
\nmeets the -axis when
\n\n
meets -axis at (0.667, 0) A1A1
\nNote: Award A1 for or seen and A1 for coordinates (, 0) given.
\n\n
METHOD 1
\nAttempt to differentiate (M1)
\n\n
when , (M1)
\nequation of the tangent is
\n\n
meets -axis at
\nA1A1
\nNote: Award A1 for or seen and A1 for coordinates (, 0) given.
\n\n
[4 marks]
\nA curve has equation .
\nFind an expression for in terms of and .
\nFind the equations of the tangents to this curve at the points where the curve intersects the line .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to differentiate implicitly M1
\nA1A1A1
\n\n
Note: Award A1 for correctly differentiating each term.
\n\n
A1
\n\n
Note: This final answer may be expressed in a number of different ways.
\n\n
[5 marks]
\nA1
\nM1
\nat the tangent is and A1
\nat the tangent is A1
\n\n
Note: These equations simplify to .
\n\n
Note: Award A0M1A1A0 if just the positive value of is considered and just one tangent is found.
\n\n
[4 marks]
\nAn aircraft’s position is given by the coordinates (, , ), where and are the aircraft’s displacement east and north of an airport, and is the height of the aircraft above the ground. All displacements are given in kilometres.
\nThe velocity of the aircraft is given as .
\nAt 13:00 it is detected at a position 30 km east and 10 km north of the airport, and at a height of 5 km. Let be the length of time in hours from 13:00.
\nIf the aircraft continued to fly with the velocity given
\nWhen the aircraft is 4 km above the ground it continues to fly on the same bearing but adjusts the angle of its descent so that it will land at the point (0, 0, 0).
\nWrite down a vector equation for the displacement, r of the aircraft in terms of .
\nverify that it would pass directly over the airport.
\nstate the height of the aircraft at this point.
\nfind the time at which it would fly directly over the airport.
\nFind the time at which the aircraft is 4 km above the ground.
\nFind the direct distance of the aircraft from the airport at this point.
\nGiven that the velocity of the aircraft, after the adjustment of the angle of descent, is , find the value of .
\nr A1A1
\n[2 marks]
\nwhen , M1
\nEITHER
\nwhen , A1
\nsince the two values of are equal the aircraft passes directly over the airport
\nOR
\n, A1
\n[2 marks]
\nheight = 5 − 0.2 × 20 = 1 km A1
\n[1 mark]
\ntime 13:12 A1
\n[1 mark]
\n(3 minutes) (M1)
\ntime 13:03 A1
\n[2 marks]
\ndisplacement is A1
\ndistance is (M1)
\n= 24.1 km A1
\n[3 marks]
\nMETHOD 1
\ntime until landing is minutes M1
\nheight to descend =
\nM1
\nA1
\n\n
METHOD 2
\nM1
\nM1
\n\n
A1
\n[3 marks]
\nLet .
\nFind
\nHence find the values of θ for which .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt at chain rule or product rule (M1)
\nA1
\n[2 marks]
\n\n
sin θ = 0 (A1)
\nθ = 0, A1
\nobtaining cos θ = sin θ (M1)
\ntan θ = 1 (M1)
\nA1
\n[5 marks]
\nConsider the functions defined for , given by and .
\nFind .
\nFind .
\nHence, or otherwise, find .
\nattempt at product rule M1
\nA1
\n[2 marks]
\nA1
\n[1 mark]
\nMETHOD 1
\nAttempt to add and (M1)
\nA1
\n(or equivalent) A1
\nNote: Condone absence of limits.
\nA1
\n\n
METHOD 2
\n\n
OR M1A1
\n\n
A1
\nA1
\n[4 marks]
\nThe following table shows the costs in US dollars (US$) of direct flights between six cities. Blank cells indicate no direct flights. The rows represent the departure cities. The columns represent the destination cities.
\nThe following table shows the least cost to travel between the cities.
\nA travelling salesman has to visit each of the cities, starting and finishing at city A.
\nShow the direct flights between the cities as a graph.
\nWrite down the adjacency matrix for this graph.
\nUsing your answer to part (b), find the number of different ways to travel from and return to city A in exactly 6 flights.
\nState whether or not it is possible to travel from and return to city A in exactly 6 flights, having visited each of the other 5 cities exactly once. Justify your answer.
\nFind the values of and .
\nUse the nearest neighbour algorithm to find an upper bound for the cost of the trip.
\nBy deleting vertex A, use the deleted vertex algorithm to find a lower bound for the cost of the trip.
\n A2
[2 marks]
\nattempt to form an adjacency matrix M1
\nA1
\n[2 marks]
\nraising the matrix to the power six (M1)
\n50 A1
\n[2 marks]
\nnot possible A1
\nbecause you must pass through B twice R1
\nNote: Do not award A1R0.
\n[2 marks]
\n, A1A1
\n[2 marks]
\nA → B → D → E → F → C → A (M1)
\n90 + 70 + 100 + 210 + 330 + 150 (A1)
\n(US$) 950 A1
\n[3 marks]
\nfinding weight of minimum spanning tree M1
\n70 + 80 + 100 + 180 = (US$) 430 A1
\nadding in two edges of minimum weight M1
\n430 + 90 + 150 = (US$) 670 A1
\n[4 marks]
\nAn earth satellite moves in a path that can be described by the curve where and are in thousands of kilometres and is time in seconds.
\nGiven that when , find the possible values of .
\nGive your answers in standard form.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1
\nsubstituting for and attempting to solve for (or vice versa) (M1)
\n(A1)
\nEITHER
\nM1A1
\nOR
\nM1A1
\nTHEN
\nattempting to find (M1)
\nA1
\n\n
Note: Award all marks except the final A1 to candidates who do not consider ±.
\n\n
METHOD 2
\nM1A1
\n(M1)(A1)
\n(M1)
\nA1
\n\n
Note: Award all marks except the final A1 to candidates who do not consider ±.
\n\n
[6 marks]
\nAn object is placed into the top of a long vertical tube, filled with a thick viscous fluid, at time seconds.
\nInitially it is thought that the resistance of the fluid would be proportional to the velocity of the object. The following model was proposed, where the object’s displacement, , from the top of the tube, measured in metres, is given by the differential equation
\n.
\nThe maximum velocity approached by the object as it falls is known as the terminal velocity.
\nAn experiment is performed in which the object is placed in the fluid on a number of occasions and its terminal velocity recorded. It is found that the terminal velocity was consistently smaller than that predicted by the model used. It was suggested that the resistance to motion is actually proportional to the velocity squared and so the following model was set up.
\n\n
At terminal velocity the acceleration of an object is equal to zero.
\nBy substituting into the equation, find an expression for the velocity of the particle at time . Give your answer in the form .
\nFrom your solution to part (a), or otherwise, find the terminal velocity of the object predicted by this model.
\nWrite down the differential equation as a system of first order differential equations.
\nUse Euler’s method, with a step length of 0.2, to find the displacement and velocity of the object when .
\nBy repeated application of Euler’s method, find an approximation for the terminal velocity, to five significant figures.
\nUse the differential equation to find the terminal velocity for the object.
\nUse your answers to parts (d), (e) and (f) to comment on the accuracy of the Euler approximation to this model.
\nM1
\nM1
\nA1
\nA1
\nA1
\nwhen , hence A1
\n\n
A1
\n[7 marks]
\neither let tend to infinity, or (M1)
\nA1
\n[2 marks]
\nM1
\nA1
\n[2 marks]
\n, (M1)(A1)
\n, (M1)A1
\n[4 marks]
\n3.3015 A1
\n[1 mark]
\nM1
\nA1
\n[2 marks]
\nthe model found the terminal velocity very accurately, so good approximation R1
\nintermediate values had object exceeding terminal velocity so not good approximation R1
\n[2 marks]
\nTwo IB schools, A and B, follow the IB Diploma Programme but have different teaching methods. A research group tested whether the different teaching methods lead to a similar final result.
\nFor the test, a group of eight students were randomly selected from each school. Both samples were given a standardized test at the start of the course and a prediction for total IB points was made based on that test; this was then compared to their points total at the end of the course.
\nPrevious results indicate that both the predictions from the standardized tests and the final IB points can be modelled by a normal distribution.
\nIt can be assumed that:
\nThe data for school A is shown in the following table.
\nFor each student, the change from the predicted points to the final points was calculated.
\nThe data for school B is shown in the following table.
\nSchool A also gives each student a score for effort in each subject. This effort score is based on a scale of 1 to 5 where 5 is regarded as outstanding effort.
\nIt is claimed that the effort put in by a student is an important factor in improving upon their predicted IB points.
\nA mathematics teacher in school A claims that the comparison between the two schools is not valid because the sample for school B contained mainly girls and that for school A, mainly boys. She believes that girls are likely to show a greater improvement from their predicted points to their final points.
\nShe collects more data from other schools, asking them to class their results into four categories as shown in the following table.
\nIdentify a test that might have been used to verify the null hypothesis that the predictions from the standardized test can be modelled by a normal distribution.
\nState why comparing only the final IB points of the students from the two schools would not be a valid test for the effectiveness of the two different teaching methods.
\nFind the mean change.
\nFind the standard deviation of the changes.
\nUse a paired -test to determine whether there is significant evidence that the students in school A have improved their IB points since the start of the course.
\nUse an appropriate test to determine whether there is evidence, at the 5 % significance level, that the students in school B have improved more than those in school A.
\nState why it was important to test that both sets of points were normally distributed.
\nPerform a test on the data from school A to show it is reasonable to assume a linear relationship between effort scores and improvements in IB points. You may assume effort scores follow a normal distribution.
\nHence, find the expected improvement between predicted and final points for an increase of one unit in effort grades, giving your answer to one decimal place.
\nUse an appropriate test to determine whether showing an improvement is independent of gender.
\nIf you were to repeat the test performed in part (e) intending to compare the quality of the teaching between the two schools, suggest two ways in which you might choose your sample to improve the validity of the test.
\n(goodness of fit) A1
\n[1 mark]
\nEITHER
\nbecause aim is to measure improvement
\nOR
\nbecause the students may be of different ability in the two schools R1
\n[1 mark]
\n0.1875 (accept 0.188, 0.19) A1
\n[1 mark]
\n2.46 (M1)A1
\nNote: Award (M1)A0 for 2.63.
\n[2 marks]
\n: there has been no improvement
\n: there has been an improvement A1
\nattempt at a one-tailed paired -test (M1)
\n-value = 0.423 A1
\nthere is no significant evidence that the students have improved R1
\nNote: If the hypotheses are not stated award a maximum of A0M1A1R0.
\n[4 marks]
\n: there is no difference between the schools
\n: school B did better than school A A1
\none-tailed 2 sample -test (M1)
\n-value = 0.0984 A1
\n0.0984 > 0.05 (not significant at the 5 % level) so do not reject the null hypothesis R1A1
\nNote: The final A1 cannot be awarded following an incorrect reason. The final R1A1 can follow through from their incorrect -value. Award a maximum of A1(M1)A0R1A1 for -value = 0.0993.
\n[5 marks]
\nsample too small for the central limit theorem to apply (and -tests assume normal distribution) R1
\n[1 mark]
\n:
\n: A1
\nNote: Allow hypotheses to be expressed in words.
\n-value = 0.00157 A1
\n(0.00157 < 0.01) there is a significant evidence of a (linear) correlation between effort and improvement (so it is reasonable to assume a linear relationship) R1
\n[3 marks]
\n(gradient of line of regression =) 6.6 A1
\n\n
[1 mark]
\n: improvement and gender are independent
\n: improvement and gender are not independent A1
\nchoice of test for independence (M1)
\ngroups first two columns as expected values in first column less than 5 M1
\nnew observed table
\n (A1)
-value = 0.581 A1
\nno significant evidence that gender and improvement are dependent R1
\n[6 marks]
\nFor example:
\nlarger samples / include data from whole school
\ntake equal numbers of boys and girls in each sample
\nhave a similar range of abilities in each sample
\n(if possible) have similar ranges of effort R1R1
\nNote: Award R1 for each reasonable suggestion to improve the validity of the test.
\n[2 marks]
\nA city has two cable companies, X and Y. Each year 20 % of the customers using company X move to company Y and 10 % of the customers using company Y move to company X. All additional losses and gains of customers by the companies may be ignored.
\nInitially company X and company Y both have 1200 customers.
\nWrite down a transition matrix T representing the movements between the two companies in a particular year.
\nFind the eigenvalues and corresponding eigenvectors of T.
\nHence write down matrices P and D such that T = PDP−1.
\nFind an expression for the number of customers company X has after years, where .
\nHence write down the number of customers that company X can expect to have in the long term.
\nM1A1
\n[2 marks]
\nM1
\nand 0.7 A1
\neigenvectors and (M1)A1
\nNote: Accept any scalar multiple of the eigenvectors.
\n[4 marks]
\nEITHER
\nP = D = A1A1
\nOR
\nP = D = A1A1
\n[2 marks]
\nP−1 = A1
\nM1A1
\nattempt to multiply matrices M1
\nso in company A, after years, A1
\n[5 marks]
\n400 × 2 = 800 A1
\n[1 mark]
\nA farmer sells bags of potatoes which he states have a mean weight of 7 kg . An inspector, however, claims that the mean weight is less than 7 kg . In order to test this claim, the inspector takes a random sample of 12 of these bags and determines the weight, kg , of each bag. He finds that You may assume that the weights of the bags of potatoes can be modelled by the normal distribution .
\nState suitable hypotheses to test the inspector’s claim.
\nFind unbiased estimates of and .
\nCarry out an appropriate test and state the -value obtained.
\nUsing a 10% significance level and justifying your answer, state your conclusion in context.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
\n[1 mark]
\nA1
\n(M1)A1
\n[3 marks]
\n(M1)(A1)
\n(A1)
\nA1
\n\n
Note: Accept any answer that rounds correctly to 0.12.
\n\n
[4 marks]
\nbecause R1
\nthe inspector’s claim is not supported (at the 10% level)
\n(or equivalent in context) A1
\n\n
Note: Only award the A1 if the R1 has been awarded
\n\n
[2 marks]
\nFind the value of .
\nIllustrate graphically the inequality .
\nHence write down a lower bound for .
\nFind an upper bound for .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)
\nNote: The above A1 for using a limit can be awarded at any stage.
Condone the use of .
Do not award this mark to candidates who use as the upper limit throughout.
\n= M1
\n\n
A1
\n[3 marks]
\n A1A1A1A1
A1 for the curve
A1 for rectangles starting at
A1 for at least three upper rectangles
A1 for at least three lower rectangles
Note: Award A0A1 for two upper rectangles and two lower rectangles.
\nsum of areas of the lower rectangles < the area under the curve < the sum of the areas of the upper rectangles so
\nAG
\n[4 marks]
\na lower bound is A1
\nNote: Allow FT from part (a).
\n[1 mark]
\nMETHOD 1
\n(M1)
\n(M1)
\n, an upper bound A1
\nNote: Allow FT from part (a).
\n\n
METHOD 2
\nchanging the lower limit in the inequality in part (b) gives
\n(A1)
\n(M1)
\n, an upper bound A1
\nNote: Condone candidates who do not use a limit.
\n[3 marks]
\nConsider the following weighted graph G.
\nState what feature of G ensures that G has an Eulerian trail.
\nState what feature of G ensures that G does not have an Eulerian circuit.
\nWrite down an Eulerian trail in G.
\nStarting and finishing at B, find a solution to the Chinese postman problem for G.
\nCalculate the total weight of the solution.
\nG has an Eulerian trail because it has (exactly) two vertices (B and F) of odd degree R1
\n[1 mark]
\nG does not have an Eulerian circuit because not all vertices are of even degree R1
\n[1 mark]
\nfor example BAEBCEFCDF A1A1
\nNote: Award A1 for start/finish at B/F, A1 for the middle vertices.
\n[2 marks]
\nwe require the Eulerian trail in (b), (weight = 65) (M1)
\nand the minimum walk FEB (15) A1
\nfor example BAEBCEFCDFEB A1
\nNote: Accept EB added to the end or FE added to the start of their answer in (b) in particular for follow through.
\n[3 marks]
\ntotal weight is (65 + 15=)80 A1
\n[1 mark]
\nThe random variables follow a bivariate normal distribution with product moment correlation coefficient .
\nA random sample of 12 observations on U, V is obtained to determine whether there is a correlation between U and V. The sample product moment correlation coefficient is denoted by r. A test to determine whether or not U, V are independent is carried out at the 1% level of significance.
\nState suitable hypotheses to investigate whether or not , are independent.
\nFind the least value of for which the test concludes that .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1A1
\n[2 marks]
\n(A1)
\n(M1)(A1)
\nwe reject if (R1)
\nattempting to solve for M1
\n\n
Note: Allow = instead of >.
\n\n
(least value of is) 0.708 (3 sf) A1
\n\n
Note: Award A1M1A0R1M1A0 to candidates who use a one-tailed test. Award A0M1A0R1M1A0 to candidates who use an incorrect number of degrees of freedom or both a one-tailed test and incorrect degrees of freedom.
\n\n
Note: Possible errors are
\n10 DF 1-tail, , least value 0.658
\n11 DF 2-tail, , least value 0.684
\n11 DF 1-tail, , least value 0.634.
\n[6 marks]
\nIn the context of graph theory, explain briefly what is meant by a circuit;
\nIn the context of graph theory, explain briefly what is meant by an Eulerian circuit.
\nThe graph has six vertices and an Eulerian circuit. Determine whether or not its complement can have an Eulerian circuit.
\nFind an example of a graph , with five vertices, such that and its complement both have an Eulerian trail but neither has an Eulerian circuit. Draw and as your solution.
\na circuit is a walk that begins and ends at the same vertex and has no repeated edges A1
\n[1 mark]
\nan Eulerian circuit is a circuit that contains every edge of a graph A1
\n[1 mark]
\nif has an Eulerian circuit all vertices are even (are of degree 2 or 4) A1
\nhence, must have all vertices odd (of degree 1 or 3) R1
\nhence, cannot have an Eulerian circuit R1
\n\n
Note: Award A1 to candidates who begin by considering a specific and (diagram). Award R1R1 to candidates who then consider a general and .
\n\n
[3 marks]
\nfor example
\n![\"M17/5/MATHL/HP3/ENG/TZ0/DM/M/03.c\"](\"../assets/uploads/tinymce_asset/asset/3523/Schermafbeelding_2017-08-10_om_10.15.08.png\"/)
A2
\nA2
\n\n
Notes: Each graph must have 3 vertices of order 2 and 2 odd vertices. Award A2 if one of the graphs satisfies that and the final A2 if the other graph is its complement.
\n\n
[4 marks]
\nConsider the functions , : defined by
\nand .
\nFind .
\nFind .
\nState with a reason whether or not and commute.
\nFind the inverse of .
\n() (M1)
\nA1A1
\n\n
[3 marks]
\n\n
\n
\n
A1A1
\n\n
[2 marks]
\nno because R1
\nNote: Accept counter example.
\n\n
[1 mark]
\n\n
(M1)
\n(M1)
\nA1
\n\n
[3 marks]
\nThe following diagram shows a circle with centre O and radius 40 cm.
\n![\"M17/5/MATME/SP2/ENG/TZ2/01\"](\"../assets/uploads/tinymce_asset/asset/3553/Schermafbeelding_2017-08-14_om_17.31.00.png\"/)
The points A, B and C are on the circumference of the circle and .
\nFind the length of arc ABC.
\nFind the perimeter of sector OABC.
\nFind the area of sector OABC.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ncorrect substitution into arc length formula (A1)
\neg
\nA1 N2
\n[2 marks]
\nvalid approach (M1)
\neg
\nA1 N2
\n[2 marks]
\ncorrect substitution into area formula (A1)
\neg
\nA1 N2
\n[2 marks]
\nA set of data comprises of five numbers which have been placed in ascending order.
\nRecalling definitions, such as the Lower Quartile is the piece of data with the data placed in order, find an expression for the Interquartile Range.
\nHence, show that a data set with only 5 numbers in it cannot have any outliers.
\nGive an example of a set of data with 7 numbers in it that does have an outlier, justify this fact by stating the Interquartile Range.
\nM1A1
\n[2 marks]
\nM1A1
\nSince due to the ascending order. R1
\nSimilarly M1A1
\nSince due to the ascending order.
\nSo there are no outliers for a data set of 5 numbers. AG
\n[5 marks]
\nFor example 1, 2, 3, 4, 5, 6, 100 where A1A1
\n[2 marks]
\nThe following diagram shows a circle, centre O and radius mm. The circle is divided into five equal sectors.
\n![\"N16/5/MATME/SP2/ENG/TZ0/03\"](\"../assets/uploads/tinymce_asset/asset/3309/Schermafbeelding_2017-03-03_om_15.31.24.png\"/)
One sector is OAB, and .
\nThe area of sector AOB is .
\nWrite down the exact value of in radians.
\nFind the value of .
\nFind AB.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nA1 N1
\n[1 mark]
\ncorrect expression for area (A1)
\neg
\nevidence of equating their expression to (M1)
\neg
\nA1 N2
\n[3 marks]
\nMETHOD 1
\nevidence of choosing cosine rule (M1)
\neg
\ncorrect substitution of their and into RHS (A1)
\neg
\n11.7557
\nA1 N2
\nMETHOD 2
\nevidence of choosing sine rule (M1)
\neg
\ncorrect substitution of their and (A1)
\neg
\n11.7557
\nA1 N2
\n[3 marks]
\nYejin plans to retire at age 60. She wants to create an annuity fund, which will pay her a monthly allowance of $4000 during her retirement. She wants to save enough money so that the payments last for 30 years. A financial advisor has told her that she can expect to earn 5% interest on her funds, compounded annually.
\nCalculate the amount Yejin needs to have saved into her annuity fund, in order to meet her retirement goal.
\nYejin has just turned 28 years old. She currently has no retirement savings. She wants to save part of her salary each month into her annuity fund.
\nCalculate the amount Yejin needs to save each month, to meet her retirement goal.
\nUse of finance solver M1
\nN = 360, I = 5%, Pmt = 4000, FV = 0, PpY = 12, CpY = 1 A1
\n$755000 (correct to 3 s.f.) A1
\n[3 marks]
\nN = 384, I = 5%, PV = 0, FV = 754638, PpY = 12, CpY = 1 M1A1
\n$817 per month (correct to 3 s.f.) A1
\n[3 marks]
\nAs part of the selection process for an engineering course at a particular university, applicants are given an exam in mathematics. This year the university has produced a new exam and they want to test if it is a valid indicator of future performance, before giving it to applicants. They randomly select 8 students in their first year of the engineering course and give them the exam. They compare the exam scores with their results in the engineering course.
\nThe results of the 8 students are shown in the table.
\nState the name of this test for validity.
\nCalculate Pearson’s product moment correlation coefficient for this data.
\nHence determine, with a reason, if the new exam is a valid indicator of future performance.
\ncriterion-related A1
\n[1 mark]
\nA2
\n[2 marks]
\nSince the value of is low (closer to 0 than +1), R1
\nThe new exam is not a valid indicator of future performance. A1
\n[2 marks]
\nOAB is a sector of the circle with centre O and radius , as shown in the following diagram.
\nThe angle AOB is radians, where .
\nThe point C lies on OA and OA is perpendicular to BC.
\nFind the area of triangle OBC in terms of and θ.
\nvalid approach (M1)
\neg , , ,
\narea (must be in terms of and θ) A1 N2
\n[2 marks]
\nSix equilateral triangles, each with side length 3 cm, are arranged to form a hexagon.
This is shown in the following diagram.
The vectors p , q and r are shown on the diagram.
\nFind p•(p + q + r).
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1 (using |p| |2q| cosθ)
\nfinding p + q + r (A1)
\neg 2q,
| p + q + r | = 2 × 3 (= 6) (seen anywhere) A1
\ncorrect angle between p and q (seen anywhere) (A1)
\n(accept 60°)
\nsubstitution of their values (M1)
\neg 3 × 6 × cos
\ncorrect value for cos (seen anywhere) (A1)
\neg
\np•(p + q + r) = 9 A1 N3
\n\n
METHOD 2 (scalar product using distributive law)
\ncorrect expression for scalar distribution (A1)
\neg p• p + p•q + p•r
\nthree correct angles between the vector pairs (seen anywhere) (A2)
\neg 0° between p and p, between p and q, between p and r
\nNote: Award A1 for only two correct angles.
\nsubstitution of their values (M1)
\neg 3.3.cos0 +3.3.cos + 3.3.cos120
\none correct value for cos0, cos or cos (seen anywhere) A1
\neg
\np•(p + q + r) = 9 A1 N3
\n\n
METHOD 3 (scalar product using relative position vectors)
\nvalid attempt to find one component of p or r (M1)
\neg sin 60 = , cos 60 = , one correct value
\none correct vector (two or three dimensions) (seen anywhere) A1
\neg
\nthree correct vectors p + q + r = 2q (A1)
\np + q + r = or (seen anywhere, including scalar product) (A1)
\ncorrect working (A1)
eg
p•(p + q + r) = 9 A1 N3
\n[6 marks]
\nSaloni wants to find a model for the temperature of a bottle of water after she removes it from the fridge. She uses a temperature probe to record the temperature of the water, every 5 minutes.
\nAfter graphing the data, Saloni believes a suitable model will be
\n, where .
\nExplain why can be modeled by an exponential function.
\nFind the equation of the least squares exponential regression curve for .
\nWrite down the coefficient of determination, .
\nInterpret what the value of tells you about the model.
\nHence predict the temperature of the water after 3 minutes.
\nRearranging the model gives A1
\nSo can be modeled by an exponential function. AG
\n[2 marks]
\n (A1)
M1A1
\n[3 marks]
\nA1
\n[1 mark]
\nSince the value of is close to +1, the model is a good fit for the data. R1
\n[1 mark]
\nminutes M1A1
\n[2 marks]
\nThe magnitudes of two vectors, u and v, are 4 and respectively. The angle between u and v is .
\nLet w = u − v. Find the magnitude of w.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1 (cosine rule)
\ndiagram including u, v and included angle of (M1)
\neg
sketch of triangle with w (does not need to be to scale) (A1)
\neg
choosing cosine rule (M1)
\neg
\ncorrect substitution A1
\neg
\n(seen anywhere) (A1)
\ncorrect working (A1)
\neg 16 + 3 − 12
\n| w | = A1 N2
\n\n
METHOD 2 (scalar product)
\nvalid approach, in terms of u and v (seen anywhere) (M1)
\neg | w |2 = (u − v)•(u − v), | w |2 = u•u − 2u•v + v•v, | w |2 = ,
\n| w | =
\ncorrect value for u•u (seen anywhere) (A1)
\neg | u |2 = 16, u•u = 16,
\ncorrect value for v•v (seen anywhere) (A1)
\neg | v |2 = 16, v•v = 3,
\n(seen anywhere) (A1)
\nu•v (= 6) (seen anywhere) A1
\ncorrect substitution into u•u − 2u•v + v•v or (2 or 3 dimensions) (A1)
\neg 16 − 2(6) + 3 (= 7)
\n| w | = A1 N2
\nLet be an obtuse angle such that .
\nLet .
\nFind the value of .
\nLine passes through the origin and has a gradient of . Find the equation of .
\nFind the derivative of .
\nThe following diagram shows the graph of for 0 ≤ ≤ 3. Line is a tangent to the graph of at point P.
\nGiven that is parallel to , find the -coordinate of P.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nevidence of valid approach (M1)
\neg sketch of triangle with sides 3 and 5,
\ncorrect working (A1)
\neg missing side is 4 (may be seen in sketch), ,
\nA2 N4
\n[4 marks]
\ncorrect substitution of either gradient or origin into equation of line (A1)
\n(do not accept )
\neg , ,
\nA2 N4
\nNote: Award A1A0 for .
\n[2 marks]
\n(seen anywhere, including answer) A1
\nchoosing product rule (M1)
\neg
\ncorrect derivatives (must be seen in a correct product rule) A1A1
\neg ,
\nA1 N5
\n[5 marks]
\nvalid approach to equate their gradients (M1)
\neg , , ,
\ncorrect equation without (A1)
\neg , ,
\ncorrect working (A1)
\neg ,
\n(do not accept ) A1 N1
\nNote: Do not award the final A1 if additional answers are given.
\n[4 marks]
\n\n
Anita is concerned that the construction of a new factory will have an adverse affect on the fish in a nearby lake. Before construction begins she catches fish at random, records their weight and returns them to the lake. After the construction is finished she collects a second, random sample of weights of fish from the lake. Her data is shown in the table.
\nAnita decides to use a t-test, at the 5% significance level, to determine if the mean weight of the fish changed after construction of the factory.
\nState an assumption that Anita is making, in order to use a t-test.
\nState the hypotheses for this t-test.
\nFind the p-value for this t-test.
\nState the conclusion of this test, in context, giving a reason.
\nEITHER
\nThe weights of the fish are distributed normally. A1
\nOR
\nThe variance of the two groups of fish is equal. A1
\n[1 mark]
\nand A1
\nwhere B and A represent the weights before and after.
\n[1 mark]
\ndf = 14, t = 0.861 (M1)
\np-value = 0.403 A2
\n[3 marks]
\nSince 0.403 > 0.05 R1
\nDo not reject H0.
\nThere is insufficient evidence, at the 5% level, of a change in weight. A1
\n[2 marks]
\nA company sends a group of employees on a training course. Afterwards, they survey these employees to gather data on the effectiveness of the training. In order to test the reliability of the survey, they design two sets of similar questions, which are given to the employees one week apart.
\nThe questions in the survey were grouped in different sections. The mean scores of the employees on the first section of each survey are given in the table.
\nState the name of this test for reliability.
\nState a possible disadvantage of using this test for reliability.
\nCalculate Pearson’s product moment correlation coefficient for this data.
\nHence determine, with a reason, if the survey is reliable.
\nParallel Forms A1
\n[1 mark]
\nEITHER
\nThe two sets of questions might not be of equal difficulty R1
\nOR
\nIt is time consuming to create two sets of questions R1
\n[1 mark]
\nA2
\n[2 marks]
\nSince the value of is close to +1, R1
\nThe survey is reliable. A1
\n[2 marks]
\nThe number of brown squirrels, , in an area of woodland can be modelled by the following differential equation.
\n, where
\nOne year conservationists notice that some black squirrels are moving into the woodland. The two species of squirrel are in competition for the same food supplies. Let be the number of black squirrels in the woodland.
\nConservationists wish to predict the likely future populations of the two species of squirrels. Research from other areas indicates that when the two populations come into contact the growth can be modelled by the following differential equations, in which is measured in tens of years.
\n, , ≥ 0
\n, , ≥ 0
\nAn equilibrium point for the populations occurs when both and .
\nWhen the two populations are small the model can be reduced to the linear system
\n\n
.
\nFor larger populations, the conservationists decide to use Euler’s method to find the long‑term outcomes for the populations. They will use Euler’s method with a step length of 2 years ().
\nFind the equilibrium population of brown squirrels suggested by this model.
\nExplain why the population of squirrels is increasing for values of less than this value.
\nVerify that , is an equilibrium point.
\nFind the other three equilibrium points.
\nBy using separation of variables, show that the general solution of is .
\nWrite down the general solution of .
\nIf both populations contain 10 squirrels at use the solutions to parts (c) (i) and (ii) to estimate the number of black and brown squirrels when . Give your answers to the nearest whole numbers.
\nWrite down the expressions for and that the conservationists will use.
\nGiven that the initial populations are , , find the populations of each species of squirrel when .
\nUse further iterations of Euler’s method to find the long-term population for each species of squirrel from these initial values.
\nUse the same method to find the long-term populations of squirrels when the initial populations are , .
\nUse Euler’s method with step length 0.2 to sketch, on the same axes, the approximate trajectories for the populations with the following initial populations.
\n(i) ,
\n(ii) ,
\nGiven that the equilibrium point at (800, 600) is a saddle point, sketch the phase portrait for ≥ 0 , ≥ 0 on the same axes used in part (e).
\n2000 (M1)A1
\n[2 marks]
\nbecause the value of is positive (for ) R1
\n[1 mark]
\nsubstitute , into both equations M1
\nboth equations equal 0 A1
\nhence an equilibrium point AG
\n[3 marks]
\n, A1
\n, , , M1A1A1
\nNote: Award M1 for an attempt at solving the system provided some values of and are found.
\n[4 marks]
\nM1
\nA1A1
\nNote: Award A1 for RHS, A1 for LHS.
\nM1
\n(where ) AG
\n[4 marks]
\nA1
\nNote: Allow any letter for the constant term, including .
\n[1 mark]
\n, (M1)A1
\n[2 marks]
\n\n
M1A1
\nNote: Accept equivalent forms.
\n[2 marks]
\n, (M1)A1A1
\n[3 marks]
\nnumber of brown squirrels go down to 0,
black squirrels to a population of 3000 A1
[1 mark]
\nnumber of brown squirrels go to 2000,
number of black squirrels goes down to 0 A1
[1 mark]
\n(i) AND (ii)
\n M1A1A1
[3 marks]
\n A1A1
Note: Award A1 for a trajectory beginning close to (0, 0) and going to (0, 3000) and A1 for a trajectory beginning close to (0, 0) and going to (2000, 0) in approximately the correct places.
\n[2 marks]
\nThe matrix A is defined by .
\nPentagon, P, which has an area of 7 cm2, is transformed by A.
\nThe matrix B is defined by .
\nB represents the combined effect of the transformation represented by a matrix X, followed by the transformation represented by A.
\nDescribe fully the geometrical transformation represented by A.
\nFind the area of the image of P.
\nFind the matrix X.
\nDescribe fully the geometrical transformation represented by X.
\nstretch A1
\nscale factor 3, A1
\ny-axis invariant (condone parallel to the x-axis) A1
\nand
\nstretch, scale factor 2, A1
\nx-axis invariant (condone parallel to the y-axis) A1
\n[5 marks]
\nA1
\nA1
\n[2 marks]
\n(A1)
\n(M1)
\nA1
\n[3 marks]
\nRotation A1
\nclockwise by 30° about the origin A1
\n[2 marks]
\nLet , for . The following diagram shows part of the graph of and the rectangle OABC, where A is on the negative -axis, B is on the graph of , and C is on the -axis.
\n![\"N17/5/MATME/SP1/ENG/TZ0/06\"](\"../assets/uploads/tinymce_asset/asset/4152/Schermafbeelding_2018-02-11_om_13.13.04.png\"/)
Find the -coordinate of A that gives the maximum area of OABC.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nattempt to find the area of OABC (M1)
\neg
\ncorrect expression for area in one variable (A1)
\neg
\nvalid approach to find maximum area (seen anywhere) (M1)
\neg
\ncorrect derivative A1
\neg
\ncorrect working (A1)
\neg
\nA2 N3
\n[7 marks]
\nNote: In this question, distance is in millimetres.
\nLet , for .
\nThe graph of passes through the origin. Let be any point on the graph of with -coordinate , where . A straight line passes through all the points .
\nDiagram 1 shows a saw. The length of the toothed edge is the distance AB.
\n![\"N17/5/MATME/SP2/ENG/TZ0/10.d_01\"](\"../assets/uploads/tinymce_asset/asset/4167/Schermafbeelding_2018-02-12_om_15.10.11.png\"/)
The toothed edge of the saw can be modelled using the graph of and the line . Diagram 2 represents this model.
\n![\"N17/5/MATME/SP2/ENG/TZ0/10.d_02\"](\"../assets/uploads/tinymce_asset/asset/4168/Schermafbeelding_2018-02-12_om_15.11.17.png\"/)
The shaded part on the graph is called a tooth. A tooth is represented by the region enclosed by the graph of and the line , between and .
\nShow that .
\nFind the coordinates of and of .
\nFind the equation of .
\nShow that the distance between the -coordinates of and is .
\nA saw has a toothed edge which is 300 mm long. Find the number of complete teeth on this saw.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
substituting M1
\neg
\n(A1)
\nA1
\nAG N0
\n[3 marks]
\nsubstituting the value of (M1)
\nA1A1 N3
\n[3 marks]
\nattempt to find the gradient (M1)
\neg
\ncorrect working (A1)
\neg
\ny = x A1 N3
\n[3 marks]
\nsubtracting -coordinates of and (in any order) (M1)
\neg
\ncorrect working (must be in correct order) A1
\neg
\ndistance is AG N0
\n[2 marks]
\nMETHOD 1
\nrecognizing the toothed-edge as the hypotenuse (M1)
\neg, sketch
\ncorrect working (using their equation of (A1)
\neg
\n(exact), 212.132 (A1)
\ndividing their value of by (M1)
\neg
\n33.7618 (A1)
\n33 (teeth) A1 N2
\nMETHOD 2
\nvertical distance of a tooth is (may be seen anywhere) (A1)
\nattempt to find the hypotenuse for one tooth (M1)
\neg
\n(exact), 8.88576 (A1)
\ndividing 300 by their value of (M1)
\neg
\n33.7618 (A1)
\n33 (teeth) A1 N2
\n[6 marks]
\nThe matrices A and B are defined by and .
\nTriangle X is mapped onto triangle Y by the transformation represented by AB. The coordinates of triangle Y are (0, 0), (−30, −20) and (−16, 32).
\nDescribe fully the geometrical transformation represented by B.
\nFind the coordinates of triangle X.
\nFind the area of triangle X.
\nHence find the area of triangle Y.
\nMatrix A represents a combination of transformations:
\nA stretch, with scale factor 3 and y-axis invariant;
Followed by a stretch, with scale factor 4 and x-axis invariant;
Followed by a transformation represented by matrix C.
Find matrix C.
\nreflection in the y-axis A1A1
\n[2 marks]
\nM1
\nEITHER
\n, so M1A1
\nOR
\nM1A1
\nTHEN
\n(A1)
\nSo the coordinates are (0, 0), (10, 0) and (0, 8). A1
\n[5 marks]
\nunits2 M1A1
\n[2 marks]
\nM1A1
\nArea units2 A1
\n[3 marks]
\nA stretch, with scale factor 3 and y-axis invariant is given by A1
\nA stretch, with scale factor 4 and x-axis invariant is given by A1
\nSo M1A1
\n[4 marks]
\nAt Grande Anse Beach the height of the water in metres is modelled by the function , where is the number of hours after 21:00 hours on 10 December 2017. The following diagram shows the graph of , for .
\n![\"M17/5/MATME/SP2/ENG/TZ1/08\"](\"../assets/uploads/tinymce_asset/asset/3544/Schermafbeelding_2017-08-14_om_10.10.26.png\"/)
The point represents the first low tide and represents the next high tide.
\nHow much time is there between the first low tide and the next high tide?
\nFind the difference in height between low tide and high tide.
\nFind the value of ;
\nFind the value of ;
\nFind the value of .
\nThere are two high tides on 12 December 2017. At what time does the second high tide occur?
\nattempt to find the difference of -values of A and B (M1)
\neg
\n6.25 (hours), (6 hours 15 minutes) A1 N2
\n[2 marks]
\nattempt to find the difference of -values of A and B (M1)
\neg
\nA1 N2
\n[2 marks]
\nvalid approach (M1)
\neg
\nA1 N2
\n[2 marks]
\nMETHOD 1
\nperiod (seen anywhere) (A1)
\nvalid approach (seen anywhere) (M1)
\neg
\n0.502654
\nA1 N2
\nMETHOD 2
\nattempt to use a coordinate to make an equation (M1)
\neg
\ncorrect substitution (A1)
\neg
\n0.502654
\nA1 N2
\n[3 marks]
\nvalid method to find (M1)
\neg
\nA1 N2
\n[2 marks]
\nMETHOD 1
\nattempt to find start or end -values for 12 December (M1)
\neg
\nfinds -value for second max (A1)
\n\n
23:00 (or 11 pm) A1 N3
\nMETHOD 2
\nvalid approach to list either the times of high tides after 21:00 or the -values of high tides after 21:00, showing at least two times (M1)
\neg
\ncorrect time of first high tide on 12 December (A1)
\neg10:30 (or 10:30 am)
\ntime of second high tide = 23:00 A1 N3
\nMETHOD 3
\nattempt to set their equal to 1.5 (M1)
\neg
\ncorrect working to find second max (A1)
\neg
\n23:00 (or 11 pm) A1 N3
\n[3 marks]
\nThe weights of the edges of a complete graph G are shown in the following table.
\nStarting at B, use Prim’s algorithm to find and draw a minimum spanning tree for G. Your solution should indicate the order in which the vertices are added. State the total weight of your tree.
\nDifferent notations may be used but the edges should be added in the following order.
\nUsing Prim’s Algorithm, (M1)
\nBD A1
\nDF A1
\nFA A1
\nFE A1
\nEC A1
\n A2
Total weight = 12 A2
\n[10 marks]
\nThe weights of the edges in a simple graph G are given in the following table.
\nUse Prim’s Algorithm, starting with vertex F, to find and draw the minimum spanning tree for G. Your solution should indicate the order in which the edges are introduced.
\nThe edges are introduced in the following order:
\nFD, FC, CB, BA, CE A2A2A2A2A2
\n A2
[12 marks]
\nA pharmaceutical company has developed a new drug to decrease cholesterol. The final stage of testing the new drug is to compare it to their current drug. They have 150 volunteers, all recently diagnosed with high cholesterol, from which they want to select a sample of size 18. They require as close as possible 20% of the sample to be below the age of 30, 30% to be between the ages of 30 and 50 and 50% to be over the age of 50.
\nHalf of the 18 volunteers are given the current drug and half are given the new drug. After six months each volunteer has their cholesterol level measured and the decrease during the six months is shown in the table.
\nCalculate the mean decrease in cholesterol for
\nThe company uses a t-test, at the 1% significance level, to determine if the new drug is more effective at decreasing cholesterol.
\nState the name for this type of sampling technique.
\nCalculate the number of volunteers in the sample under the age of 30.
\nThe new drug.
\nThe current drug.
\nState an assumption that the company is making, in order to use a t-test.
\nState the hypotheses for this t-test.
\nFind the p-value for this t-test.
\nState the conclusion of this test, in context, giving a reason.
\nstratified sampling A1
\n[1 mark]
\nM1A1
\nso 4 volunteers need to be chosen A1
\n[3 marks]
\n34.8 mg/dL A1
\n[1 mark]
\n24.7 mg/dL A1
\n[1 mark]
\nEITHER
\nThe decreases in cholesterol are distributed normally A1
\nOR
\nThe variance of the two groups of volunteers is equal. A1
\n[1 mark]
\nand A1
\nwhere N and C represent the decreases of the new and current drug
\n[1 mark]
\ndf = 16, t = 2.77 (M1)
\np-value = 0.00683 A2
\n[3 marks]
\nSince 0.00683 < 0.01 R1
\nReject H0. There is evidence, at the 1% level, that the new drug is more effective. A1
\n[2 marks]
\nThis question explores methods to determine the area bounded by an unknown curve.
\nThe curve is shown in the graph, for .
\nThe curve passes through the following points.
\nIt is required to find the area bounded by the curve, the -axis, the -axis and the line .
\nOne possible model for the curve is a cubic function.
\nA second possible model for the curve is an exponential function, , where .
\nUse the trapezoidal rule to find an estimate for the area.
\nWith reference to the shape of the graph, explain whether your answer to part (a)(i) will be an over-estimate or an underestimate of the area.
\nUse all the coordinates in the table to find the equation of the least squares cubic regression curve.
\nWrite down the coefficient of determination.
\nWrite down an expression for the area enclosed by the cubic function, the -axis, the -axis and the line .
\nFind the value of this area.
\nShow that .
\nHence explain how a straight line graph could be drawn using the coordinates in the table.
\nBy finding the equation of a suitable regression line, show that and .
\nHence find the area enclosed by the exponential function, the -axis, the -axis and the line .
\nArea M1A1
\nArea = 156 units2 A1
\n[3 marks]
\nThe graph is concave up, R1
\nso the trapezoidal rule will give an overestimate. A1
\n[2 marks]
\nM1A2
\n[3 marks]
\nA1
\n[1 mark]
\nArea A1A1
\n[2 marks]
\nArea = 145 units2 (Condone 143–145 units2, using rounded values.) A2
\n[2 marks]
\nM1
\nA1
\nAG
\n[2 marks]
\nPlot against . R1
\n[1 mark]
\nRegression line is M1A1
\nSo gradient = 0.986 R1
\nM1A1
\n[5 marks]
\nArea units2 M1A1
\n[2 marks]
\nGiven the matrix A = find the values of the real number for which where
\n\n
\n
(M1)
\n(M1)
\n\n
(A2) (C4)
\n[4 marks]
\nFind the values of and given that the matrix is the inverse of the matrix .
\nFor the values of and found in part (a), solve the system of linear equations
\n\n
AB = I
\n(AB)11 = 1 ⇒ – 12 + 6 = 1, giving = 7 (A1) (C1)
\n(AB)22 = 1 ⇒ –16 + 5 + 7 = 1, giving = 2 (A1) (C1)
\n[2 marks]
\nthe system is where .
\nThen, (M1)
\nThus , , (A1) (C2)
\n[2 marks]
\nLet , where .
\nFind in terms of .
\nIf is equal to , find the value of .
\nUsing this value of , find and hence solve the system of equations:
\n\n
\n
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n\n
(A1)(A1)(A1)(A1)
\n[4 marks]
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n\n
(A1)
\nSubstituting: (A1)
\nNote: Candidates may solve to give , and then show that only satisfies .
\n[2 marks]
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n\n
(M1)
\nor (A1)
\n\n
\n
(M1)(M1)
\n(A1)
\n(A1)
\nie
\n\n
Note: The solution must use matrices. Award no marks for solutions using other methods.
\n[6 marks]
\nFind the values of the real number for which the determinant of the matrix is equal to zero.
\n\n
(M1)
\n(M1)
\n\n
or (A1) (C3)
\n[3 marks]
\nThis question explores methods to analyse the scores in an exam.
\nA random sample of 149 scores for a university exam are given in the table.
\nThe university wants to know if the scores follow a normal distribution, with the mean and variance found in part (a).
\nThe expected frequencies are given in the table.
\nThe university assigns a pass grade to students whose scores are in the top 80%.
\nThe university also wants to know if the exam is gender neutral. They obtain random samples of scores for male and female students. The mean, sample variance and sample size are shown in the table.
\nThe university awards a distinction to students who achieve high scores in the exam. Typically, 15% of students achieve a distinction. A new exam is trialed with a random selection of students on the course. 5 out of 20 students achieve a distinction.
\nA different exam is trialed with 16 students. Let be the percentage of students achieving a distinction. It is desired to test the hypotheses
\nagainst
\nIt is decided to reject the null hypothesis if the number of students achieving a distinction is greater than 3.
\nFind unbiased estimates for the population mean.
\nFind unbiased estimates for the population Variance.
\nShow that the expected frequency for 20 < ≤ 4 is 31.5 correct to 1 decimal place.
\nPerform a suitable test, at the 5% significance level, to determine if the scores follow a normal distribution, with the mean and variance found in part (a). You should clearly state your hypotheses, the degrees of freedom, the p-value and your conclusion.
\nUse the normal distribution model to find the score required to pass.
\nPerform a suitable test, at the 5% significance level, to determine if there is a difference between the mean scores of males and females. You should clearly state your hypotheses, the p-value and your conclusion.
\nPerform a suitable test, at the 5% significance level, to determine if it is easier to achieve a distinction on the new exam. You should clearly state your hypotheses, the critical region and your conclusion.
\nFind the probability of making a Type I error.
\nGiven that find the probability of making a Type II error.
\n52.8 A1
\n[1 mark]
\nM1A1
\n[2 marks]
\nM1A1
\nM1
\n= 31.5 AG
\n[3 marks]
\nuse of a goodness of fit test M1
\n and A1A1
A1
\np-value = 0.569 A2
\nSince 0.569 > 0.05 R1
\nInsufficient evidence to reject . The scores follow a normal distribution. A1
\n[8 marks]
\nM1A1
\n[2 marks]
\nuse of a t-test M1
\nand A1
\np-value = 0.180 A2
\nSince 0.180 > 0.05 R1
\nInsufficient evidence to reject . There is no difference between males and females. A1
\n[6 marks]
\nuse of test for proportion using Binomial distribution M1
\nand A1
\nand M1
\nSo the critical region is A1
\nSince 5 < 7 R1
\nInsufficient evidence to reject . It is not easier to achieve a distinction on the new exam. A1
\n[6 marks]
\nusing M1
\nM1A1
\n[3 marks]
\nusing M1
\nM1A1
\n[3 marks]
\nIf and det , find the possible values of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(M1)(A1)
\n\n
(A1)
\nor (A1) (C4)
\nNote: Both answers are required for the final (A1).
\n[4 marks]
\nConsider the system of equations A where A and .
\nFind det A.
\nFind the set of values of for which the system has a unique solution.
\nAttempting to find det A (M1)
\ndet A A1 N2
\n[2 marks]
\nSystem has a unique solution provided det A ≠ 0 (R1)
\n(A1)
\nSolving or equivalent for M1
\nA1 N3
\n[4 marks]
\nand are 2 × 2 matrices, where and . Find
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(M1)
\n(M1)
\n(A1)
\n(A1)
\nOR
\n(M1)
\n\n
, (A1)
\n\n
, (A1)
\n(A1) (C4)
\nNote: Correct solution with inversion (ie AB instead of BA) earns FT marks, (maximum [3 marks]).
\n[4 marks]
\nLet A = , D = , and C = .
\nGiven matrices A, B, C for which AB = C and det A ≠ 0, express B in terms of A and C.
\nFind the matrix DA.
\nFind B if AB = C.
\nFind the coordinates of the point of intersection of the planes , , .
\nSince det A ≠ 0, A–1 exists. (M1)
\nHence AB = C ⇒ B = A–1C (C1)
\n[2 marks]
\nDA = (A1)
\n[1 mark]
\nB = A–1C = DC (M1)
\n(A1)
\n[2 marks]
\nThe system of equations is
\nor A C (M1)
\nThe required point = (1, –1, 2). (A1)
\n[2 marks]
\nThe matrices A, B, X are given by
\nA = , B = , X = , , , , .
\nGiven that AX + X = Β, find the exact values of , , and .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nX + X =
\nX = (M1)
\nPre-multiply by inverse of (M1)
\nX = (A1)(A1)
\nNote: Award (A1) for determinant, (A1) for matrix .
\n(A1)(A1)(A1)(A1)
\n\n
\n
OR
\n(A1)
\n(A1)
\n\n
(A1)
\n\n
(A1)
\nNotes: Award (A1) for each pair of equations.
Allow ft from their equations.
(A1)(A1)(A1)(A1)
\nNote: Award (A0)(A0)(A1)(A1) if the final answers are given as decimals ie 0.848, 1.79, 0.606, 0.848.
\n[8 marks]
\nLet .
\nWrite down the value of .
\nFind the value of .
\nLet .
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n= 5 A1 N1
\n[1 mark]
\n+ 9 = 4 (M1)
\n= −5 A1 N2
\n[2 marks]
\nComparing elements 3(2) − 5() = −9 M1
\n= 3 A2 N2
\n[3 marks]
\nMr Burke teaches a mathematics class with 15 students. In this class there are 6 female students and 9 male students.
\nEach day Mr Burke randomly chooses one student to answer a homework question.
\nIn the first month, Mr Burke will teach his class 20 times.
\nFind the probability that on any given day Mr Burke chooses a female student to answer a question.
\nFind the probability he will choose a female student 8 times.
\nFind the probability he will choose a male student at most 9 times.
\nA1
\n[1 mark]
\nP(X = 8) (M1)
\nNote: Award (M1) for evidence of recognizing binomial probability. eg P(X = 8), X ∼ B.
\n0.180 (0.179705…) A1
\n[2 marks]
\nP(male) = A1
\nP(X ≤ 9) = 0.128 (0.127521…) (M1)A1
\nNote: Award (M1) for evidence of correct approach eg, P(X ≤ 9).
\n[3 marks]
\nConsider the matrix A .
\nB, C and X are also 2 × 2 matrices.
\nWrite down the inverse, A–1.
\nGiven that XA + B = C, express X in terms of A–1, B and C.
\nGiven that B , and C , find X.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ndet A = 5(1) − 7(−2) = 19
\nA–1 (A2)
\nNote: Award (A1) for , (A1) for dividing by 19.
\nOR
\nA–1 (G2)
\n[2 marks]
\nXA + B = C ⇒ XA = C – Β (M1)
\nX = (C – Β)Α–1 (A1)
\nOR
\nX = (C – B)A–1 (A2)
\n[2 marks]
\n(C – Β)Α–1 = (A1)
\n⇒ X = (A1)
\nOR
\nX = (G2)
\nNote: If premultiplication by A–1 is used, award (M1)(M0) in part (i) but award (A2) for in part (ii).
\n[2 marks]
\nIn this question you will explore possible models for the spread of an infectious disease
\nAn infectious disease has begun spreading in a country. The National Disease Control Centre (NDCC) has compiled the following data after receiving alerts from hospitals.
\nA graph of against is shown below.
\nThe NDCC want to find a model to predict the total number of people infected, so they can plan for medicine and hospital facilities. After looking at the data, they think an exponential function in the form could be used as a model.
\nUse your answer to part (a) to predict
\nThe NDCC want to verify the accuracy of these predictions. They decide to perform a goodness of fit test.
\nThe predictions given by the model for the first five days are shown in the table.
\nIn fact, the first day when the total number of people infected is greater than 1000 is day 14, when a total of 1015 people are infected.
\nBased on this new data, the NDCC decide to try a logistic model in the form .
\nUse the data from days 1–5, together with day 14, to find the value of
\nUse an exponential regression to find the value of and of , correct to 4 decimal places.
\nthe number of new people infected on day 6.
\nthe day when the total number of people infected will be greater than 1000.
\nUse your answer to part (a) to show that the model predicts 16.7 people will be infected on the first day.
\nExplain why the number of degrees of freedom is 2.
\nPerform a goodness of fit test at the 5% significance level. You should clearly state your hypotheses, the p-value, and your conclusion.
\nGive two reasons why the prediction in part (b)(ii) might be lower than 14.
\n.
\n.
\n.
\nHence predict the total number of people infected by this disease after several months.
\nUse the logistic model to find the day when the rate of increase of people infected is greatest.
\nM1A1A1
\n[3 marks]
\nA1
\nnumber of new people infected = 247 – 140 = 107 M1A1
\n[3 marks]
\nuse of graph or table M1
\nday 9 A1
\n[2 marks]
\n9.7782(1.7125)1 M1
\n= 16.7 people AG
\n[1 mark]
\n2 parameters ( and ) were estimated from the data. R1
\nM1
\n= 2 AG
\n[2 marks]
\ndata is modeled by and data is not modeled by A1
\np-value = 0.893 A2
\nSince 0.893 > 0.05 R1
\nInsufficient evidence to reject . So data is modeled by A1
\n[5 marks]
\nvaccine or medicine might slow down rate of infection R1
\nPeople become more aware of disease and take precautions to avoid infection R1
\nAccept other valid reasons
\n[2 marks]
\n1060 M1A1
\n[2 marks]
\n108 A1
\n[1 mark]
\n0.560 A1
\n[1 mark]
\nAs M1
\nA1
\n[2 marks]
\nsketch of or solve M1
\nA1
\nDay 8 A1
\n[3 marks]
\nLet A .
\nLet A2 + A + I = O where , and O = .
\nFind the values of for which the matrix (A − I) is singular.
\nFind the value of and of .
\nHence show that I = A (6I – A).
\nUse the result from part (b) (ii) to explain why A is non-singular.
\nUse the values from part (b) (i) to express A4 in the form A+ I where , .
\nA − I = A1
\nIf A − I is singular then det (A − I) = 0 (R1)
\ndet (A − I) (A1)
\nAttempting to solve or equivalent for M1
\n= 1, 5 A1 N2
\nNote: Candidates need both values of for the final A1.
\n[5 marks]
\nA1
\n(A1)
\nForming any two independent equations M1
\n(eg , or equivalent)
\nNote: Accept equations in matrix form.
\nSolving these two equations (M1)
\nand A1 N2
\n[5 marks]
\nA2 − 6A + 5I = O (M1)
\n5I = 6A − A2 A1
\n= A(6I − A) A1A1
\nNote: Award A1 for A and A1 for (6I − A).
\nI = A(6I − A) AG N0
\nSpecial Case: Award M1A0A0A0 only for candidates following alternative methods.
\n[5 marks]
\nMETHOD 1
\nI = A(6I − A) = A × (6I − A) M1
\nHence by definition (6I − A) is the inverse of A. R1
\nHence A−1 exists and so A is non-singular R1 N0
\n\n
METHOD 2
\nAs det I = 1 (≠ 0), then R1
\ndet A(6I − A) = det A × det (6I − A) (≠ 0) M1
\n⇒ det A ≠ 0 and so A is non-singular. R1 N0
\n\n
[3 marks]
\nMETHOD 1
\nA2 = 6A − 5I (A1)
\nA4 = (6A − 5I)2 M1
\n= 36A2 − 60AI + 25I2 A1
\n= 36(6A − 5I) − 60A + 25I M1
\n= 156A − 155I ( = 156, = −155) A1 N0
\n\n
METHOD 2
\nA2 = 6A − 5I (A1)
\nA3 = 6A2 − 5A where A2 = 6A − 5I M1
\n= 31A − 30I A1
\nA4 = 31A2 − 30A where A2 = 6A − 5I M1
\n= 156A − 155I ( = 156, = −155) A1 N0
\n\n
Note: Do not accept methods that evaluate A4 directly from A.
\n\n
[5 marks]
\nLet A = , B = , and X = .
\nConsider the equation AX = B.
\nWrite down the inverse matrix A−1.
\nExpress X in terms of A−1 and B.
\nHence, solve for X.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nA−1 = or A2 N2
\n[2 marks]
\nX = A−1B A1 N1
\n[1 mark]
\nX = A3 N3
\n[3 marks]
\nIf A = and B = , find 2 values of and , given that AB = BA.
\nAB = (A1)
\nBA = (A1)
\nAB = BA ⇒ 8 + 16 = 24 and 4 + 8 = 40
\nThis gives and . (A1) (C3)
\n[3 marks]
\nGiven that A = and I = , find the values of for which (A – I) is a singular matrix.
\nsingular matrix ⇒ det = 0 (R1)
\n(A1)
\n(M1)
\n(A1)
\nor 6 (A1)(A1) (C6)
\nNote: Award (C2) for one correct answer with no working.
\n[6 marks]
\nLet be the sum of the first terms of the arithmetic series 2 + 4 + 6 + ….
\nLet M = .
\nIt may now be assumed that M = , for ≥ 4. The sum T is defined by
\nT = M1 + M2 + M3 + ... + M.
\nFind 4.
\nFind 100.
\nFind M2.
\nShow that M3 = .
\nWrite down M4.
\nFind T4.
\nUsing your results from part (a) (ii), find T100.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n4 = 20 A1 N1
\n[1 mark]
\n1 = 2, = 2 (A1)
\nAttempting to use formula for M1
\n100 = 10100 A1 N2
\n[3 marks]
\nM2 = A2 N2
\n[2 marks]
\nFor writing M3 as M2 × M or M × M2 M1
\nM3 = A2
\nM3 = AG N0
\n[3 marks]
\nM4 = A1 N1
\n[1 mark]
\nT4 = (M1)
\n= A1A1 N3
\n[3 marks]
\nT100 = (M1)
\nA1A1 N3
\n[3 marks]
\nThe matrices A, B, C and X are all non-singular 3 × 3 matrices.
\nGiven that A–1XB = C, express X in terms of the other matrices.
\nAA–1XB = ΑC (M1)(A1)
\nIXBB–1 = ACB–1 (M1)(A1)
\nX = ACB–1 (M1)(A1) (C6)
\n[6 marks]
\nLet A = and B = . Giving your answers in terms of , , , and ,
\nwrite down A + B.
\nfind AB.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nA + B =
\nA2
\n[2 marks]
\nAB = A1A1A1A1
\nNote: Award N2 for finding BA = .
\n[4 marks]
\nThe matrix A = has inverse A−1 = .
\nConsider the simultaneous equations
\n\n
\n
\n
Write down the value of .
\nWrite down the value of .
\nWrite these equations as a matrix equation.
\nSolve the matrix equation.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n= 4 A1 N1
\n[1 mark]
\n= 7 A1 N1
\n[1 mark]
\nEITHER
\nA A1 N1
\nOR
\nA1 N1
\n[1 mark]
\n= A−1 (accept algebraic method) (M1)
\n(accept = −3, = 5, = 4) A2 N3
\n[3 marks]
\nThe hens on a farm lay either white or brown eggs. The eggs are put into boxes of six. The farmer claims that the number of brown eggs in a box can be modelled by the binomial distribution, B(6, ). By inspecting the contents of 150 boxes of eggs she obtains the following data.
\nShow that this data leads to an estimated value of .
\nStating null and alternative hypotheses, carry out an appropriate test at the 5 % level to decide whether the farmer’s claim can be justified.
\nfrom the sample, the probability of a brown egg is
\nA1
\nAG
\n[1 mark]
\nif the data can be modelled by a binomial distribution with , the expected frequencies of boxes are given in the table
\n A3
Notes: Deduct one mark for each error or omission.
Accept any rounding to at least one decimal place.
null hypothesis: the distribution is binomial A1
\nalternative hypothesis: the distribution is not binomial A1
\nfor a chi-squared test the last two columns should be combined R1
\n(Accept 6.06) (M1)A1
\ndegrees of freedom = 4 A1
\ncritical value = 9.488 A1
\nOr use of -value
\nwe conclude that the farmer’s claim can be justified R1
\n[11 marks]
\nThe square matrix X is such that X3 = 0. Show that the inverse of the matrix (I – X) is I + X + X2.
\nFor multiplying (I – X)(I + X + X2) M1
\n= I2 + IX + IX2 – XI – X2 – X3 = I + X + X2 – X – X2 – X3 (A1)(A1)
\n= I – X3 A1
\n= I A1
\nAB = I ⇒ A–1 = B (R1)
\n(I – X)(I + X + X2) = I ⇒ (I – X)–1 = I + X + X2 AG N0
\n[6 marks]
\nJim writes a computer program to generate 500 values of a variable Z. He obtains the following table from his results.
\nIn this situation, state briefly what is meant by
\nUse a chi-squared goodness of fit test to investigate whether or not, at the 5 % level of significance, the N(0, 1) distribution can be used to model these results.
\na Type I error.
\na Type II error.
\n (A1)(A1)(A1)(A1)(A1)(A1)
(M1)
\n= 7.94 A1
\nDegrees of freedom = 5 A1
\nCritical value = 11.07 A1
\nOr use of p-value
\nWe conclude that the data fit the N(0, 1) distribution. R1
\nat the 5% level of significance A1
\n[12 marks]
\nType I error concluding that the data do not fit N(0, 1) when in fact they do. R2
\n[2 marks]
\nType II error concluding that data fit N(0, 1) when in fact they do not. R2
\n[2 marks]
\nGiven that A = and B = , find X if BX = A – AB.
\n\n
METHOD 1
\nA – AB = (M1)(A1)
\nX = B–1(A – AB) = B–1 (M1)
\n(A1)
\n(A2) (C6)
\n\n
METHOD 2
\nAttempting to set up a matrix equation (M2)
\nX = B–1(A – AB) (A2)
\n(from GDC) (A2) (C6)
\n\n
[6 marks]
\nFind the inverse of the matrix .
\nHence solve the system of equations
\n\n
\n
\n
A2 N2
\n[2 marks]
\nIn matrix form A = B or = A−1 B M1
\n, , A1A1A1 N0
\n[4 marks]
\nLet C = and D = .
\nThe 2 × 2 matrix Q is such that 3Q = 2C – D
\nFind Q.
\nFind CD.
\nFind D–1.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n3Q = (A1)
\n3Q = (A1)
\nQ = (A1) (N3)
\n[3 marks]
\nCD =
\n(A1)(A1)(A1)(A1) (N4)
\n[4 marks]
\ndet D = 5 + 2 (may be implied) (A1)
\nD–1 = (A1) (N2)
\n[2 marks]
\nIn an effort to study the level of intelligence of students entering college, a psychologist collected data from 4000 students who were given a standard test. The predictive norms for this particular test were computed from a very large population of scores having a normal distribution with mean 100 and standard deviation of 10. The psychologist wishes to determine whether the 4000 test scores he obtained also came from a normal distribution with mean 100 and standard deviation 10. He prepared the following table (expected frequencies are rounded to the nearest integer):
\n\n
Copy and complete the table, showing how you arrived at your answers.
\nTest the hypothesis at the 5% level of significance.
\nTo calculate expected frequencies, we multiply 4000 by the probability of each cell:
\n(M1)
\n\n
\n
\n
Therefore, the expected frequency (M1)
\n(A1)
\nSimilarly:
\n\n
Frequency
\n(A1)
\nAnd
\n\n
Frequency
\n(A1)
\n[5 marks]
\nTo test the goodness of fit of the normal distribution, we use the distribution. Since the last cell has an expected frequency less than 5, it is combined with the cell preceding it. There are therefore 7 – 1 = 6 degrees of freedom. (C1)
\n(M1)
\n= 53.03 (A1)
\nH0: Distribution is Normal with and .
\nH1: Distribution is not Normal with and . (M1)
\n\n
Since , we reject H0 (A1)
\nOr use of p-value
\nTherefore, we have enough evidence to suggest that the normal distribution with mean 100 and standard deviation 10 does not fit the data well. (R1)
\nNote: If a candidate has not combined the last 2 cells, award (C0)(M1)(A0)(M1)(A1)(R1) (or as appropriate).
\n[6 marks]
\nSix coins are tossed simultaneously 320 times, with the following results.
\nAt the 5% level of significance, test the hypothesis that all the coins are fair.
\nLet H0 be the hypothesis that all coins are fair, (C1)
\nand let H1 be the hypothesis that not all coins are fair. (C1)
\nLet be the number of tails obtained, is binomially distributed. (M1)
\n (A3)
Notes: Award (A2) if one entry on the third row is incorrect. Award (A1) if two entries on the third row are incorrect. Award (A0) if three or more entries on the third row are incorrect.
\n\n
(A1)
\nAlso (A1)
\nSince 7.24 < 12.592, H0 cannot be rejected. (R1)
\n[9 marks]
\nA solid right circular cone has a base radius of 21 cm and a slant height of 35 cm.
A smaller right circular cone has a height of 12 cm and a slant height of 15 cm, and is removed from the top of the larger cone, as shown in the diagram.
Calculate the radius of the base of the cone which has been removed.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\nNote: Award (M1) for correct substitution into Pythagoras theorem.
\nOR
\n(M1)
\nNote: Award (M1) for a correct equation.
\n= 9 (cm) (A1) (C2)
\n[2 marks]
\nA pan, in which to cook a pizza, is in the shape of a cylinder. The pan has a diameter of 35 cm and a height of 0.5 cm.
\n![\"M17/5/MATSD/SP2/ENG/TZ1/04\"](\"../assets/uploads/tinymce_asset/asset/3600/Schermafbeelding_2017-08-16_om_11.14.51.png\"/)
A chef had enough pizza dough to exactly fill the pan. The dough was in the shape of a sphere.
\nThe pizza was cooked in a hot oven. Once taken out of the oven, the pizza was placed in a dining room.
\nThe temperature, , of the pizza, in degrees Celsius, °C, can be modelled by
\n\n
where is a constant and is the time, in minutes, since the pizza was taken out of the oven.
\nWhen the pizza was taken out of the oven its temperature was 230 °C.
\nThe pizza can be eaten once its temperature drops to 45 °C.
\nCalculate the volume of this pan.
\nFind the radius of the sphere in cm, correct to one decimal place.
\nFind the value of .
\nFind the temperature that the pizza will be 5 minutes after it is taken out of the oven.
\nCalculate, to the nearest second, the time since the pizza was taken out of the oven until it can be eaten.
\nIn the context of this model, state what the value of 19 represents.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(M1)
\n\n
Notes: Award (A1) for 17.5 (or equivalent) seen.
\nAward (M1) for correct substitutions into volume of a cylinder formula.
\n\n
(A1)(G2)
\n[3 marks]
\n(M1)
\n\n
Note: Award (M1) for equating their answer to part (a) to the volume of sphere.
\n\n
(M1)
\n\n
Note: Award (M1) for correctly rearranging so is the subject.
\n\n
(A1)(ft)(G2)
\n\n
Note: Award (A1) for correct unrounded answer seen. Follow through from part (a).
\n\n
(A1)(ft)(G3)
\n\n
Note: The final (A1)(ft) is awarded for rounding their unrounded answer to one decimal place.
\n\n
[4 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution.
\n\n
(A1)(G2)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution into the function, . Follow through from part (c). The negative sign in the exponent is required for correct substitution.
\n\n
(°C) (°C)) (A1)(ft)(G2)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for equating 45 to the exponential equation and for correct substitution (follow through for their in part (c)).
\n\n
(A1)(ft)(G1)
\n(A1)(ft)(G2)
\n\n
Note: Award final (A1)(ft) for converting their minutes into seconds.
\n\n
[3 marks]
\nthe temperature of the (dining) room (A1)
\nOR
\nthe lowest final temperature to which the pizza will cool (A1)
\n[1 mark]
\nGive your answers to four significant figures.
\nA die is thrown 120 times with the following results.
\nShowing all steps clearly, test whether the die is fair
\n(i) at the 5% level of significance;
\n(ii) at the 1% level of significance.
\nExplain what is meant by “level of significance” in part (a).
\nThe data can be described by the following table
\n (A1)
(C1)
\n\n
(A1)
\n\n
(i) (A1)
\nHence, since 11.3 > 11.07 at the 5% level we must accept H1. (R1)
\n\n
(ii) (A1)
\nHence, since 11.3 < 15.086, at the 1% level, there is not enough evidence to conclude that the die is not fair (and hence we accept H0). (R1)
\n\n
[7 marks]
\nLet denote the significance level. If is greater than then it means that the probability of obtaining the results obtained is less than if H0 is correct. (R3)
\nNote: Award (R3) for any correct explanation. Use discretion to award (R2) or (R1).
\n[3 marks]
\nMatrices A, B and C are defined by
\nA = B = C = .
\nLet X be an unknown 2 × 2 matrix satisfying the equation
\nAX + B = C.
\nThis equation may be solved for X by rewriting it in the form
\nX = A−1 D.
\nwhere D is a 2 × 2 matrix.
\nWrite down A−1.
\nFind D.
\nFind X.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nA−1 = or or (A1)(A1)(N2)
\n[2 marks]
\nAX = C − B (may be implied) (A1)
\nX = A−1 (C−B) (A1)
\nD = C − B
\n(A1) (N3)
\n[3 marks]
\nX = (A2) (N2)
\n[2 marks]
\nA calculator generates a random sequence of digits. A sample of 200 digits is randomly selected from the first 100 000 digits of the sequence. The following table gives the number of times each digit occurs in this sample.
\nIt is claimed that all digits have the same probability of appearing in the sequence.
\nTest this claim at the 5% level of significance.
\nExplain what is meant by the 5% level of significance.
\nH0: The sequence contains equal numbers of each digit. (A1)
\nH1: The sequence does not contain equal numbers of each digit. (A1)
\n(M1)(A1)
\nThe number of degrees of freedom is 9. (A1)
\n(A1)
\n. Hence H0 is accepted. (A1)
\n[7 marks]
\nThe probability of rejecting H0 when it is true (A1)
\nis 0.05. (A1)
\nNote: Award (A1)(A1) for “the probability of a type I error is 0.05.”
\n[2 marks]
\nA zoologist believes that the number of eggs laid in the Spring by female birds of a certain breed follows a Poisson law. She observes 100 birds during this period and she produces the following table.
\nThe zoologist wishes to determine whether or not a Poisson law provides a suitable model.
\nCalculate the mean number of eggs laid by these birds.
\nWrite down appropriate hypotheses.
\nCarry out a test at the 1% significance level, and state your conclusion.
\nMean (M1)
\nA1 N2
\n[2 marks]
\nH0 : Poisson law provides a suitable model A1
\nH1 : Poisson law does not provide a suitable model A1
\n[2 marks]
\nThe expected frequencies are
\n A1A1A1A1A1A1
Note: Accept expected frequencies rounded to a minimum of three significant figures.
\n(M1)(A2)
\n(accept 5.33 and 5.34) A2
\n(6 cells − 2 restrictions) A1
\nNote: If candidates have combined rows allow FT on their value of .
\nCritical value
\nBecause 5.35 < 13.277, the Poisson law does provide a suitable model. R1 N0
\n[14 marks]
\nThe number of cars passing a certain point in a road was recorded during 80 equal time intervals and summarized in the table below.
\nCarry out a goodness of fit test at the 5% significance level to decide if the above data can be modelled by a Poisson distribution.
\nH0 : The data can be modeled by a Poisson distribution.
\nH1 : The data cannot be modeled by a Poisson distribution.
\nA1
\nTheoretical frequencies are
\n(M1)(A1)
\nA1
\n\n
\n
A1
\nNote: Award A1 for , , .
\n(5 or more) A1
\n\n
\n
(accept 1.82) (M1)(A1)
\n(six frequencies and two restrictions) (A1)
\nat the 5% level. A1
\nSince 1.83 < 9.488 we accept H0 and conclude that the distribution can be modeled by a Poisson distribution. R1 N0
\n[11 marks]
\nEggs at a farm are sold in boxes of six. Each egg is either brown or white. The owner believes that the number of brown eggs in a box can be modelled by a binomial distribution. He examines 100 boxes and obtains the following data.
\nCalculate the mean number of brown eggs in a box.
\nHence estimate , the probability that a randomly chosen egg is brown.
\nBy calculating an appropriate statistic, test, at the 5% significance level, whether or not the binomial distribution gives a good fit to these data.
\nNote: Candidates may obtain slightly different numerical answers depending on the calculator and approach used. Use discretion in marking.
\nMean (A1)
\n[1 mark]
\nNote: Candidates may obtain slightly different numerical answers depending on the calculator and approach used. Use discretion in marking.
\n(A1)
\n[1 mark]
\nNote: Candidates may obtain slightly different numerical answers depending on the calculator and approach used. Use discretion in marking.
\nThe calculated values are
\n
10 9.046 0.910
29 26.732 5.14 (M1)
31 32.917 3.675 (A1)
18 21.617 13.083 (A1)
12 9.688 5.345 (A1)
Note: Award (M1) for the attempt to calculate expected values, (A1) for correct expected values, (A1) for correct values, (A1) for combining cells.
\n(A1)
\nOR
\n(G5)
\nDegrees of freedom = 3; Critical value = 7.815
\n(or p-value = 0.668 (or 0.669)) (A1)(A1)
\nWe conclude that the binomial distribution does provide a good fit. (R1)
\n[8 marks]
\nA water container is made in the shape of a cylinder with internal height cm and internal base radius cm.
\n![\"N16/5/MATSD/SP2/ENG/TZ0/06\"](\"../assets/uploads/tinymce_asset/asset/3342/Schermafbeelding_2017-03-07_om_08.31.01.png\"/)
The water container has no top. The inner surfaces of the container are to be coated with a water-resistant material.
\nThe volume of the water container is .
\nThe water container is designed so that the area to be coated is minimized.
\nOne can of water-resistant material coats a surface area of .
\nWrite down a formula for , the surface area to be coated.
\nExpress this volume in .
\nWrite down, in terms of and , an equation for the volume of this water container.
\nShow that .
\nFind .
\nUsing your answer to part (e), find the value of which minimizes .
\nFind the value of this minimum area.
\nFind the least number of cans of water-resistant material that will coat the area in part (g).
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(A1)(A1)
\n\n
Note: Award (A1) for either OR seen. Award (A1) for two correct terms added together.
\n\n
[2 marks]
\n(A1)
\n\n
Notes: Units not required.
\n\n
[1 mark]
\n(A1)(ft)
\n\n
Notes: Award (A1)(ft) for equating to their part (b).
\nDo not accept unless is explicitly defined as their part (b).
\n\n
[1 mark]
\n(A1)(ft)(M1)
\n\n
Note: Award (A1)(ft) for their seen.
\nAward (M1) for correctly substituting only into a correct part (a).
\nAward (A1)(ft)(M1) for rearranging part (c) to and substituting for in expression for .
\n\n
(AG)
\n\n
Notes: The conclusion, , must be consistent with their working seen for the (A1) to be awarded.
\nAccept as equivalent to .
\n\n
[2 marks]
\n(A1)(A1)(A1)
\n\n
Note: Award (A1) for , (A1) for or , (A1) for .
\n\n
[3 marks]
\n(M1)
\n\n
Note: Award (M1) for equating their part (e) to zero.
\n\n
OR (M1)
\n\n
Note: Award (M1) for isolating .
\n\n
OR
\nsketch of derivative function (M1)
\nwith its zero indicated (M1)
\n(A1)(ft)(G2)
\n[3 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution of their part (f) into the given equation.
\n\n
(A1)(ft)(G2)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for dividing their part (g) by 2000.
\n\n
(A1)(ft)
\n\n
Notes: Follow through from part (g).
\n\n
14 (cans) (A1)(ft)(G3)
\n\n
Notes: Final (A1) awarded for rounding up their to the next integer.
\n\n
[3 marks]
\nLet A = and B = . Find, in terms of ,
\n2A − B.
\ndet (2A − B).
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n\n
2A = (A1)
\n2A − B = A2 N3
\n[3 marks]
\nEvidence of using the definition of determinant (M1)
\nCorrect substitution (A1)
\neg 4(5) − 2(2 − 1), 20 − 2(2 − 1), 20 − 4 + 2
\ndet (2A − B) = 22 − 4 A1 N3
\n[3 marks]
\nThe number of telephone calls received by a helpline over 80 one-minute periods are summarized in the table below.
\nFind the exact value of the mean of this distribution.
\nTest, at the 5% level of significance, whether or not the data can be modelled by a Poisson distribution.
\nMean (M1)
\nA1
\nNote: Do not accept 2.73.
\n[2 marks]
\nH0: the data can be modelled by a Poisson distribution A1
\nH1: the data cannot be modelled by a Poisson distribution A1
\n A3
Note: Award A2 for one error, A1 for two errors, A0 for three or more errors.
\nCombining last two columns (M1)
\nNote: Allow FT from not combining the last two columns and / or getting 2.98 for the last expected frequency.
\nEITHER
\n(M1)(A1)
\n= 8.804 (accept 8.8) A1
\n, A1A1
\nHence 8.804 is not significant since 8.804 < 9.488 and we accept H0 R1
\nOR
\np-value = 0.0662 (accept 0.066) which is not significant since A5
\n0.0662 > 0.05 and we accept H0 R1 N0
\n[12 marks]
\nIn a reforested area of pine trees, heights of trees planted in a specific year seem to follow a normal distribution. A sample of 100 such trees is selected to test the validity of this hypothesis. The results of measuring tree heights, to the nearest centimetre, are recorded in the first two columns of the table below.
\nDescribe what is meant by
\na goodness of fit test (a complete explanation required);
\nthe level of significance of a hypothesis test.
\nFind the mean and standard deviation of the sample data in the table above. Show how you arrived at your answers.
\nMost of the expected frequencies have been calculated in the third column. (Frequencies have been rounded to the nearest integer, and frequencies in the first and last classes have been extended to include the rest of the data beyond 15 and 225. Find the values of , and and show how you arrived at your answers.
\nIn order to test for the goodness of fit, the test statistic was calculated to be 1.0847. Show how this was done.
\nState your hypotheses, critical number, decision rule and conclusion (using a 5% level of significance).
\nA goodness of fit test is a statistical test of the hypothesis that a set of observed counts of cells of a certain large population is consistent with a set of theoretical counts. (R1)
\nThe test statistic has a distribution with degrees of freedom. One degree of freedom is lost for every parameter that has to be estimated from the sample. (R1)
\n[2 marks]
\nThe level of significance of a hypothesis test is the maximal probability that we reject a true null hypothesis. (R1)
\n[1 mark]
\nWe use the class midpoints in the calculation of the mean and standard deviation.
\n(M1)
\n= 133.5 (A1)
\n(M1)
\n= 56.345 (= 56.3 to 3 sf) (A1)
\n[4 marks]
\nEvery frequency is the product of the number of observations and the probability of a number in each class. Since by hypothesis we have a normal distribution, the probabilities can be read from a normal table with mean 133.5 and standard deviation 56.345 (M1)
\nE1 = 100 × P(45 ≤ ≤ 75) ≈ 9 so = 9 (A1)
\nE2 = 100 × P(135 ≤ ≤ 165) ≈ 20 so b = 20 (A1)
\nE3 = 100 × P(195 ≤ ≤ 225) ≈ 9 so = 9 (A1)
\n[4 marks]
\nThe test statistic is a variable. Hence (M1)
\n(M1)
\n= 1.0847 (A1)
\n[3 marks]
\nH0: The distribution of tree heights is normally distributed
\nH1: The distribution is not normal (M1)
\nSince the mean and standard deviation were estimated from the sample, the number of degrees of freedom is 8 – 1 – 2 = 5 (A1)
\nThe critical number is = 110705
\nIf > 11.0705 we reject H0 (A1)
\nSince = 1.0847 < 11.0705, we fail to reject H0 (R1)
\nConclusion: we do not have enough evidence to claim that the distribution of tree heights is not normal (R1)
\n[5 marks]
\nThe random variable X is thought to follow a binomial distribution B (4, ). In order to investigate this belief, a random sample of 100 observations on X was taken with the following results.
\nAn automatic machine is used to fill bottles of water. The amount delivered, ml, may be assumed to be normally distributed with mean ml and standard deviation 8 ml. Initially, the machine is adjusted so that the value of is 500. In order to check that the value of remains equal to 500, a random sample of 10 bottles is selected at regular intervals, and the mean amount of water, , in these bottles is calculated. The following hypotheses are set up.
\nH0: = 500; H1: ≠ 500
\nThe critical region is defined to be .
\nState suitable hypotheses for testing this belief.
\nCalculate the mean of these data and hence estimate the value of .
\nCalculate an appropriate value of and state your conclusion, using a 1% significance level.
\nFind the significance level of this procedure.
\nSome time later, the actual value of is 503. Find the probability of a Type II error.
\nH0 : The data are B (4, ); H1 : The data are not B (4, ) A1
\n[1 mark]
\nMean M1A1
\n= 1.8 A1
\nM1A1
\n[5 marks]
\nThe expected frequencies are
\n A1A1A1A1A1
The last two classes must be combined because the expected frequency for = 4 is less than 5. R1
\nM2
\n= 18.0 A2
\nDF = 2 (A1)
\nCritical value = 9.21 A1
\nWe conclude, at the 1% significance level, that X does not fit a binomial model. R1
\n\n
Special case: award the following marks to candidates who do not combine classes.
\nM2
\n= 39.6 A0
\nDF = 3 (A1)
\nCritical value = 11.345 A1
\nWe conclude, at the 1% significance level, that X does not fit a binomial model. R1
\n\n
[13 marks]
\nUnder H0, the distribution of is N (500, 6.4). (A1)
\nSignificance level = P < 495 or > 505 | H0 M2
\n= 2 × 0.02405 (A1)
\n= 0.0481 A1 N5
\n[5 marks]
\nThe distribution of is now N (503, 6.4). (A1)
\nP(Type ΙΙ error) = P(495 < < 505) (M1)
\n= 0.785 A1 N3
\n[3 marks]
\nA horse breeder records the number of births for each of 100 horses during the past eight years. The results are summarized in the following table:
\nStating null and alternative hypotheses carry out an appropriate test at the 5% significance level to decide whether the results can be modelled by B (6, 0.5).
\nWithout doing any further calculations, explain briefly how you would carry out a test, at the 5% significance level, to decide if the data can be modelled by B(6, ), where is unspecified.
\nA different horse breeder collected data on the time and outcome of births. The data are summarized in the following table:
\nCarry out an appropriate test at the 5% significance level to decide whether there is an association between time and outcome.
\nMETHOD 1
\nH0: distribution is B(6, 0.5); H1: distribution is not B(6, 0.5) A1
\nA3
\nCombining the first two columns and the last two columns: A1
\n\n
(M1)
\n= 5.22 A1
\n= 4, so critical value of A1A1
\nSince 5.22 < 9.488 the result is not significant and we accept H0 R1
\n\n
METHOD 2
\nH0: distribution is B(6, 0.5); H1: distribution is not B(6, 0.5) A1
\nBy GDC, A8
\nSince 0.266 > 0.05 the result is not significant and we accept H0 R1
\n\n
[10 marks]
\nEstimate from the data which would entail the loss of one degree of freedom A1A1
\n[2 marks]
\nH0: there is no association H1: there is an association A1
\n A2
(M1)
\n= 15.7
\nA1A1
\nSince 15.7 > 7.815 we reject H0 R1
\n\n
METHOD 2
\nH0: there is no association H1: there is an association A1
\nBy GDC, = 0.00129 A6
\nSince 0.00129 < 0.05 we reject H0. R1
\n\n
[8 marks]
\nThe heights, metres, of the 241 new entrants to a men’s college were measured and the following statistics calculated.
\n\n
The Head of Mathematics decided to use a test to determine whether or not these heights could be modelled by a normal distribution. He therefore divided the data into classes as follows.
\nCalculate unbiased estimates of the population mean and the population variance.
\nState suitable hypotheses.
\nCalculate the value of the statistic and state your conclusion using a 10% level of significance.
\nA1
\nM1A1
\n[3 marks]
\nH0: Data can be modelled by a normal distribution
\nH1: Data cannot be modelled by a normal distribution A1
\n[1 mark]
\nThe expected frequencies are
\n A1A1A1A1A1A1
M1A1
\nDegrees of freedom = 3 A1
\nCritical value = 6.251 or p-value = 0.35 A1
\nThe data can be modelled by a normal distribution. R1
\n[11 marks]
\nA right circular cone of radius is inscribed in a sphere with centre O and radius as shown in the following diagram. The perpendicular height of the cone is , X denotes the centre of its base and B a point where the cone touches the sphere.
\nShow that the volume of the cone may be expressed by .
\nGiven that there is one inscribed cone having a maximum volume, show that the volume of this cone is .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to use Pythagoras in triangle OXB M1
\nA1
\nsubstitution of their into formula for volume of cone M1
\n\n
A1
\nNote: This A mark is independent and may be seen anywhere for the correct expansion of .
\n\n
AG
\n[4 marks]
\nat max, R1
\n\n
\n
(since ) A1
\nEITHER
\nfrom part (a)
\nA1
\nA1
\nOR
\n\n
A1
\n\n
A1
\nTHEN
\nAG
\n[4 marks]
\nConsider the function .
\nExpress in the form .
\nFactorize .
\nSketch the graph of , indicating on it the equations of the asymptotes, the coordinates of the -intercept and the local maximum.
\nShow that .
\nHence find the value of if .
\nSketch the graph of .
\nDetermine the area of the region enclosed between the graph of , the -axis and the lines with equations and .
\nA1
\n[1 mark]
\nA1
\n[1 mark]
\n![\"M17/5/MATHL/HP1/ENG/TZ1/B11.b/M\"](\"../assets/uploads/tinymce_asset/asset/3504/Schermafbeelding_2017-08-08_om_13.58.40.png\"/)
A1 for the shape
\nA1 for the equation
\nA1 for asymptotes and
\nA1 for coordinates
\nA1 -intercept
\n[5 marks]
\nM1
\nAG
\n[1 mark]
\n\n
A1
\nM1
\nM1A1
\n\n
[4 marks]
\n![\"M17/5/MATHL/HP1/ENG/TZ1/B11.e/M\"](\"../assets/uploads/tinymce_asset/asset/3505/Schermafbeelding_2017-08-08_om_14.20.03.png\"/)
symmetry about the -axis M1
\ncorrect shape A1
\n\n
Note: Allow FT from part (b).
\n\n
[2 marks]
\n(M1)(A1)
\nA1
\n\n
Note: Do not award FT from part (e).
\n\n
[3 marks]
\nLet .
\nThe graph of has a local maximum at A. Find the coordinates of A.
\nShow that there is exactly one point of inflexion, B, on the graph of .
\nThe coordinates of B can be expressed in the form B where a, b. Find the value of a and the value of b.
\nSketch the graph of showing clearly the position of the points A and B.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to differentiate (M1)
\nA1
\nNote: Award M1 for using quotient or product rule award A1 if correct derivative seen even in unsimplified form, for example .
\nM1
\nA1
\nA1
\n[5 marks]
\nM1
\nA1
\nNote: Award A1 for correct derivative seen even if not simplified.
\nA1
\nhence (at most) one point of inflexion R1
\nNote: This mark is independent of the two A1 marks above. If they have shown or stated their equation has only one solution this mark can be awarded.
\nchanges sign at R1
\nso exactly one point of inflexion
\n[5 marks]
\nA1
\n(M1)A1
\nNote: Award M1 for the substitution of their value for into .
\n[3 marks]
\nA1A1A1A1
A1 for shape for x < 0
A1 for shape for x > 0
A1 for maximum at A
A1 for POI at B.
Note: Only award last two A1s if A and B are placed in the correct quadrants, allowing for follow through.
\n[4 marks]
\nScientists have developed a type of corn whose protein quality may help chickens gain weight faster than the present type used. To test this new type, 20 one-day-old chicks were fed a ration that contained the new corn while another control group of 20 chicks was fed the ordinary corn. The data below gives the weight gains in grams, for each group after three weeks.
\nThe scientists wish to investigate the claim that Group B gain weight faster than Group A. Test this claim at the 5% level of significance, noting which hypothesis test you are using. You may assume that the weight gain for each group is normally distributed, with the same variance, and independent from each other.
\nThe data from the two samples above are combined to form a single set of data. The following frequency table gives the observed frequencies for the combined sample. The data has been divided into five intervals.
\nTest, at the 5% level, whether the combined data can be considered to be a sample from a normal population with a mean of 380.
\nThis is a t-test of the difference of two means. Our assumptions are that the two populations are approximately normal, samples are random, and they are independent from each other. (R1)
\nH0: μ1 − μ2 = 0
\nH1: μ1 − μ2 < 0 (A1)
\nt = −2.460, (A1)
\ndegrees of freedom = 38 (A1)
\nSince the value of critical t = −1.686 we reject H0. (A1)
\nHence group B grows faster. (R1)
\n[6 marks]
\nThis is a goodness-of-fit test.
\nTo finish the table, the frequencies of the respective cells have to be calculated. Since the standard deviation is not given, it has to be estimated using the data itself. s = 49.59, eg the third expected frequency is 40 × 0.308 = 12.32, since P(350.5 < W < 390.5) = 0.3078...
\nThe table of observed and expected frequencies is:
\n (M1)(A2)
Since the first expected frequency is 3.22, we combine the two cells, so that the first two rows become one row, that is,
\n (M1)
Number of degrees of freedom is 4 – 1 – 1 = 2 (C1)
\nH0: The distribution is normal with mean 380
\nH1: The distribution is not normal with mean 380 (A1)
\nThe test statistic is
\n\n
= 3.469 (A1)
\nWith 2 degrees of freedom, the critical number is = 5.99 (A2)
\nSo, we do not have enough evidence to reject the null hypothesis. Therefore, there is no evidence to say that the distribution is not normal with mean 380. (R1)
\n[10 marks]
\nThe speed of light is kilometres per second. The average distance from the Sun to the Earth is 149.6 million km.
\nCalculate the time, in minutes, it takes for light from the Sun to reach the Earth.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)(M1)
\n\n
Note: Award (M1) for dividing the correct numerator (which can be presented in a different form such as or ) by and (M1) for dividing by 60.
\n\n
(A1) (C3)
\n[3 marks]
\nLittle Green island originally had no turtles. After 55 turtles were introduced to the island, their population is modelled by
\n\n
where is a constant and is the time in years since the turtles were introduced.
\n\n
Find the value of .
\nFind the time, in years, for the population to decrease to 20 turtles.
\nThere is a number beyond which the turtle population will not decrease.
\nFind the value of . Justify your answer.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\nNote: Award (M1) for correct substitution of zero and 55 into the function.
\n45 (A1) (C2)
\n[2 marks]
\n(M1)
\nNote: Award (M1) for comparing correct expression involving 20 and their 45. Accept an equation.
\n(2.16992…) (A1)(ft) (C2)
\nNote: Follow through from their part (a), but only if positive.
Answer must be in years; do not accept months for the final (A1).
[2 marks]
\n10 (A1)
\nbecause as the number of years increases the number of turtles approaches 10 (R1) (C2)
\nNote: Award (R1) for a sketch with an asymptote at approximately ,
OR for table with values such as 10.003 and 10.001 for and , for example,
OR when approaches large numbers approaches 10. Do not award (A1)(R0).
[2 marks]
\nA factory packages coconut water in cone-shaped containers with a base radius of 5.2 cm and a height of 13 cm.
\nThe factory designers are currently investigating whether a cone-shaped container can be replaced with a cylinder-shaped container with the same radius and the same total surface area.
\nFind the slant height of the cone-shaped container.
\nFind the slant height of the cone-shaped container.
\nShow that the total surface area of the cone-shaped container is 314 cm2, correct to three significant figures.
\nFind the height, , of this cylinder-shaped container.
\nThe factory director wants to increase the volume of coconut water sold per container.
\nState whether or not they should replace the cone-shaped containers with cylinder‑shaped containers. Justify your conclusion.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(M1)
\nNote: Award (M1) for correct substitution in the volume formula for cone.
\n368 (368.110…) cm3 (A1)(G2)
\nNote: Accept 117.173… cm3 or cm3.
\n[2 marks]
\n(slant height2) = (5.2)2 + 132 (M1)
\nNote: Award (M1) for correct substitution into the formula.
\n14.0 (14.0014…) (cm) (A1)(G2)
\n[2 marks]
\n14.0014… × (5.2) × + (5.2)2 × (M1)(M1)
\nNote: Award (M1) for their correct substitution in the curved surface area formula for cone; (M1) for adding the correct area of the base. The addition must be explicitly seen for the second (M1) to be awarded. Do not accept rounded values here as may come from working backwards.
\n313.679… (cm2) (A1)
\nNote: Use of 3 sf value 14.0 gives an unrounded answer of 313.656….
\n314 (cm2) (AG)
\nNote: Both the unrounded and rounded answers must be seen for the final (A1) to be awarded.
\n[3 marks]
\n2 × × (5.2) × + 2 × × (5.2)2 = 314 (M1)(M1)(M1)
\nNote: Award (M1) for correct substitution in the curved surface area formula for cylinder; (M1) for adding two correct base areas of the cylinder; (M1) for equating their total cylinder surface area to 314 (313.679…). For this mark to be awarded the areas of the two bases must be added to the cylinder curved surface area and equated to 314. Award at most (M1)(M0)(M0) for cylinder curved surface area equated to 314.
\n( =) 4.41 (4.41051…) (cm) (A1)(G3)
\n[4 marks]
\n× (5.2)2 × 4.41051… (M1)
\nNote: Award (M1) for correct substitution in the volume formula for cylinder.
\n375 (374.666…) (cm3) (A1)(ft)(G2)
\nNote: Follow through from part (d).
\n375 (cm3) > 368 (cm3) (R1)(ft)
\nOR
\n“volume of cylinder is larger than volume of cone” or similar (R1)(ft)
\nNote: Follow through from their answer to part (a). The verbal statement should be consistent with their answers from parts (e) and (a) for the (R1) to be awarded.
\nreplace with the cylinder containers (A1)(ft)
\nNote: Do not award (A1)(ft)(R0). Follow through from their incorrect volume for the cylinder in this question part but only if substitution in the volume formula shown.
\n[4 marks]
\nA solid glass paperweight consists of a hemisphere of diameter 6 cm on top of a cuboid with a square base of length 6 cm, as shown in the diagram.
\nThe height of the cuboid, x cm, is equal to the height of the hemisphere.
\nWrite down the value of x.
\nCalculate the volume of the paperweight.
\n1 cm3 of glass has a mass of 2.56 grams.
\nCalculate the mass, in grams, of the paperweight.
\n3 (cm) (A1) (C1)
\n\n
[1 mark]
\nunits are required in part (a)(ii)
\n\n
(M1)(M1)
\nNote: Award (M1) for their correct substitution in volume of sphere formula divided by 2, (M1) for adding their correctly substituted volume of the cuboid.
\n\n
= 165 cm3 (164.548…) (A1)(ft) (C3)
\nNote: The answer is 165 cm3; the units are required. Follow through from part (a)(i).
\n\n
[3 marks]
\ntheir 164.548… × 2.56 (M1)
\nNote: Award (M1) for multiplying their part (a)(ii) by 2.56.
\n\n
= 421 (g) (421.244…(g)) (A1)(ft) (C2)
\nNote: Follow through from part (a)(ii).
\n\n
[2 marks]
\nA window is made in the shape of a rectangle with a semicircle of radius metres on top, as shown in the diagram. The perimeter of the window is a constant P metres.
\n![\"M17/5/MATHL/HP1/ENG/TZ2/10\"](\"../assets/uploads/tinymce_asset/asset/3508/Schermafbeelding_2017-08-08_om_17.46.34.png\"/)
Find the area of the window in terms of P and .
\nFind the width of the window in terms of P when the area is a maximum, justifying that this is a maximum.
\nShow that in this case the height of the rectangle is equal to the radius of the semicircle.
\nthe width of the rectangle is and let the height of the rectangle be
\n(A1)
\n(A1)
\n\n
M1A1
\n[4 marks]
\nA1
\nM1
\n(A1)
\nhence the width is A1
\nR1
\nhence maximum AG
\n[5 marks]
\nEITHER
\n\n
M1
\nA1
\nAG
\nOR
\n\n
M1
\nA1
\nAG
\n[2 marks]
\nAn archaeological site is to be made accessible for viewing by the public. To do this, archaeologists built two straight paths from point A to point B and from point B to point C as shown in the following diagram. The length of path AB is 185 m, the length of path BC is 250 m, and angle is 125°.
\nThe archaeologists plan to build two more straight paths, AD and DC. For the paths to go around the site, angle is to be made equal to 85° and angle is to be made equal to 70° as shown in the following diagram.
\nFind the size of angle .
\nFind the size of angle .
\n(CAD =) 53.1° (53.0521…°) (A1)(ft)
\nNote: Follow through from their part (b)(i) only if working seen.
\n[1 mark]
\n(ACD = ) 70° − (180° − 125° − 31.9478°…) (M1)
\nNote: Award (M1) for subtracting their angle from 70°.
\nOR
\n(ADC =) 360 − (85 + 70 + 125) = 80
\n(ACD =) 180 − 80 − 53.0521... (M1)
\n46.9° (46.9478…°) (A1)(ft)(G2)
\nNote: Follow through from part (b)(i).
\n[2 marks]
\nIn this question, give all answers correct to 2 decimal places.
\nJose travelled from Buenos Aires to Sydney. He used Argentine pesos, ARS, to buy 350 Australian dollars, AUD, at a bank. The exchange rate was 1 ARS = 0.1559 AUD.
\nThe bank charged Jose a commission of 2%.
\nJose used his credit card to pay his hotel bill in Sydney. The bill was 585 AUD. The value the credit card company charged for this payment was 4228.38 ARS. The exchange rate used by the credit card company was 1 AUD = ARS. No commission was charged.
\nUse this exchange rate to calculate the amount of ARS that is equal to 350 AUD.
\nCalculate the total amount of ARS Jose paid to get 350 AUD.
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
Note: In this question, the first time an answer is not to 2 dp the final (A1) is not awarded.
\n\n
(M1)
\n\n
Note: Award (M1) for dividing 350 by 0.1559.
\n\n
(A1) (C2)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for multiplying their answer to part (a) by 1.02.
\n\n
(A1)(ft) (C2)
\nOR
\n(M1)
\n\n
Note: Award (M1) for multiplying their answer to part (a) by 0.02.
\n\n
\n
\n
(A1)(ft) (C2)
\n\n
Note: Follow through from part (a).
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for dividing 4228.38 by 585.
\n\n
(A1) (C2)
\n[2 marks]
\nThree points in three-dimensional space have coordinates A(0, 0, 2), B(0, 2, 0) and C(3, 1, 0).
\nFind the vector .
\nFind the vector .
\nHence or otherwise, find the area of the triangle ABC.
\nA1
\nNote: Accept row vectors or equivalent.
\n[1 mark]
\nA1
\nNote: Accept row vectors or equivalent.
\n[1 mark]
\nMETHOD 1
\nattempt at vector product using and . (M1)
\n±(2i + 6j +6k) A1
\nattempt to use area M1
\nA1
\n\n
METHOD 2
\nattempt to use M1
\n\n
A1
\n\n
attempt to use area M1
\n\n
A1
\n[4 marks]
\nConsider the lines and defined by
\nr and where is a constant.
\nGiven that the lines and intersect at a point P,
\nfind the value of ;
\ndetermine the coordinates of the point of intersection P.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nr M1
\nM1A1
\nA1
\nMETHOD 2
\nM1
\nattempt to solve M1
\nA1
\nA1
\n[4 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nConsider the curve .
\nFind the x-coordinates of the points on the curve where the gradient is zero.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid attempt to find M1
\nA1A1
\nattempt to solve M1
\nA1A1
\n[6 marks]
\nWillow finds that she receives approximately 70 emails per working day.
\nShe decides to model the number of emails received per working day using the random variable , where follows a Poisson distribution with mean 70.
\nIn order to test her model, Willow records the number of emails she receives per working day over a period of 6 months. The results are shown in the following table.
\nFrom the table, calculate
\nArchie works for a different company and knows that he receives emails according to a Poisson distribution, with a mean of emails per day.
\nUsing this distribution model, find .
\nUsing this distribution model, find the standard deviation of .
\nan estimate for the mean number of emails received per working day.
\nan estimate for the standard deviation of the number of emails received per working day.
\nGive one piece of evidence that suggests Willow’s Poisson distribution model is not a good fit.
\nSuppose that the probability of Archie receiving more than 10 emails in total on any one day is 0.99. Find the value of λ.
\nNow suppose that Archie received exactly 20 emails in total in a consecutive two day period. Show that the probability that he received exactly 10 of them on the first day is independent of λ.
\n\n
(M1)
\n= 0.102 A1
\n\n
[2 marks]
\nstandard deviation = (= 8.37) (M1)A1
\n\n
[2 marks]
\nuse of midpoints (accept consistent use of 45, 55 etc.) (M1)
\n(M1)
\nA1
\nNote: If 45, 55, etc. are used consistently instead of midpoints (implied by the answer 71.58…) award M1M1A0.
\n\n
[3 marks]
\n13.9 (M1)A1
\n\n
[2 marks]
\nvalid reason given to include the examples below R1
\nvariance is 192 which is not close to the mean (accept not equal to) standard deviation too high (using parts (a)(ii) and (b)(ii))
\nrelative frequency of ≤ 59 is 0.142 which is too high (using part (a)(i))
\nPoisson would give a frequency of roughly 14 for 80 ≤ ≤ 89
\nNote: Reasons which do not use values found in previous parts must be backed up with numerical evidence.
\n\n
[1 mark]
\n\n
\n
(M1)
\nattempt to solve a correct equation (M1)
\nλ = 20.1 A1
\n\n
[3 marks]
\nin 1 day, no of emails is X ~ Po(λ)
\nin 2 days, no of emails is Y ~ Po(2λ) (A1)
\nP(10 on first day | 20 in 2 days) (M1)
\n(M1)
\nA1
\nA1
\n\n
which is independent of λ AG
\n\n
[5 marks]
\nGive your answers to parts (b), (c) and (d) to the nearest whole number.
\nHarinder has 14 000 US Dollars (USD) to invest for a period of five years. He has two options of how to invest the money.
\nOption A: Invest the full amount, in USD, in a fixed deposit account in an American bank.
\nThe account pays a nominal annual interest rate of r % , compounded yearly, for the five years. The bank manager says that this will give Harinder a return of 17 500 USD.
\nOption B: Invest the full amount, in Indian Rupees (INR), in a fixed deposit account in an Indian bank. The money must be converted from USD to INR before it is invested.
\nThe exchange rate is 1 USD = 66.91 INR.
\nThe account in the Indian bank pays a nominal annual interest rate of 5.2 % compounded monthly.
\nCalculate the value of r.
\nCalculate 14 000 USD in INR.
\nCalculate the amount of this investment, in INR, in this account after five years.
\nHarinder chose option B. At the end of five years, Harinder converted this investment back to USD. The exchange rate, at that time, was 1 USD = 67.16 INR.
\nCalculate how much more money, in USD, Harinder earned by choosing option B instead of option A.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)(A1)
\nNote: Award (M1) for substitution into the compound interest formula, (A1) for correct substitution. Award at most (M1)(A0) if not equated to 17500.
\nOR
\nN = 5
\nPV = ±14000
\nFV = 17500
\nP/Y = 1
\nC/Y = 1 (A1)(M1)
\nNote: Award (A1) for C/Y = 1 seen, (M1) for all other correct entries. FV and PV must have opposite signs.
\n= 4.56 (%) (4.56395… (%)) (A1) (G3)
\n[3 marks]
\n\n
14000 × 66.91 (M1)
\nNote: Award (M1) for multiplying 14000 by 66.91.
\n936740 (INR) (A1) (G2)
\nNote: Answer must be given to the nearest whole number.
\n[2 marks]
\n(M1)(A1)(ft)
\nNote: Award (M1) for substitution into the compound interest formula, (A1)(ft) for their correct substitution.
\nOR
\nN = 60
\nI% = 5.2
\nPV = ±936740
\nP/Y= 12
\nC/Y= 12 (A1)(M1)
\nNote: Award (A1) for C/Y = 12 seen, (M1) for all other correct entries.
\nOR
\nN = 5
\nI% = 5.2
\nPV = ±936740
\nP/Y= 1
\nC/Y= 12 (A1)(M1)
\nNote: Award (A1) for C/Y = 12 seen, (M1) for all other correct entries
\n= 1214204 (INR) (A1)(ft) (G3)
\nNote: Follow through from part (b). Answer must be given to the nearest whole number.
\n[3 marks]
\n(M1)
\nNote: Award (M1) for dividing their (c) by 67.16.
\n(USD) (M1)(A1)(ft) (G3)
\nNote: Award (M1) for finding the difference between their conversion and 17500. Answer must be given to the nearest whole number. Follow through from part (c).
\n[3 marks]
\nThe discrete random variable X has the following probability distribution, where p is a constant.
\nFind the value of p.
\nFind μ, the expected value of X.
\nFind P(X > μ).
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
equating sum of probabilities to 1 (p + 0.5 − p + 0.25 + 0.125 + p3 = 1) M1
\np3 = 0.125 =
\np= 0.5 A1
\n[2 marks]
\nμ = 0 × 0.5 + 1 × 0 + 2 × 0.25 + 3 × 0.125 + 4 × 0.125 M1
\n= 1.375 A1
\n[2 marks]
\nP(X > μ) = P(X = 2) + P(X = 3) + P(X = 4) (M1)
\n= 0.5 A1
\nNote: Do not award follow through A marks in (b)(i) from an incorrect value of p.
\nNote: Award M marks in both (b)(i) and (b)(ii) provided no negative probabilities, and provided a numerical value for μ has been found.
\n[2 marks]
\nA Chocolate Shop advertises free gifts to customers that collect three vouchers. The vouchers are placed at random into 10% of all chocolate bars sold at this shop. Kati buys some of these bars and she opens them one at a time to see if they contain a voucher. Let be the probability that Kati obtains her third voucher on the bar opened.
\n(It is assumed that the probability that a chocolate bar contains a voucher stays at 10% throughout the question.)
\nIt is given that for .
\nKati’s mother goes to the shop and buys chocolate bars. She takes the bars home for Kati to open.
\nShow that and .
\nFind the values of the constants and .
\nDeduce that for .
\n(i) Hence show that has two modes and .
\n(ii) State the values of and .
\nDetermine the minimum value of such that the probability Kati receives at least one free gift is greater than 0.5.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
\nAG
\n(M1)
\n(or equivalent) A1
\nAG
\n[3 marks]
\nMETHOD 1
\nattempting to form equations in and M1
\nA1
\nA1
\nattempting to solve simultaneously (M1)
\nA1
\nMETHOD 2
\nM1
\n(M1)A1
\nA1
\nA1
\n\n
Note: Condone the absence of in the determination of the values of and .
\n\n
[5 marks]
\nMETHOD 1
\nEITHER
\n(M1)
\nOR
\n(M1)
\nTHEN
\nA1
\nA1
\nA1
\nAG
\nMETHOD 2
\n(M1)
\nA1A1
\n\n
Note: Award A1 for a correct numerator and A1 for a correct denominator.
\n\n
A1
\nAG
\n[4 marks]
\n(i) attempting to solve for M1
\nA1
\nR1
\nR1
\nhas two modes AG
\n\n
Note: Award R1R1 for a clearly labelled graphical representation of the two inequalities (using ).
\n\n
(ii) the modes are 20 and 21 A1
\n[5 marks]
\nMETHOD 1
\n(A1)
\nattempting to solve (or equivalent eg ) for (M1)
\n\n
Note: Award (M1) for attempting to solve an equality (obtaining ).
\n\n
A1
\nMETHOD 2
\n(A1)
\nattempting to solve for (M1)
\nA1
\n[3 marks]
\nThe vectors a and b are defined by a = , b = , where .
\nFind and simplify an expression for a • b in terms of .
\nHence or otherwise, find the values of for which the angle between a and b is obtuse .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
a • b = (M1)
\n= A1
\n\n
[2 marks]
\nrecognition that a • b = |a||b|cos θ (M1)
\na • b < 0 or < 0 or cos θ < 0 R1
\nNote: Allow ≤ for R1.
\n\n
attempt to solve using sketch or sign diagram (M1)
\nA1
\n\n
[4 marks]
\nLet a = and b = , .
\nGiven that a and b are perpendicular, find the possible values of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
a • b =
\nA1
\na • b = 0 (M1)
\n\n
attempt at solving their quadratic equation (M1)
\n\n
A1
\nNote: Attempt at solving using |a||b| = |a × b| will be M1A0A0A0 if neither answer found M1(A1)A1A0
for one correct answer and M1(A1)A1A1 for two correct answers.
[4 marks]
\nIn this question, give all answers to two decimal places.
\nKarl invests 1000 US dollars (USD) in an account that pays a nominal annual interest of 3.5%, compounded quarterly. He leaves the money in the account for 5 years.
\nCalculate the amount of money he has in the account after 5 years.
\nWrite down the amount of interest he earned after 5 years.
\nKarl decides to donate this interest to a charity in France. The charity receives 170 euros (EUR). The exchange rate is 1 USD = t EUR.
\nCalculate the value of t.
\n(M1)(A1)
\nNote: Award (M1) for substitution in compound interest formula, (A1) for correct substitution.
\nOR
\nN = 5
\nI = 3.5
\nPV = 1000
\nP/Y = 1
\nC/Y = 4
\nNote: Award (A1) for C/Y = 4 seen, (M1) for other correct entries.
\nOR
\nN = 5 × 4
\nI = 3.5
\nPV = 1000
\nP/Y = 1
\nC/Y = 4
\nNote: Award (A1) for C/Y = 4 seen, (M1) for other correct entries.
\n= 1190.34 (USD) (A1)
\nNote: Award (M1) for substitution in compound interest formula, (A1) for correct substitution.
\n[3 marks]
\n190.34 (USD) (A1)(ft) (C4)
\nNote: Award (A1)(ft) for subtraction of 1000 from their part (a)(i). Follow through from (a)(i).
\n[1 mark]
\n(M1)
\nNote: Award (M1) for division of 170 by their part (a)(ii).
\n= 0.89 (A1)(ft) (C2)
\nNote: Follow through from their part (a)(ii).
\n[2 marks]
\nSketch the graphs and on the following axes for 0 < ≤ 9.
\nHence solve in the range 0 < ≤ 9.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1A1
Note: Award A1 for each correct curve, showing all local max & mins.
\nNote: Award A0A0 for the curves drawn in degrees.
\n[2 marks]
\n= 1.35, 4.35, 6.64 (M1)
\nNote: Award M1 for attempt to find points of intersections between two curves.
\n0 < < 1.35 A1
\nNote: Accept < 1.35.
\n4.35 < < 6.64 A1A1
\nNote: Award A1 for correct endpoints, A1 for correct inequalities.
\nNote: Award M1FTA1FTA0FTA0FT for 0 < < 7.31.
\nNote: Accept < 7.31.
\n[4 marks]
\nJohn likes to go sailing every day in July. To help him make a decision on whether it is safe to go sailing he classifies each day in July as windy or calm. Given that a day in July is calm, the probability that the next day is calm is 0.9. Given that a day in July is windy, the probability that the next day is calm is 0.3. The weather forecast for the 1st July predicts that the probability that it will be calm is 0.8.
\nDraw a tree diagram to represent this information for the first three days of July.
\nFind the probability that the 3rd July is calm.
\nFind the probability that the 1st July was calm given that the 3rd July is windy.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"M17/5/MATHL/HP2/ENG/TZ2/05.a/M\"](\"../assets/uploads/tinymce_asset/asset/3519/Schermafbeelding_2017-08-09_om_18.23.32.png\"/)
M1A2
\n
Note: Award M1 for 3 stage tree-diagram, A2 for 0.8, 0.9, 0.3 probabilities correctly placed.
\n\n
[3 marks]
\n(M1)A1
\n[2 marks]
\n(M1)
\n\n
OR
\nOR (A1)(A1)
\n\n
Note: Award A1 for correct numerator, A1 for correct denominator.
\n\n
A1
\n[4 marks]
\nThere are 75 players in a golf club who take part in a golf tournament. The scores obtained on the 18th hole are as shown in the following table.
\n![\"M17/5/MATHL/HP2/ENG/TZ2/01\"](\"../assets/uploads/tinymce_asset/asset/3517/Schermafbeelding_2017-08-09_om_16.43.55.png\"/)
One of the players is chosen at random. Find the probability that this player’s score was 5 or more.
\nCalculate the mean score.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)A1
\n[2 marks]
\n(M1)
\nA1
\n[2 marks]
\nSketch the graphs of and on the following axes.
\nSolve the equation .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
straight line graph with correct axis intercepts A1
\nmodulus graph: V shape in upper half plane A1
\nmodulus graph having correct vertex and y-intercept A1
\n[3 marks]
\nMETHOD 1
\nattempt to solve (M1)
\nA1
\nNote: Accept using the graph.
\nattempt to solve (algebraically) M1
\nA1
\n[4 marks]
\n\n
\n
METHOD 2
\nM1
\n\n
\n
\n
attempt to factorise (or equivalent) M1
\n\n
A1
\nA1
\n[4 marks]
\nConsider the graphs of and , where .
\nSketch the graphs on the same set of axes.
\nGiven that the graphs enclose a region of area 18 square units, find the value of b.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"M17/5/MATHL/HP1/ENG/TZ1/A6.a/M\"](\"../assets/uploads/tinymce_asset/asset/3503/Schermafbeelding_2017-08-08_om_11.15.31.png\"/)
graphs sketched correctly (condone missing b) A1A1
\n[2 marks]
\n(M1)A1
\nA1
\n[3 marks]
\nThe marks achieved by eight students in a class test are given in the following list.
\nThe teacher increases all the marks by 2. Write down the new value for
\nthe standard deviation.
\n2.22 A1
\n[1 mark]
\nIqbal attempts three practice papers in mathematics. The probability that he passes the first paper is 0.6. Whenever he gains a pass in a paper, his confidence increases so that the probability of him passing the next paper increases by 0.1. Whenever he fails a paper the probability of him passing the next paper is 0.6.
\nComplete the given probability tree diagram for Iqbal’s three attempts, labelling each branch with the correct probability.
\nCalculate the probability that Iqbal passes at least two of the papers he attempts.
\nFind the probability that Iqbal passes his third paper, given that he passed only one previous paper.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1A1A1
Note: Award A1 for each correct column of probabilities.
\n[3 marks]
\nprobability (at least twice) =
\nEITHER
\n(M1)
\nOR
\n(M1)
\nNote: Award M1 for summing all required probabilities.
\nTHEN
\n= 0.696 A1
\n[2 marks]
\nP(passes third paper given only one paper passed before)
\n(M1)
\nA1
\n= 0.657 A1
\n[3 marks]
\nWhen carpet is manufactured, small faults occur at random. The number of faults in Premium carpets can be modelled by a Poisson distribution with mean 0.5 faults per 20m2. Mr Jones chooses Premium carpets to replace the carpets in his office building. The office building has 10 rooms, each with the area of 80m2.
\nFind the probability that the carpet laid in the first room has fewer than three faults.
\nFind the probability that exactly seven rooms will have fewer than three faults in the carpet.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(M1)
\n(A1)
\nA1
\n[3 marks]
\n(M1)(A1)
\nA1
\n\n
Note: Award M1 for clear recognition of binomial distribution.
\n\n
[3 marks]
\nEvents and are such that and .
\nFind .
\nFind .
\nHence show that events and are independent.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
(M1)
\nA1
\n[2 marks]
\n\n
(M1)
\nA1
\n[2 marks]
\nMETHOD 1
\nA1
\nR1
\nhence and are independent AG
\n\n
Note: If there is evidence that the student has calculated by assuming independence in the first place, award A0R0.
\n\n
METHOD 2
\nEITHER
\nA1
\nOR
\nA1
\nTHEN
\nand are independent R1
\nhence and are independent AG
\nMETHOD 3
\nA1
\nR1
\nhence and are independent AG
\n[2 marks]
\nDaniela is going for a holiday to South America. She flies from the US to Argentina stopping in Peru on the way.
\nIn Peru she exchanges 85 United States dollars (USD) for Peruvian nuevo sol (PEN). The exchange rate is 1 USD = 3.25 PEN and a flat fee of 5 USD commission is charged.
\nAt the end of Daniela’s holiday she has 370 Argentinean peso (ARS). She converts this back to USD at a bank that charges a 4% commission on the exchange. The exchange rate is 1 USD = 9.60 ARS.
\nCalculate the amount of PEN she receives.
\nCalculate the amount of USD she receives.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)(M1)
\n\n
Note: Award (M1) for subtracting 5 from 85, (M1) for multiplying by 3.25.
\nAward (M1) for , (M1) for subtracting .
\n\n
(A1) (C3)
\n[3 marks]
\n(M1)(M1)
\n\n
Note: Award (M1) for multiplying by 0.96 (or equivalent), (M1) for dividing by 9.6. If division by 3.25 seen in part (a), condone multiplication by 9.6 in part (b).
\n\n
(A1) (C3)
\n[3 marks]
\nA rational function is defined by where the parameters and . The following diagram represents the graph of .
\n![\"N16/5/MATHL/HP1/ENG/TZ0/03\"](\"../assets/uploads/tinymce_asset/asset/3290/Schermafbeelding_2017-02-28_om_09.42.27.png\"/)
Using the information on the graph,
\nstate the value of and the value of ;
\nfind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
\nA1
\n[2 marks]
\nuse the coordinates of on the graph M1
\nA1
\n[2 marks]
\nConsider two events and such that and .
\nCalculate ;
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
use of M1
\nA1
\n\n
A1
\n\n
Note: Do not award the final A1 if two solutions are given.
\n\n
[3 marks]
\nuse of or alternative (M1)
\n(A1)
\nA1
\n[3 marks]
\nClaudia travels from Buenos Aires to Barcelona. She exchanges 8000 Argentine Pesos (ARS) into Euros (EUR).
\nThe exchange rate is 1 ARS = 0.09819 EUR. The bank charges a 2% commission on the exchange.
\nWhen Claudia returns to Buenos Aires she has 85 EUR left and exchanges this money back into ARS. The exchange rate is 1 ARS = 0.08753 EUR. The bank charges % commission. The commission charged on this exchange is 14.57 ARS.
\nFind the amount of Euros that Claudia receives. Give your answer correct to two decimal places.
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)(M1)
\n\n
Note: Award (M1) for multiplying 8000 by 0.09819, (M1) for multiplying by 0.98 (or equivalent).
\n\n
769.81 (EUR) (A1) (C3)
\n[3 marks]
\n(M1)(M1)
\n\n
Note: Award (M1) for dividing 85 by 0.08753, and (M1) for multiplying their by and equating to 14.57.
\n\n
OR
\n(M1)
\n\n
Note: Award (M1) for dividing 85 by 0.08753.
\n\n
OR (M1)
\n\n
Note: Award (M1) for dividing 14.57 by 9.71095… or equivalent.
\n\n
(A1) (C3)
\n[3 marks]
\nConsider the function defined by where .
\nSketch the graph of indicating clearly any intercepts with the axes and the coordinates of any local maximum or minimum points.
\nState the range of .
\nSolve the inequality .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"N16/5/MATHL/HP2/ENG/TZ0/05.a/M\"](\"../assets/uploads/tinymce_asset/asset/3296/Schermafbeelding_2017-03-01_om_06.12.12.png\"/)
correct shape passing through the origin and correct domain A1
\n\n
Note: Endpoint coordinates are not required. The domain can be indicated by and 1 marked on the axis.
\nA1
\ntwo correct intercepts (coordinates not required) A1
\n\n
Note: A graph passing through the origin is sufficient for .
\n\n
[3 marks]
\nA1A1
\n\n
Note: Award A1A0 for open or semi-open intervals with correct endpoints. Award A1A0 for closed intervals with one correct endpoint.
\n\n
[2 marks]
\nattempting to solve either (or equivalent) or (or equivalent) (eg. graphically) (M1)
\n![\"N16/5/MATHL/HP2/ENG/TZ0/05.c/M\"](\"../assets/uploads/tinymce_asset/asset/3297/Schermafbeelding_2017-03-01_om_06.22.47.png\"/)
(A1)
\nA1A1
\n\n
Note: Award A0 for .
\n\n
[4 marks]
\nJohn purchases a new bicycle for 880 US dollars (USD) and pays for it with a Canadian credit card. There is a transaction fee of 4.2 % charged to John by the credit card company to convert this purchase into Canadian dollars (CAD).
\nThe exchange rate is 1 USD = 1.25 CAD.
\nJohn insures his bicycle with a US company. The insurance company produces the following table for the bicycle’s value during each year.
\nThe values of the bicycle form a geometric sequence.
\nDuring the 1st year John pays 120 USD to insure his bicycle. Each year the amount he pays to insure his bicycle is reduced by 3.50 USD.
\nCalculate, in CAD, the total amount John pays for the bicycle.
\nFind the value of the bicycle during the 5th year. Give your answer to two decimal places.
\nCalculate, in years, when the bicycle value will be less than 50 USD.
\nFind the total amount John has paid to insure his bicycle for the first 5 years.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
1.042 × 880 × 1.25 OR (880 + 0.042 × 880) × 1.25 (M1)(M1)
\nNote: Award (M1) for multiplying 880 by 1.042 and (M1) for multiplying 880 by 1.25.
\n1150 (CAD) (1146.20 (CAD)) (A1)(G2)
\nNote: Accept 1146.2 (CAD)
\n[3 marks]
\nOR (M1)
\nNote: Award (M1) for correctly dividing sequential terms to find the common ratio, or 0.8 seen.
\n880(0.8)5−1 (M1)
\nNote: Award (M1) for correct substitution into geometric sequence formula.
\n360.45 (USD) (A1)(G3)
\nNote: Do not award the final (A1) if the answer is not correct to 2 decimal places. Award at most (M0)(M1)(A0) if .
\n[3 marks]
\n(M1)
\nNote: Award (M1) for correct substitution into geometric sequence formula and (in)equating to 50. Accept weak or strict inequalities. Accept an equation. Follow through from their common ratio in part (b). Accept a sketch of their GP with as a valid method.
\nOR
\nAND (M1)
\nNote: Award (M1) for their and both seen. If the student states , without seen, this is not sufficient to award (M1).
\n14 or “14th year” or “after the 13th year” (A1)(ft)(G2)
\nNote: The context of the question requires the final answer to be an integer. Award at most (M1)(A0) for a final answer of 13.9 years. Follow through from their 0.8 in part (b).
\n[2 marks]
\n(M1)(A1)
\nNote: Award (M1) for substitution into arithmetic series formula, (A1) for correct substitution.
\n565 (USD) (A1)(G2)
\n[3 marks]
\nEach of the 25 students in a class are asked how many pets they own. Two students own three pets and no students own more than three pets. The mean and standard deviation of the number of pets owned by students in the class are and respectively.
\nFind the number of students in the class who do not own a pet.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nlet p have no pets, q have one pet and r have two pets (M1)
\np + q + r + 2 = 25 (A1)
\n0p + 1q + 2r + 6 = 18 A1
\nNote: Accept a statement that there are a total of 12 pets.
\nattempt to use variance equation, or evidence of trial and error (M1)
\n(A1)
\nattempt to solve a system of linear equations (M1)
\np = 14 A1
\n\n
METHOD 2
\n (M1)
(A1)
\nA1
\n(M1)(A1)
\n(M1)
\nA1
\nso 14 have no pets
\n[7 marks]
\nConsider two events, and , such that and .
\nBy drawing a Venn diagram, or otherwise, find .
\nShow that the events and are not independent.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
Note: Award M1 for a Venn diagram with at least one probability in the correct region.
\n\n
EITHER
\n(A1)
\nA1
\nOR
\n(A1)
\nA1
\n\n
[3 marks]
\nMETHOD 1
\n(M1)
\n= 0.2 A1
\nstatement that their R1
\nNote: Award R1 for correct reasoning from their value.
\n⇒ , not independent AG
\n\n
METHOD 2
\n(M1)
\n= 0.2 A1
\nstatement that their R1
\nNote: Award R1 for correct reasoning from their value.
\n⇒ , not independent AG
\nNote: Accept equivalent argument using .
\n\n
[3 marks]
\nHarry travelled from the USA to Mexico and changed 700 dollars (USD) into pesos (MXN).
\nThe exchange rate was 1 USD = 18.86 MXN.
\nOn his return, Harry had 2400 MXN to change back into USD.
\nThere was a 3.5 % commission to be paid on the exchange.
\nCalculate the amount of MXN Harry received.
\nCalculate the value of the commission, in MXN, that Harry paid.
\nThe exchange rate for this exchange was 1 USD = 17.24 MXN.
\nCalculate the amount of USD Harry received. Give your answer correct to the nearest cent.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
700 × 18.86 (M1)
\nNote: Award (M1) for multiplication by 18.86.
\n= 13 200 (13 202) (MXN) (A1) (C2)
\n\n
[2 marks]
\n2400 × 0.035 (M1)
\nNote: Award (M1) for multiplication by 0.035.
\n= 84 (MXN) (A1) (C2)
\n\n
[2 marks]
\n(M1)
\nNote: Award (M1) for dividing 2400 minus their part (b), by 17.24. Follow through from part (b).
\n= 134.34 (USD) (A1)(ft) (C2)
\nNote: Award at most (M1)(A0) if final answer is not given to nearest cent.
\n\n
[2 marks]
\nConsider two events and defined in the same sample space.
\nGiven that and ,
\nShow that .
\n(i) show that ;
\n(ii) hence find .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nM1
\nM1A1
\nAG
\nMETHOD 2
\nM1
\nM1
\n\n
A1
\nAG
\n[3 marks]
\n(i) use and (M1)
\nA1
\nM1
\nAG
\n(ii) METHOD 1
\nM1
\nM1
\nA1
\nMETHOD 2
\nM1
\nM1
\nA1
\n[6 marks]
\nIn this question give all answers correct to two decimal places.
\nJavier takes 5000 US dollars (USD) on a business trip to Venezuela. He exchanges 3000 USD into Venezuelan bolívars (VEF).
\nThe exchange rate is 1 USD 6.3021 VEF.
\nDuring his time in Venezuela, Javier spends 1250 USD and 12 000 VEF. On his return home, Javier exchanges his remaining VEF into USD.
\nThe exchange rate is 1 USD 8.7268 VEF.
\nCalculate the amount of VEF that Javier receives.
\nCalculate the total amount, in USD, that Javier has remaining from his 5000 USD after his trip to Venezuela.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
The first answer not given correct to two decimal places is not awarded the final (A1).
\nIncorrect rounding is not penalized thereafter.
\n(M1)
\n\n
Note: Award (M1) for multiplying 3000 by 6.3021.
\n\n
(A1) (C2)
\n[2 marks]
\n(M1)(M1)(M1)
\n\n
Note: Award (M1) for subtracting 12 000 from their answer to part (a) OR for 6906.30 seen, (M1) for dividing their amount by 8.7268 (can be implied if 791.389… seen) and (M1) for OR 750 seen.
\n\n
(A1)(ft) (C4)
\n\n
Note: Follow through from part (a).
\n\n
[4 marks]
\nA city hired 160 employees to work at a festival. The following cumulative frequency curve shows the number of hours employees worked during the festival.
\n![\"M17/5/MATME/SP1/ENG/TZ2/08.a.ii\"](\"../assets/uploads/tinymce_asset/asset/3541/Schermafbeelding_2017-08-14_om_07.01.49.png\"/)
The city paid each of the employees £8 per hour for the first 40 hours worked, and £10 per hour for each hour they worked after the first 40 hours.
\nWrite down the number of employees who worked 50 hours or less.
\nFind the amount of money an employee earned for working 40 hours;
\n130 employees A1 N1
\n[1 mark]
\n£320 A1 N1
\n[1 mark]
\nAdam is a beekeeper who collected data about monthly honey production in his bee hives. The data for six of his hives is shown in the following table.
\n![\"N17/5/MATME/SP2/ENG/TZ0/08\"](\"../assets/uploads/tinymce_asset/asset/4160/Schermafbeelding_2018-02-12_om_10.46.13.png\"/)
The relationship between the variables is modelled by the regression line with equation .
\nAdam has 200 hives in total. He collects data on the monthly honey production of all the hives. This data is shown in the following cumulative frequency graph.
\n![\"N17/5/MATME/SP2/ENG/TZ0/08.c.d.e\"](\"../assets/uploads/tinymce_asset/asset/4161/Schermafbeelding_2018-02-12_om_10.49.33.png\"/)
Adam’s hives are labelled as low, regular or high production, as defined in the following table.
\n![\"N17/5/MATME/SP2/ENG/TZ0/08.c.d.e_02\"](\"../assets/uploads/tinymce_asset/asset/4162/Schermafbeelding_2018-02-12_om_10.51.25.png\"/)
Adam knows that 128 of his hives have a regular production.
\nWrite down the value of and of .
\nUse this regression line to estimate the monthly honey production from a hive that has 270 bees.
\nWrite down the number of low production hives.
\nFind the value of ;
\nFind the number of hives that have a high production.
\nAdam decides to increase the number of bees in each low production hive. Research suggests that there is a probability of 0.75 that a low production hive becomes a regular production hive. Calculate the probability that 30 low production hives become regular production hives.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of setup (M1)
\negcorrect value for or
\n\n
A1A1 N3
\n[3 marks]
\nsubstituting into their equation (M1)
\neg
\n1424.67
\nA1 N2
\n[2 marks]
\n40 (hives) A1 N1
\n[1 mark]
\nvalid approach (M1)
\neg
\n168 hives have a production less than (A1)
\nA1 N3
\n[3 marks]
\nvalid approach (M1)
\neg
\n32 (hives) A1 N2
\n[2 marks]
\nrecognize binomial distribution (seen anywhere) (M1)
\neg
\ncorrect values (A1)
\neg (check FT) and and
\n0.144364
\n0.144 A1 N2
\n[3 marks]
\nThe following table shows the mean weight, y kg , of children who are x years old.
\nThe relationship between the variables is modelled by the regression line with equation .
\nFind the value of a and of b.
\nWrite down the correlation coefficient.
\nUse your equation to estimate the mean weight of a child that is 1.95 years old.
\nvalid approach (M1)
\neg correct value for a or b (or for r seen in (ii))
\na = 1.91966 b = 7.97717
\na = 1.92, b = 7.98 A1A1 N3
\n[3 marks]
\n0.984674
\nr = 0.985 A1 N1
\n[1 mark]
\ncorrect substitution into their equation (A1)
eg 1.92 × 1.95 + 7.98
11.7205
\n11.7 (kg) A1 N2
\n[2 marks]
\nLet , be a periodic function with
\nThe following diagram shows the graph of .
\nThere is a maximum point at A. The minimum value of is −13 .
\nA ball on a spring is attached to a fixed point O. The ball is then pulled down and released, so that it moves back and forth vertically.
\nThe distance, d centimetres, of the centre of the ball from O at time t seconds, is given by
\n\n
Find the coordinates of A.
\nFor the graph of , write down the amplitude.
\nFor the graph of , write down the period.
\nHence, write in the form .
\nFind the maximum speed of the ball.
\nFind the first time when the ball’s speed is changing at a rate of 2 cm s−2.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n−0.394791,13
\nA(−0.395, 13) A1A1 N2
\n[2 marks]
\n13 A1 N1
\n[1 mark]
\n, 6.28 A1 N1
\n[1 mark]
\nvalid approach (M1)
\neg recognizing that amplitude is p or shift is r
\n(accept p = 13, r = 0.395) A1A1 N3
\nNote: Accept any value of r of the form
\n[3 marks]
\nrecognizing need for d ′(t) (M1)
\neg −12 sin(t) − 5 cos(t)
\ncorrect approach (accept any variable for t) (A1)
\neg −13 sin(t + 0.395), sketch of d′, (1.18, −13), t = 4.32
\nmaximum speed = 13 (cms−1) A1 N2
\n[3 marks]
\nrecognizing that acceleration is needed (M1)
\neg a(t), d \"(t)
\ncorrect equation (accept any variable for t) (A1)
\neg
\nvalid attempt to solve their equation (M1)
\neg sketch, 1.33
\n1.02154
\n1.02 A2 N3
\n[5 marks]
\nThe maximum temperature , in degrees Celsius, in a park on six randomly selected days is shown in the following table. The table also shows the number of visitors, , to the park on each of those six days.
\n![\"M17/5/MATME/SP2/ENG/TZ2/02\"](\"../assets/uploads/tinymce_asset/asset/3554/Schermafbeelding_2017-08-14_om_17.34.22.png\"/)
The relationship between the variables can be modelled by the regression equation .
\nFind the value of and of .
\nWrite down the value of .
\nUse the regression equation to estimate the number of visitors on a day when the maximum temperature is 15 °C.
\nevidence of set up (M1)
\negcorrect value for or
\n0.667315, 22.2117
\nA1A1 N3
\n[3 marks]
\n0.922958
\nA1 N1
\n[1 marks]
\nvalid approach (M1)
\neg
\n32.2214 (A1)
\n32 (visitors) (must be an integer) A1 N2
\n[3 marks]
\nJim heated a liquid until it boiled. He measured the temperature of the liquid as it cooled. The following table shows its temperature, degrees Celsius, minutes after it boiled.
\n![\"M17/5/MATME/SP1/ENG/TZ1/04\"](\"../assets/uploads/tinymce_asset/asset/3526/Schermafbeelding_2017-08-11_om_08.45.05.png\"/)
Jim believes that the relationship between and can be modelled by a linear regression equation.
\nWrite down the independent variable.
\nWrite down the boiling temperature of the liquid.
\nJim describes the correlation as very strong. Circle the value below which best represents the correlation coefficient.
\n\n
Jim’s model is , for . Use his model to predict the decrease in temperature for any 2 minute interval.
\nA1 N1
\n[1 mark]
\n105 A1 N1
\n[1 mark]
\nA2 N2
\n[2 marks]
\nvalid approach (M1)
\neg
\nfinding where
\n4.48 (degrees) A1 N2
\n\n
Notes: Award no marks for answers that directly use the table to find the decrease in temperature for 2 minutes eg .
\n\n
[2 marks]
\nThe following table shows the hand lengths and the heights of five athletes on a sports team.
\nThe relationship between x and y can be modelled by the regression line with equation y = ax + b.
\nAnother athlete on this sports team has a hand length of 21.5 cm. Use the regression equation to estimate the height of this athlete.
\nsubstituting x = 21.5 into their equation (M1)
\neg 9.91(21.5) − 31.3
\n181.755
\n182 (cm) A1 N2
\n\n
[2 marks]
\nA bag contains 5 green balls and 3 white balls. Two balls are selected at random without replacement.
\nComplete the following tree diagram.
\n![\"N17/5/MATME/SP1/ENG/TZ0/01.a\"](\"../assets/uploads/tinymce_asset/asset/4147/Schermafbeelding_2018-02-11_om_09.35.12.png\"/)
Find the probability that exactly one of the selected balls is green.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct probabilities
\n![\"N17/5/MATME/SP1/ENG/TZ0/01.a/M\"](\"../assets/uploads/tinymce_asset/asset/4149/Schermafbeelding_2018-02-11_om_09.51.13.png\"/)
A1A1A1 N3
\n
Note: Award A1 for each correct bold answer.
\n\n
[3 marks]
\nmultiplying along branches (M1)
\neg
\nadding probabilities of correct mutually exclusive paths (A1)
\neg
\nA1 N2
\n[3 marks]
\nA group of 7 adult men wanted to see if there was a relationship between their Body Mass Index (BMI) and their waist size. Their waist sizes, in centimetres, were recorded and their BMI calculated. The following table shows the results.
\nThe relationship between and can be modelled by the regression equation .
\nWrite down the value of and of .
\nFind the correlation coefficient.
\nUse the regression equation to estimate the BMI of an adult man whose waist size is 95 cm.
\nvalid approach (M1)
\neg correct value for or (or for correct or = 0.955631 seen in (ii))
\n0.141120, 11.1424
\n= 0.141, = 11.1 A1A1 N3
\n[3 marks]
\n0.977563
\n= 0.978 A1 N1
\n[1 mark]
\ncorrect substitution into their regression equation (A1)
\neg 0.141(95) + 11.1
\n24.5488
\n24.5 A1 N2
\n[2 marks]
\nConsider the function , .
\nThe graph of is translated two units to the left to form the function .
\nExpress in the form where , , , , .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1
\nattempt to expand M1
\n(A1)
\nA1
\n\n
A1
\nNote: For correct expansion of award max M0M1(A1)A0A1.
\n[5 marks]
\nAt an amusement park, a Ferris wheel with diameter 111 metres rotates at a constant speed. The bottom of the wheel is k metres above the ground. A seat starts at the bottom of the wheel.
\nThe wheel completes one revolution in 16 minutes.
\nAfter t minutes, the height of the seat above ground is given by , for 0 ≤ t ≤ 32.
\nFind when the seat is 30 m above the ground for the third time.
\nvalid approach (M1)
eg sketch of h and
18.4630
\nt = 18.5 (minutes) A1 N3
\n[3 marks]
\nTwo events A and B are such that P(A) = 0.62 and P = 0.18.
\nFind P(A ∩ B′ ).
\nGiven that P((A ∪ B)′ ) = 0.19, find P(A | B′ ).
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach
\neg Venn diagram, P(A) − P (A ∩ B), 0.62 − 0.18 (M1)
\nP(A ∩ B' ) = 0.44 A1 N2
\n[2 marks]
\nvalid approach to find either P(B′ ) or P(B) (M1)
\neg (seen anywhere), 1 − P(A ∩ B′ ) − P((A ∪ B)′ )
correct calculation for P(B′ ) or P(B) (A1)
\neg 0.44 + 0.19, 0.81 − 0.62 + 0.18
\ncorrect substitution into (A1)
\neg
\n0.698412
\nP(A | B′ ) = (exact), 0.698 A1 N3
\n[4 marks]
\nThe depth of water in a port is modelled by the function , for , where is the number of hours after high tide.
\nAt high tide, the depth is 9.7 metres.
\nAt low tide, which is 7 hours later, the depth is 5.3 metres.
\nFind the value of .
\nFind the value of .
\nUse the model to find the depth of the water 10 hours after high tide.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg, sketch of graph,
\nA1 N2
\n[2 marks]
\nvalid approach (M1)
\neg, period is
\n0.448798
\n, (do not accept degrees) A1 N2
\n[2 marks]
\nvalid approach (M1)
\neg
\n7.01045
\n7.01 (m) A1 N2
\n[2 marks]
\nLet for .
\nLet .
\nThe function can be written in the form .
\nThe range of is ≤ ≤ . Find and .
\nFind the range of .
\nFind the value of and of .
\nFind the period of .
\nThe equation has two solutions where ≤ ≤ . Find both solutions.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nvalid attempt to find range (M1)
\neg , max = 6 min = 2,
and , and ,
\n, A1A1 N3
\n[3 marks]
\n10 ≤ ≤ 30 A2 N2
\n[2 marks]
\nevidence of substitution (may be seen in part (b)) (M1)
\neg ,
\n, (accept ) A1A1 N3
\nNote: If no working shown, award N2 for one correct value.
\n[3 marks]
\ncorrect working (A1)
\neg
\n1.04719
\n, 1.05 A1 N2
\n[2 marks]
\nvalid approach (M1)
\neg ,
Note: Award M1 for any correct value for or which lies outside the domain of .
\n3.81974, 4.03424
\n, (do not accept answers in degrees) A1A1 N3
\n[3 marks]
\nThe following diagram shows the graph of , for .
\n![\"N16/5/MATME/SP2/ENG/TZ0/10\"](\"../assets/uploads/tinymce_asset/asset/3313/Schermafbeelding_2017-03-03_om_16.53.31.png\"/)
The graph of has a minimum point at and a maximum point at .
\nThe graph of is obtained from the graph of by a translation of . The maximum point on the graph of has coordinates .
\nThe graph of changes from concave-up to concave-down when .
\n(i) Find the value of .
\n(ii) Show that .
\n(iii) Find the value of .
\n(i) Write down the value of .
\n(ii) Find .
\n(i) Find .
\n(ii) Hence or otherwise, find the maximum positive rate of change of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(i) valid approach (M1)
\neg
\nA1 N2
\n(ii) valid approach (M1)
\negperiod is 12, per
\nA1
\nAG N0
\n(iii) METHOD 1
\nvalid approach (M1)
\neg, substitution of points
\nA1 N2
\nMETHOD 2
\nvalid approach (M1)
\neg, amplitude is 6
\nA1 N2
\n[6 marks]
\n(i) A1 N1
\n(ii) A2 N2
\n[3 marks]
\n(i) METHOD 1 Using
\nrecognizing that a point of inflexion is required M1
\negsketch, recognizing change in concavity
\nevidence of valid approach (M1)
\neg, sketch, coordinates of max/min on
\n(exact) A1 N2
\nMETHOD 2 Using
\nrecognizing that a point of inflexion is required M1
\negsketch, recognizing change in concavity
\nevidence of valid approach involving translation (M1)
\neg, sketch,
\n(exact) A1 N2
\n(ii) valid approach involving the derivative of or (seen anywhere) (M1)
\neg, max on derivative, sketch of derivative
\nattempt to find max value on derivative M1
\neg, dot on max of sketch
\n3.14159
\nmax rate of change (exact), 3.14 A1 N2
\n[6 marks]
\nThe following diagram shows quadrilateral ABCD.
\n\n
Find DB.
\nFind DC.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nevidence of choosing sine rule (M1)
\neg
\ncorrect substitution (A1)
eg
9.57429
\nDB = 9.57 (cm) A1 N2
\n[3 marks]
\nevidence of choosing cosine rule (M1)
\neg
\ncorrect substitution into RHS (A1)
eg
10.5677
\nDC = 10.6 (cm) A1 N2
\n[3 marks]
\nA ship is sailing north from a point A towards point D. Point C is 175 km north of A. Point D is 60 km north of C. There is an island at E. The bearing of E from A is 055°. The bearing of E from C is 134°. This is shown in the following diagram.
\n![\"M17/5/MATME/SP2/ENG/TZ2/09\"](\"../assets/uploads/tinymce_asset/asset/3557/Schermafbeelding_2017-08-15_om_06.15.07.png\"/)
When the ship reaches D, it changes direction and travels directly to the island at 50 km per hour. At the same time as the ship changes direction, a boat starts travelling to the island from a point B. This point B lies on (AC), between A and C, and is the closest point to the island. The ship and the boat arrive at the island at the same time. Find the speed of the boat.
\nvalid approach for locating B (M1)
\negBE is perpendicular to ship’s path, angle
\ncorrect working for BE (A1)
\neg
\nvalid approach for expressing time (M1)
\neg
\ncorrect working equating time (A1)
\neg
\n27.2694
\n27.3 (km per hour) A1 N3
\n[5 marks]
\nTwo points P and Q have coordinates (3, 2, 5) and (7, 4, 9) respectively.
\nLet = 6i − j + 3k.
\nFind .
\nFind .
\nFind the angle between PQ and PR.
\nFind the area of triangle PQR.
\nHence or otherwise find the shortest distance from R to the line through P and Q.
\nvalid approach (M1)
\neg (7, 4, 9) − (3, 2, 5) A − B
\n4i + 2j + 4k A1 N2
\n[2 marks]
\ncorrect substitution into magnitude formula (A1)
eg
A1 N2
\n[2 marks]
\nfinding scalar product and magnitudes (A1)(A1)
\nscalar product = (4 × 6) + (2 × (−1) + (4 × 3) (= 34)
\nmagnitude of PR =
\ncorrect substitution of their values to find cos M1
\neg cos
\n0.581746
\n= 0.582 radians or = 33.3° A1 N3
\n[4 marks]
\ncorrect substitution (A1)
eg
area is 11.2 (sq. units) A1 N2
\n[2 marks]
\nrecognizing shortest distance is perpendicular distance from R to line through P and Q (M1)
\neg sketch, height of triangle with base
\ncorrect working (A1)
\neg
\n3.72677
\ndistance = 3.73 (units) A1 N2
\n[3 marks]
\nA communication tower, T, produces a signal that can reach cellular phones within a radius of 32 km. A straight road passes through the area covered by the tower’s signal.
\nThe following diagram shows a line representing the road and a circle representing the area covered by the tower’s signal. Point R is on the circumference of the circle and points S and R are on the road. Point S is 38 km from the tower and RŜT = 43˚.
\nLet SR = . Use the cosine rule to show that .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nrecognizing TR =32 (seen anywhere, including diagram) A1
\ncorrect working A1
\neg ,
\nAG N0
\n\n
[2 marks]
\nConsider the points A(−3, 4, 2) and B(8, −1, 5).
\nA line L has vector equation . The point C (5, , 1) lies on line L.
\nFind .
\nFind the value of .
\nShow that .
\nFind the angle between and .
\nFind the area of triangle ABC.
\nNote: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.
\n\n
correct substitution into formula (A1)
\neg
\n12.4498
\n(exact), 12.4 A1 N2
\n\n
[2 marks]
\nNote: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.
\n\n
valid approach to find (M1)
\neg , ,
\n(seen anywhere) (A1)
\nattempt to substitute their parameter into the vector equation (M1)
\neg ,
\nA1 N2
\n\n
[3 marks]
\nNote: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.
\n\n
correct approach A1
\neg , AO + OC,
\nAG N0
\nNote: Do not award A1 in part (ii) unless answer in part (i) is correct and does not result from working backwards.
\n\n
[2 marks]
\nNote: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.
\n\n
finding scalar product and magnitude (A1)(A1)
\nscalar product = 11 × 8 + −5 × −10 + 3 × −1 (=135)
\n\n
evidence of substitution into formula (M1)
\neg
\ncorrect substitution (A1)
\neg , ,
\n\n
0.565795, 32.4177°
\n= 0.566, 32.4° A1 N3
\n\n
[5 marks]
\nNote: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.
\n\n
correct substitution into area formula (A1)
\neg ,
\n42.8660
\narea = 42.9 A1 N2
\n\n
[2 marks]
\nLet .
\nFind .
\nLet . Find .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ncorrect substitution (A1)
\neg
\n4.58257
\n(exact), 4.58 A1 N2
\n[2 marks]
\nfinding scalar product and (A1)(A1)
\nscalar product
\n\n
substituting their values into cosine formula (M1)
\neg cos BC
\n0.509739 (29.2059°)
\n(29.2°) A1 N2
\n[4 marks]
\nA line passes through points and .
\nThe line also passes through the point .
\nShow that .
\nFind a vector equation for .
\nFind the value of .
\nThe point D has coordinates . Given that is perpendicular to , find the possible values of .
\ncorrect approach A1
\n\n
eg
\n\n
AG N0
\n[1 mark]
\nany correct equation in the form (any parameter for )
\n\n
where is or and is a scalar multiple of A2 N2
\n\n
eg
\n\n
Note: Award A1 for the form , A1 for the form , A0 for the form .
\n\n
[2 marks]
\nMETHOD 1 – finding value of parameter
\nvalid approach (M1)
\n\n
eg
\n\n
one correct equation (not involving ) (A1)
\neg
\ncorrect parameter from their equation (may be seen in substitution) A1
\neg
\ncorrect substitution (A1)
\n\n
eg
\n\n
A1 N2
\n\n
METHOD 2 – eliminating parameter
\nvalid approach (M1)
\n\n
eg
\n\n
one correct equation (not involving ) (A1)
\neg
\ncorrect equation (with ) A1
\neg
\ncorrect working to solve for (A1)
\neg
\n\n
A1 N2
\n\n
[5 marks]
\nvalid approach to find or (M1)
\n\n
eg
\n\n
correct vector for or (may be seen in scalar product) A1
\n\n
eg
\n\n
recognizing scalar product of or with direction vector of is zero (seen anywhere) (M1)
\n\n
eg
\n\n
correct scalar product in terms of only A1
\neg
\ncorrect working to solve quadratic (A1)
\neg
\nA1A1 N3
\n\n
[7 marks]
\nPoint A has coordinates (−4, −12, 1) and point B has coordinates (2, −4, −4).
\nThe line L passes through A and B.
\nShow that
\nFind a vector equation for L.
\nPoint C (k , 12 , −k) is on L. Show that k = 14.
\nFind .
\nWrite down the value of angle OBA.
\nPoint D is also on L and has coordinates (8, 4, −9).
\nFind the area of triangle OCD.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ncorrect approach A1
\neg
\nAG N0
\n[1 mark]
\nany correct equation in the form r = a + tb (any parameter for t) A2 N2
\nwhere a is or and b is a scalar multiple of
\neg r r
\nNote: Award A1 for the form a + tb, A1 for the form L = a + tb, A0 for the form r = b + ta.
\n[2 marks]
\nMETHOD 1 (solving for t)
\nvalid approach (M1)
\neg
\none correct equation A1
\neg −4 + 8t = 12, −12 + 8t = 12
\ncorrect value for t (A1)
\neg t = 2 or 3
\ncorrect substitution A1
\neg 2 + 6(2), −4 + 6(3), −[1 + 3(−5)]
\nk = 14 AG N0
\n\n
METHOD 2 (solving simultaneously)
\nvalid approach (M1)
\neg
\ntwo correct equations in A1
\neg k = −4 + 6t, −k = 1 −5t
\nEITHER (eliminating k)
\ncorrect value for t (A1)
\neg t = 2 or 3
\ncorrect substitution A1
\neg 2 + 6(2), −4 + 6(3)
\nOR (eliminating t)
\ncorrect equation(s) (A1)
\neg 5k + 20 = 30t and −6k − 6 = 30t, −k = 1 − 5
\ncorrect working clearly leading to k = 14 A1
\neg −k + 14 = 0, −6k = 6 −5k − 20, 5k = −20 + 6(1 + k)
\nTHEN
\nk = 14 AG N0
\n[4 marks]
\n\n
correct substitution into scalar product A1
\neg (2)(6) − (4)(8) − (4)(−5), 12 − 32 + 20
\n= 0 A1 N0
\n[2 marks]
\n\n
A1 N1
\n[1 marks]
\nMETHOD 1 ( × height × CD)
\nrecognizing that OB is altitude of triangle with base CD (seen anywhere) M1
\neg sketch showing right angle at B
\nor (seen anywhere) (A1)
\ncorrect magnitudes (seen anywhere) (A1)(A1)
\n\n
\n
correct substitution into A1
\neg
\narea A1 N3
\n\n
METHOD 2 (subtracting triangles)
\nrecognizing that OB is altitude of either ΔOBD or ΔOBC(seen anywhere) M1
\neg sketch of triangle showing right angle at B
\none correct vector or or or (seen anywhere) (A1)
\neg ,
\n(seen anywhere) (A1)
\none correct magnitude of a base (seen anywhere) (A1)
\n\n
correct working A1
\neg
\narea A1 N3
\n\n
METHOD 3 (using ab sin C with ΔOCD)
\ntwo correct side lengths (seen anywhere) (A1)(A1)
\n\n
attempt to find cosine ratio (seen anywhere) M1
eg
correct working for sine ratio A1
\neg
\ncorrect substitution into A1
\neg
\narea A1 N3
\n[6 marks]
\nLet and , where O is the origin. L1 is the line that passes through A and B.
\nFind a vector equation for L1.
\nThe vector is perpendicular to . Find the value of p.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
any correct equation in the form r = a + tb (accept any parameter for t)
\nwhere a is , and b is a scalar multiple of A2 N2
\neg r = , r = 2i + j + 3k + s(i + 3j + k)
\nNote: Award A1 for the form a + tb, A1 for the form L = a + tb, A0 for the form r = b + ta.
\n[2 marks]
\nMETHOD 1
\ncorrect scalar product (A1)
\neg (1 × 2) + (3 × p) + (1 × 0), 2 + 3p
\nevidence of equating their scalar product to zero (M1)
\neg a•b = 0, 2 + 3p = 0, 3p = −2
\nA1 N3
\n\n
METHOD 2
\nvalid attempt to find angle between vectors (M1)
\ncorrect substitution into numerator and/or angle (A1)
\neg
\nA1 N3
\n[3 marks]
\nEvents and are independent with and .
\nFind .
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid interpretation (may be seen on a Venn diagram) (M1)
\neg
\nA1 N2
\n[2 marks]
\nvalid attempt to find (M1)
\neg
\ncorrect working for (A1)
\neg
\ncorrect working for (A1)
\neg
\nA1 N3
\n[4 marks]
\nA line, , has equation . Point lies on .
\nFind .
\nA second line, , is parallel to and passes through (1, 2, 3).
\nWrite down a vector equation for .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ncorrect equation (A1)
\neg ,
\n(A1)
\nsubstitute their value into component (M1)
\neg ,
\nA1 N3
\n[4 marks]
\n(=(i + 2j + 3k) + (6i + 2k)) A2 N2
\nNote: Accept any scalar multiple of for the direction vector.
\nAward A1 for , A1 for , A0 for .
\n[2 marks]
\nThe position vectors of points P and Q are i 2 j k and 7i 3j 4k respectively.
\nFind a vector equation of the line that passes through P and Q.
\nThe line through P and Q is perpendicular to the vector 2i nk. Find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid attempt to find direction vector (M1)
\neg
\ncorrect direction vector (or multiple of) (A1)
\neg6i j 3k
\nany correct equation in the form r a tb (any parameter for ) A2 N3
\nwhere a is i 2j k or 7i 3j 4k , and b is a scalar multiple of 6i j 3k
\negr 7i 3j 4k t(6i j 3k), r
\n\n
Notes: Award A1 for the form a tb, A1 for the form L a tb, A0 for the form r b ta.
\n\n
[4 marks]
\ncorrect expression for scalar product (A1)
\neg
\nsetting scalar product equal to zero (seen anywhere) (M1)
\negu v
\nA1 N2
\n[3 marks]
\nThe function is defined by where .
\nFind the remainder when is divided by .
\nFind the remainder when is divided by .
\nProve that has only one real zero.
\nWrite down the transformation that will transform the graph of onto the graph of .
\nThe random variable follows a Poisson distribution with a mean of and .
\nFind the value of .
\n(M1)
\nNote: Award M1 for a valid attempt at remainder theorem or polynomial division.
\n= −12 A1
\nremainder = −12
\n[2 marks]
\n= 0 A1
\nremainder = 0
\n[1 mark]
\n(is a zero) A1
\nNote: Can be seen anywhere.
\nEITHER
\nfactorise to get (M1)A1
\n(for ) (or equivalent statement) R1
\nNote: Award R1 if correct two complex roots are given.
\nOR
\nA1
\nattempting to show M1
\neg discriminant = 36 – 96 < 0, completing the square
\nno turning points R1
\nTHEN
\nonly one real zero (as the curve is continuous) AG
\n[4 marks]
\nnew graph is (M1)
\nstretch parallel to the -axis (with invariant), scale factor 0.5 A1
\nNote: Accept “horizontal” instead of “parallel to the -axis”.
\n[2 marks]
\nM1A1
\nNote: Allow factorials in the denominator for A1.
\nA1
\nNote: Accept any correct cubic equation without factorials and .
\nEITHER
\n(M1)
\n(A1)
\nOR
\n(M1)(A1)
\nTHEN
\n= 1.5 A1
\n[6 marks]
\nThe following table shows a probability distribution for the random variable , where .
\n![\"M17/5/MATME/SP2/ENG/TZ2/10\"](\"../assets/uploads/tinymce_asset/asset/3558/Schermafbeelding_2017-08-15_om_06.18.09.png\"/)
A bag contains white and blue marbles, with at least three of each colour. Three marbles are drawn from the bag, without replacement. The number of blue marbles drawn is given by the random variable .
\nA game is played in which three marbles are drawn from the bag of ten marbles, without replacement. A player wins a prize if three white marbles are drawn.
\nFind .
\nFind .
\nWrite down the probability of drawing three blue marbles.
\nExplain why the probability of drawing three white marbles is .
\nThe bag contains a total of ten marbles of which are white. Find .
\nGrant plays the game until he wins two prizes. Find the probability that he wins his second prize on his eighth attempt.
\ncorrect substitution into formula (A1)
\neg
\n, 0.0333 A1 N2
\n[2 marks]
\nevidence of summing probabilities to 1 (M1)
\neg
\nA1 N2
\n[2 marks]
\nA1 N1
\n[1 mark]
\nvalid reasoning R1
\neg
\nAG N0
\n[1 mark]
\nvalid method (M1)
\neg
\ncorrect equation A1
\neg
\nA1 N2
\n[3 marks]
\nrecognizing one prize in first seven attempts (M1)
\neg
\ncorrect working (A1)
\neg
\ncorrect approach (A1)
\neg
\n0.065119
\n0.0651 A1 N2
\n[4 marks]
\nPablo drives to work. The probability that he leaves home before 07:00 is .
\nIf he leaves home before 07:00 the probability he will be late for work is .
\nIf he leaves home at 07:00 or later the probability he will be late for work is .
\nCopy and complete the following tree diagram.
\nFind the probability that Pablo leaves home before 07:00 and is late for work.
\nFind the probability that Pablo is late for work.
\nGiven that Pablo is late for work, find the probability that he left home before 07:00.
\nTwo days next week Pablo will drive to work. Find the probability that he will be late at least once.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1A1A1 N3
Note: Award A1 for each bold fraction.
\n[3 marks]
\nmultiplying along correct branches (A1)
eg
P(leaves before 07:00 ∩ late) = A1 N2
\n[2 marks]
\n\n
multiplying along other “late” branch (M1)
eg
adding probabilities of two mutually exclusive late paths (A1)
eg
A1 N2
\n[3 marks]
\nrecognizing conditional probability (seen anywhere) (M1)
eg
correct substitution of their values into formula (A1)
eg
A1 N2
\n[3 marks]
\nvalid approach (M1)
eg 1 − P(not late twice), P(late once) + P(late twice)
correct working (A1)
eg
A1 N2
\n[3 marks]
\nThe weights, in grams, of oranges grown in an orchard, are normally distributed with a mean of 297 g. It is known that 79 % of the oranges weigh more than 289 g and 9.5 % of the oranges weigh more than 310 g.
\nThe weights of the oranges have a standard deviation of σ.
\nThe grocer at a local grocery store will buy the oranges whose weights exceed the 35th percentile.
\nThe orchard packs oranges in boxes of 36.
\nFind the probability that an orange weighs between 289 g and 310 g.
\nFind the standardized value for 289 g.
\nHence, find the value of σ.
\nTo the nearest gram, find the minimum weight of an orange that the grocer will buy.
\nFind the probability that the grocer buys more than half the oranges in a box selected at random.
\nThe grocer selects two boxes at random.
\nFind the probability that the grocer buys more than half the oranges in each box.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct approach indicating subtraction (A1)
\neg 0.79 − 0.095, appropriate shading in diagram
\nP(289 < w < 310) = 0.695 (exact), 69.5 % A1 N2
\n[2 marks]
\nMETHOD 1
\nvalid approach (M1)
\neg 1 − p, 21
\n−0.806421
\nz = −0.806 A1 N2
\n\n
METHOD 2
\n(i) & (ii)
\ncorrect expression for z (seen anywhere) (A1)
\neg
\nvalid approach (M1)
\neg 1 − p, 21
\n−0.806421
\nz = −0.806 (seen anywhere) A1 N2
\n\n
[2 marks]
\nMETHOD 1
\nattempt to standardize (M1)
\neg
\ncorrect substitution with their z (do not accept a probability) A1
\neg
\n9.92037
\nσ = 9.92 A1 N2
\n\n
METHOD 2
\n(i) & (ii)
\ncorrect expression for z (seen anywhere) (A1)
\neg
\nvalid approach (M1)
\neg 1 − p, 21
\n−0.806421
\nz = −0.806 (seen anywhere) A1 N2
\nvalid attempt to set up an equation with their z (do not accept a probability) (M1)
\neg
\n9.92037
\nσ = 9.92 A1 N2
\n[3 marks]
\nvalid approach (M1)
\neg P(W < w) = 0.35, −0.338520 (accept 0.385320), diagram showing values in a standard normal distribution
\ncorrect score at the 35th percentile (A1)
\neg 293.177
\n294 (g) A1 N2
\nNote: If working shown, award (M1)(A1)A0 for 293.
If no working shown, award N1 for 293.177, N1 for 293.
Exception to the FT rule: If the score is incorrect, and working shown, award A1FT for correctly finding their minimum weight (by rounding up)
\n[3 marks]
\nevidence of recognizing binomial (seen anywhere) (M1)
\neg
\ncorrect probability (seen anywhere) (A1)
\neg 0.65
\nEITHER
\nfinding P(X ≤ 18) from GDC (A1)
\neg 0.045720
\nevidence of using complement (M1)
\neg 1−P(X ≤ 18)
\n0.954279
\nP(X > 18) = 0.954 A1 N2
\nOR
\nrecognizing P(X > 18) = P(X ≥ 19) (M1)
\nsumming terms from 19 to 36 (A1)
\neg P(X = 19) + P(X = 20) + … + P(X = 36)
\n0.954279
\nP(X > 18) = 0.954 A1 N2
\n[5 marks]
\ncorrect calculation (A1)
\n\n
0.910650
\n0.911 A1 N2
\n[2 marks]
\nThe following box-and-whisker plot shows the number of text messages sent by students in a school on a particular day.
\nFind the value of the interquartile range.
\nOne student sent k text messages, where k > 11 . Given that k is an outlier, find the least value of k.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognizing Q1 or Q3 (seen anywhere) (M1)
\neg 4,11 , indicated on diagram
\nIQR = 7 A1 N2
\n[2 marks]
\nrecognizing the need to find 1.5 IQR (M1)
\neg 1.5 × IQR, 1.5 × 7
\nvalid approach to find k (M1)
\neg 10.5 + 11, 1.5 × IQR + Q3
\n21.5 (A1)
\nk = 22 A1 N3
\nNote: If no working shown, award N2 for an answer of 21.5.
\n[4 marks]
\nA random variable is normally distributed with mean, . In the following diagram, the shaded region between 9 and represents 30% of the distribution.
\n![\"M17/5/MATME/SP2/ENG/TZ1/09\"](\"../assets/uploads/tinymce_asset/asset/3545/Schermafbeelding_2017-08-14_om_10.15.49.png\"/)
The standard deviation of is 2.1.
\nThe random variable is normally distributed with mean and standard deviation 3.5. The events and are independent, and .
\nFind .
\nFind the value of .
\nFind .
\nGiven that , find .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg
\n(exact) A1 N2
\n[2 marks]
\n(may be seen in equation) (A1)
\nvalid attempt to set up an equation with their (M1)
\neg
\n10.7674
\nA1 N3
\n[3 marks]
\n(seen anywhere) (A1)
\nvalid approach (M1)
\neg
\ncorrect equation (A1)
\neg
\nA1
\nA1 N3
\n[5 marks]
\nfinding (seen anywhere) (A2)
\nrecognizing conditional probability (M1)
\neg
\ncorrect working (A1)
\neg
\n0.746901
\n0.747 A1 N3
\n[5 marks]
\nConsider the vectors a = and b = .
\nFind the value of for which a and b are
\nparallel.
\nperpendicular.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nvalid approach (M1)
\neg b = 2a, a = b, cos θ = 1, a•b = −|a||b|, 2 = 18
\n= 9 A1 N2
\n[2 marks]
\nevidence of scalar product (M1)
\neg a•b, (0)(0) + (3)(6) + (18)
\nrecognizing a•b = 0 (seen anywhere) (M1)
\ncorrect working (A1)
\neg 18 + 18 = 0, 18 = −18 (A1)
\n= −1 A1 N3
\n[4 marks]
\nConsider the function .
\nSketch the graph of y = f (x), for −4 ≤ x ≤ 3 and −50 ≤ y ≤ 100.
\nUse your graphic display calculator to find the zero of f (x).
\nUse your graphic display calculator to find the coordinates of the local minimum point.
\nUse your graphic display calculator to find the equation of the tangent to the graph of y = f (x) at the point (–2, 38.75).
\nGive your answer in the form y = mx + c.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(A1)(A1)
\n
Note: Award (A1) for axis labels and some indication of scale; accept y or f(x).
\nUse of graph paper is not required. If no scale is given, assume the given window for zero and minimum point.
\nAward (A1) for smooth curve with correct general shape.
\nAward (A1) for x-intercept closer to y-axis than to end of sketch.
\nAward (A1) for correct local minimum with x-coordinate closer to y-axis than end of sketch and y-coordinate less than half way to top of sketch.
\nAward at most (A1)(A0)(A1)(A1) if the sketch intersects the y-axis or if the sketch curves away from the y-axis as x approaches zero.
\n\n
[4 marks]
\n1.19 (1.19055…) (A1)
\n\n
Note: Accept an answer of (1.19, 0).
\nDo not follow through from an incorrect sketch.
\n\n
[1 mark]
\n(−1.5, 36) (A1)(A1)
\nNote: Award (A0)(A1) if parentheses are omitted.
\nAccept x = −1.5, y = 36.
\n\n
[2 marks]
\ny = −9.25x + 20.3 (y = −9.25x + 20.25) (A1)(A1)
\nNote: Award (A1) for −9.25x, award (A1) for +20.25, award a maximum of (A0)(A1) if answer is not an equation.
\n\n
[2 marks]
\nThere are three fair six-sided dice. Each die has two green faces, two yellow faces and two red faces.
\nAll three dice are rolled.
\nTed plays a game using these dice. The rules are:
\nThe random variable ($) represents how much is added to his winnings after a turn.
\nThe following table shows the distribution for , where $ represents his winnings in the game so far.
\nFind the probability of rolling exactly one red face.
\nFind the probability of rolling two or more red faces.
\nShow that, after a turn, the probability that Ted adds exactly $10 to his winnings is .
\nWrite down the value of .
\nHence, find the value of .
\nTed will always have another turn if he expects an increase to his winnings.
\nFind the least value of for which Ted should end the game instead of having another turn.
\nvalid approach to find P(one red) (M1)
eg , , ,
\nlisting all possible cases for exactly one red (may be indicated on tree diagram)
\nP(1 red) = 0.444 [0.444, 0.445] A1 N2
\n[3 marks] [5 maximum for parts (a.i) and (a.ii)]
\nvalid approach (M1)
eg P() + P(), 1 − P( ≤ 1), binomcdf
\ncorrect working (A1)
\neg , 0.222 + 0.037 ,
\n0.259259
\nP(at least two red) = 0.259 A1 N3
\n[3 marks] [5 maximum for parts (a.i) and (a.ii)]
\nrecognition that winning $10 means rolling exactly one green (M1)
\nrecognition that winning $10 also means rolling at most 1 red (M1)
\neg “cannot have 2 or more reds”
\ncorrect approach A1
\neg P(1G ∩ 0R) + P(1G ∩ 1R), P(1G) − P(1G ∩ 2R),
\n“one green and two yellows or one of each colour”
\nNote: Because this is a “show that” question, do not award this A1 for purely numerical expressions.
\none correct probability for their approach (A1)
\neg , , , ,
\ncorrect working leading to A1
\neg , ,
\nprobability = AG N0
\n[5 marks]
\n, 0.259 (check FT from (a)(ii)) A1 N1
\n[1 mark]
\nevidence of summing probabilities to 1 (M1)
\neg , ,
\n0.148147 (0.148407 if working with their value to 3 sf)
\n(exact), 0.148 A1 N2
\n[2 marks]
\ncorrect substitution into the formula for expected value (A1)
\neg
\ncorrect critical value (accept inequality) A1
\neg = 34.2857 , > 34.2857
\n$40 A1 N2
\n[3 marks]
\nThe voltage in a circuit is given by the equation
\n, where is measured in seconds.
\nThe current in this circuit is given by the equation
\n.
\nThe power in this circuit is given by .
\nThe average power in this circuit from to is given by the equation
\n, where .
\nWrite down the maximum and minimum value of .
\nWrite down two transformations that will transform the graph of onto the graph of .
\nSketch the graph of for 0 ≤ ≤ 0.02 , showing clearly the coordinates of the first maximum and the first minimum.
\nFind the total time in the interval 0 ≤ ≤ 0.02 for which ≥ 3.
\n\n
Find (0.007).
\n\n
With reference to your graph of explain why > 0 for all > 0.
\n\n
Given that can be written as where , , , > 0, use your graph to find the values of , , and .
\n\n
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
3, −3 A1A1
\n[2 marks]
\nstretch parallel to the -axis (with -axis invariant), scale factor A1
\ntranslation of (shift to the left by 0.003) A1
\nNote: Can be done in either order.
\n[2 marks]
\ncorrect shape over correct domain with correct endpoints A1
first maximum at (0.0035, 4.76) A1
first minimum at (0.0085, −1.24) A1
[3 marks]
\n≥ 3 between = 0.0016762 and 0.0053238 and = 0.011676 and 0.015324 (M1)(A1)
\nNote: Award M1A1 for either interval.
\n= 0.00730 A1
\n[3 marks]
\n(M1)
\n= 2.87 A1
\n[2 marks]
\nin each cycle the area under the axis is smaller than area above the axis R1
\nthe curve begins with the positive part of the cycle R1
\n[2 marks]
\n(M1)
\nA1
\n\n
A1
\n\n
A1
\n(M1)
\nA1
\n[6 marks]
\nA random variable Z is normally distributed with mean 0 and standard deviation 1. It is known that P( < −1.6) = and P( > 2.4) = . This is shown in the following diagram.
\nA second random variable is normally distributed with mean and standard deviation .
\nIt is known that P( < 1) = .
\nFind P(−1.6 < < 2.4). Write your answer in terms of and .
\nGiven that > −1.6, find the probability that z < 2.4 . Write your answer in terms of and .
\nWrite down the standardized value for .
\nIt is also known that P( > 2) = .
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognizing area under curve = 1 (M1)
\neg , ,
\nA1 N2
\n[2 marks]
\n(seen anywhere) (A1)
\nrecognizing conditional probability (M1)
\neg ,
\ncorrect working (A1)
\neg ,
\nA1 N4
\nNote: Do not award the final A1 if correct answer is seen followed by incorrect simplification.
\n[4 marks]
\n(may be seen in part (d)) A1 N1
\nNote: Depending on the candidate’s interpretation of the question, they may give as the answer to part (c). Such answers should be awarded the first (M1) in part (d), even when part (d) is left blank. If the candidate goes on to show as part of their working in part (d), the A1 in part (c) may be awarded.
\n[1 mark]
\nattempt to standardize (do not accept ) (M1)
\neg (may be seen in part (c)), ,
\ncorrect equation with each -value (A1)(A1)
\neg , ,
\nvalid approach (to set up equation in one variable) M1
\neg ,
\ncorrect working (A1)
\neg , ,
\nA1 N2
\n[6 marks]
\nThe vector equation of line is given by r .
\nPoint P is the point on that is closest to the origin. Find the coordinates of P.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1 (Distance between the origin and P)
\ncorrect position vector for OP (A1)
\neg ,
\ncorrect expression for OP or OP2 (seen anywhere) A1
\neg ,
\nvalid attempt to find the minimum of OP (M1)
\neg , root on sketch of , min indicated on sketch of
\n(A1)
\nsubstitute their value of into (only award if there is working to find ) (M1)
\neg one correct coordinate,
\n\n
A1 N2
\n\n
METHOD 2 (Perpendicular vectors)
\nrecognizing that closest implies perpendicular (M1)
\neg (may be seen on sketch),
\nvalid approach involving (M1)
\neg
\ncorrect scalar product A1
\neg , ,
\n(A1)
\nsubstitute their value of into or (only award if scalar product used to find ) (M1)
\neg one correct coordinate,
\n\n
A1 N2
\n\n
[6 marks]
\nThe following table shows a probability distribution for the random variable , where .
\n
A bag contains white and blue marbles, with at least three of each colour. Three marbles are drawn from the bag, without replacement. The number of blue marbles drawn is given by the random variable .
\nA game is played in which three marbles are drawn from the bag of ten marbles, without replacement. A player wins a prize if three white marbles are drawn.
\nJill plays the game nine times. Find the probability that she wins exactly two prizes.
\nvalid approach (M1)
\neg
\n0.279081
\n0.279 A1 N2
\n[2 marks]
\nA discrete random variable has the following probability distribution.
\n![\"N17/5/MATME/SP2/ENG/TZ0/04\"](\"../assets/uploads/tinymce_asset/asset/4159/Schermafbeelding_2018-02-12_om_10.34.18.png\"/)
Find the value of .
\nWrite down .
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\negtotal probability = 1
\ncorrect equation (A1)
\neg
\nA2 N3
\n[4 marks]
\nA1 N1
\n[1 mark]
\nvalid approach for finding (M1)
\neg
\ncorrect substitution into formula for conditional probability (A1)
\neg
\n0.0476190
\n(exact), 0.0476 A1 N2
\n[3 marks]
\nRosa joins a club to prepare to run a marathon. During the first training session Rosa runs a distance of 3000 metres. Each training session she increases the distance she runs by 400 metres.
\nA marathon is 42.195 kilometres.
\nIn the th training session Rosa will run further than a marathon for the first time.
\nCarlos joins the club to lose weight. He runs 7500 metres during the first month. The distance he runs increases by 20% each month.
\nWrite down the distance Rosa runs in the third training session;
\nWrite down the distance Rosa runs in the th training session.
\nFind the value of .
\nCalculate the total distance, in kilometres, Rosa runs in the first 50 training sessions.
\nFind the distance Carlos runs in the fifth month of training.
\nCalculate the total distance Carlos runs in the first year.
\n3800 m (A1)
\n[1 mark]
\nOR (M1)(A1)
\n\n
Note: Award (M1) for substitution into arithmetic sequence formula, (A1) for correct substitution.
\n\n
[2 marks]
\n(M1)
\n\n
Notes: Award (M1) for their correct inequality. Accept .
\nAccept OR . Award (M0) for .
\n\n
(A1)(ft)(G2)
\n\n
Note: Follow through from part (a)(ii), but only if is a positive integer.
\n\n
[2 marks]
\n(M1)(A1)(ft)
\n\n
Note: Award (M1) for substitution into sum of an arithmetic series formula, (A1)(ft) for correct substitution.
\n\n
(A1)
\n\n
Note: Award (A1) for their seen.
\n\n
(A1)(ft)(G3)
\n\n
Note: Award (A1)(ft) for correctly converting their answer in metres to km; this can be awarded independently from previous marks.
\n\n
OR
\n(M1)(A1)(ft)(A1)
\n\n
Note: Award (M1) for substitution into sum of an arithmetic series formula, (A1)(ft) for correct substitution, (A1) for correctly converting 3000 m and 400 m into km.
\n\n
(A1)(G3)
\n[4 marks]
\n(M1)(A1)
\n\n
Note: Award (M1) for substitution into geometric series formula, (A1) for correct substitutions.
\n\n
(A1)(G3)
\nOR
\n(M1)(A1)
\n\n
Note: Award (M1) for substitution into geometric series formula, (A1) for correct substitutions.
\n\n
(A1)(G3)
\n[3 marks]
\n(M1)(A1)
\n\n
Notes: Award (M1) for substitution into sum of a geometric series formula, (A1) for correct substitutions. Follow through from their ratio () in part (d). If (distance does not increase) or the final answer is unrealistic (eg ), do not award the final (A1).
\n\n
(A1)(G2)
\n[3 marks]
\nSergei is training to be a weightlifter. Each day he trains at the local gym by lifting a metal bar that has heavy weights attached. He carries out successive lifts. After each lift, the same amount of weight is added to the bar to increase the weight to be lifted.
\nThe weights of each of Sergei’s lifts form an arithmetic sequence.
\nSergei’s friend, Yuri, records the weight of each lift. Unfortunately, last Monday, Yuri misplaced all but two of the recordings of Sergei’s lifts.
\nOn that day, Sergei lifted 21 kg on the third lift and 46 kg on the eighth lift.
\nFor that day find how much weight was added after each lift.
\nFor that day find the weight of Sergei’s first lift.
\nOn that day, Sergei made 12 successive lifts. Find the total combined weight of these lifts.
\n5d = 46 − 21 OR u1 + 2d = 21 and u1 + 7d = 46 (M1)
\nNote: Award (M1) for a correct equation in d or for two correct equations in u1 and d.
\n(d =) 5 (kg) (A1) (C2)
\n[2 marks]
\nu1 + 2 × 5 = 21 (M1)
\nOR
\nu1 + 7 × 5 = 46 (M1)
\nNote: Award (M1) for substitution of their d into either of the two equations.
\n(u1 =) 11 (kg) (A1)(ft) (C2)
\nNote: Follow through from part (a)(i).
\n[2 marks]
\n(M1)
\nNote: Award (M1) for correct substitution into arithmetic series formula.
\n= 462 (kg) (A1)(ft) (C2)
\nNote: Follow through from parts (a) and (b).
\n[2 marks]
\nA new café opened and during the first week their profit was $60.
\nThe café’s profit increases by $10 every week.
\nA new tea-shop opened at the same time as the café. During the first week their profit was also $60.
\nThe tea-shop’s profit increases by 10 % every week.
\nFind the café’s profit during the 11th week.
\nCalculate the café’s total profit for the first 12 weeks.
\nFind the tea-shop’s profit during the 11th week.
\nCalculate the tea-shop’s total profit for the first 12 weeks.
\nIn the mth week the tea-shop’s total profit exceeds the café’s total profit, for the first time since they both opened.
\nFind the value of m.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
60 + 10 × 10 (M1)(A1)
\nNote: Award (M1) for substitution into the arithmetic sequence formula, (A1) for correct substitution.
\n= ($) 160 (A1)(G3)
\n[3 marks]
\n(M1)(A1)(ft)
\nNote: Award (M1) for substituting the arithmetic series formula, (A1)(ft) for correct substitution. Follow through from their first term and common difference in part (a).
\n= ($) 1380 (A1)(ft)(G2)
\n[3 marks]
\n\n
60 × 1.110 (M1)(A1)
\nNote: Award (M1) for substituting the geometric progression nth term formula, (A1) for correct substitution.
\n= ($) 156 (155.624…) (A1)(G3)
\nNote: Accept the answer if it rounds correctly to 3 sf, as per the accuracy instructions.
\n[3 marks]
\n\n
(M1)(A1)(ft)
\nNote: Award (M1) for substituting the geometric series formula, (A1)(ft) for correct substitution. Follow through from part (c) for their first term and common ratio.
\n= ($)1280 (1283.05…) (A1)(ft)(G2)
\n[3 marks]
\n(M1)(M1)
\nNote: Award (M1) for correctly substituted geometric and arithmetic series formula with n (accept other variable for “n”), (M1) for comparing their expressions consistent with their part (b) and part (d).
\nOR
\n (M1)(M1)
Note: Award (M1) for two curves with approximately correct shape drawn in the first quadrant, (M1) for one point of intersection with approximate correct position.
\nAccept alternative correct sketches, such as
\nAward (M1) for a curve with approximate correct shape drawn in the 1st (or 4th) quadrant and all above (or below) the x-axis, (M1) for one point of intersection with the x-axis with approximate correct position.
\n17 (A2)(ft)(G3)
\nNote: Follow through from parts (b) and (d).
An answer of 16 is incorrect. Award at most (M1)(M1)(A0)(A0) with working seen. Award (G0) if final answer is 16 without working seen.
[4 marks]
\nAt Penna Airport the probability, P(A), that all passengers arrive on time for a flight is 0.70. The probability, P(D), that a flight departs on time is 0.85. The probability that all passengers arrive on time for a flight and it departs on time is 0.65.
\nThe number of hours that pilots fly per week is normally distributed with a mean of 25 hours and a standard deviation . 90 % of pilots fly less than 28 hours in a week.
\nShow that event A and event D are not independent.
\nFind .
\nGiven that all passengers for a flight arrive on time, find the probability that the flight does not depart on time.
\nFind the value of .
\nAll flights have two pilots. Find the percentage of flights where both pilots flew more than 30 hours last week.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nmultiplication of P(A) and P(D) (A1)
\neg 0.70 × 0.85, 0.595
\ncorrect reasoning for their probabilities R1
\neg ,
\nA and D are not independent AG N0
\n\n
METHOD 2
\ncalculation of (A1)
\neg , 0.928
\ncorrect reasoning for their probabilities R1
\neg ,
\nA and D are not independent AG N0
\n[2 marks]
\ncorrect working (A1)
\neg , 0.7 − 0.65 , correct shading and/or value on Venn diagram
\nA1 N2
\n[2 marks]
\n\n
recognizing conditional probability (seen anywhere) (M1)
\neg ,
\ncorrect working (A1)
\neg
\n0.071428
\n, 0.0714 A1 N2
\n[3 marks]
\nfinding standardized value for 28 hours (seen anywhere) (A1)
\neg
\ncorrect working to find (A1)
eg ,
\n2.34091
\nA1 N2
\n[3 marks]
\n(A1)
\nvalid approach (seen anywhere) (M1)
\neg , (0.01634)2 , B(2, 0.0163429) , 2.67E-4 , 2.66E-4
\n0.0267090
\n0.0267 % A2 N3
\n[4 marks]
\nOllie has installed security lights on the side of his house that are activated by a sensor. The sensor is located at point C directly above point D. The area covered by the sensor is shown by the shaded region enclosed by triangle ABC. The distance from A to B is 4.5 m and the distance from B to C is 6 m. Angle AĈB is 15°.
\nFind CÂB.
\nPoint B on the ground is 5 m from point E at the entrance to Ollie’s house. He is 1.8 m tall and is standing at point D, below the sensor. He walks towards point B.
\nFind the distance Ollie is from the entrance to his house when he first activates the sensor.
\n(M1)(A1)
\nCÂB = 20.2º (20.187415…) A1
\nNote: Award (M1) for substituted sine rule formula and award (A1) for correct substitutions.
\n[3 marks]
\nA1
\n(let X be the point on BD where Ollie activates the sensor)
\n(M1)
\nNote: Award A1 for their correct angle . Award M1 for correctly substituted trigonometric formula.
\nA1
\n(M1)
\n= 2.45 (m) (2.44714…) A1
\n[5 marks]
\nA nationwide study on reaction time is conducted on participants in two age groups. The participants in Group X are less than 40 years old. Their reaction times are normally distributed with mean 0.489 seconds and standard deviation 0.07 seconds.
\nThe participants in Group Y are 40 years or older. Their reaction times are normally distributed with mean 0.592 seconds and standard deviation σ seconds.
\nIn the study, 38 % of the participants are in Group X.
\nA randomly selected participant has a reaction time greater than 0.65 seconds. Find the probability that the participant is in Group X.
\nTen of the participants with reaction times greater than 0.65 are selected at random. Find the probability that at least two of them are in Group X.
\ncorrect work for P(group X and > 0.65) or P(group Y and > 0.65) (may be seen anywhere) (A1)
\neg , ,
\n\n
recognizing conditional probability (seen anywhere) (M1)
\neg ,
\nvalid approach to find (M1)
\neg ,
correct work for (A1)
\neg 0.0107 × 0.38 + 0.396 × 0.62, 0.249595
\ncorrect substitution into conditional probability formula A1
\neg ,
\n0.016327
\nA1 N3
\n\n
[6 marks]
\nrecognizing binomial probability (M1)
\neg , , (0.016327)2(0.983672)8,
\nvalid approach (M1)
\neg , , summing terms from 2 to 10 (accept binomcdf(10, 0.0163, 2, 10))
\n0.010994
\nA1 N2
\n\n
[3 marks]
\nA biased four-sided die is rolled. The following table gives the probability of each score.
\nFind the value of k.
\nCalculate the expected value of the score.
\nThe die is rolled 80 times. On how many rolls would you expect to obtain a three?
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of summing to 1 (M1)
\neg 0.28 + k + 1.5 + 0.3 = 1, 0.73 + k = 1
\nk = 0.27 A1 N2
\n[2 marks]
\ncorrect substitution into formula for E (X) (A1)
eg 1 × 0.28 + 2 × k + 3 × 0.15 + 4 × 0.3
E (X) = 2.47 (exact) A1 N2
\n[2 marks]
\nvalid approach (M1)
\neg np, 80 × 0.15
\n12 A1 N2
\n[2 marks]
\nHelen is building a cabin using cylindrical logs of length 2.4 m and radius 8.4 cm. A wedge is cut from one log and the cross-section of this log is illustrated in the following diagram.
\nFind 50° in radians.
\nFind the volume of this log.
\nA1
\n[1 mark]
\nvolume M1M1M1
\nNote: Award M1 240 × area, award M1 for correctly substituting area sector formula, award M1 for subtraction of the angles or their areas.
\n= 45800 (= 45811.96071) A1
\n[4 marks]
\nThe mass M of apples in grams is normally distributed with mean μ. The following table shows probabilities for values of M.
\nThe apples are packed in bags of ten.
\nAny apples with a mass less than 95 g are classified as small.
\nWrite down the value of k.
\nShow that μ = 106.
\nFind P(M < 95) .
\nFind the probability that a bag of apples selected at random contains at most one small apple.
\nFind the expected number of bags in this crate that contain at most one small apple.
\nFind the probability that at least 48 bags in this crate contain at most one small apple.
\nevidence of using (M1)
\neg k + 0.98 + 0.01 = 1
\nk = 0.01 A1 N2
\n[2 marks]
\nrecognizing that 93 and 119 are symmetrical about μ (M1)
\neg μ is midpoint of 93 and 119
\ncorrect working to find μ A1
\n\n
μ = 106 AG N0
\n[2 marks]
\nfinding standardized value for 93 or 119 (A1)
eg z = −2.32634, z = 2.32634
correct substitution using their z value (A1)
eg
σ = 5.58815 (A1)
\n0.024508
\nP(X < 95) = 0.0245 A2 N3
\n[5 marks]
\nevidence of recognizing binomial (M1)
\neg 10, ananaCpqn−=××and 0.024B(5,,)pnp=
\nvalid approach (M1)
\neg P(1),P(0)P(1)XXX≤=+=
\n0.976285
\n0.976 A1 N2
\n[3 marks]
\nrecognizing new binomial probability (M1)
eg B(50, 0.976)
correct substitution (A1)
eg E(X) = 50 (0.976285)
48.81425
\n48.8 A1 N2
\n[3 marks]
\nvalid approach (M1)
\neg P(X ≥ 48), 1 − P(X ≤ 47)
\n0.884688
\n0.885 A1 N2
\n[2 marks]
\nA jar contains 5 red discs, 10 blue discs and green discs. A disc is selected at random and replaced. This process is performed four times.
\nWrite down the probability that the first disc selected is red.
\nLet be the number of red discs selected. Find the smallest value of for which .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1 N1
\n[1 mark]
\nrecognizing binomial distribution (M1)
\neg
\ncorrect value for the complement of their (seen anywhere) A1
\neg
\ncorrect substitution into (A1)
\neg
\n(A1)
\nA1 N3
\n[5 marks]
\nA particle, A, moves so that its velocity ( ms−1) at time is given by = 2 sin , ≥ 0.
\nThe kinetic energy () of the particle A is measured in joules (J) and is given by = 52.
\nWrite down an expression for as a function of time.
\nHence find .
\nHence or otherwise find the first time at which the kinetic energy is changing at a rate of 5 J s−1.
\nA1
\n[1 mark]
\n(M1)A1
\n[2 marks]
\n= 0.126 (M1)A1
\n[2 marks]
\nThe random variable is normally distributed with a mean of 100. The following diagram shows the normal curve for .
\n![\"M17/5/MATME/SP1/ENG/TZ2/03\"](\"../assets/uploads/tinymce_asset/asset/3534/Schermafbeelding_2017-08-11_om_16.32.27.png\"/)
Let be the shaded region under the curve, to the right of 107. The area of is 0.24.
\nWrite down .
\nFind .
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1 N1
\n[1 mark]
\nvalid approach (M1)
\neg
\ncorrect working (A1)
\neg
\nA1 N2
\n[3 marks]
\nvalid approach (M1)
\neg
\nA1 N2
\n[2 marks]
\nIn a large university the probability that a student is left handed is 0.08. A sample of 150 students is randomly selected from the university. Let be the expected number of left-handed students in this sample.
\nFind .
\nHence, find the probability that exactly students are left handed;
\nHence, find the probability that fewer than students are left handed.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of binomial distribution (may be seen in part (b)) (M1)
\neg
\nA1 N2
\n[2 marks]
\n(A1)
\n0.119231
\nprobability A1 N2
\n[2 marks]
\nrecognition that (M1)
\n0.456800
\nA1 N2
\n[2 marks]
\nThe weights, , of newborn babies in Australia are normally distributed with a mean 3.41 kg and standard deviation 0.57 kg. A newborn baby has a low birth weight if it weighs less than kg.
\nGiven that 5.3% of newborn babies have a low birth weight, find .
\nA newborn baby has a low birth weight.
\nFind the probability that the baby weighs at least 2.15 kg.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg, ![\"N16/5/MATME/SP2/ENG/TZ0/05.a/M\"](\"../assets/uploads/tinymce_asset/asset/3317/Schermafbeelding_2017-03-06_om_06.21.13.png\"/)
2.48863
\nA2 N3
\n[3 marks]
\ncorrect value or expression (seen anywhere)
\neg (A1)
\nevidence of conditional probability (M1)
\neg
\n0.744631
\n0.745 A1 N2
\n[3 marks]
\nIn a group of 20 girls, 13 take history and 8 take economics. Three girls take both history and economics, as shown in the following Venn diagram. The values and represent numbers of girls.
\n![\"M17/5/MATME/SP1/ENG/TZ1/01\"](\"../assets/uploads/tinymce_asset/asset/3525/Schermafbeelding_2017-08-11_om_08.39.12.png\"/)
Find the value of ;
\nFind the value of .
\nA girl is selected at random. Find the probability that she takes economics but not history.
\nvalid approach (M1)
\neg
\nA1 N2
\n[2 marks]
\nvalid approach (M1)
\neg
\nA1 N2
\n[2 marks]
\nvalid approach (M1)
\neg
\nA1 N2
\n[2 marks]
\nThe heights of adult males in a country are normally distributed with a mean of 180 cm and a standard deviation of . 17% of these men are shorter than 168 cm. 80% of them have heights between and 192 cm.
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
finding the -value for 0.17 (A1)
\neg
\nsetting up equation to find , (M1)
\neg
\n(A1)
\nEITHER (Properties of the Normal curve)
\ncorrect value (seen anywhere) (A1)
\neg
\ncorrect working (A1)
\neg
\n\n
correct equation in
\neg (A1)
\n35.6536
\nA1 N3
\nOR (Trial and error using different values of h)
\ntwo correct probabilities whose 2 sf will round up and down, respectively, to 0.8 A2
\neg
\n\n
A2
\n[7 marks]
\nThe following diagram shows the graph of , the derivative of .
\n![\"M17/5/MATME/SP1/ENG/TZ1/06\"](\"../assets/uploads/tinymce_asset/asset/3527/Schermafbeelding_2017-08-11_om_08.50.59.png\"/)
The graph of has a local minimum at A, a local maximum at B and passes through .
\nThe point lies on the graph of the function, .
\nWrite down the gradient of the curve of at P.
\nFind the equation of the normal to the curve of at P.
\nDetermine the concavity of the graph of when and justify your answer.
\nA1 N1
\n[1 mark]
\ngradient of normal (A1)
\nattempt to substitute their normal gradient and coordinates of P (in any order) (M1)
\neg
\nA1 N3
\n[3 marks]
\ncorrect answer and valid reasoning A2 N2
\nanswer: eg graph of is concave up, concavity is positive (between )
\nreason: eg slope of is positive, is increasing, ,
\nsign chart (must clearly be for and show A and B)
\n![\"M17/5/MATME/SP1/ENG/TZ1/06.b/M\"](\"../assets/uploads/tinymce_asset/asset/3532/Schermafbeelding_2017-08-11_om_10.53.43.png\"/)
\n
Note: The reason given must refer to a specific function/graph. Referring to “the graph” or “it” is not sufficient.
\n\n
[2 marks]
\nA manager wishes to check the mean mass of flour put into bags in his factory. He randomly samples 10 bags and finds the mean mass is 1.478 kg and the standard deviation of the sample is 0.0196 kg.
\nFind for this sample.
\nFind a 95 % confidence interval for the population mean, giving your answer to 4 significant figures.
\nThe bags are labelled as being 1.5 kg mass. Comment on this claim with reference to your answer in part (b).
\n(M1)A1
\n[2 marks]
\n(1.463, 1.493) (M1)A1
\nNote: If used answer is (1.464, 1.492), award M1A0.
\n[2 marks]
\n95 % of the time these results would be produced by a population with mean of less than 1.5 kg, so it is likely the mean mass is less than 1.5 kg R1
\n[1 mark]
\nA quadratic function can be written in the form . The graph of has axis of symmetry and -intercept at
\nFind the value of .
\nFind the value of .
\nThe line is a tangent to the curve of . Find the values of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1 (using x-intercept)
\ndetermining that 3 is an -intercept (M1)
\neg, ![\"M17/5/MATME/SP1/ENG/TZ1/09.a/M\"](\"../assets/uploads/tinymce_asset/asset/3533/Schermafbeelding_2017-08-11_om_13.55.43.png\"/)
valid approach (M1)
\neg
\nA1 N2
\nMETHOD 2 (expanding f (x))
\ncorrect expansion (accept absence of ) (A1)
\neg
\nvalid approach involving equation of axis of symmetry (M1)
\neg
\nA1 N2
\nMETHOD 3 (using derivative)
\ncorrect derivative (accept absence of ) (A1)
\neg
\nvalid approach (M1)
\neg
\nA1 N2
\n[3 marks]
\nattempt to substitute (M1)
\neg
\ncorrect working (A1)
\neg
\nA1 N2
\n[3 marks]
\nMETHOD 1 (using discriminant)
\nrecognizing tangent intersects curve once (M1)
\nrecognizing one solution when discriminant = 0 M1
\nattempt to set up equation (M1)
\neg
\nrearranging their equation to equal zero (M1)
\neg
\ncorrect discriminant (if seen explicitly, not just in quadratic formula) A1
\neg
\ncorrect working (A1)
\neg
\nA1A1 N0
\nMETHOD 2 (using derivatives)
\nattempt to set up equation (M1)
\neg
\nrecognizing derivative/slope are equal (M1)
\neg
\ncorrect derivative of (A1)
\neg
\nattempt to set up equation in terms of either or M1
\neg
\nrearranging their equation to equal zero (M1)
\neg
\ncorrect working (A1)
\neg
\nA1A1 N0
\n[8 marks]
\nLet . The following diagram shows part of the graph of .
\n![\"M17/5/MATME/SP2/ENG/TZ2/08\"](\"../assets/uploads/tinymce_asset/asset/3556/Schermafbeelding_2017-08-15_om_06.09.00.png\"/)
\n
There are -intercepts at and at . There is a maximum at A where , and a point of inflexion at B where .
\nFind the value of .
\nWrite down the coordinates of A.
\nWrite down the rate of change of at A.
\nFind the coordinates of B.
\nFind the the rate of change of at B.
\nLet be the region enclosed by the graph of , the -axis, the line and the line . The region is rotated 360° about the -axis. Find the volume of the solid formed.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of valid approach (M1)
\neg
\n2.73205
\nA1 N2
\n[2 marks]
\n1.87938, 8.11721
\nA2 N2
\n[2 marks]
\nrate of change is 0 (do not accept decimals) A1 N1
\n[1 marks]
\nMETHOD 1 (using GDC)
\nvalid approach M1
\neg, max/min on
\nsketch of either or , with max/min or root (respectively) (A1)
\nA1 N1
\nSubstituting their value into (M1)
\neg
\nA1 N1
\nMETHOD 2 (analytical)
\nA1
\nsetting (M1)
\nA1 N1
\nsubstituting their value into (M1)
\neg
\nA1 N1
\n[4 marks]
\nrecognizing rate of change is (M1)
\neg
\nrate of change is 6 A1 N2
\n[3 marks]
\nattempt to substitute either limits or the function into formula (M1)
\ninvolving (accept absence of and/or )
\neg
\n128.890
\nA2 N3
\n[3 marks]
\nIn a coffee shop, the time it takes to serve a customer can be modelled by a normal distribution with a mean of 1.5 minutes and a standard deviation of 0.4 minutes.
\nTwo customers enter the shop together. They are served one at a time.
\nFind the probability that the total time taken to serve both customers will be less than 4 minutes.
\nClearly state any assumptions you have made.
\nlet be the time to serve both customers and the time to serve the th customer
\nassuming independence of and R1
\nis normally distributed and (M1)
\nA1
\nM1A1
\nA1
\n[6 marks]
\nTomás is playing with sticks and he forms the first three diagrams of a pattern. These diagrams are shown below.
\n![\"M17/5/MATSD/SP1/ENG/TZ2/05\"](\"../assets/uploads/tinymce_asset/asset/3582/Schermafbeelding_2017-08-15_om_17.25.44.png\"/)
Tomás continues forming diagrams following this pattern.
\nTomás forms a total of 24 diagrams.
\nDiagram is formed with 52 sticks. Find the value of .
\nFind the total number of sticks used by Tomás for all 24 diagrams.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)(A1)
\n\n
Note: Award (M1) for substitution into the formula of the th term of an arithmetic sequence, (A1) for correct substitution.
\n\n
(A1) (C3)
\n[3 marks]
\nOR (M1)(A1)(ft)
\n\n
Notes: Award (M1) for substitution into the sum of the first terms of an arithmetic sequence formula, (A1)(ft) for their correct substitution, consistent with part (a).
\n\n
924 (A1)(ft) (C3)
\n\n
Note: Follow through from part (a).
\n\n
[3 marks]
\nA particle P moves with velocity v = in a magnetic field, B = , .
\nGiven that v is perpendicular to B, find the value of .
\nThe force, F, produced by P moving in the magnetic field is given by the vector equation F = v × B, .
\nGiven that | F | = 14, find the value of .
\n(M1)
\nA1
\n[2 marks]
\n(M1)
\nA1
\nmagnitude is M1
\nA1
\n[4 marks]
\nConsider f(x), g(x) and h(x), for x∈ where h(x) = (x).
\nGiven that g(3) = 7 , g′ (3) = 4 and f ′ (7) = −5 , find the gradient of the normal to the curve of h at x = 3.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognizing the need to find h′ (M1)
\nrecognizing the need to find h′ (3) (seen anywhere) (M1)
\nevidence of choosing chain rule (M1)
\neg
\ncorrect working (A1)
\neg
\n(A1)
\nevidence of taking their negative reciprocal for normal (M1)
\neg
\ngradient of normal is A1 N4
\n[7 marks]
\nThe company Snakezen’s Ladders makes ladders of different lengths. All the ladders that the company makes have the same design such that:
\nthe first rung is 30 cm from the base of the ladder,
\nthe second rung is 57 cm from the base of the ladder,
\nthe distance between the first and second rung is equal to the distance between all adjacent rungs on the ladder.
\nThe ladder in the diagram was made by this company and has eleven equally spaced rungs.
\n![\"M17/5/MATSD/SP1/ENG/TZ1/05\"](\"../assets/uploads/tinymce_asset/asset/3566/Schermafbeelding_2017-08-15_om_10.59.54.png\"/)
Find the distance from the base of this ladder to the top rung.
\nThe company also makes a ladder that is 1050 cm long.
\nFind the maximum number of rungs in this 1050 cm long ladder.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)(A1)
\n\n
Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions.
\n\n
(A1) (C3)
\n\n
Note: Units are not required.
\n\n
[3 marks]
\n(M1)(A1)(ft)
\n\n
Note: Award (M1) for substituted arithmetic sequence formula , accept an equation, (A1) for correct substitutions.
\n\n
(A1)(ft) (C3)
\n\n
Note: Follow through from their 27 in part (a). The answer must be an integer and rounded down.
\n\n
[3 marks]
\nLet . Part of the graph of is shown in the following diagram.
\nThe graph of crosses the -axis at the point P. The line L is tangent to the graph of at P.
\nFind the coordinates of P.
\nFind .
\nHence, find the equation of L in terms of .
\nThe graph of has a local minimum at the point Q. The line L passes through Q.
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg , , ,
\n(0, 6) (accept = 0 and = 6) A1 N2
\n\n
[2 marks]
\nA2 N2
\n\n
[2 marks]
\nvalid approach (M1)
\neg
\ncorrect working (A1)
\neg , slope = ,
\nattempt to substitute gradient and coordinates into linear equation (M1)
\neg , , , L
\ncorrect equation A1 N3
\neg , ,
\n\n
[4 marks]
\nvalid approach to find intersection (M1)
\neg
\ncorrect equation (A1)
\neg
\ncorrect working (A1)
\neg ,
\nat Q (A1)
\n\n
valid approach to find minimum (M1)
\neg
\ncorrect equation (A1)
\neg
\nsubstitution of their value of at Q into their equation (M1)
\neg ,
\n= −4 A1 N0
\n\n
[8 marks]
\nProduct research leads a company to believe that the revenue () made by selling its goods at a price () can be modelled by the equation.
\n, ,
\nThere are two competing models, A and B with different values for the parameters and .
\nModel A has = 3, = −0.5 and model B has = 2.5, = −0.6.
\nThe company experiments by selling the goods at three different prices in three similar areas and the results are shown in the following table.
\nThe company will choose the model with the smallest value for the sum of square residuals.
\nDetermine which model the company chose.
\n(Model A)
\nM1
\npredicted values
\n (A1)
(M1)
\n= 0.5263… A1
\n\n
(Model B)
\n\n
predicted values
\n (A1)
0.170576… A1
\nchose model B A1
\nNote: Method marks can be awarded if seen for either model A or model B. Award final A1 if it is a correct deduction from their calculated values for A and B.
\n[7 marks]
\n\n
Consider the geometric sequence .
\nWrite down the common ratio of the sequence.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1) (C1)
\n[1 mark]
\nGabriella purchases a new car.
\nThe car’s value in dollars, , is modelled by the function
\n\n
where is the number of years since the car was purchased and is a constant.
\nAfter two years, the car’s value is $9143.20.
\nThis model is defined for . At years the car’s value will be zero dollars.
\nWrite down, and simplify, an expression for the car’s value when Gabriella purchased it.
\nFind the value of .
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(M1)
\n\n
Note: Award (M1) for correct substitution into .
\n\n
(A1) (C2)
\n\n
Note: Accept OR 9790 for a final answer.
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution into .
\n\n
(A1) (C2)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution into .
\n\n
OR
\n![\"N16/5/MATSD/SP1/ENG/TZ0/15.c/M\"](\"../assets/uploads/tinymce_asset/asset/3338/Schermafbeelding_2017-03-07_om_07.33.35.png\"/)
(M1)
\n
Note: Award (M1) for a correctly shaped curve with some indication of scale on the vertical axis.
\n\n
(A1)(ft) (C2)
\n\n
Note: Follow through from part (b).
\n\n
[2 marks]
\nThe rates of change of the area covered by two types of fungi, X and Y, on a particular tree are given by the following equations, where is the area covered by X and is the area covered by Y.
\n\n
\n
The matrix has eigenvalues of 2 and −1 with corresponding eigenvectors and .
\nInitially = 8 cm2 and = 10 cm2.
\nFind the value of when .
\nOn the following axes, sketch a possible trajectory for the growth of the two fungi, making clear any asymptotic behaviour.
\nM1
\n= −1 A1
\n[2 marks]
\nasymptote of trajectory along r M1A1
\nNote: Award M1A0 if asymptote along .
\ntrajectory begins at (8, 10) with negative gradient A1A1
\n[4 marks]
\nThe lengths of trout in a fisherman’s catch were recorded over one month, and are represented in the following histogram.
\n![\"M17/5/MATSD/SP1/ENG/TZ1/01\"](\"../assets/uploads/tinymce_asset/asset/3564/Schermafbeelding_2017-08-15_om_10.45.21.png\"/)
Complete the following table.
\n![\"M17/5/MATSD/SP1/ENG/TZ1/01\"](\"../assets/uploads/tinymce_asset/asset/3572/Schermafbeelding_2017-08-15_om_12.36.53.png\"/)
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"M17/5/MATSD/SP1/ENG/TZ1/01.a/M\"](\"../assets/uploads/tinymce_asset/asset/3574/Schermafbeelding_2017-08-15_om_12.38.42.png\"/)
(A2) (C2)
\n
Note: Award (A2) for all correct entries, (A1) for 3 correct entries.
\n\n
[2 marks]
\nAll lengths in this question are in metres.
\n\n
Consider the function , for −2 ≤ ≤ 2. In the following diagram, the shaded region is enclosed by the graph of and the -axis.
\nA container can be modelled by rotating this region by 360˚ about the -axis.
\nWater can flow in and out of the container.
\nThe volume of water in the container is given by the function , for 0 ≤ ≤ 4 , where is measured in hours and is measured in m3. The rate of change of the volume of water in the container is given by .
\nThe volume of water in the container is increasing only when < < .
\nFind the volume of the container.
\nFind the value of and of .
\nDuring the interval < < , he volume of water in the container increases by m3. Find the value of .
\nWhen = 0, the volume of water in the container is 2.3 m3. It is known that the container is never completely full of water during the 4 hour period.
\n\n
Find the minimum volume of empty space in the container during the 4 hour period.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to substitute correct limits or the function into formula involving (M1)
\neg ,
\n4.18879
\nvolume = 4.19, (exact) (m3) A2 N3
\nNote: If candidates have their GDC incorrectly set in degrees, award M marks where appropriate, but no A marks may be awarded. Answers from degrees are p = 13.1243 and q = 26.9768 in (b)(i) and 12.3130 or 28.3505 in (b)(ii).
\n\n
[3 marks]
\n\n
\n
recognizing the volume increases when is positive (M1)
\neg > 0, sketch of graph of indicating correct interval
\n1.73387, 3.56393
\n= 1.73, = 3.56 A1A1 N3
\n\n
[3 marks]
\n\n
\n
valid approach to find change in volume (M1)
\neg ,
\n3.74541
\ntotal amount = 3.75 (m3) A2 N3
\n\n
[3 marks]
\nNote: There may be slight differences in the final answer, depending on which values candidates carry through from previous parts. Accept answers that are consistent with correct working.
\n\n
recognizing when the volume of water is a maximum (M1)
\neg maximum when ,
\nvalid approach to find maximum volume of water (M1)
\neg , , 3.85745
\ncorrect expression for the difference between volume of container and maximum value (A1)
\neg , 4.19 − 3.85745
\n0.331334
\n0.331 (m3) A2 N3
\n\n
[5 marks]
\nOn one day 180 flights arrived at a particular airport. The distance travelled and the arrival status for each incoming flight was recorded. The flight was then classified as on time, slightly delayed, or heavily delayed.
\nThe results are shown in the following table.
\nA χ2 test is carried out at the 10 % significance level to determine whether the arrival status of incoming flights is independent of the distance travelled.
\nThe critical value for this test is 7.779.
\nA flight is chosen at random from the 180 recorded flights.
\nState the alternative hypothesis.
\nCalculate the expected frequency of flights travelling at most 500 km and arriving slightly delayed.
\nWrite down the number of degrees of freedom.
\nWrite down the χ2 statistic.
\nWrite down the associated p-value.
\nState, with a reason, whether you would reject the null hypothesis.
\nWrite down the probability that this flight arrived on time.
\nGiven that this flight was not heavily delayed, find the probability that it travelled between 500 km and 5000 km.
\nTwo flights are chosen at random from those which were slightly delayed.
\nFind the probability that each of these flights travelled at least 5000 km.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nThe arrival status is dependent on the distance travelled by the incoming flight (A1)
\nNote: Accept “associated” or “not independent”.
\n[1 mark]
\nOR (M1)
\nNote: Award (M1) for correct substitution into expected value formula.
\n= 15 (A1) (G2)
\n[2 marks]
\n4 (A1)
\nNote: Award (A0) if “2 + 2 = 4” is seen.
\n[1 mark]
\n9.55 (9.54671…) (G2)
\nNote: Award (G1) for an answer of 9.54.
\n[2 marks]
\n0.0488 (0.0487961…) (G1)
\n[1 mark]
\nReject the Null Hypothesis (A1)(ft)
\nNote: Follow through from their hypothesis in part (a).
\n9.55 (9.54671…) > 7.779 (R1)(ft)
\nOR
\n0.0488 (0.0487961…) < 0.1 (R1)(ft)
\nNote: Do not award (A1)(ft)(R0)(ft). Follow through from part (d). Award (R1)(ft) for a correct comparison, (A1)(ft) for a consistent conclusion with the answers to parts (a) and (d). Award (R1)(ft) for χ2calc > χ2crit , provided the calculated value is explicitly seen in part (d)(i).
\n[2 marks]
\n(A1)(A1) (G2)
\nNote: Award (A1) for correct numerator, (A1) for correct denominator.
\n[2 marks]
\n(A1)(A1) (G2)
\nNote: Award (A1) for correct numerator, (A1) for correct denominator.
\n[2 marks]
\n(A1)(M1)
\nNote: Award (A1) for two correct fractions and (M1) for multiplying their two fractions.
\n(A1) (G2)
\n[3 marks]
\nConsider the function , .
\nFind .
\nThe graph of has a horizontal tangent line at and at . Find .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nchoosing product rule (M1)
\neg ,
\ncorrect derivatives (must be seen in the rule) A1A1
\neg ,
\nA1 N4
\n[4 marks]
\nvalid method (M1)
\neg , ,
(accept ) A1 N2
\n[2 marks]
\nA bag contains marbles, two of which are blue. Hayley plays a game in which she randomly draws marbles out of the bag, one after another, without replacement. The game ends when Hayley draws a blue marble.
\nLet = 5. Find the probability that the game will end on her
\nFind the probability, in terms of , that the game will end on her first draw.
\nFind the probability, in terms of , that the game will end on her second draw.
\nthird draw.
\nfourth draw.
\nHayley plays the game when = 5. She pays $20 to play and can earn money back depending on the number of draws it takes to obtain a blue marble. She earns no money back if she obtains a blue marble on her first draw. Let M be the amount of money that she earns back playing the game. This information is shown in the following table.
\nFind the value of so that this is a fair game.
\nA1 N1
\n\n
[1 mark]
\ncorrect probability for one of the draws A1
\neg P(not blue first) = , blue second =
\nvalid approach (M1)
\neg recognizing loss on first in order to win on second, P(B' then B), P(B') × P(B | B'), tree diagram
\ncorrect expression in terms of A1 N3
\neg , ,
\n\n
[3 marks]
\ncorrect working (A1)
\neg
\nA1 N2
\n\n
[2 marks]
\ncorrect working (A1)
\neg
\nA1 N2
\n\n
[2 marks]
\ncorrect probabilities (seen anywhere) (A1)(A1)
\neg , (may be seen on tree diagram)
\nvalid approach to find E (M) or expected winnings using their probabilities (M1)
\neg ,
\n\n
correct working to find E (M) or expected winnings (A1)
\neg ,
\n\n
correct equation for fair game A1
\neg ,
\ncorrect working to combine terms in (A1)
\neg , ,
\n= 5 A1 N0
\nNote: Do not award the final A1 if the candidate’s FT probabilities do not sum to 1.
\n\n
[7 marks]
\nThe following Venn diagram shows the events and , where . The values shown are probabilities.
\nFind the value of .
\nFind the value of .
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg 0.30 − 0.1, + 0.1 = 0.3
\n= 0.2 A1 N2
\n[2 marks]
\nvalid approach (M1)
\neg 1 − (0.3 + 0.4), 1 − 0.4 − 0.1 −
\n= 0.3 A1 N2
\n[2 marks]
\nvalid approach (M1)
\neg , , ,
\nA1 N2
\n[2 marks]
\nLet .
\nConsider the functions and , for ≥ 0.
\nThe graphs of and are shown in the following diagram.
\nThe shaded region is enclosed by the graphs of , , the -axis and .
\nFind .
\nHence find .
\nWrite down an expression for the area of .
\nHence find the exact area of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nevidence of choosing chain rule (M1)
\neg , ,
\nA2 N3
\n[3 marks]
\nintegrating by inspection from (a) or by substitution (M1)
\neg , , , ,
\ncorrect integrated expression in terms of A2 N3
\neg ,
\n[3 marks]
\n\n
\n
integrating and subtracting functions (in any order) (M1)
\neg ,
\ncorrect integral (including limits, accept absence of ) A1 N2
\neg , ,
\n[2 marks]
\nrecognizing is a common factor (seen anywhere, may be seen in part (c)) (M1)
\neg , ,
\ncorrect integration (A1)(A1)
\neg
\nNote: Award A1 for and award A1 for .
\nsubstituting limits into their integrated function and subtracting (in any order) (M1)
\neg ,
\ncorrect working (A1)
\neg ,
\narea of A1 N3
\n[6 marks]
\nA particle P moves along a straight line. The velocity v m s−1 of P after t seconds is given by v (t) = 7 cos t − 5t cos t, for 0 ≤ t ≤ 7.
\nThe following diagram shows the graph of v.
\nFind the initial velocity of P.
\nFind the maximum speed of P.
\nWrite down the number of times that the acceleration of P is 0 m s−2 .
\nFind the acceleration of P when it changes direction.
\nFind the total distance travelled by P.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ninitial velocity when t = 0 (M1)
\neg v(0)
\nv = 17 (m s−1) A1 N2
\n[2 marks]
\nrecognizing maximum speed when is greatest (M1)
\neg minimum, maximum, v' = 0
\none correct coordinate for minimum (A1)
\neg 6.37896, −24.6571
\n24.7 (ms−1) A1 N2
\n[3 marks]
\nrecognizing a = v ′ (M1)
\neg , correct derivative of first term
\nidentifying when a = 0 (M1)
\neg turning points of v, t-intercepts of v ′
\n3 A1 N3
\n[3 marks]
\nrecognizing P changes direction when v = 0 (M1)
\nt = 0.863851 (A1)
\n−9.24689
\na = −9.25 (ms−2) A2 N3
\n[4 marks]
\ncorrect substitution of limits or function into formula (A1)
eg
63.8874
\n63.9 (metres) A2 N3
\n[3 marks]
\nConsider the following diagram.
\nThe sides of the equilateral triangle ABC have lengths 1 m. The midpoint of [AB] is denoted by P. The circular arc AB has centre, M, the midpoint of [CP].
\nFind AM.
\nFind the area of the shaded region.
\nMETHOD 1
\nPC or 0.8660 (M1)
\nPM PC or 0.4330 (A1)
\nAM
\nor 0.661 (m) A1
\n\n
METHOD 2
\nusing the cosine rule
\nAM2 M1A1
\nAM or 0.661 (m) A1
\n[3 marks]
\nEITHER
\n(M1)A1
\nOR
\n(M1)A1
\n= 0.158(m2) A1
\nNote: Award M1 for attempting to calculate area of a sector minus area of a triangle.
\n[3 marks]
\nThe diagram shows two circles with centres at the points A and B and radii and , respectively. The point B lies on the circle with centre A. The circles intersect at the points C and D.
\n![\"N16/5/MATHL/HP2/ENG/TZ0/09\"](\"../assets/uploads/tinymce_asset/asset/3295/Schermafbeelding_2017-02-28_om_17.29.37.png\"/)
Let be the measure of the angle CAD and be the measure of the angle CBD in radians.
\nFind an expression for the shaded area in terms of , and .
\nShow that .
\nHence find the value of given that the shaded area is equal to 4.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nM1A1A1
\n\n
Note: Award M1A1A1 for alternative correct expressions eg. .
\n\n
[3 marks]
\nMETHOD 1
\nconsider for example triangle ADM where M is the midpoint of BD M1
\nA1
\n\n
AG
\nMETHOD 2
\nattempting to use the cosine rule (to obtain ) M1
\n(obtained from ) A1
\n\n
AG
\nMETHOD 3
\nwhere
\nM1
\n\n
Note: Award M1 either for use of the double angle formula or the conversion from sine to cosine.
\n\n
A1
\n\n
AG
\n[2 marks]
\n(from triangle ADM), A1
\nattempting to solve
\nwith and for (M1)
\nA1
\n[3 marks]
\n160 students attend a dual language school in which the students are taught only in Spanish or taught only in English.
\nA survey was conducted in order to analyse the number of students studying Biology or Mathematics. The results are shown in the Venn diagram.
\n\n
Set S represents those students who are taught in Spanish.
\nSet B represents those students who study Biology.
\nSet M represents those students who study Mathematics.
\n\n
A student from the school is chosen at random.
\nFind the number of students in the school that are taught in Spanish.
\nFind the number of students in the school that study Mathematics in English.
\nFind the number of students in the school that study both Biology and Mathematics.
\nWrite down .
\nWrite down .
\nFind the probability that this student studies Mathematics.
\nFind the probability that this student studies neither Biology nor Mathematics.
\nFind the probability that this student is taught in Spanish, given that the student studies Biology.
\n10 + 40 + 28 + 17 (M1)
\n= 95 (A1)(G2)
\n\n
Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17) seen.
\n[2 marks]
\n20 + 12 (M1)
\n= 32 (A1)(G2)
\n\n
Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17) seen.
\n[2 marks]
\n12 + 40 (M1)
\n= 52 (A1)(G2)
\n\n
Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17) seen.
\n[2 marks]
\n78 (A1)
\n\n
[1 mark]
\n12 (A1)
\n\n
[1 mark]
\n(A1)(A1) (G2)
\n\n
Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct denominator. All answers must be probabilities to award (A1).
\n\n
[2 marks]
\n(A1)(A1) (G2)
\n\n
Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct denominator. All answers must be probabilities to award (A1).
\n\n
[2 marks]
\n(A1)(A1) (G2)
\n\n
Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct denominator. All answers must be probabilities to award (A1).
\n\n
[2 marks]
\nIn an international competition, participants can answer questions in only one of the three following languages: Portuguese, Mandarin or Hindi. 80 participants took part in the competition. The number of participants answering in Portuguese, Mandarin or Hindi is shown in the table.
\nA boy is chosen at random.
\nState the number of boys who answered questions in Portuguese.
\nFind the probability that the boy answered questions in Hindi.
\nTwo girls are selected at random.
\nCalculate the probability that one girl answered questions in Mandarin and the other answered questions in Hindi.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n20 (A1) (C1)
\n[1 mark]
\nnull (A1)(A1) (C2)
\nNote: Award (A1) for correct numerator, (A1) for correct denominator.
\n[2 marks]
\n(A1)(M1)
\nNote: Award (A1) for first or second correct product seen, (M1) for adding their two products or for multiplying their product by two.
\n(A1) (C3)
\n[3 marks]
\nA sector of a circle with radius cm , where > 0, is shown on the following diagram.
The sector has an angle of 1 radian at the centre.
Let the area of the sector be cm2 and the perimeter be cm. Given that , find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n\n
use of the correct formula for area and arc length (M1)
\nperimeter is (A1)
\nNote: A1 independent of previous M1.
\nA1
\n\n
(as > 0) A1
\nNote: Do not award final A1 if is included.
\n[4 marks]
\nA tetrahedral (four-sided) die has written on it the numbers 1, 2, 3 and 4. The die is rolled many times and the scores are noted. The table below shows the resulting frequency distribution.
\n![\"M17/5/MATSD/SP1/ENG/TZ2/07\"](\"../assets/uploads/tinymce_asset/asset/3584/Schermafbeelding_2017-08-16_om_05.55.47.png\"/)
The die was rolled a total of 100 times.
\nThe mean score is 2.71.
\nWrite down an equation, in terms of and , for the total number of times the die was rolled.
\nUsing the mean score, write down a second equation in terms of and .
\nFind the value of and of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
or equivalent (A1) (C1)
\n[1 mark]
\nor equivalent (M1)(A1) (C2)
\n\n
Note: Award (M1) for a sum including and , divided by 100 and equated to 2.71, (A1) for a correct equation.
\n\n
[2 marks]
\nand (M1)
\n\n
Note: Award (M1) for obtaining a correct linear equation in one variable from their (a) and their (b).
\nThis may be implied if seen in part (a) or part (b).
\n\n
(A1)(ft)(A1)(ft) (C3)
\n\n
Notes: Follow through from parts (a) and (b), irrespective of working seen provided the answers are positive integers.
\n\n
[3 marks]
\nA water trough which is 10 metres long has a uniform cross-section in the shape of a semicircle with radius 0.5 metres. It is partly filled with water as shown in the following diagram of the cross-section. The centre of the circle is O and the angle KOL is radians.
\n![\"M17/5/MATHL/HP2/ENG/TZ1/08\"](\"../assets/uploads/tinymce_asset/asset/3515/Schermafbeelding_2017-08-09_om_11.09.30.png\"/)
The volume of water is increasing at a constant rate of .
\nFind an expression for the volume of water in the trough in terms of .
\nCalculate when .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
area of segment M1A1
\n\n
A1
\n[3 marks]
\nMETHOD 1
\nM1A1
\n(M1)
\nA1
\nMETHOD 2
\n(M1)
\nA1
\n(M1)
\nA1
\n[4 marks]
\nThe histogram shows the time, t, in minutes, that it takes the customers of a restaurant to eat their lunch on one particular day. Each customer took less than 25 minutes.
\nThe histogram is incomplete, and only shows data for 0 ≤ t < 20.
\nThe mean time it took all customers to eat their lunch was estimated to be 12 minutes.
\nIt was found that k customers took between 20 and 25 minutes to eat their lunch.
\nWrite down the mid-interval value for 10 ≤ t < 15.
\nWrite down the total number of customers in terms of k.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
12.5 (A1) (C1)
\n[1 mark]
\n33 + k OR 10 + 8 + 5 + 10 + k (A1)
\nNote: Award (A1) for “number of customers = 33 + k”.
\n[1 mark]
\nA florist sells bouquets of roses. The florist recorded, in Table 1, the number of roses in each bouquet sold to customers.
\nTable 1
\nThe roses can be arranged into bouquets of size small, medium or large. The data from Table 1 has been organized into a cumulative frequency table, Table 2.
\nTable 2
\nComplete the cumulative frequency table.
\nWrite down the probability that a bouquet of roses sold is not small.
\nA customer buys a large bouquet.
\nFind the probability that there are 12 roses in this bouquet.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(ft) (C2)
Note: Award (A1) for 10; (A1)(ft) for the last column all correct. Follow through from their 10 for their 50 in the last column.
\n[2 marks]
\n(A1)(ft)(A1)(ft) (C2)
\nNote: Award (A1)(ft) for their numerator being 25 + their 10, and (A1)(ft) for their denominator being their 50. Follow through from part (a).
\n[2 marks]
\n(A1)(A1)(ft) (C2)
\nNote: Award (A1) for a numerator of 4 and (A1)(ft) for their 10 as denominator. Follow through from part (a).
\n[2 marks]
\nA group of 800 students answered 40 questions on a category of their choice out of History, Science and Literature.
\nFor each student the category and the number of correct answers, , was recorded. The results obtained are represented in the following table.
\n![\"N17/5/MATSD/SP2/ENG/TZ0/01\"](\"../assets/uploads/tinymce_asset/asset/4185/Schermafbeelding_2018-02-13_om_14.11.54.png\"/)
A test at the 5% significance level is carried out on the results. The critical value for this test is 12.592.
\nState whether is a discrete or a continuous variable.
\nWrite down, for , the modal class;
\nWrite down, for , the mid-interval value of the modal class.
\nUse your graphic display calculator to estimate the mean of ;
\nUse your graphic display calculator to estimate the standard deviation of .
\nFind the expected frequency of students choosing the Science category and obtaining 31 to 40 correct answers.
\nWrite down the null hypothesis for this test;
\nWrite down the number of degrees of freedom.
\nWrite down the -value for the test;
\nWrite down the statistic.
\nState the result of the test. Give a reason for your answer.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
discrete (A1)
\n[1 mark]
\n(A1)
\n[1 mark]
\n15.5 (A1)(ft)
\n\n
Note: Follow through from part (b)(i).
\n\n
[1 mark]
\n(G2)
\n[2 marks]
\n(G1)
\n[1 marks]
\nOR (M1)
\n\n
Note: Award (M1) for correct substitution into expected frequency formula.
\n\n
(A1)(G2)
\n[2 marks]
\nchoice of category and number of correct answers are independent (A1)
\n\n
Notes: Accept “no association” between (choice of) category and number of correct answers. Do not accept “not related” or “not correlated” or “influenced”.
\n\n
[1 mark]
\n6 (A1)
\n[1 mark]
\n\n
(G1)
\n[1 mark]
\n(G2)
\n[2 marks]
\nthe null hypothesis is not rejected (the null hypothesis is accepted) (A1)(ft)
\nOR
\n(choice of) category and number of correct answers are independent (A1)(ft)
\nas OR (R1)
\n\n
Notes: Award (R1) for a correct comparison of either their statistic to the critical value or their -value to the significance level. Award (A1)(ft) from that comparison.
\nFollow through from part (f). Do not award (A1)(ft)(R0).
\n\n
[2 marks]
\nSolve the equation .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nuse of M1
\n\n
(M1)
\nA1
\nA1A1
\nMETHOD 2
\nM1
\n\n
M1A1
\n\n
A1A1
\n\n
Note: Award A1A0 if extra solutions given or if solutions given in degrees (or both).
\n\n
[5 marks]
\nFind the value of .
\nShow that where .
\nUse the principle of mathematical induction to prove that
\nwhere .
\nHence or otherwise solve the equation in the interval .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)A1
\n\n
Note: Award M1 for 5 equal terms with \\) + \\) or signs.
\n\n
[2 marks]
\nM1
\nA1
\nAG
\n[2 marks]
\nlet
\nif
\nwhich is true (as proved in part (b)) R1
\nassume true, M1
\n\n
Notes: Only award M1 if the words “assume” and “true” appear. Do not award M1 for “let ” only. Subsequent marks are independent of this M1.
\n\n
consider :
\n\n
M1
\nA1
\n\n
M1
\nM1
\nA1
\nA1
\n\n
so if true for , then also true for
\nas true for then true for all R1
\n\n
Note: Accept answers using transformation formula for product of sines if steps are shown clearly.
\n\n
Note: Award R1 only if candidate is awarded at least 5 marks in the previous steps.
\n\n
[9 marks]
\nEITHER
\nM1
\nA1
\nM1
\nM1
\nor A1
\nand
\nOR
\nM1A1
\nM1A1
\nof A1
\nand
\nTHEN
\nand A1
\n\n
Note: Do not award the final A1 if extra solutions are seen.
\n\n
[6 marks]
\nLet f(x) = ln x − 5x , for x > 0 .
\nFind f '(x).
\nFind f \"(x).
\nSolve f '(x) = f \"(x).
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1A1 N2
\n[2 marks]
\nf \"(x) = −x−2 A1 N1
\n[1 mark]
\nMETHOD 1 (using GDC)
\nvalid approach (M1)
\neg
0.558257
\nx = 0.558 A1 N2
\nNote: Do not award A1 if additional answers given.
\n\n
METHOD 2 (analytical)
\nattempt to solve their equation f '(x) = f \"(x) (do not accept ) (M1)
\neg
\n0.558257
\nx = 0.558 A1 N2
\nNote: Do not award A1 if additional answers given.
\n[2 marks]
\nIn a high school, 160 students completed a questionnaire which asked for the number of people they are following on a social media website. The results were recorded in the following box-and-whisker diagram.
\nThe following incomplete table shows the distribution of the responses from these 160 students.
\nWrite down the mid-interval value for the 100 < x ≤ 150 group.
\n125 (accept 125.5) (A1)
\nLet . Find the term in in the expansion of the derivative, .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nderivative of A2
\n\n
recognizing need to find term in (seen anywhere) R1
\neg
\nvalid approach to find the terms in (M1)
\neg, Pascal’s triangle to 6th row
\nidentifying correct term (may be indicated in expansion) (A1)
\neg
\ncorrect working (may be seen in expansion) (A1)
\neg
\nA1 N3
\nMETHOD 2
\nrecognition of need to find in (seen anywhere) R1
\nvalid approach to find the terms in (M1)
\neg, Pascal’s triangle to 7th row
\nidentifying correct term (may be indicated in expansion) (A1)
\neg6th term,
\ncorrect working (may be seen in expansion) (A1)
\neg
\ncorrect term (A1)
\n\n
differentiating their term in (M1)
\neg
\nA1 N3
\n[7 marks]
\nLet , .
\nThe following diagram shows part of the graph of .
\nRectangle PQRS is drawn with P and Q on the -axis and R and S on the graph of .
\nLet OP = .
\nConsider another function , .
\nFind the -intercepts of the graph of .
\nShow that the area of PQRS is .
\nHence find the value of such that the area of PQRS is a maximum.
\nShow that when the graphs of and intersect, .
\nGiven that the graphs of and intersect only once, find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nvalid approach (M1)
\neg , , one correct solution
\n, 3 (accept (3, 0), (−3, 0)) A1 N2
\n[2 marks]
\nvalid approach (M1)
\neg height = , base = 2(OP) or , ,
\ncorrect working that clearly leads to given answer A1
\neg
\nNote: Do not accept sloppy notation eg .
\narea = AG N0
\n[2 marks]
\nsetting derivative = 0 (seen anywhere) (M1)
\neg ,
\ncorrect derivative (must be in terms of only) (seen anywhere) A2
\neg ,
\ncorrect working (A1)
\neg ,
\nA1 N3
\n[5 marks]
\nvalid approach (M1)
\neg ,
\ncorrect working (A1)
\neg ,
\nAG N0
\n[2 marks]
\nMETHOD 1 (discriminant)
\nrecognizing to use discriminant (seen anywhere) (M1)
\neg Δ,
\ndiscriminant = 0 (seen anywhere) M1
\ncorrect substitution into discriminant (do not accept only in quadratic formula) (A1)
\neg ,
\ncorrect working (A1)
\neg ,
\nA1 N2
\n\n
METHOD 2 (completing the square)
\nvalid approach to complete the square (M1)
\neg ,
\ncorrect working (A1)
\neg ,
\nrecognizing condition for one solution M1
\neg ,
\ncorrect working (A1)
\neg ,
\nA1 N2
\n\n
METHOD 3 (using vertex)
valid approach to find vertex (seen anywhere) M1
\neg ,
\ncorrect working (A1)
\neg ,
\n(A1)
\ncorrect substitution (A1)
\neg
\nA1 N2
\n\n
[5 marks]
\nIn triangle and .
\nConsider the possible triangles with .
\nConsider the case where , the length of QR is not fixed at 8 cm.
\nUse the cosine rule to show that .
\nCalculate the two corresponding values of PQ.
\nHence, find the area of the smaller triangle.
\nDetermine the range of values of for which it is possible to form two triangles.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1
\nAG
\n[2 marks]
\nEITHER
\n(M1)
\nOR
\nusing the sine rule (M1)
\nTHEN
\nor A1
\nA1
\n[3 marks]
\nM1A1
\nA1
\n[3 marks]
\nMETHOD 1
\nEITHER
\n\n
discriminant M1
\nA1
\nM1
\nA1
\nOR
\nconstruction of a right angle triangle (M1)
\nM1(A1)
\nhence for two triangles R1
\nTHEN
\nA1
\nto ensure two positive solutions or valid geometric argument R1
\nA1
\nMETHOD 2
\ndiagram showing two triangles (M1)
\nM1A1
\none right angled triangle when (A1)
\nfor two triangles R1
\nfor two triangles A1
\nA1
\n[7 marks]
\nIn triangle ABC, AB = 5, BC = 14 and AC = 11.
\nFind all the interior angles of the triangle. Give your answers in degrees to one decimal place.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nattempt to apply cosine rule M1
\n\n
\n
A1
\nattempt to apply sine rule or cosine rule: M1
\n\n
\n
A1
\n\n
A1
\nNote: Candidates may attempt to find angles in any order of their choosing.
\n[5 marks]
\nA survey was carried out to investigate the relationship between a person’s age in years ( ) and the number of hours they watch television per week (). The scatter diagram represents the results of the survey.
\n![\"N17/5/MATSD/SP1/ENG/TZ0/05\"](\"../assets/uploads/tinymce_asset/asset/4170/Schermafbeelding_2018-02-12_om_18.00.45.png\"/)
The mean age of the people surveyed was 50.
\nFor these results, the equation of the regression line on is .
\nFind the mean number of hours that the people surveyed watch television per week.
\nDraw the regression line on the scatter diagram.
\nBy placing a tick (✔) in the correct box, determine which of the following statements is true:
\n![\"N17/5/MATSD/SP1/ENG/TZ0/05.c\"](\"../assets/uploads/tinymce_asset/asset/4178/Schermafbeelding_2018-02-13_om_07.09.18.png\"/)
Diogo is 18 years old. Give a reason why the regression line should not be used to estimate the number of hours Diogo watches television per week.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\n\n
Note: Award (M1) for correct substitution of 50 into equation of the regression line.
\n\n
(A1) (C2)
\nOR
\n(M1)
\n\n
Note: Award (M1) for correctly summing the values of the points, and dividing by 25.
\n\n
(A1) (C2)
\n[2 marks]
\nline through and (A1)(ft)(A1) (C2)
\n\n
Note: Award (A1)(ft) for a straight line through (50, their ), and (A1) for the line intercepting the -axis at ; this may need to be extrapolated. Follow through from part (a). Award at most (A0)(A1) if the line is not drawn with a ruler.
\n\n
[2 marks]
\n![\"N17/5/MATSD/SP1/ENG/TZ0/05.c/M\"](\"../assets/uploads/tinymce_asset/asset/4180/Schermafbeelding_2018-02-13_om_07.53.00.png\"/)
(A1) (C1)
\n
Note: Award (A0) if more than one tick (✔) is seen.
\n\n
[1 mark]
\n18 is less than the lowest age in the survey OR extrapolation. (A1) (C1)
\n\n
Note: Accept equivalent statements.
\n\n
[1 mark]
\nA closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20 cm3.
\nThe material for the base and top of the can costs 10 cents per cm2 and the material for the curved side costs 8 cents per cm2. The total cost of the material, in cents, is C.
\nExpress h in terms of r.
\nShow that .
\nGiven that there is a minimum value for C, find this minimum value in terms of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct equation for volume (A1)
eg
A1 N2
\n[2 marks]
\n\n
attempt to find formula for cost of parts (M1)
eg 10 × two circles, 8 × curved side
correct expression for cost of two circles in terms of r (seen anywhere) A1
eg
correct expression for cost of curved side (seen anywhere) (A1)
eg
correct expression for cost of curved side in terms of r A1
eg
AG N0
\n[4 marks]
\nrecognize at minimum (R1)
eg
correct differentiation (may be seen in equation)
\nA1A1
\ncorrect equation A1
eg
correct working (A1)
eg
r = 2 (m) A1
\nattempt to substitute their value of r into C
eg (M1)
correct working
eg (A1)
(cents) A1 N3
\nNote: Do not accept 753.6, 753.98 or 754, even if 240 is seen.
\n[9 marks]
\nBoat A is situated 10km away from boat B, and each boat has a marine radio transmitter on board. The range of the transmitter on boat A is 7km, and the range of the transmitter on boat B is 5km. The region in which both transmitters can be detected is represented by the shaded region in the following diagram. Find the area of this region.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nuse of cosine rule (M1)
\nCÂB = arccos (A1)
\nCA = arccos (A1)
\nattempt to subtract triangle area from sector area (M1)
\narea
\n= 3.5079… + 5.3385… (A1)
\nNote: Award this A1 for either of these two values.
\n= 8.85 (km2) A1
\nNote: Accept all answers that round to 8.8 or 8.9.
\n\n
[6 marks]
\nIn a triangle and .
\nUse the cosine rule to find the two possible values for AC.
\nFind the difference between the areas of the two possible triangles ABC.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1
\nlet
\nM1A1
\nattempting to solve for (M1)
\nA1A1
\nMETHOD 2
\nlet
\nusing the sine rule to find a value of M1
\n(M1)A1
\n(M1)A1
\nMETHOD 3
\nlet
\nusing the sine rule to find a value of and a value of M1
\nobtaining and A1
\nand
\nattempting to find a value of using the cosine rule (M1)
\nA1A1
\n\n
Note: Award M1A0(M1)A1A0 for one correct value of
\n\n
[5 marks]
\nand (A1)
\n( and )
\nlet be the difference between the two areas
\n(M1)
\n\n
A1
\n[3 marks]
\nThe lengths of two of the sides in a triangle are 4 cm and 5 cm. Let θ be the angle between the two given sides. The triangle has an area of cm2.
\nShow that .
\nFind the two possible values for the length of the third side.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
EITHER
\nA1
\nOR
\nheight of triangle is if using 4 as the base or if using 5 as the base A1
\nTHEN
\nAG
\n[1 mark]
\nlet the third side be
\nM1
\nvalid attempt to find (M1)
\nNote: Do not accept writing as a valid method.
\n\n
A1A1
\n\n
or A1A1
\n[6 marks]
\nThe manager of a folder factory recorded the number of folders produced by the factory (in thousands) and the production costs (in thousand Euros), for six consecutive months.
\n![\"M17/5/MATSD/SP2/ENG/TZ2/03\"](\"../assets/uploads/tinymce_asset/asset/3608/Schermafbeelding_2017-08-16_om_17.30.09.png\"/)
Every month the factory sells all the folders produced. Each folder is sold for 2.99 Euros.
\nDraw a scatter diagram for this data. Use a scale of 2 cm for 5000 folders on the horizontal axis and 2 cm for 10 000 Euros on the vertical axis.
\nWrite down, for this set of data the mean number of folders produced, ;
\nWrite down, for this set of data the mean production cost, .
\nLabel the point on the scatter diagram.
\nState a reason why the regression line on is appropriate to model the relationship between these variables.
\nUse your graphic display calculator to find the equation of the regression line on .
\nDraw the regression line on on the scatter diagram.
\nUse the equation of the regression line to estimate the least number of folders that the factory needs to sell in a month to exceed its production cost for that month.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"M17/5/MATSD/SP2/ENG/TZ2/03.a/M\"](\"../assets/uploads/tinymce_asset/asset/3616/Schermafbeelding_2017-08-17_om_06.54.04.png\"/)
(A4)
\n
Notes: Award (A1) for correct scales and labels. Award (A0) if axes are reversed and follow through for their points.
\nAward (A3) for all six points correctly plotted, (A2) for four or five points correctly plotted, (A1) for two or three points correctly plotted.
\nIf graph paper has not been used, award at most (A1)(A0)(A0)(A0). If accuracy cannot be determined award (A0)(A0)(A0)(A0).
\n\n
[4 marks]
\n(A1)(G1)
\n[1 mark]
\n(A1)(G1)
\n\n
Note: Accept (i) 21000 and (ii) 55000 seen.
\n\n
[1 mark]
\ntheir mean point M labelled on diagram (A1)(ft)(G1)
\n\n
Note: Follow through from part (b).
\nAward (A1)(ft) if their part (b) is correct and their attempt at plotting in part (a) is labelled M.
\nIf graph paper not used, award (A1) if is labelled. If their answer from part (b) is incorrect and accuracy cannot be determined, award (A0).
\n\n
[1 mark]
\nthe correlation coefficient/r is (very) close to 1 (R1)(ft)
\nOR
\nthe correlation is (very) strong (R1)(ft)
\n\n
Note: Follow through from their answer to part (d).
\n\n
OR
\nthe position of the data points on the scatter graphs suggests that the tendency is linear (R1)(ft)
\n\n
Note: Follow through from their scatter graph in part (a).
\n[1 mark]
\n(G2)
\n\n
Notes: Award (G1) for , (G1) for 14.2.
\nAward a maximum of (G0)(G1) if the answer is not an equation.
\nAward (G0)(G1)(ft) if gradient and -intercept are swapped in the equation.
\n\n
[2 marks]
\nstraight line through their (A1)(ft)
\n-intercept of the line (or extension of line) passing through (A1)(ft)
\n\n
Notes: Follow through from part (f). In the event that the regression line is not straight (ruler not used), award (A0)(A1)(ft) if line passes through both their and , otherwise award (A0)(A0). The line must pass through the midpoint, not near this point. If it is not clear award (A0).
\nIf graph paper is not used, award at most (A1)(ft)(A0).
\n\n
[2 marks]
\n(M1)(M1)
\n\n
Note: Award (M1) for seen and (M1) for equating to their equation of the regression line. Accept an inequality sign.
\nAccept a correct graphical method involving their part (f) and .
\nAccept drawn on their scatter graph.
\n\n
(this step may be implied by their final answer) (A1)(ft)(G2)
\n(A1)(ft)(G3)
\n\n
Note: Follow through from their answer to (f). Use of 3 sf gives an answer of .
\nAward (G2) for or 13.524 or a value which rounds to 13500 seen without workings.
\nAward the last (A1)(ft) for correct multiplication by 1000 and an answer satisfying revenue > their production cost.
\nAccept 13.6 thousand (folders).
\n\n
[4 marks]
\nThe values of the functions and and their derivatives for and are shown in the following table.
\n![\"M17/5/MATME/SP1/ENG/TZ2/06\"](\"../assets/uploads/tinymce_asset/asset/3537/Schermafbeelding_2017-08-11_om_16.42.43.png\"/)
Let .
\nFind .
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
expressing as a product of and (A1)
\neg
\nA1 N2
\n[2 marks]
\nattempt to use product rule (do not accept ) (M1)
\neg
\ncorrect substitution of values into product rule (A1)
\neg
\nA1 N2
\n[3 marks]
\nFind the set of values of that satisfy the inequality .
\nThe triangle ABC is shown in the following diagram. Given that , find the range of possible values for AB.
\n![\"M17/5/MATHL/HP2/ENG/TZ2/04.b\"](\"../assets/uploads/tinymce_asset/asset/3518/Schermafbeelding_2017-08-09_om_18.13.24.png\"/)
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
(M1)
\nA1
\n[2 marks]
\nM1
\nA1
\n\n
from result in (a)
\nor (A1)
\nbut AB must be at least 2
\nA1
\n\n
Note: Allow for either of the final two A marks.
\n\n
[4 marks]
\nLet .
\nLet , where .
\nLet and .
\n(i) Find the first four derivatives of .
\n(ii) Find .
\n(i) Find the first three derivatives of .
\n(ii) Given that , find .
\n(i) Find .
\n(ii) Hence, show that .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(i) A2 N2
\n(ii) valid approach (M1)
\negrecognizing that 19 is one less than a multiple of 4,
\nA1 N2
\n[4 marks]
\n(i)
\nA1A1 N2
\n(ii) METHOD 1
\ncorrect working that leads to the correct answer, involving the correct expression for the 19th derivative A2
\neg
\n(accept ) A1 N1
\nMETHOD 2
\ncorrect working involving recognizing patterns in coefficients of first three derivatives (may be seen in part (b)(i)) leading to a general rule for 19th coefficient A2
\neg
\n\n
(accept ) A1 N1
\n[5 marks]
\n(i) valid approach using product rule (M1)
\neg
\ncorrect 20th derivatives (must be seen in product rule) (A1)(A1)
\neg
\nA1 N3
\n(ii) substituting (seen anywhere) (A1)
\neg
\nevidence of one correct value for or (seen anywhere) (A1)
\neg
\nevidence of correct values substituted into A1
\neg
\n\n
Note: If candidates write only the first line followed by the answer, award A1A0A0.
\n\n
AG N0
\n[7 marks]
\nIn the following diagram, = a, = b. C is the midpoint of [OA] and .
\n![\"N17/5/MATHL/HP1/ENG/TZ0/09\"](\"../assets/uploads/tinymce_asset/asset/4113/Schermafbeelding_2018-02-07_om_14.26.10.png\"/)
It is given also that and , where .
\nFind, in terms of a and b .
\nFind, in terms of a and b .
\nFind an expression for in terms of a, b and ;
\nFind an expression for in terms of a, b and .
\nShow that , and find the value of .
\nDeduce an expression for in terms of a and b only.
\nGiven that area , find the value of .
\nb A1
\n[1 mark]
\n(M1)
\nb – a A1
\n[2 marks]
\na M1A1
\n[2 marks]
\na M1A1
\n[2 marks]
\nequating coefficients: M1
\nA1
\nsolving simultaneously: M1
\nA1AG
\n[4 marks]
\n\n
M1A1
\n[2 marks]
\nMETHOD 1
\n(M1)
\n(M1)
\nA1
\n(M1)
\nA1
\n\n
METHOD 2
\nA1
\nor M1
\n\n
(M1)
\nA1
\n\n
\n
A1
\n[5 marks]
\nA group of 60 sports enthusiasts visited the PyeongChang 2018 Winter Olympic games to watch a variety of sporting events.
\nThe most popular sports were snowboarding (S), figure skating (F) and ice hockey (H).
\nFor this group of 60 people:
\n4 did not watch any of the most popular sports,
x watched all three of the most popular sports,
9 watched snowboarding only,
11 watched figure skating only,
15 watched ice hockey only,
7 watched snowboarding and figure skating,
13 watched figure skating and ice hockey,
11 watched snowboarding and ice hockey.
Complete the Venn diagram using the given information.
\nFind the value of x.
\nWrite down the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(A1) (C3)
Note: Award (A1) for 4 in correct place.
\nAward (A1) for 9, 11, 15 in correct place.
\nAward (A1) for 7 − x, 13 − x, 11 − x in correct place.
\nAccept 2, 8 and 6 in place of 7 − x, 13 − x, 11 − x.
\n[3 marks]
\n(M1)
\nNote: Award (M1) for equating the sum of at least seven of the entries in their Venn diagram to 60.
\n(A1)(ft) (C2)
\nNote: Follow through from part (a), but only if answer is positive.
\n[2 marks]
\n34 (A1)(ft) (C1)
\nNote: Follow through from their Venn diagram.
\n[1 mark]
\nA school café sells three flavours of smoothies: mango (), kiwi fruit () and banana ().
85 students were surveyed about which of these three flavours they like.
35 students liked mango, 37 liked banana, and 26 liked kiwi fruit
2 liked all three flavours
20 liked both mango and banana
14 liked mango and kiwi fruit
3 liked banana and kiwi fruit
Using the given information, complete the following Venn diagram.
\nFind the number of surveyed students who did not like any of the three flavours.
\nA student is chosen at random from the surveyed students.
\nFind the probability that this student likes kiwi fruit smoothies given that they like mango smoothies.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n (A1)(A1) (C2)
Note: Award (A1) for 18, 12 and 1 in correct place on Venn diagram, (A1) for 3, 16 and 11 in correct place on Venn diagram.
\n[2 marks]
\n85 − (3 + 16 + 11 + 18 + 12 + 1 + 2) (M1)
\nNote: Award (M1) for subtracting the sum of their values from 85.
\n22 (A1)(ft) (C2)
\nNote: Follow through from their Venn diagram in part (a).
If any numbers that are being subtracted are negative award (M1)(A0).
[2 marks]
\n(A1)(ft)(A1)(ft) (C2)
\nNote: Award (A1) for correct numerator; (A1) for correct denominator. Follow through from their Venn diagram.
\n[2 marks]
\nConsider the following Venn diagrams.
\nWrite down an expression, in set notation, for the shaded region represented by Diagram 1.
\nWrite down an expression, in set notation, for the shaded region represented by Diagram 2.
\nWrite down an expression, in set notation, for the shaded region represented by Diagram 3.
\nShade, on the Venn diagram, the region represented by the set .
\n\n
Shade, on the Venn diagram, the region represented by the set .
\nA' (A1)
Note: Accept alternative set notation for complement such as U − A.
\n[1 mark]
\nOR (A1)
\nNote: Accept alternative set notation for complement.
\n[1 mark]
\nOR (A2) (C4)
\nNote: Accept equivalent answers, for example .
\n[2 marks]
\n\n
(A1)
[1 mark]
\n(A1) (C2)
[1 mark]
The amount of yeast, g grams, in a sugar solution can be modelled by the function,
\ng(t) = 10 − k(c−t) for t ≥ 0
\nwhere t is the time in minutes.
\nThe graph of g(t) is shown.
\nThe initial amount of yeast in this solution is 2 grams.
\nThe amount of yeast in this solution after 3 minutes is 9 grams.
\nWrite down the maximum amount of yeast in this solution.
\n10 (grams) (A1) (C1)
\n[1 mark]
\nIn a company it is found that 25 % of the employees encountered traffic on their way to work. From those who encountered traffic the probability of being late for work is 80 %.
\nFrom those who did not encounter traffic, the probability of being late for work is 15 %.
\nThe tree diagram illustrates the information.
\nThe company investigates the different means of transport used by their employees in the past year to travel to work. It was found that the three most common means of transport used to travel to work were public transportation (P ), car (C ) and bicycle (B ).
\nThe company finds that 20 employees travelled by car, 28 travelled by bicycle and 19 travelled by public transportation in the last year.
\nSome of the information is shown in the Venn diagram.
\nThere are 54 employees in the company.
\nWrite down the value of a.
\nWrite down the value of b.
\nUse the tree diagram to find the probability that an employee encountered traffic and was late for work.
\nUse the tree diagram to find the probability that an employee was late for work.
\nUse the tree diagram to find the probability that an employee encountered traffic given that they were late for work.
\nFind the value of x.
\nFind the value of y.
\nFind the number of employees who, in the last year, did not travel to work by car, bicycle or public transportation.
\nFind .
\na = 0.2 (A1)
\n[1 mark]
\nb = 0.85 (A1)
\n[1 mark]
\n0.25 × 0.8 (M1)
\nNote: Award (M1) for a correct product.
\n(A1)(G2)
\n[2 marks]
\n0.25 × 0.8 + 0.75 × 0.15 (A1)(ft)(M1)
\nNote: Award (A1)(ft) for their (0.25 × 0.8) and (0.75 × 0.15), (M1) for adding two products.
\n(A1)(ft)(G3)
\nNote: Award the final (A1)(ft) only if answer does not exceed 1. Follow through from part (b)(i).
\n[3 marks]
\n\n
\n
\n
\n
(A1)(ft)(A1)(ft)
\nNote: Award (A1)(ft) for a correct numerator (their part (b)(i)), (A1)(ft) for a correct denominator (their part (b)(ii)). Follow through from parts (b)(i) and (b)(ii).
\n(A1)(ft)(G3)
\nNote: Award final (A1)(ft) only if answer does not exceed 1.
\n[3 marks]
\n(x =) 3 (A1)
\n[1 Mark]
\n(y =) 10 (A1)(ft)
\nNote: Following through from part (c)(i) but only if their x is less than or equal to 13.
\n[1 Mark]
\n54 − (10 + 3 + 4 + 2 + 6 + 8 + 13) (M1)
\nNote: Award (M1) for subtracting their correct sum from 54. Follow through from their part (c).
\n= 8 (A1)(ft)(G2)
\nNote: Award (A1)(ft) only if their sum does not exceed 54. Follow through from their part (c).
\n[2 marks]
\n6 + 8 + 13 (M1)
\nNote: Award (M1) for summing 6, 8 and 13.
\n27 (A1)(G2)
\n[2 marks]
\nA function is given by .
\nThe graph of the function intersects the graph of .
\nFind the exact value of each of the zeros of .
\nExpand the expression for .
\nFind .
\nUse your answer to part (b)(ii) to find the values of for which is increasing.
\nDraw the graph of for and . Use a scale of 2 cm to represent 1 unit on the -axis and 1 cm to represent 5 units on the -axis.
\nWrite down the coordinates of the point of intersection.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(A1)
\n\n
Note: Award (A1) for –1 and each exact value seen. Award at most (A1)(A0)(A1) for use of 2.23606… instead of .
\n\n
[3 marks]
\n(A1)
\n\n
Notes: The expansion may be seen in part (b)(ii).
\n\n
[1 mark]
\n(A1)(ft)(A1)(ft)(A1)(ft)
\n\n
Notes: Follow through from part (b)(i). Award (A1)(ft) for each correct term. Award at most (A1)(ft)(A1)(ft)(A0) if extra terms are seen.
\n\n
[3 marks]
\n(M1)
\n\n
Notes: Award (M1) for their . Accept equality or weak inequality.
\n\n
(A1)(ft)(A1)(ft)(G2)
\n\n
Notes: Award (A1)(ft) for correct endpoints, (A1)(ft) for correct weak or strict inequalities. Follow through from part (b)(ii). Do not award any marks if there is no answer in part (b)(ii).
\n\n
[3 marks]
\n![\"N17/5/MATSD/SP2/ENG/TZ0/05.d/M\"](\"../assets/uploads/tinymce_asset/asset/4193/Schermafbeelding_2018-02-14_om_06.23.51.png\"/)
(A1)(A1)(ft)(A1)(ft)(A1)
\n
Notes: Award (A1) for correct scale; axes labelled and drawn with a ruler.
\nAward (A1)(ft) for their correct -intercepts in approximately correct location.
\nAward (A1) for correct minimum and maximum points in approximately correct location.
\nAward (A1) for a smooth continuous curve with approximate correct shape. The curve should be in the given domain.
\nFollow through from part (a) for the -intercepts.
\n\n
[4 marks]
\n(G1)(ft)(G1)(ft)
\n\n
Notes: Award (G1) for 1.49 and (G1) for 13.9 written as a coordinate pair. Award at most (G0)(G1) if parentheses are missing. Accept and . Follow through from part (b)(i).
\n\n
[2 marks]
\nConsider , for , where .
\nThe equation has exactly one solution. Find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1 – using discriminant
\ncorrect equation without logs (A1)
\neg
\nvalid approach (M1)
\neg
\nrecognizing discriminant must be zero (seen anywhere) M1
\neg
\ncorrect discriminant (A1)
\neg
\ncorrect working (A1)
\neg
\nA2 N2
\nMETHOD 2 – completing the square
\ncorrect equation without logs (A1)
\neg
\nvalid approach to complete the square (M1)
\neg
\ncorrect working (A1)
\neg
\nrecognizing conditions for one solution M1
\neg
\ncorrect working (A1)
\neg
\nA2 N2
\n[7 marks]
\nLet , for 0 ≤ ≤ 1.
\nSketch the graph of on the grid below:
\nFind the -coordinates of the points of inflexion of the graph of .
\nHence find the values of for which the graph of is concave-down.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1A1A1 N3
Note: Only if the shape is approximately correct with exactly 2 maximums and 1 minimum on the interval 0 ≤ ≤ 0, award the following:
A1 for correct domain with both endpoints within circle and oval.
A1 for passing through the other -intercepts within the circles.
A1 for passing through the three turning points within circles (ignore -intercepts and extrema outside of the domain).
[3 marks]
\nevidence of reasoning (may be seen on graph) (M1)
\neg , (0.524, 0), (0.785, 0)
\n0.523598, 0.785398
\n, A1A1 N3
\nNote: Award M1A1A0 if any solution outside domain (eg ) is also included.
\n[3 marks]
\nA2 N2
\nNote: Award A1 if any correct interval outside domain also included, unless additional solutions already penalized in (b).
Award A0 if any incorrect intervals are also included.
[2 marks]
\nTwo submarines A and B have their routes planned so that their positions at time t hours, 0 ≤ t < 20 , would be defined by the position vectors rA and rB relative to a fixed point on the surface of the ocean (all lengths are in kilometres).
\nTo avoid the collision submarine B adjusts its velocity so that its position vector is now given by
\nrB .
\nShow that the two submarines would collide at a point P and write down the coordinates of P.
\nFind the value of t when submarine B passes through P.
\nFind the value of t when the two submarines are closest together.
\nFind the distance between the two submarines at this time.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nrA = rB (M1)
\n2 − t = − 0.5t ⇒ t = 4 A1
\nchecking t = 4 satisfies 4 + t = 3.2 + 1.2t and − 1 − 0.15t = − 2 + 0.1t R1
\nP(−2, 8, −1.6) A1
\nNote: Do not award final A1 if answer given as column vector.
\n[4 marks]
\nM1
\nNote: The M1 can be awarded for any one of the resultant equations.
\nA1
\n[2 marks]
\nminimum when (M1)
\nt = 3.83 A1
\n[2 marks]
\n0.511 (km) A1
\n[1 mark]
\nThe population of fish in a lake is modelled by the function
\n, 0 ≤ ≤ 30 , where is measured in months.
\nFind the population of fish at = 10.
\nFind the rate at which the population of fish is increasing at = 10.
\nFind the value of for which the population of fish is increasing most rapidly.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg (10)
\n235.402
\n235 (fish) (must be an integer) A1 N2
\n[2 marks]
\nrecognizing rate of change is derivative (M1)
\neg rate = , (10) , sketch of , 35 (fish per month)
\n35.9976
\n36.0 (fish per month) A1 N2
\n[2 marks]
\nvalid approach (M1)
\neg maximum of , = 0
\n15.890
\n15.9 (months) A1 N2
\n[2 marks]
\nThe following scatter diagram shows the scores obtained by seven students in their mathematics test, m, and their physics test, p.
\nThe mean point, M, for these data is (40, 16).
\nPlot and label the point M on the scatter diagram.
\nDraw the line of best fit, by eye, on the scatter diagram.
\nUsing your line of best fit, estimate the physics test score for a student with a score of 20 in their mathematics test.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1) (C2)
Note: Award (A1) for mean point plotted and (A1) for labelled M.
\n[2 marks]
\nstraight line through their mean point crossing the p-axis at 5±2 (A1)(ft)(A1)(ft) (C2)
\nNote: Award (A1)(ft) for a straight line through their mean point. Award (A1)(ft) for a correct p-intercept if line is extended.
\n[2 marks]
\npoint on line where m = 20 identified and an attempt to identify y-coordinate (M1)
\n10.5 (A1)(ft) (C2)
\nNote: Follow through from their line in part (b).
\n[2 marks]
\nLet . The graph of is shown in the following diagram.
\nFind .
\nFind the area of the region enclosed by the graph of , the x-axis and the lines x = 1 and x = 2 .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1A1 N2
\nNotes: Award A1A0 for both correct terms if +c is omitted.
Award A1A0 for one correct term eg .
Award A1A0 if both terms are correct, but candidate attempts further working to solve for c.
[2 marks]
\nsubstitution of limits or function (A1)
\neg
\nsubstituting limits into their integrated function and subtracting (M1)
\neg
\nNote: Award M0 if substituted into original function.
\ncorrect working (A1)
\neg
\nA1 N3
\n[4 marks]
\nConsider the graph of the function .
\nWrite down the equation of the vertical asymptote.
\nWrite down the equation of the horizontal asymptote.
\nCalculate the value of x for which f(x) = 0 .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
x = 0 (A1)(A1) (C2)
\nNote: Award (A1) for x = “a constant” (A1) for = 0. Award (A0)(A0) for an answer of “0”.
\n[2 marks]
\nf(x) = −2 (y = −2) (A1)(A1) (C2)
\nNote: Award (A1) for y = “a constant” (A1) for = −2. Award (A0)(A0) for an answer of “−2”.
\n[2 marks]
\n(M1)
\nNote: Award (M1) for equating f(x) to 0.
\n(A1) (C2)
\n[2 marks]
\nAll the children in a summer camp play at least one sport, from a choice of football () or basketball (). 15 children play both sports.
\nThe number of children who play only football is double the number of children who play only basketball.
\nLet be the number of children who play only football.
\nThere are 120 children in the summer camp.
\nWrite down an expression, in terms of , for the number of children who play only basketball.
\nComplete the Venn diagram using the above information.
\nFind the number of children who play only football.
\nWrite down the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1) (C1)
\n[1 mark]
\n (A1)(A1)(ft) (C2)
\n
Notes: Award (A1) for 15 placed in the correct position, award (A1)(ft) for and their placed in the correct positions of diagram. Do not penalize the absence of 0 inside the rectangle and award at most (A1)(A0) if any value other than 0 is seen outside the circles. Award at most (A1)(A0) if 35 and 70 are seen instead of and their .
\n\n
[2 marks]
\nor equivalent (M1)
\n\n
Note: Award (M1) for adding the values in their Venn and equating to 120 (or equivalent).
\n\n
(A1)(ft) (C2)
\n\n
Note: Follow through from their Venn diagram, but only if the answer is a positive integer and is seen in their Venn diagram.
\n\n
[2 marks]
\n85 (A1)(ft) (C1)
\n\n
Note: Follow through from their Venn diagram and their answer to part (c), but only if the answer is a positive integer and less than 120.
\n\n
[1 mark]
\nA particle P moves along a straight line. Its velocity after seconds is given by , for . The following diagram shows the graph of .
\n![\"M17/5/MATME/SP2/ENG/TZ1/07\"](\"../assets/uploads/tinymce_asset/asset/3543/Schermafbeelding_2017-08-14_om_10.04.21.png\"/)
Write down the first value of at which P changes direction.
\nFind the total distance travelled by P, for .
\nA second particle Q also moves along a straight line. Its velocity, after seconds is given by for . After seconds Q has travelled the same total distance as P.
\nFind .
\nA1 N1
\n[1 mark]
\nsubstitution of limits or function into formula or correct sum (A1)
\neg
\n9.64782
\ndistance A1 N2
\n[2 marks]
\ncorrect approach (A1)
\neg
\ncorrect integration (A1)
\neg
\nequating their expression to the distance travelled by their P (M1)
\neg
\n5.93855
\n5.94 (seconds) A1 N3
\n[4 marks]
\nConsider the function , where is a constant. Part of the graph of is shown below.
\n![\"M17/5/MATSD/SP2/ENG/TZ2/06\"](\"../assets/uploads/tinymce_asset/asset/3611/Schermafbeelding_2017-08-16_om_17.47.40.png\"/)
It is known that at the point where the tangent to the graph of is horizontal.
\nThere are two other points on the graph of at which the tangent is horizontal.
\nWrite down the -intercept of the graph.
\nFind .
\nShow that .
\nFind .
\nWrite down the -coordinates of these two points;
\nWrite down the intervals where the gradient of the graph of is positive.
\nWrite down the range of .
\nWrite down the number of possible solutions to the equation .
\nThe equation , where , has four solutions. Find the possible values of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
5 (A1)
\n\n
Note: Accept an answer of .
\n\n
[1 mark]
\n(A1)(A1)
\n\n
Note: Award (A1) for and (A1) for . Award at most (A1)(A0) if extra terms are seen.
\n\n
[2 marks]
\n(M1)(M1)
\n\n
Note: Award (M1) for substitution of into their derivative, (M1) for equating their derivative, written in terms of , to 0 leading to a correct answer (note, the 8 does not need to be seen).
\n\n
(AG)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution of and into the formula of the function.
\n\n
21 (A1)(G2)
\n[2 marks]
\n(A1)(A1)
\n\n
Note: Award (A1) for each correct solution. Award at most (A0)(A1)(ft) if answers are given as and or and .
\n\n
[2 marks]
\n(A1)(ft)(A1)(ft)
\n\n
Note: Award (A1)(ft) for , follow through from part (d)(i) provided their value is negative.
\nAward (A1)(ft) for , follow through only from their 0 from part (d)(i); 2 must be the upper limit.
\nAccept interval notation.
\n\n
[2 marks]
\n(A1)(ft)(A1)
\n\n
Notes: Award (A1)(ft) for 21 seen in an interval or an inequality, (A1) for “”.
\nAccept interval notation.
\nAccept or .
\nFollow through from their answer to part (c)(ii). Award at most (A1)(ft)(A0) if is seen instead of . Do not award the second (A1) if a (finite) lower limit is seen.
\n\n
[2 marks]
\n3 (solutions) (A1)
\n[1 mark]
\nor equivalent (A1)(ft)(A1)
\n\n
Note: Award (A1)(ft) for 5 and 21 seen in an interval or an inequality, (A1) for correct strict inequalities. Follow through from their answers to parts (a) and (c)(ii).
\nAccept interval notation.
\n\n
[2 marks]
\nFind .
\nFind , given that and .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach to set up integration by substitution/inspection (M1)
\neg
\ncorrect expression (A1)
\neg
\nA2 N4
\n\n
Notes: Award A1 if missing “”.
\n\n
[4 marks]
\nsubstituting into their answer from (a) (M1)
\neg
\ncorrect working (A1)
\neg
\nA1 N2
\n[3 marks]
\nThe following Venn diagram shows the sets , , and .
\nis an element of .
\n![\"N16/5/MATSD/SP1/ENG/TZ0/03\"](\"../assets/uploads/tinymce_asset/asset/3319/Schermafbeelding_2017-03-06_om_09.16.47.png\"/)
In the table indicate whether the given statements are True or False.
\n![\"N16/5/MATSD/SP1/ENG/TZ0/03.a\"](\"../assets/uploads/tinymce_asset/asset/3325/Schermafbeelding_2017-03-06_om_12.54.23.png\"/)
On the Venn diagram, shade the region .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
![\"N16/5/MATSD/SP1/ENG/TZ0/03.a/M\"](\"../assets/uploads/tinymce_asset/asset/3327/Schermafbeelding_2017-03-06_om_12.57.08.png\"/)
(A1)(A1)(A1)(A1)(A1) (C5)
[5 marks]
\n![\"N16/5/MATSD/SP1/ENG/TZ0/03.b/M\"](\"../assets/uploads/tinymce_asset/asset/3328/Schermafbeelding_2017-03-06_om_13.00.17.png\"/)
(A1) (C1)
[1 mark]
\nA company performs an experiment on the efficiency of a liquid that is used to detect a nut allergy.
\nA group of 60 people took part in the experiment. In this group 26 are allergic to nuts. One person from the group is chosen at random.
\nA second person is chosen from the group.
\nWhen the liquid is added to a person’s blood sample, it is expected to turn blue if the person is allergic to nuts and to turn red if the person is not allergic to nuts.
\nThe company claims that the probability that the test result is correct is 98% for people who are allergic to nuts and 95% for people who are not allergic to nuts.
\nIt is known that 6 in every 1000 adults are allergic to nuts.
\nThis information can be represented in a tree diagram.
\n![\"N17/5/MATSD/SP2/ENG/TZ0/04.c.d.e.f.g\"](\"../assets/uploads/tinymce_asset/asset/4187/Schermafbeelding_2018-02-13_om_14.31.34.png\"/)
An adult, who was not part of the original group of 60, is chosen at random and tested using this liquid.
\nThe liquid is used in an office to identify employees who might be allergic to nuts. The liquid turned blue for 38 employees.
\nFind the probability that this person is not allergic to nuts.
\nFind the probability that both people chosen are not allergic to nuts.
\nCopy and complete the tree diagram.
\nFind the probability that this adult is allergic to nuts and the liquid turns blue.
\nFind the probability that the liquid turns blue.
\nFind the probability that the tested adult is allergic to nuts given that the liquid turned blue.
\nEstimate the number of employees, from this 38, who are allergic to nuts.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)
\n\n
Note: Award (A1) for correct numerator, (A1) for correct denominator.
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for their correct product.
\n\n
(A1)(ft)(G2)
\n\n
Note: Follow through from part (a).
\n\n
[2 marks]
\n![\"N17/5/MATSD/SP2/ENG/TZ0/04.c/M\"](\"../assets/uploads/tinymce_asset/asset/4192/Schermafbeelding_2018-02-14_om_05.54.09.png\"/)
(A1)(A1)(A1)
\n
Note: Award (A1) for each correct pair of branches.
\n\n
[3 marks]
\n(M1)
\n\n
Note: Award (M1) for multiplying 0.006 by 0.98.
\n\n
(A1)(G2)
\n[2 marks]
\n(A1)(ft)(M1)
\n\n
Note: Award (A1)(ft) for their two correct products, (M1) for adding two products.
\n\n
(A1)(ft)(G3)
\n\n
Note: Follow through from parts (c) and (d).
\n\n
[3 marks]
\n(M1)(M1)
\n\n
Note: Award (M1) for their correct numerator, (M1) for their correct denominator.
\n\n
(A1)(ft)(G3)
\n\n
Note: Follow through from parts (d) and (e).
\n\n
[3 marks]
\n(M1)
\n\n
Note: Award (M1) for multiplying 38 by their answer to part (f).
\n\n
(A1)(ft)(G2)
\n\n
Notes: Follow through from part (f). Use of 3 sf result from part (f) results in an answer of 4.03 (4.028).
\n\n
[2 marks]
\nDune Canyon High School organizes its school year into three trimesters: fall/autumn (), winter () and spring (). The school offers a variety of sporting activities during and outside the school year.
\nThe activities offered by the school are summarized in the following Venn diagram.
\n![\"M17/5/MATSD/SP1/ENG/TZ1/04\"](\"../assets/uploads/tinymce_asset/asset/3565/Schermafbeelding_2017-08-15_om_10.56.10.png\"/)
Write down the number of sporting activities offered by the school during its school year.
\nDetermine whether rock-climbing is offered by the school in the fall/autumn trimester.
\nWrite down the elements of the set ;
\nWrite down .
\nWrite down, in terms of , and , an expression for the set which contains only archery, baseball, kayaking and surfing.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
15 (A1) (C1)
\n[1 mark]
\nno (A1) (C1)
\n\n
Note: Accept “it is only offered in Winter and Spring”.
\n\n
[1 mark]
\nvolleyball, golf, cycling (A1) (C1)
\n\n
Note: Responses must list all three sports for the (A1) to be awarded.
\n\n
[1 mark]
\n4 (A1) (C1)
\n[1 mark]
\nOR (or equivalent) (A2) (C2)
\n[2 marks]
\nConsider the curve y = 5x3 − 3x.
\nThe curve has a tangent at the point P(−1, −2).
\nFind .
\nFind the gradient of this tangent at point P.
\nFind the equation of this tangent. Give your answer in the form y = mx + c.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
15x2 − 3 (A1)(A1) (C2)
\nNote: Award (A1) for 15x2, (A1) for −3. Award at most (A1)(A0) if additional terms are seen.
\n\n
[2 marks]
\n15 (−1)2 − 3 (M1)
\nNote: Award (M1) for substituting −1 into their .
\n\n
= 12 (A1)(ft) (C2)
\nNote: Follow through from part (a).
\n\n
[2 marks]
\n(y − (−2)) = 12 (x − (−1)) (M1)
\nOR
\n−2 = 12(−1) + c (M1)
\nNote: Award (M1) for point and their gradient substituted into the equation of a line.
\n\n
y = 12x + 10 (A1)(ft) (C2)
\nNote: Follow through from part (b).
\n\n
[2 marks]
\nA function is given by .
\nWrite down the derivative of .
\nFind the point on the graph of at which the gradient of the tangent is equal to 6.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
or equivalent (A1)(A1)(A1) (C3)
\n\n
Note: Award (A1) for , (A1) for and (A1) for or . Award at most (A1)(A1)(A0) if additional terms seen.
\n\n
[3 marks]
\n(M1)
\n\n
Note: Award (M1) for equating their derivative to 6.
\n\n
OR (A1)(ft)(A1)(ft) (C3)
\n\n
Note: A frequent wrong answer seen in scripts is for this answer with correct working award (M1)(A0)(A1) and if there is no working award (C1).
\n\n
[3 marks]
\nContestants in a TV gameshow try to get through three walls by passing through doors without falling into a trap. Contestants choose doors at random.
If they avoid a trap they progress to the next wall.
If a contestant falls into a trap they exit the game before the next contestant plays.
Contestants are not allowed to watch each other attempt the game.
The first wall has four doors with a trap behind one door.
\nAyako is a contestant.
\nNatsuko is the second contestant.
\nThe second wall has five doors with a trap behind two of the doors.
\nThe third wall has six doors with a trap behind three of the doors.
\nThe following diagram shows the branches of a probability tree diagram for a contestant in the game.
\nWrite down the probability that Ayako avoids the trap in this wall.
\nFind the probability that only one of Ayako and Natsuko falls into a trap while attempting to pass through a door in the first wall.
\nCopy the probability tree diagram and write down the relevant probabilities along the branches.
\nA contestant is chosen at random. Find the probability that this contestant fell into a trap while attempting to pass through a door in the second wall.
\nA contestant is chosen at random. Find the probability that this contestant fell into a trap.
\n120 contestants attempted this game.
\nFind the expected number of contestants who fell into a trap while attempting to pass through a door in the third wall.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(0.75, 75%) (A1)
\n[1 mark]
\nOR (M1)(M1)
\nNote: Award (M1) for their product seen, and (M1) for adding their two products or multiplying their product by 2.
\n(A1)(ft) (G3)
\nNote: Follow through from part (a), but only if the sum of their two fractions is 1.
\n[3 marks]
\n(A1)(ft)(A1)(A1)
Note: Award (A1) for each correct pair of branches. Follow through from part (a).
\n[3 marks]
\n(M1)
\nNote: Award (M1) for correct probabilities multiplied together.
\n(A1)(ft) (G2)
\nNote: Follow through from their tree diagram or part (a).
\n[2 marks]
\nOR (M1)(M1)
\nNote: Award (M1) for and (M1) for subtracting their correct probability from 1, or adding to their .
\n(A1)(ft) (G2)
\nNote: Follow through from their tree diagram.
\n[3 marks]
\n(M1)(M1)
\nNote: Award (M1) for and (M1) for multiplying by 120.
\n= 27 (A1)(ft) (G3)
\nNote: Follow through from their tree diagram or their from their calculation in part (d)(ii).
\n[3 marks]
\nA potter sells vases per month.
\nHis monthly profit in Australian dollars (AUD) can be modelled by
\n\n
Find the value of if no vases are sold.
\nDifferentiate .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
−120 (AUD) (A1) (C1)
\n[1 mark]
\n(A1)(A1) (C2)
\nNote: Award (A1) for each correct term. Award at most (A1)(A0) for extra terms seen.
\n[2 marks]
\nLet . Given that , find .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg
\ncorrect integration by substitution/inspection A2
\neg
\ncorrect substitution into their integrated function (must include ) M1
\neg
\n\n
Note: Award M0 if candidates substitute into or .
\n\n
(A1)
\nA1 N4
\n[6 marks]
\nLet . The following diagram shows part of the graph of .
\nThe region R is enclosed by the graph of , the -axis, and the -axis. Find the area of R.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1 (limits in terms of )
\nvalid approach to find -intercept (M1)
\neg , ,
\n-intercept is 3 (A1)
\nvalid approach using substitution or inspection (M1)
\neg , , , , ,
\n, , ,
\n(A2)
\nsubstituting both of their limits into their integrated function and subtracting (M1)
\neg ,
\nNote: Award M0 if they substitute into original or differentiated function. Do not accept only “– 0” as evidence of substituting lower limit.
\n\n
correct working (A1)
\neg ,
\narea = 2 A1 N2
\n\n
\n
METHOD 2 (limits in terms of )
\nvalid approach to find -intercept (M1)
\neg , ,
\n-intercept is 3 (A1)
\nvalid approach using substitution or inspection (M1)
\neg , , , ,
\n, ,
\ncorrect integration (A2)
\neg ,
\nboth correct limits for (A1)
\neg = 16 and = 25, , , = 4 and = 5, ,
\nsubstituting both of their limits for (do not accept 0 and 3) into their integrated function and subtracting (M1)
\neg ,
\nNote: Award M0 if they substitute into original or differentiated function, or if they have not attempted to find limits for .
\n\n
area = 2 A1 N2
\n\n
[8 marks]
\nLet , for .
\nFind .
\nPart of the graph of f is shown in the following diagram.
\nThe shaded region R is enclosed by the graph of f, the x-axis, and the lines x = 1 and x = 9 . Find the volume of the solid formed when R is revolved 360° about the x-axis.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct working (A1)
\neg
\nA2 N3
\nNote: Award A1 for .
\n[3 marks]
\nattempt to substitute either limits or the function into formula involving f 2 (accept absence of / dx) (M1)
\neg
\nsubstituting limits into their integral and subtracting (in any order) (M1)
\neg
\ncorrect working involving calculating a log value or using log law (A1)
\neg
\nA1 N3
\nNote: Full FT may be awarded as normal, from their incorrect answer in part (a), however, do not award the final two A marks unless they involve logarithms.
\n[4 marks]
\nThe point A has coordinates (4 , −8) and the point B has coordinates (−2 , 4).
\nThe point D has coordinates (−3 , 1).
\nWrite down the coordinates of C, the midpoint of line segment AB.
\nFind the gradient of the line DC.
\nFind the equation of the line DC. Write your answer in the form ax + by + d = 0 where a , b and d are integers.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(1, −2) (A1)(A1) (C2)
Note: Award (A1) for 1 and (A1) for −2, seen as a coordinate pair.
Accept x = 1, y = −2. Award (A1)(A0) if x and y coordinates are reversed.
\n[2 marks]
\n(M1)
\nNote: Award (M1) for correct substitution, of their part (a), into gradient formula.
\n(A1)(ft) (C2)
\nNote: Follow through from part (a).
\n[2 marks]
\n\n
OR OR (M1)
\nNote: Award (M1) for correct substitution of their part (b) and a given point.
\nOR
\nOR (M1)
\nNote: Award (M1) for correct substitution of their part (b) and a given point.
\n(accept any integer multiple, including negative multiples) (A1)(ft) (C2)
\nNote: Follow through from parts (a) and (b). Where the gradient in part (b) is found to be , award at most (M1)(A0) for either or .
\n[2 marks]
\n\n
\n
A group of 66 people went on holiday to Hawaii. During their stay, three trips were arranged: a boat trip (), a coach trip () and a helicopter trip ().
\nFrom this group of people:
\n| 3 | \nwent on all three trips; | \n
| 16 | \nwent on the coach trip only; | \n
| 13 | \nwent on the boat trip only; | \n
| 5 | \nwent on the helicopter trip only; | \n
| x | \nwent on the coach trip and the helicopter trip but not the boat trip; | \n
| 2x \n | \nwent on the boat trip and the helicopter trip but not the coach trip; | \n
| 4x \n | \nwent on the boat trip and the coach trip but not the helicopter trip; | \n
| 8 | \ndid not go on any of the trips. | \n
One person in the group is selected at random.
\nDraw a Venn diagram to represent the given information, using sets labelled , and .
\nShow that .
\nWrite down the value of .
\nFind the probability that this person
\n(i) went on at most one trip;
\n(ii) went on the coach trip, given that this person also went on both the helicopter trip and the boat trip.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n![\"N16/5/MATSD/SP2/ENG/TZ0/02.a/M\"](\"../assets/uploads/tinymce_asset/asset/3344/Schermafbeelding_2017-03-07_om_10.03.03.png\"/)
(A5)
\n
Notes: Award (A1) for rectangle and three labelled intersecting circles (U need not be seen),
\n(A1) for 3 in the correct region,
\n(A1) for 8 in the correct region,
\n(A1) for 5, 13 and 16 in the correct regions,
\n(A1) for , and in the correct regions.
\n\n
[5 marks]
\n(M1)
\n\n
Note: Award (M1) for either a completely correct equation or adding all the terms from their diagram in part (a) and equating to 66.
\nAward (M0)(A0) if their equation has no .
\n\n
OR (A1)
\n\n
Note: Award (A1) for adding their like terms correctly, but only when the solution to their equation is equal to 3 and is consistent with their original equation.
\n\n
(AG)
\n\n
Note: The conclusion must be seen for the (A1) to be awarded.
\n\n
[2 marks]
\n15 (A1)(ft)
\n\n
Note: Follow through from part (a). The answer must be an integer.
\n\n
[1 mark]
\n(i) (A1)(ft)(A1)(G2)
\n\n
Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn diagram.
\n\n
(ii) (A1)(A1)(ft)(G2)
\n\n
Note: Award (A1) for numerator, (A1)(ft) for denominator. Follow through from their Venn diagram.
\n\n
[4 marks]
\nLet . Find , given that .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of integration (M1)
\neg
\ncorrect integration (accept missing ) (A2)
\neg
\nsubstituting initial condition into their integrated expression (must have ) M1
\neg
\n\n
Note: Award M0 if they substitute into the original or differentiated function.
\n\n
recognizing (A1)
\neg
\n(A1)
\nA1 N5
\n[7 marks]
\nA farmer owns a plot of land in the shape of a quadrilateral ABCD.
\nand angle .
\n![\"N16/5/MATSD/SP2/ENG/TZ0/05\"](\"../assets/uploads/tinymce_asset/asset/3341/Schermafbeelding_2017-03-07_om_08.23.38.png\"/)
The farmer wants to divide the land into two equal areas. He builds a fence in a straight line from point B to point P on AD, so that the area of PAB is equal to the area of PBCD.
\nCalculate
\nthe length of BD;
\nthe size of angle DAB;
\nthe area of triangle ABD;
\nthe area of quadrilateral ABCD;
\nthe length of AP;
\nthe length of the fence, BP.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\n\n
Note: Award (M1) for correct substitution into Pythagoras’ theorem.
\n\n
(A1)(G2)
\n[2 marks]
\n(M1)(A1)(ft)
\n\n
Note: Award (M1) for substitution into cosine rule, (A1)(ft) for their correct substitutions. Follow through from part (a).
\n\n
(A1)(ft)(G2)
\n\n
Note: If their 103 used, the answer is
\n\n
[3 marks]
\n(M1)(A1)(ft)
\n\n
Notes: Award (M1) for substitution into the trig form of the area of a triangle formula.
\nAward (A1)(ft) for their correct substitutions.
\nFollow through from part (b).
\nIf 68.8° is used the area .
\n\n
(A1)(ft)(G2)
\n[3 marks]
\n(M1)
\n\n
Note: Award (M1) for correctly substituted area of triangle formula added to their answer to part (c).
\n\n
(A1)(ft)(G2)
\n[2 marks]
\n(M1)(M1)
\n\n
Notes: Award (M1) for the correct substitution into triangle formula.
\nAward (M1) for equating their triangle area to half their part (d).
\n\n
(A1)(ft)(G2)
\n\n
Notes: Follow through from parts (b) and (d).
\n\n
[3 marks]
\n(M1)(A1)(ft)
\n\n
Notes: Award (M1) for substituted cosine rule formula.
\nAward (A1)(ft) for their correct substitutions. Accept the exact fraction in place of .
\nFollow through from parts (b) and (e).
\n\n
(A1)(ft)(G2)
\n\n
Notes: If 54.4 and are used the answer is
\n\n
[3 marks]
\nThe diagram shows part of the graph of a function . The graph passes through point .
\n![\"M17/5/MATSD/SP1/ENG/TZ2/13\"](\"../assets/uploads/tinymce_asset/asset/3587/Schermafbeelding_2017-08-16_om_06.22.37.png\"/)
The tangent to the graph of at A has equation . Let be the normal to the graph of at A.
\nWrite down the value of .
\nFind the equation of . Give your answer in the form where , , .
\nDraw the line on the diagram above.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
3 (A1) (C1)
\n\n
Notes: Accept
\n\n
[1 mark]
\nOR (A1)(A1)
\n\n
Note: Award (A1) for correct gradient, (A1) for correct substitution of in the equation of line.
\n\n
or any integer multiple (A1)(ft) (C3)
\n\n
Note: Award (A1)(ft) for their equation correctly rearranged in the indicated form.
\nThe candidate’s answer must be an equation for this mark.
\n\n
[3 marks]
\n![\"M17/5/MATSD/SP1/ENG/TZ2/13.c/M\"](\"../assets/uploads/tinymce_asset/asset/3594/Schermafbeelding_2017-08-16_om_08.26.38.png\"/)
(M1)(A1)(ft) (C2)
\n
Note: Award M1) for a straight line, with positive gradient, passing through , (A1)(ft) for line (or extension of their line) passing approximately through 2.5 or their intercept with the -axis.
\n\n
[2 marks]
\nThe derivative of a function is given by . The graph of passes through .
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognizing to integrate (M1)
\neg , ,
\ncorrect integral (do not penalize for missing +) (A2)
eg
\nsubstituting (in any order) into their integrated expression (must have +) M1
\neg
\nNote: Award M0 if they substitute into original or differentiated function.
\n(or any equivalent form, eg ) A1 N4
\n[5 marks]
\nA restaurant serves desserts in glasses in the shape of a cone and in the shape of a hemisphere. The diameter of a cone shaped glass is 7.2 cm and the height of the cone is 11.8 cm as shown.
\n![\"N17/5/MATSD/SP2/ENG/TZ0/06\"](\"../assets/uploads/tinymce_asset/asset/4188/Schermafbeelding_2018-02-13_om_14.40.46.png\"/)
The volume of a hemisphere shaped glass is .
\nThe restaurant offers two types of dessert.
\nThe regular dessert is a hemisphere shaped glass completely filled with chocolate mousse. The cost, to the restaurant, of the chocolate mousse for one regular dessert is $1.89.
\nThe special dessert is a cone shaped glass filled with two ingredients. It is first filled with orange paste to half of its height and then with chocolate mousse for the remaining volume.
\n![\"N17/5/MATSD/SP2/ENG/TZ0/06.d.e.f\"](\"../assets/uploads/tinymce_asset/asset/4189/Schermafbeelding_2018-02-13_om_14.44.32.png\"/)
The cost, to the restaurant, of of orange paste is $7.42.
\nA chef at the restaurant prepares 50 desserts; regular desserts and special desserts. The cost of the ingredients for the 50 desserts is $111.44.
\nShow that the volume of a cone shaped glass is , correct to 3 significant figures.
\nCalculate the radius, , of a hemisphere shaped glass.
\nFind the cost of of chocolate mousse.
\nShow that there is of orange paste in each special dessert.
\nFind the total cost of the ingredients of one special dessert.
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\n\n
Note: Award (M1) for correct substitution into volume of a cone formula.
\n\n
(A1)
\n(AG)
\n\n
Note: Both rounded and unrounded answers must be seen for the final (A1) to be awarded.
\n\n
[2 marks]
\n(M1)(A1)
\n\n
Notes: Award (M1) for multiplying volume of sphere formula by (or equivalent).
\nAward (A1) for equating the volume of hemisphere formula to 225.
\n\n
OR
\n(A1)(M1)
\n\n
Notes: Award (A1) for 450 seen, (M1) for equating the volume of sphere formula to 450.
\n\n
(A1)(G2)
\n[3 marks]
\n(M1)
\n\n
Note: Award (M1) for dividing 1.89 by 2.25, or equivalent.
\n\n
(A1)(G2)
\n\n
Note: Accept 84 cents if the units are explicit.
\n\n
[2 marks]
\n(A1)
\n(M1)
\n\n
Note: Award (M1) for correct substitution into volume of a cone formula, but only if the result rounds to 20.
\n\n
(AG)
\nOR
\n(A1)
\n(M1)
\n\n
Notes: Award (M1) for multiplying 160 by . Award (A0)(M1) for if is not seen.
\n\n
(AG)
\n\n
Notes: Do not award any marks if the response substitutes in the known value to find the radius of the cone.
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for the sum of two correct products.
\n\n
$ 2.66 (A1)(ft)(G2)
\n\n
Note: Follow through from part (c).
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for correct equation.
\n\n
(M1)
\n\n
Note: Award (M1) for setting up correct equation, including their 2.66 from part (e).
\n\n
(A1)(ft)(G3)
\n\n
Note: Follow through from part (e), but only if their answer for is rounded to the nearest positive integer, where .
\nAward at most (M1)(M1)(A0) for a final answer of “28, 22”, where the -value is not clearly defined.
\n\n
[3 marks]
\nA healthy human body temperature is 37.0 °C. Eight people were medically examined and the difference in their body temperature (°C), from 37.0 °C, was recorded. Their heartbeat (beats per minute) was also recorded.
\nWrite down, for this set of data the mean temperature difference from 37 °C, .
\nWrite down, for this set of data the mean number of heartbeats per minute, .
\nPlot and label the point M(, ) on the scatter diagram.
\nUse your graphic display calculator to find the Pearson’s product–moment correlation coefficient, .
\nHence describe the correlation between temperature difference from 37 °C and heartbeat.
\nDraw the regression line on on the scatter diagram.
\n0.025 (A1)
\n[1 mark]
\n74 (A1)
\n[1 mark]
\nthe point M labelled, correctly plotted on their diagram (A1)(A1)(ft)
\nNote: Award (A1) for labelled M. Do not accept any other label. Award (A1)(ft) for their point M correctly plotted. Follow through from part (b).
\n[2 marks]
\n0.807 (0.806797…) (G2)
\n[2 marks]
\n(moderately) strong, positive (A1)(ft)(A1)(ft)
\nNote: Award (A1) for (moderately) strong, (A1) for positive. Follow through from part (d)(i). If there is no answer to part (d)(i), award at most (A0)(A1).
\n[2 marks]
\ntheir regression line correctly drawn on scatter diagram (A1)(ft)(A1)(ft)
\nNote: Award (A1)(ft) for a straight line, using a ruler, intercepting their mean point, and (A1)(ft) for intercepting the -axis at their 73.5 and the gradient of the line is positive. If graph paper is not used, award at most (A1)(A0). Follow through from part (e).
\n[2 marks]
\nIn this question distance is in centimetres and time is in seconds.
\nParticle A is moving along a straight line such that its displacement from a point P, after seconds, is given by , 0 ≤ ≤ 25. This is shown in the following diagram.
\nAnother particle, B, moves along the same line, starting at the same time as particle A. The velocity of particle B is given by , 0 ≤ ≤ 25.
\nFind the initial displacement of particle A from point P.
\nFind the value of when particle A first reaches point P.
\nFind the value of when particle A first changes direction.
\nFind the total distance travelled by particle A in the first 3 seconds.
\nGiven that particles A and B start at the same point, find the displacement function for particle B.
\nFind the other value of when particles A and B meet.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nvalid approach (M1)
\neg
\n15 (cm) A1 N2
\n[2 marks]
\nvalid approach (M1)
\neg
\n2.46941
\n= 2.47 (seconds) A1 N2
\n[2 marks]
\nrecognizing when change in direction occurs (M1)
\neg slope of changes sign, , minimum point, 10.0144, (4.08, −4.66)
\n4.07702
\n= 4.08 (seconds) A1 N2
\n[2 marks]
\nMETHOD 1 (using displacement)
\ncorrect displacement or distance from P at (seen anywhere) (A1)
\neg −2.69630, 2.69630
\nvalid approach (M1)
\neg 15 + 2.69630, , −17.6963
\n17.6963
\n17.7 (cm) A1 N2
\n\n
METHOD 2 (using velocity)
\nattempt to substitute either limits or the velocity function into distance formula involving (M1)
\neg ,
\n17.6963
\n17.7 (cm) A1 N2
\n[3 marks]
\nrecognize the need to integrate velocity (M1)
\neg
\n(accept instead of and missing ) (A2)
\nsubstituting initial condition into their integrated expression (must have ) (M1)
\neg ,
\nA1 N3
\n[5 marks]
\nvalid approach (M1)
\neg , sketch, (9.30404, 2.86710)
\n9.30404
\n(seconds) A1 N2
\nNote: If candidates obtain in part (e)(i), there are 2 solutions for part (e)(ii), 1.32463 and 7.79009. Award the last A1 in part (e)(ii) only if both solutions are given.
\n[2 marks]
\nHaruka has an eco-friendly bag in the shape of a cuboid with width 12 cm, length 36 cm and height of 9 cm. The bag is made from five rectangular pieces of cloth and is open at the top.
\n\n
Nanako decides to make her own eco-friendly bag in the shape of a cuboid such that the surface area is minimized.
\nThe width of Nanako’s bag is x cm, its length is three times its width and its height is y cm.
\n\n
The volume of Nanako’s bag is 3888 cm3.
\nCalculate the area of cloth, in cm2, needed to make Haruka’s bag.
\nCalculate the volume, in cm3, of the bag.
\nUse this value to write down, and simplify, the equation in x and y for the volume of Nanako’s bag.
\nWrite down and simplify an expression in x and y for the area of cloth, A, used to make Nanako’s bag.
\nUse your answers to parts (c) and (d) to show that
\n.
\nFind .
\nUse your answer to part (f) to show that the width of Nanako’s bag is 12 cm.
\nThe cloth used to make Nanako’s bag costs 4 Japanese Yen (JPY) per cm2.
\nFind the cost of the cloth used to make Nanako’s bag.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
36 × 12 + 2(9 ×12) + 2(9 × 36) (M1)
\nNote: Award (M1) for correct substitution into surface area of cuboid formula.
\n\n
= 1300 (cm2) (1296 (cm2)) (A1)(G2)
\n\n
[2 marks]
\n36 × 9 ×12 (M1)
\nNote: Award (M1) for correct substitution into volume of cuboid formula.
\n\n
= 3890 (cm3) (3888 (cm3)) (A1)(G2)
\n\n
[2 marks]
\n3 × × = 3888 (M1)
\nNote: Award (M1) for correct substitution into volume of cuboid formula and equated to 3888.
\n\n
2 = 1296 (A1)(G2)
\nNote: Award (A1) for correct fully simplified volume of cuboid.
\nAccept .
\n\n
[2 marks]
\n(A =) 3x2 + 2(xy) + 2(3xy) (M1)
\nNote: Award (M1) for correct substitution into surface area of cuboid formula.
\n\n
(A =) 3x2 + 8xy (A1)(G2)
\nNote: Award (A1) for correct simplified surface area of cuboid formula.
\n\n
\n
[2 marks]
\n(A1)(ft)(M1)
\nNote: Award (A1)(ft) for correct rearrangement of their part (c) seen (rearrangement may be seen in part(c)), award (M1) for substitution of their part (c) into their part (d) but only if this leads to the given answer, which must be shown.
\n\n
(AG)
\n\n
[2 marks]
\n(A1)(A1)(A1)
\nNote: Award (A1) for , (A1) for −10368, (A1) for . Award a maximum of (A1)(A1)(A0) if any extra terms seen.
\n\n
[3 marks]
\n(M1)
\nNote: Award (M1) for equating their to zero.
\n\n
OR OR (M1)
\nNote: Award (M1) for correctly rearranging their equation so that fractions are removed.
\n\n
(A1)
\n(cm) (AG)
\nNote: The (AG) line must be seen for the final (A1) to be awarded. Substituting invalidates the method, award a maximum of (M1)(M0)(A0).
\n\n
[3 marks]
\n(M1)
\n\n
Note: Award (M1) for substituting 12 into the area formula and for multiplying the area formula by 4.
\n\n
= 5180 (JPY) (5184 (JPY)) (A1)(G2)
\n\n
[2 marks]
\nThe function has a local maximum and a local minimum. The local maximum is at .
\nShow that .
\nFind the coordinates of the local minimum.
\nWrite down the interval where the gradient of the graph of is negative.
\nDetermine the equation of the normal at in the form .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(A1)
\nNote: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if additional terms are seen or for an answer . If their derivative is seen in parts (b), (c) or (d) and not in part (a), award at most (A1)(A1)(A0).
\n(M1)(M1)
\nNote: Award (M1) for substituting in into their derivative and (M1) for setting it equal to zero. Substituting invalidates the process, award at most (A1)(A1)(A1)(M0)(M0).
\n(AG)
\nNote: For the final (M1) to be awarded, no incorrect working must be seen, and must lead to the conclusion . The final (AG) must be seen.
\n[5 marks]
\n(2, −2.33) OR (A1)(A1)
\nNote: Award (A1) for each correct coordinate. Award (A0)(A1) if parentheses are missing. Accept , . Award (M1)(A0) for their derivative, a quadratic expression with –6 substituted for , equated to zero but leading to an incorrect answer.
\n[2 marks]
\n(A1)(ft)(A1)
\nNote: Award (A1) for , (A1)(ft) for . Follow through for their \"2\" in part (b). It is possible to award (A0)(A1). For award (A1)(A0). Accept equivalent notation such as (−3, 2). Award (A0)(A1)(ft) for .
\n[2 marks]
\n−4 (A1)(ft)
\nNote: Award (A1)(ft) for the gradient of the tangent seen. If an incorrect derivative was used in part (a), then working for their must be seen. Follow through from their derivative in part (a).
\ngradient of normal is (A1)(ft)
\nNote: Award (A1)(ft) for the negative reciprocal of their gradient of tangent. Follow through within this part. Award (G2) for an unsupported gradient of the normal.
\n(A1)
\nNote: Award (A1) for (16.3333…) seen.
\nOR (M1)
\nNote: Award (M1) for substituting their normal gradient into equation of line formula.
\nOR (A1)(ft)(G4)
\nNote: Award (G4) for the correct equation of line in correct form without any prior working. The final (A1)(ft) is contingent on and .
\n[5 marks]
\nMoney boxes are coin containers used by children and come in a variety of shapes. The money box shown is in the shape of a cylinder. It has a radius of 4.43 cm and a height of 12.2 cm.
\nFind the volume of the money box.
\nA second money box is in the shape of a sphere and has the same volume as the cylindrical money box.
\nFind the diameter of the second money box.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(V =) (M1)(A1)
\nNote: Award (M1) for substitution into volume of a cylinder formula, (A1) for correct substitution.
\n752 cm3 (752.171…cm3) (A1)(C3)
\n[3 marks]
\n(M1)
\nNote: Award (M1) for equating their volume to the volume of a sphere formula.
\n5.64169…cm (A1)(ft)
\nNote: Follow through from part (a).
\n11.3 cm (11.2833…cm) (A1)(ft) (C3)
\n[3 marks]
\nThe equation of a curve is .
\nThe gradient of the tangent to the curve at a point P is .
\nFind .
\nFind the coordinates of P.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1) (C2)
\n\n
Note: Award (A1) for , award (A1) for .
\nAward at most (A1)(A0) if there are any extra terms.
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for equating their answer to part (a) to .
\n\n
(A1)(ft)
\n\n
Note: Follow through from part (a). Award (M0)(A0) for seen without working.
\n\n
(M1)
\n\n
Note: Award (M1) substituting their into the original function.
\n\n
(A1)(ft) (C4)
\n\n
Note: Accept .
\n\n
[4 marks]
\nThe matrix M is given by M .
\nGiven that M3 can be written as a quadratic expression in M in the form aM2 + bM + cI , determine the values of the constants a, b and c.
\nShow that M4 = 19M2 + 40M + 30I.
\nUsing mathematical induction, prove that Mn can be written as a quadratic expression in M for all positive integers n ≥ 3.
\nFind a quadratic expression in M for the inverse matrix M–1.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nM2 ; M3 (A1)(A1)
\nlet (M1)
\nthen, for example
\n11a + b + c = 53
\n10a + 2b = 50 M1A1
\n6a + 2b = 38
\nthe solution is a = 3, b =10, c =10 (M1)A1
\n(M3 = 3M2 +10M +10I)
\n\n
METHOD 2
\ndet(M − I) = 0 (M1)
\nM1A1
\nM1A1
\napplying the Cayley – Hamilton theorem (M1)
\nM3 = 3M2 +10M +10I and so a = 3, b =10, c =10 A1
\n\n
[7 marks]
\nM4 = 3M3 + 10M2 + 10M M1
\n= 3(3M2 + 10M + 10I) + 10M2 + 10M M1
\n=19M2 + 40M +30I AG
\n[2 marks]
\nthe statement is true for n = 3 as shown in part (a) A1
\nassume true for n = k, ie Mk = pM2 + qM + rI M1
\nNote: Subsequent marks after this M1 are independent and can be awarded.
\nMk+1 = pM3 + qM2 + rM M1
\n= p(3M2 +10M +10I) + qM2 + rM M1
\n= (3p +q)M2 + (10p + r)M + 10pI A1
\nhence true for n = k ⇒ true for n = k +1 and since true for n = 3, the statement is proved by induction R1
\nNote: Award R1 provided at least four of the previous marks are gained.
\n[6 marks]
\nM2 = 3M + 10I + 10M–1 M1
\nM–1 = (M2 − 3M − 10I) A1
\n[2 marks]
\nNote: In this question, distance is in metres and time is in seconds.
\nA particle P moves in a straight line for five seconds. Its acceleration at time is given by , for .
\nWhen , the velocity of P is .
\nWrite down the values of when .
\nHence or otherwise, find all possible values of for which the velocity of P is decreasing.
\nFind an expression for the velocity of P at time .
\nFind the total distance travelled by P when its velocity is increasing.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nA1A1 N2
\n[2 marks]
\nrecognizing that is decreasing when is negative (M1)
\neg, sketch of
\ncorrect interval A1 N2
\neg
\n[2 marks]
\nvalid approach (do not accept a definite integral) (M1)
\neg
\ncorrect integration (accept missing ) (A1)(A1)(A1)
\n\n
substituting , (must have ) (M1)
\neg
\nA1 N6
\n[6 marks]
\nrecognizing that increases outside the interval found in part (b) (M1)
\neg, diagram
\none correct substitution into distance formula (A1)
\neg
\none correct pair (A1)
\neg3.13580 and 11.0833, 20.9906 and 35.2097
\n14.2191 A1 N2
\n\n
[4 marks]
\nLet A2 = 2A + I where A is a 2 × 2 matrix.
\nShow that A4 = 12A + 5I.
\nLet B = .
\nGiven that B2 – B – 4I = , find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
A4 = 4A2 + 4AI + I2 or equivalent M1A1
= 4(2A + I) + 4A + I A1
= 8A + 4I + 4A + I
= 12A + 5I AG
[3 marks]
\nMETHOD 2
A3 = A(2A + I) = 2A2 + AI = 2(2A + I) + A(= 5A + 2I) M1A1
A4 = A(5A + 2I) A1
= 5A2 + 2A = 5(2A + I) + 2A
= 12A + 5I AG
[3 marks]
\nB2 = (A1)
\n(A1)
\nA1
\n[3 marks]
\nThe function f : M → M where M is the set of 2 × 2 matrices, is given by f(X) = AX where A is a 2 × 2 matrix.
\nGiven that A is non-singular, prove that f is a bijection.
\nIt is now given that A is singular.
\nBy considering appropriate determinants, prove that f is not a bijection.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
suppose f(X) = f(Y) , ie AX = AY (M1)
\nthen A−1AX = A−1AY A1
\nX = Y A1
\nsince f(X) = f(Y) ⇒ X = Y, f is an injection R1
\nnow suppose C ∈ M and consider f(D) = C , ie AD = C M1
\nthen D = A−1 C (A−1 exists since A is non- singular) A1
\nsince given C ∈ M, there exists D ∈ M such that f(D) = C , f is a surjection R1
\ntherefore f is a bijection AG
\n[7 marks]
\nsuppose f(X) = Y, ie AX = Y (M1)
\nthen det(A)det(X) = det(Y) A1
\nsince det(A) = 0, it follows that det(Y) = 0 A1
\nit follows that f is not surjective since the function cannot reach non-singular matrices R1
\ntherefore f is not a bijection AG
\n[4 marks]
\nA cylinder with radius and height is shown in the following diagram.
\nThe sum of and for this cylinder is 12 cm.
\nWrite down an equation for the area, , of the curved surface in terms of .
\nFind .
\nFind the value of when the area of the curved surface is maximized.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
OR (A1)(M1) (C2)
\nNote: Award (A1) for or seen. Award (M1) for correctly substituting into curved surface area of a cylinder. Accept OR .
\n[2 marks]
\n(A1)(ft)(A1)(ft) (C2)
\nNote: Award (A1)(ft) for and (A1)(ft) for . Follow through from part (a). Award at most (A1)(ft)(A0) if additional terms are seen.
\n[2 marks]
\n(M1)
\nNote: Award (M1) for setting their part (b) equal to zero.
\n6 (cm) (A1)(ft) (C2)
\nNote: Follow through from part (b).
\n[2 marks]
\nThe matrix A is given by A .
\nThe matrix B is given by B .
\nShow that the eigenvalues of A are real if .
\nDeduce that the eigenvalues are real if A is symmetric.
\nDetermine the eigenvalues of B.
\nDetermine the corresponding eigenvectors.
\nthe eigenvalues satisfy
\nM1
\nA1
\nA1
\nthe condition for real roots is
\nM1
\nAG
\n[4 marks]
\nif the matrix is symmetric, b = c. In this case, M1
\n\n
because each square term is non-negative R1AG
\n[2 marks]
\nthe characteristic equation is
\nM1
\nA1
\n[2 marks]
\ntaking
\nM1
\ngiving eigenvector A1
\n\n
taking
\nM1
\ngiving eigenvector A1
\n[4 marks]
\nA driver needs to make deliveries to five shops , , , and . The driver starts and finishes his journey at the warehouse . The driver wants to find the shortest route to visit all the shops and return to the warehouse. The distances, in kilometres, between the locations are given in the following table.
\nBy deleting , use the deleted vertex algorithm to find a lower bound for the length of a route that visits every shop, starting and finishing at .
\nStarting from , use the nearest-neighbour algorithm to find a route which gives an upper bound for this problem and calculate its length.
\ndeleting and its adjacent edges, the minimal spanning tree is
\n A1A1A1A1
Note: Award the A1’s for either the edges or their weights.
\nthe minimum spanning tree has weight = 54
\nNote: Accept a correct drawing of the minimal spanning tree.
\nadding in the weights of 2 deleted edges of least weight and (M1)
lower bound = 54 + 36 + 39
= 129 A1
\n[6 marks]
\nattempt at the nearest-neighbour algorithm M1
\n
A1
Note: Award M1 for a route that begins with and then .
\nupper bound = 36 + 11 + 15 + 12 + 22 + 39 = 135 (M1)A1
\n[4 marks]
\nis a simple, connected graph with eight vertices.
\nis a connected, planar graph, with vertices, edges and faces. Every face in is bounded by exactly edges.
\nWrite down the minimum number of edges in .
\nFind the maximum number of edges in .
\nFind the maximum number of edges in , given that contains an Eulerian circuit.
\nExplain why .
\nFind the value of when and .
\nFind the possible values of when .
\n7 (a tree) A1
\n[1 mark]
\n(a complete graph) (M1)
\nA1
\n[2 marks]
\n(since every vertex must be of degree 6) (M1)
\nA1
\n[2 marks]
\ncounting the edges around every face gives edges A1
\nbut as every edge is counted in 2 faces R1
\nAG
\n[2 marks]
\nusing with M1
\nEITHER
\nsubstituting into (M1)
\nOR
\nsubstituting into (M1)
\nTHEN
\n\n
A1
\n[3 marks]
\n(or equivalent) M1
\nwhen
\nor A1
\nEITHER
\nM1
\nOR
\nsubstituting at least two of into (or equivalent) M1
\nTHEN
\n(since ) A1
\n[4 marks]
\nA balloon in the shape of a sphere is filled with helium until the radius is 6 cm.
\nThe volume of the balloon is increased by 40%.
\nCalculate the volume of the balloon.
\nCalculate the radius of the balloon following this increase.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
Units are required in parts (a) and (b).
\n(M1)
\n\n
Note: Award (M1) for correct substitution into volume of sphere formula.
\n\n
(A1) (C2)
\n\n
Note: Answers derived from the use of approximations of (3.14; 22/7) are awarded (A0).
\n\n
[2 marks]
\nUnits are required in parts (a) and (b).
\nOR OR (M1)(M1)
\n\n
Note: Award (M1) for multiplying their part (a) by 1.4 or equivalent, (M1) for equating to the volume of a sphere formula.
\n\n
OR OR OR (M1)
\n\n
Note: Award (M1) for isolating .
\n\n
(A1)(ft) (C4)
\n\n
Note: Follow through from part (a).
\n\n
[4 marks]
\nConsider the function , where x > 0 and k is a constant.
\nThe graph of the function passes through the point with coordinates (4 , 2).
\nP is the minimum point of the graph of f (x).
\nFind the value of k.
\nUsing your value of k , find f ′(x).
\nUse your answer to part (b) to show that the minimum value of f(x) is −22 .
\nSketch the graph of y = f (x) for 0 < x ≤ 6 and −30 ≤ y ≤ 60.
Clearly indicate the minimum point P and the x-intercepts on your graph.
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
Note: Award (M1) for correct substitution of x = 4 and y = 2 into the function.
k = 3 (A1) (G2)
\n[2 marks]
\n(A1)(A1)(A1)(ft) (G3)
\nNote: Award (A1) for −48 , (A1) for x−2, (A1)(ft) for their 6x. Follow through from part (a). Award at most (A1)(A1)(A0) if additional terms are seen.
\n[3 marks]
\n(M1)
\nNote: Award (M1) for equating their part (b) to zero.
\nx = 2 (A1)(ft)
\nNote: Follow through from part (b). Award (M1)(A1) for seen.
\nAward (M0)(A0) for x = 2 seen either from a graphical method or without working.
\n(M1)
\nNote: Award (M1) for substituting their 2 into their function, but only if the final answer is −22. Substitution of the known result invalidates the process; award (M0)(A0)(M0).
\n−22 (AG)
\n[3 marks]
\n(A1)(A1)(ft)(A1)(ft)(A1)(ft)
Note: Award (A1) for correct window. Axes must be labelled.
(A1)(ft) for a smooth curve with correct shape and zeros in approximately correct positions relative to each other.
(A1)(ft) for point P indicated in approximately the correct position. Follow through from their x-coordinate in part (c). (A1)(ft) for two x-intercepts identified on the graph and curve reflecting asymptotic properties.
[4 marks]
\nPeter, the Principal of a college, believes that there is an association between the score in a Mathematics test, , and the time taken to run 500 m, seconds, of his students. The following paired data are collected.
\nIt can be assumed that follow a bivariate normal distribution with product moment correlation coefficient .
\nState suitable hypotheses and to test Peter’s claim, using a two-tailed test.
\nCarry out a suitable test at the 5 % significance level. With reference to the -value, state your conclusion in the context of Peter’s claim.
\nPeter uses the regression line of on as and calculates that a student with a Mathematics test score of 73 will have a running time of 101 seconds. Comment on the validity of his calculation.
\nA1
\nNote: It must be .
\n[1 mark]
\nA2
\nNote: Accept anything that rounds to 0.65
\n0.649 > 0.05 R1
\nhence, we accept and conclude that Peter’s claim is wrong A1
\nNote: The A mark depends on the R mark and the answer must be given in context. Follow through the -value in part (b).
\n[4 marks]
\na statement along along the lines of ‘(we have accepted that) the two variables are independent’ or ‘the two variables are weakly correlated’ R1
\na statement along the lines of ‘the use of the regression line is invalid’ or ‘it would give an inaccurate result’ R1
\nNote: Award the second R1 only if the first R1 is awarded.
\nNote: FT the conclusion in(a)(ii). If a candidate concludes that the claim is correct, mark as follows: (as we have accepted H1) the 2 variables are dependent and 73 lies in the range of values R1, hence the use of the regression line is valid R1.
\n[2 marks]
\nLet g(x) = −(x − 1)2 + 5.
\nLet f(x) = x2. The following diagram shows part of the graph of f.
\nThe graph of g intersects the graph of f at x = −1 and x = 2.
\nWrite down the coordinates of the vertex of the graph of g.
\nOn the grid above, sketch the graph of g for −2 ≤ x ≤ 4.
\nFind the area of the region enclosed by the graphs of f and g.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(1,5) (exact) A1 N1
\n[1 mark]
\n A1A1A1 N3
Note: The shape must be a concave-down parabola.
Only if the shape is correct, award the following for points in circles:
A1 for vertex,
A1 for correct intersection points,
A1 for correct endpoints.
[3 marks]
\nintegrating and subtracting functions (in any order) (M1)
eg
correct substitution of limits or functions (accept missing dx, but do not accept any errors, including extra bits) (A1)
eg
area = 9 (exact) A1 N2
\n[3 marks]
\nIn this question, give your answers to the nearest whole number.
\n
Criselda travelled to Kota Kinabalu in Malaysia. At the airport, she saw the following information at the Currency Exchange counter.
This means the Currency Exchange counter would buy from a traveller and in exchange return at a rate of . There is no commission charged.
\nCriselda changed to .
\nCalculate the amount of that Criselda received.
\nWhile in Kota Kinabalu, Criselda spent . She returned to the Currency Exchange counter and changed the remainder of her into .
\nCalculate the amount of she received.
\n (A1)(M1)
Note: Award (A1) for selecting as the exchange rate, (M1) for multiplying by an exchange rate from the table.
(A1) (C3)
Note: Do not award the final (A1) if the answer is to the wrong level of accuracy.
\n[3 marks]
\n (M1)(M1)
Note: Award (M1) for their correct subtraction or for or their correct difference seen. Award (M1) for dividing by . Follow through from part (a).
(A1)(ft) (C3)
Note: Do not award the final (A1) if the answer is to the wrong level of accuracy.
\n[3 marks]
\nA random variable has a distribution with mean and variance 4. A random sample of size 100 is to be taken from the distribution of .
\nJosie takes a different random sample of size 100 to test the null hypothesis that against the alternative hypothesis that at the 5 % level.
\nState the central limit theorem as applied to a random sample of size , taken from a distribution with mean and variance .
\nJack takes a random sample of size 100 and calculates that . Find an approximate 90 % confidence interval for .
\nFind the critical region for Josie’s test, giving your answer correct to two decimal places.
\nWrite down the probability that Josie makes a Type I error.
\nGiven that the probability that Josie makes a Type II error is 0.25, find the value of , giving your answer correct to three significant figures.
\nfor (sufficiently) large the sample mean approximately A1
\nA1
\nNote: Award the first A1 for large and reference to the sample mean , the second A1 is for normal and the two parameters.
\nNote: Award the second A1 only if the first A1 is awarded.
\nNote: Allow ‘ tends to infinity’ or ‘ ≥ 30’ in place of ‘large’.
\n[2 marks]
\n[59.9, 60.5] A1A1
\nNote: Accept answers which round to the correct 3sf answers.
\n[2 marks]
\nunder , (A1)
\nrequired to find such that (M1)
\nuse of any valid method, eg GDC Inv(Normal) or (M1)
\nhence critical region is A1
\n[4 marks]
\n0.05 A1
\n[1 mark]
\n(Type II error) = ( is accepted / is false) (R1)
\nNote: Accept Type II error means is accepted given is false.
\nwhen (M1)
\n(M1)
\nwhere
\n(A1)
\n\n
A1
\n[5 marks]
\nLet for 0 ≤ ≤ 1.5. The following diagram shows the graph of .
\nFind the x-intercept of the graph of .
\nThe region enclosed by the graph of , the y-axis and the x-axis is rotated 360° about the x-axis.
\nFind the volume of the solid formed.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
eg or 0…
1.14472
\n(exact), 1.14 A1 N2
\n[2 marks]
\nattempt to substitute either their limits or the function into formula involving . (M1)
\neg
\n2.49799
\nvolume = 2.50 A2 N3
\n[3 marks]
\nElvis Presley is an extremely popular singer. Although he passed away in , many of his fans continue to pay tribute by dressing like Elvis and singing his songs.
\nThe number of Elvis impersonators, , can be modelled by the function
\n\nwhere , is the number of years since .
\nWrite down the number of Elvis impersonators in .
\nCalculate the time taken for the number of Elvis impersonators to reach .
\nCalculate the number of Elvis impersonators when .
\nThe world population in is projected to be people.
\nUse this information to explain why the model for the number of Elvis impersonators is unrealistic.
\n (A1) (C1)
[1 mark]
\n (M1)
Note: Award (M1) for equating to the exponential function.
\n( (years)) (A1) (C2)
\n[2 marks]
\n (M1)
Note: Award (M1) for correct substitution in the function .
\n(A1) (C2)
\n[2 marks]
\nThe number of Elvis impersonators in , is greater than the world population. (R1) (C1)
\nOR
\n(R1) (C1)
\nNote: Award (R1) for a correct comparison of their number of impersonators with the world population. Follow through from part (c) if a reasonable argument can be made that the model is unrealistic.
Award (R0) if the number of impersonators is not explicitly seen in part (c) or in part (d).
[1 mark]
\nNote: In this question, distance is in metres and time is in seconds.
\n\n
A particle moves along a horizontal line starting at a fixed point A. The velocity of the particle, at time , is given by , for . The following diagram shows the graph of
\n![\"M17/5/MATME/SP2/ENG/TZ2/07\"](\"../assets/uploads/tinymce_asset/asset/3563/Schermafbeelding_2017-08-15_om_08.18.11.png\"/)
There are -intercepts at and .
\nFind the maximum distance of the particle from A during the time and justify your answer.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1 (displacement)
\nrecognizing (M1)
\nconsideration of displacement at and (seen anywhere) M1
\neg and
\n\n
Note: Must have both for any further marks.
\n\n
correct displacement at and (seen anywhere) A1A1
\n(accept 2.28318), 1.55513
\nvalid reasoning comparing correct displacements R1
\neg, more left than right
\n2.28 (m) A1 N1
\n\n
Note: Do not award the final A1 without the R1.
\n\n
METHOD 2 (distance travelled)
\nrecognizing distance (M1)
\nconsideration of distance travelled from to 2 and to 5 (seen anywhere) M1
\neg and
\n\n
Note: Must have both for any further marks
\n\n
correct distances travelled (seen anywhere) A1A1
\n2.28318, (accept ), 3.83832
\nvalid reasoning comparing correct distance values R1
\neg
\n2.28 (m) A1 N1
\n\n
Note: Do not award the final A1 without the R1.
\n\n
[6 marks]
\nThe weights, in grams, of individual packets of coffee can be modelled by a normal distribution, with mean and standard deviation .
\nFind the probability that a randomly selected packet has a weight less than .
\nThe probability that a randomly selected packet has a weight greater than grams is . Find the value of .
\nA packet is randomly selected. Given that the packet has a weight greater than , find the probability that it has a weight greater than .
\nFrom a random sample of packets, determine the number of packets that would be expected to have a weight lying within standard deviations of the mean.
\nPackets are delivered to supermarkets in batches of . Determine the probability that at least packets from a randomly selected batch have a weight less than .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(M1)A1
\n
[2 marks]
(M1)
\nA1
\n
[2 marks]
(M1)
\n(A1)
\n\nA1
\n
[3 marks]
EITHER
\n
(A1)
OR
(A1)
THEN
(M1)
A1
\n
[3 marks]
(A1)
\nrecognising (M1)
\nnow using (M1)
\nA1
\n
[4 marks]
Let , for . The graph of passes through the point , where .
\nFind the value of .
\nThe following diagram shows part of the graph of .
\n![\"N17/5/MATME/SP2/ENG/TZ0/05.b\"](\"../assets/uploads/tinymce_asset/asset/4166/Schermafbeelding_2018-02-12_om_13.30.18.png\"/)
The region enclosed by the graph of , the -axis and the lines and is rotated 360° about the -axis. Find the volume of the solid formed.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg, intersection with
\n2.32143
\n(exact), 2.32 A1 N2
\n[2 marks]
\nattempt to substitute either their limits or the function into volume formula (must involve , accept reversed limits and absence of and/or , but do not accept any other errors) (M1)
\neg
\n331.989
\nA2 N3
\n[3 marks]
\nA particle P starts from a point A and moves along a horizontal straight line. Its velocity after seconds is given by
\n\n
The following diagram shows the graph of .
\n![\"N16/5/MATME/SP2/ENG/TZ0/09\"](\"../assets/uploads/tinymce_asset/asset/3312/Schermafbeelding_2017-03-03_om_16.40.29.png\"/)
P is at rest when and .
\nWhen , the acceleration of P is zero.
\nFind the initial velocity of .
\nFind the value of .
\n(i) Find the value of .
\n(ii) Hence, find the speed of P when .
\n(i) Find the total distance travelled by P between and .
\n(ii) Hence or otherwise, find the displacement of P from A when .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nvalid attempt to substitute into the correct function (M1)
\neg
\n2 A1 N2
\n[2 marks]
\nrecognizing when P is at rest (M1)
\n5.21834
\nA1 N2
\n[2 marks]
\n(i) recognizing that (M1)
\neg, minimum on graph
\n1.95343
\nA1 N2
\n(ii) valid approach to find their minimum (M1)
\neg, reference to min on graph
\n1.75879
\nspeed A1 N2
\n[4 marks]
\n(i) substitution of correct into distance formula, (A1)
\neg
\n4.45368
\ndistance A1 N2
\n(ii) displacement from to (seen anywhere) (A1)
\neg
\ndisplacement from to (A1)
\neg
\nvalid approach to find displacement for M1
\neg
\n\n
displacement A1 N2
\n[6 marks]
\nThe number of marathons that Audrey runs in any given year can be modelled by a Poisson distribution with mean 1.3 .
\nCalculate the probability that Audrey will run at least two marathons in a particular year.
\nFind the probability that she will run at least two marathons in exactly four out of the following five years.
\n\n
(M1)A1
\n[2 marks]
\n(M1)A1
\nNote: Award (M1) for recognition of binomial or equivalent, A1 for correct parameters.
\n(M1)A1
\n[4 marks]
\nChristine and her friends live in Winnipeg, Canada. The weighted graph shows the location of their houses and the time, in minutes, to travel between each house.
\nChristine’s house is located at vertex .
\nUse Dijkstra’s algorithm to find the shortest time to travel from to , clearly showing how the algorithm has been applied.
\nHence write down the shortest path from to .
\nA new road is constructed that allows Christine to travel from to in minutes. If Christine starts from home and uses this new road her minimum travel time to is reduced, but her minimum travel time to remains the same.
\nFind the possible values of .
\nattempts to construct a graph or table to represent Dijkstra’s algorithm M1
\nEITHER
\nOR
\na clear attempt at Step 1 ( and considered) M1
\nSteps 2 and 3 correctly completed A1
\nStep 4 () correctly completed A1
\nSteps 5 and 6 ( and or or ) correctly completed A1
\nshortest time (mins) A1
\n
[6 marks]
A1
\nNote: Award A1 only if it is clear that Dijkstra’s algorithm has been attempted in part (a) (i). This A1 can be awarded if the candidate attempts to use Dijkstra’s algorithm but neglects to state (mins).
\n
[1 mark]
minimum travel time from to is reduced
\nis now (mins) (M1)
\nis still (mins)
\nso (A1)
\n
Note: Condone .
travel time from to remains the same ( mins)
is now (mins) (M1)
\n(A1)
\nso A1
\n
Note: Accept .
[5 marks]
Consider the curve y = 2x3 − 9x2 + 12x + 2, for −1 < x < 3
\nSketch the curve for −1 < x < 3 and −2 < y < 12.
\nA teacher asks her students to make some observations about the curve.
\nThree students responded.
Nadia said “The x-intercept of the curve is between −1 and zero”.
Rick said “The curve is decreasing when x < 1 ”.
Paula said “The gradient of the curve is less than zero between x = 1 and x = 2 ”.
State the name of the student who made an incorrect observation.
\nFind .
\nGiven that y = 2x3 − 9x2 + 12x + 2 = k has three solutions, find the possible values of k.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(A1)(A1)
Note: Award (A1) for correct window (condone a window which is slightly off) and axes labels. An indication of window is necessary. −1 to 3 on the x-axis and −2 to 12 on the y-axis and a graph in that window.
(A1) for correct shape (curve having cubic shape and must be smooth).
(A1) for both stationary points in the 1st quadrant with approximate correct position,
(A1) for intercepts (negative x-intercept and positive y intercept) with approximate correct position.
[4 marks]
\nRick (A1)
\nNote: Award (A0) if extra names stated.
\n[1 mark]
\n6x2 − 18x + 12 (A1)(A1)(A1)
\nNote: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if extra terms seen.
\n[3 marks]
\n6 < k < 7 (A1)(A1)(ft)(A1)
\nNote: Award (A1) for an inequality with 6, award (A1)(ft) for an inequality with 7 from their part (c) provided it is greater than 6, (A1) for their correct strict inequalities. Accept ]6, 7[ or (6, 7).
\n[3 marks]
\nComplete the following table by placing ticks (✓) to show which of the number sets , , , and these numbers belong to. The first row has been completed as an example.
\nNote: Award (A1) for each completely correct row, (A0) otherwise.
\n[6 marks]
\nA particle P starts from point O and moves along a straight line. The graph of its velocity, ms−1 after seconds, for 0 ≤ ≤ 6 , is shown in the following diagram.
\nThe graph of has -intercepts when = 0, 2 and 4.
\nThe function represents the displacement of P from O after seconds.
\nIt is known that P travels a distance of 15 metres in the first 2 seconds. It is also known that and .
\nFind the value of .
\nFind the total distance travelled in the first 5 seconds.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nrecognizing relationship between and (M1)
\neg ,
\nA1 N2
\n[2 marks]
\ncorrectly interpreting distance travelled in first 2 seconds (seen anywhere, including part (a) or the area of 15 indicated on diagram) (A1)
\neg ,
\nvalid approach to find total distance travelled (M1)
\neg sum of 3 areas, , shaded areas in diagram between 0 and 5
\nNote: Award M0 if only is seen.
\ncorrect working towards finding distance travelled between 2 and 5 (seen anywhere including within total area expression or on diagram) (A1)
\neg , , , ,
\nequal areas
correct working using (A1)
\neg , , , ,
\ntotal distance travelled = 33 (m) A1 N2
\n[5 marks]
\nJashanti is saving money to buy a car. The price of the car, in US Dollars (USD), can be modelled by the equation
\n\n
Jashanti’s savings, in USD, can be modelled by the equation
\n\n
In both equations is the time in months since Jashanti started saving for the car.
\nJashanti does not want to wait too long and wants to buy the car two months after she started saving. She decides to ask her parents for the extra money that she needs.
\nWrite down the amount of money Jashanti saves per month.
\nUse your graphic display calculator to find how long it will take for Jashanti to have saved enough money to buy the car.
\nCalculate how much extra money Jashanti needs.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
400 (USD) (A1) (C1)
\n[1 mark]
\n(M1)
\n\n
Note: Award (M1) for equating to or for comparing the difference between the two expressions to zero or for showing a sketch of both functions.
\n\n
(A1) (C2)
\n\n
Note: Accept 9 months.
\n\n
[2 marks]
\n(M1)(M1)
\n\n
Note: Award (M1) for correct substitution of into equation for , (M1) for finding the difference between a value/expression for and a value/expression for . The first (M1) is implied if 7671.25 seen.
\n\n
4870 (USD) (4871.25) (A1) (C3)
\n\n
Note: Accept 4871.3.
\n\n
[3 marks]
\nSila High School has 110 students. They each take exactly one language class from a choice of English, Spanish or Chinese. The following table shows the number of female and male students in the three different language classes.
\nA test was carried out at the 5 % significance level to analyse the relationship between gender and student choice of language class.
\nUse your graphic display calculator to write down
\nThe critical value at the 5 % significance level for this test is 5.99.
\nOne student is chosen at random from this school.
\nAnother student is chosen at random from this school.
\nWrite down the null hypothesis, H0 , for this test.
\nState the number of degrees of freedom.
\nthe expected frequency of female students who chose to take the Chinese class.
\nState whether or not H0 should be rejected. Justify your statement.
\nFind the probability that the student does not take the Spanish class.
\nFind the probability that neither of the two students take the Spanish class.
\nFind the probability that at least one of the two students is female.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(H0:) (choice of) language is independent of gender (A1)
\nNote: Accept “there is no association between language (choice) and gender”. Accept “language (choice) is not dependent on gender”. Do not accept “not related” or “not correlated” or “not influenced”.
\n[1 mark]
\n2 (AG)
\n[1 mark]
\n16.4 (16.4181…) (G1)
\n[1 mark]
\n(we) reject the null hypothesis (A1)(ft)
\n8.68507… > 5.99 (R1)(ft)
\nNote: Follow through from part (c)(ii). Accept “do not accept” in place of “reject.” Do not award (A1)(ft)(R0).
\nOR
\n(we) reject the null hypothesis (A1)
\n0.0130034 < 0.05 (R1)
\nNote: Accept “do not accept” in place of “reject.” Do not award (A1)(ft)(R0).
\n[2 marks]
\n(A1)(A1)(G2)
\nNote: Award (A1) for correct numerator, (A1) for correct denominator.
\n\n
[2 marks]
\n(M1)(M1)
\nNote: Award (M1) for multiplying two fractions. Award (M1) for multiplying their correct fractions.
\nOR
\n(M1)(M1)
\nNote: Award (M1) for correct products; (M1) for adding 4 products.
\n(A1)(ft)(G2)
\nNote: Follow through from their answer to part (e)(i).
\n[3 marks]
\n(M1)(M1)
\nNote: Award (M1) for multiplying two correct fractions. Award (M1) for subtracting their product of two fractions from 1.
\nOR
\n(M1)(M1)
\nNote: Award (M1) for correct products; (M1) for adding three products.
\n(A1)(G2)
\n[3 marks]
\nLet . The following diagram shows part of the graph of .
\nFind the -intercept of the graph of .
\nThe region enclosed by the graph of , the -axis and the -axis is rotated 360º about the -axis. Find the volume of the solid formed.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg ,
\n0.693147
\n= ln 2 (exact), 0.693 A1 N2
\n[2 marks]
\nattempt to substitute either their correct limits or the function into formula (M1)
\ninvolving
\neg , ,
\n3.42545
\nvolume = 3.43 A2 N3
\n[3 marks]
\nThe function is of the form , where , and are positive integers.
\nPart of the graph of is shown on the axes below. The graph of the function has its local maximum at and its local minimum at .
\n![\"M17/5/MATSD/SP1/ENG/TZ1/12\"](\"../assets/uploads/tinymce_asset/asset/3570/Schermafbeelding_2017-08-15_om_11.28.21.png\"/)
Write down the domain of the function.
\nDraw the line on the axes.
\nWrite down the number of solutions to .
\nFind the range of values of for which has no solution.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A2) (C2)
\n\n
Note: Accept equivalent notation. Award (A1)(A0) for .
\nAward (A1) for a clear statement that demonstrates understanding of the meaning of domain. For example, should be awarded (A1)(A0).
\n\n
[2 marks]
\n![\"M17/5/MATSD/SP1/ENG/TZ1/21.b.i/M\"](\"../assets/uploads/tinymce_asset/asset/3579/Schermafbeelding_2017-08-15_om_15.32.51.png\"/)
(A1) (C1)
\n
Note: The command term “Draw” states: “A ruler (straight edge) should be used for straight lines”; do not accept a freehand line.
\n\n
[1 mark]
\n2 (A1)(ft) (C1)
\n\n
Note: Follow through from part (b)(i).
\n\n
[1 mark]
\n(A1)(A1) (C2)
\n\n
Note: Award (A1) for both end points correct and (A1) for correct strict inequalities.
\nAward at most (A1)(A0) if the stated variable is different from or for example is (A1)(A0).
\n\n
[2 marks]
\nA particle moves along a straight line so that its velocity, m s−1, after seconds is given by , for 0 ≤ ≤ 5.
\nFind when the particle is at rest.
\nFind the acceleration of the particle when .
\nFind the total distance travelled by the particle.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg , sketch of graph
\n2.95195
\n(exact), (s) A1 N2
\n\n
[2 marks]
\nvalid approach (M1)
\neg ,
\n0.659485
\n= 1.96 ln 1.4 (exact), = 0.659 (m s−2) A1 N2
\n\n
[2 marks]
\ncorrect approach (A1)
\neg ,
\n5.3479
\ndistance = 5.35 (m) A2 N3
\n\n
[3 marks]
\nConsider the function ,
\nThe function , , models the path of a river, as shown on the following map, where both axes represent distance and are measured in kilometres. On the same map, the location of a highway is defined by the function .
\nThe origin, O(0, 0) , is the location of the centre of a town called Orangeton.
\nA straight footpath, , is built to connect the centre of Orangeton to the river at the point where .
\nBridges are located where the highway crosses the river.
\nA straight road is built from the centre of Orangeton, due north, to connect the town to the highway.
\nState the domain of .
\nFind the distance from the centre of Orangeton to the point at which the road meets the highway.
\nThis straight road crosses the highway and then carries on due north.
\nState whether the straight road will ever cross the river. Justify your answer.
\n0 < < (A1)(A1)
\nNote: Award (A1) for both endpoints correct, (A1) for correct mathematical notation indicating an interval with two endpoints. Accept weak inequalities. Award at most (A1)(A0) for incorrect notation such as 0 − 0.5 or a written description of the domain with correct endpoints. Award at most (A1)(A0) for 0 < < .
\n[2 marks]
\n(M1)
\n1.5 (km) (A1)(G2)
\n[2 marks]
\ndomain given as (but equation of road is ) (R1)
\nOR
\n(equation of road is ) the function of the river is asymptotic to (R1)
\nso it does not meet the river (A1)
\nNote: Award the (R1) for a correct mathematical statement about the equation of the river (and the equation of the road). Justification must be based on mathematical reasoning. Do not award (R0)(A1).
\n[2 marks]
\nConsider the function .
\nFind .
\nFind the gradient of the graph of at .
\nFind the equation of the tangent line to the graph of at . Give the equation in the form where, , , and .
\n(A1)(A1)(A1)
\nNote: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if there are extra terms.
\n[3 marks]
\n(M1)
\nNote: Award (M1) for correct substitution of 2 in their derivative of the function.
\n6 (A1)(ft)(G2)
\nNote: Follow through from part (d).
\n[2 marks]
\n(M1)
\nNote: Award (M1) for 2, their part (a) and their part (e) substituted into equation of a straight line.
\n\n
OR
\n(M1)
\nNote: Award (M1) for 2, their part (a) and their part (e) substituted into equation of a straight line.
\nOR
\n(M1)
\nNote: Award (M1) for their answer to (e) and intercept substituted in the gradient-intercept line equation.
\n(accept integer multiples) (A1)(ft)(G2)
\nNote: Follow through from parts (a) and (e).
\n[2 marks]
\nFarmer Brown has built a new barn, on horizontal ground, on his farm. The barn has a cuboid base and a triangular prism roof, as shown in the diagram.
\nThe cuboid has a width of 10 m, a length of 16 m and a height of 5 m.
The roof has two sloping faces and two vertical and identical sides, ADE and GLF.
The face DEFL slopes at an angle of 15° to the horizontal and ED = 7 m .
The roof was built using metal supports. Each support is made from five lengths of metal AE, ED, AD, EM and MN, and the design is shown in the following diagram.
\nED = 7 m , AD = 10 m and angle ADE = 15° .
M is the midpoint of AD.
N is the point on ED such that MN is at right angles to ED.
Farmer Brown believes that N is the midpoint of ED.
\nCalculate the area of triangle EAD.
\nCalculate the total volume of the barn.
\nCalculate the length of MN.
\nCalculate the length of AE.
\nShow that Farmer Brown is incorrect.
\nCalculate the total length of metal required for one support.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(Area of EAD =) (M1)(A1)
\nNote: Award (M1) for substitution into area of a triangle formula, (A1) for correct substitution. Award (M0)(A0)(A0) if EAD or AED is considered to be a right-angled triangle.
\n= 9.06 m2 (9.05866… m2) (A1) (G3)
\n[3 marks]
\n(10 × 5 × 16) + (9.05866… × 16) (M1)(M1)
\nNote: Award (M1) for correct substitution into volume of a cuboid, (M1) for adding the correctly substituted volume of their triangular prism.
\n= 945 m3 (944.938… m3) (A1)(ft) (G3)
\nNote: Follow through from part (a).
\n[3 marks]
\n(M1)
\nNote: Award (M1) for correct substitution into trigonometric equation.
\n(MN =) 1.29(m) (1.29409… (m)) (A1) (G2)
\n[2 marks]
\n(AE2 =) 102 + 72 − 2 × 10 × 7 × cos 15 (M1)(A1)
\nNote: Award (M1) for substitution into cosine rule formula, and (A1) for correct substitution.
\n(AE =) 3.71(m) (3.71084… (m)) (A1) (G2)
\n[3 marks]
\nND2 = 52 − (1.29409…)2 (M1)
\nNote: Award (M1) for correct substitution into Pythagoras theorem.
\n(ND =) 4.83 (4.82962…) (A1)(ft)
\nNote: Follow through from part (c).
\nOR
\n(M1)
\nNote: Award (M1) for correct substitution into tangent.
\n(ND =) 4.83 (4.82962…) (A1)(ft)
\nNote: Follow through from part (c).
OR
\n(M1)
\nNote: Award (M1) for correct substitution into cosine.
\n(ND =) 4.83 (4.82962…) (A1)(ft)
\nNote: Follow through from part (c).
\nOR
\nND2 = 1.29409…2 + 52 − 2 × 1.29409… × 5 × cos 75° (M1)
\nNote: Award (M1) for correct substitution into cosine rule.
\n(ND =) 4.83 (4.82962…) (A1)(ft)
\nNote: Follow through from part (c).
\n4.82962… ≠ 3.5 (ND ≠ 3.5) (R1)(ft)
\nOR
\n4.82962… ≠ 2.17038… (ND ≠ NE) (R1)(ft)
\n(hence Farmer Brown is incorrect)
\nNote: Do not award (M0)(A0)(R1)(ft). Award (M0)(A0)(R0) for a correct conclusion without any working seen.
\n[3 marks]
\n(EM2 =) 1.29409…2 + (7 − 4.82962…)2 (M1)
\nNote: Award (M1) for their correct substitution into Pythagoras theorem.
\nOR
\n(EM2 =) 52 + 72 − 2 × 5 × 7 × cos 15 (M1)
\nNote: Award (M1) for correct substitution into cosine rule formula.
\n(EM =) 2.53(m) (2.52689...(m)) (A1)(ft) (G2)(ft)
\nNote: Follow through from parts (c), (d) and (e).
\n(Total length =) 2.52689… + 3.71084… + 1.29409… +10 + 7 (M1)
\nNote: Award (M1) for adding their EM, their parts (c) and (d), and 10 and 7.
\n= 24.5 (m) (24.5318… (m)) (A1)(ft) (G4)
\nNote: Follow through from parts (c) and (d).
\n[4 marks]
\nThe diagram shows a circular horizontal board divided into six equal sectors. The sectors are labelled white (W), yellow (Y) and blue (B).
\nA pointer is pinned to the centre of the board. The pointer is to be spun and when it stops the colour of the sector on which the pointer stops is recorded. The pointer is equally likely to stop on any of the six sectors.
\nEva will spin the pointer twice. The following tree diagram shows all the possible outcomes.
\nFind the probability that both spins are yellow.
\nFind the probability that at least one of the spins is yellow.
\nWrite down the probability that the second spin is yellow, given that the first spin is blue.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
OR (M1)
\nNote: Award (M1) for multiplying correct probabilities.
\n(0.111, 0.111111…, 11.1%) (A1) (C2)
\n[2 marks]
\n(M1)(M1)
\nNote: Award (M1) for and or equivalent, and (M1) for and adding only the three correct probabilities.
\nOR
\n(M1)(M1)
\nNote: Award (M1) for seen and (M1) for subtracting from 1. This may be shown in a tree diagram with “yellow” and “not yellow” branches.
\n(0.556, 0.555555…, 55.6%) (A1)(ft) (C3)
\nNote: Follow through marks may be awarded if their answer to part (a) is used in a correct calculation.
\n[3 marks]
\n\n
(0.333, 0.333333…, 33.3%) (A1) (C1)
\n[1 mark]
\nThe graph of the quadratic function intersects the -axis at the point and has its vertex at the point .
\n![\"N16/5/MATSD/SP1/ENG/TZ0/09\"](\"../assets/uploads/tinymce_asset/asset/3321/Schermafbeelding_2017-03-06_om_09.57.03.png\"/)
Find the value of .
\n(M1)
\n\n
Note: Award (M1) for correct substitution into axis of symmetry formula.
\n\n
OR
\n(M1)
\n\n
Note: Award (M1) for correctly differentiating and equating to zero.
\n\n
OR
\n(or equivalent)
\n(or equivalent) (M1)
\n\n
Note: Award (M1) for correct substitution of and in the original quadratic function.
\n\n
(A1)(ft) (C2)
\n\n
Note: Follow through from part (a).
\n\n
[2 marks]
\nThe size of a computer screen is the length of its diagonal. Zuzana buys a rectangular computer screen with a size of 68 cm, a height of cm and a width of cm, as shown in the diagram.
\n![\"N17/5/MATSD/SP1/ENG/TZ0/06\"](\"../assets/uploads/tinymce_asset/asset/4171/Schermafbeelding_2018-02-12_om_18.05.15.png\")
The ratio between the height and the width of the screen is 3:4.
\nUse this information to write down an equation involving and .
\nUse this ratio to write down in terms of .
\nFind the value of and of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(or 4624 or equivalent) (A1) (C1)
\n[1 mark]
\n(M1)
\n\n
Note: Award (M1) for a correct equation.
\n\n
(A1) (C2)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution of their expression for into their answer to part (a). Accept correct substitution of in terms of .
\n\n
(A1)(ft)(A1)(ft) (C3)
\n\n
Note: Follow through from parts (a) and (b) as long as and .
\n\n
[3 marks]
\nThe Malthouse Charity Run is a kilometre race. The time taken for each runner to complete the race was recorded. The data was found to be normally distributed with a mean time of minutes and a standard deviation of minutes.
\nA runner who completed the race is chosen at random.
\nWrite down the probability that the runner completed the race in more than minutes.
\nCalculate the probability that the runner completed the race in less than minutes.
\nIt is known that of the runners took more than minutes and less than minutes to complete the race.
\nFind the value of .
\n(A1) (C1)
\n[1 mark]
\n(M1)
\n\n
Note: Award (M1) for a correct mathematical statement.
\nOR
Award (M1) for a diagram that shows the value labelled to the left of the mean and the correct shaded region.
\n
(A1) (C2)
\n[2 marks]
\nOR (seen) (A1)
\nNote: Award (A1) for or seen.
\n\n
OR (M1)
\nNote: Award (M1) for a correct mathematical statement.
OR
Award (M1) for a diagram that shows greater than the mean and shading in the region below , above , or between and the mean.
(minutes) (A1) (C3)
\nNote: Accept “ minutes and seconds” or (from sf value) “ minutes and seconds”.
\n[3 marks]
\nConsider the straight lines L1 and L2 . R is the point of intersection of these lines.
\nThe equation of line L1 is y = ax + 5.
\nThe equation of line L2 is y = −2x + 3.
\nFind the value of a.
\nFind the coordinates of R.
\nLine L3 is parallel to line L2 and passes through the point (2, 3).
\nFind the equation of line L3. Give your answer in the form y = mx + c.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
0 = 10a + 5 (M1)
\nNote: Award (M1) for correctly substituting any point from L1 into the equation.
\nOR
\n(M1)
\nNote: Award (M1) for correctly substituting any two points on L1 into the gradient formula.
\n(A1) (C2)
\n[2 marks]
\n(A1)(ft)(A1)(ft) (C2)
\nNote: Award (A1) for x-coordinate and (A1) for y-coordinate. Follow through from their part (a). Award (A1)(A0) if brackets are missing. Accept x = −1.33, y = 5.67.
\n[2 marks]
\n3 = −2(2) + c (M1)
\nNote: Award (M1) for correctly substituting –2 and the given point into the equation of a line.
\ny = −2x + 7 (A1) (C2)
\nNote: Award (A0) if the equation is not written in the form y = mx + c.
\n[2 marks]
\nConsider the function .
\nConsider a second function, .
\nCalculate .
\nSketch the graph of for and .
\nWrite down the equation of the vertical asymptote.
\nWrite down the coordinates of the -intercept.
\nWrite down the possible values of for which and .
\nFind the solution of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\n\n
Note: Award (M1) for correct substitution into function.
\n\n
(A1)(G2)
\n[2 marks]
\n![\"M17/5/MATSD/SP2/ENG/TZ1/03.b/M\"](\"../assets/uploads/tinymce_asset/asset/3603/Schermafbeelding_2017-08-16_om_14.01.31.png\"/)
(A1)(A1)(A1)(A1)
\n
Note: Award (A1) for indication of correct window and labelled axes.
\nAward (A1) for correct shape and position for (with the local maximum, local minimum and -intercept in relative approximate location in quadrant).
\nAward (A1) for correct shape and position for (with the local minimum in relative approximate location in quadrant).
\nAward (A1) for smooth curve with indication of asymptote (graph should not touch -axis and should not curve away from the -axis). The asymptote is only assessed in this mark.
\n\n
[4 marks]
\n(A2)
\n\n
Note: Award (A1) for “” and (A1) for “”.
\nThe answer must be an equation.
\n\n
[2 marks]
\n(A1)(A1)
\n\n
Note: Award (A1) for each correct coordinate. Award (A0)(A1) if parentheses are missing.
\n\n
[2 marks]
\n(A1)(A1)
\n\n
Note: Award (A1) for both correct end points, (A1) for strict inequalities used with 2 endpoints.
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for equating the expressions for and or for the line sketched (positive gradient, negative -intercept) on their graph from part (a).
\n\n
(A1)(G2)
\n\n
Note: Award a maximum of (M1)(A0) or (G1) for coordinate pair seen as final answer.
\n\n
[2 marks]
\nA manufacturer makes trash cans in the form of a cylinder with a hemispherical top. The trash can has a height of 70 cm. The base radius of both the cylinder and the hemispherical top is 20 cm.
\nA designer is asked to produce a new trash can.
\nThe new trash can will also be in the form of a cylinder with a hemispherical top.
\nThis trash can will have a height of H cm and a base radius of r cm.
\nThere is a design constraint such that H + 2r = 110 cm.
\nThe designer has to maximize the volume of the trash can.
\nWrite down the height of the cylinder.
\nFind the total volume of the trash can.
\nFind the height of the cylinder, h , of the new trash can, in terms of r.
\nShow that the volume, V cm3 , of the new trash can is given by
\n.
\nUsing your graphic display calculator, find the value of r which maximizes the value of V.
\nThe designer claims that the new trash can has a capacity that is at least 40% greater than the capacity of the original trash can.
\nState whether the designer’s claim is correct. Justify your answer.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n50 (cm) (A1)
\n[1 mark]
\n(M1)(M1)(M1)
\nNote: Award (M1) for their correctly substituted volume of cylinder, (M1) for correctly substituted volume of sphere formula, (M1) for halving the substituted volume of sphere formula. Award at most (M1)(M1)(M0) if there is no addition of the volumes.
\n(A1)(ft) (G3)
\nNote: Follow through from part (a).
\n[4 marks]
\nh = H − r (or equivalent) OR H = 110 − 2r (M1)
\nNote: Award (M1) for writing h in terms of H and r or for writing H in terms of r.
\n(h =) 110 − 3r (A1) (G2)
\n[2 marks]
\n(M1)(M1)(M1)
\nNote: Award (M1) for volume of hemisphere, (M1) for correct substitution of their h into the volume of a cylinder, (M1) for addition of two correctly substituted volumes leading to the given answer. Award at most (M1)(M1)(M0) for subsequent working that does not lead to the given answer. Award at most (M1)(M1)(M0) for substituting H = 110 − 2r as their h.
\n(AG)
\n[3 marks]
\n(r =) 31.4 (cm) (31.4285… (cm)) (G2)
\nOR
\n(M1)
\nNote: Award (M1) for setting the correct derivative equal to zero.
\n(r =) 31.4 (cm) (31.4285… (cm)) (A1)
\n[2 marks]
\n(M1)
\nNote: Award (M1) for correct substitution of their 31.4285… into the given equation.
\n= 114000 (113781…) (A1)(ft)
\nNote: Follow through from part (e).
\n(increase in capacity =) (R1)(ft)
\nNote: Award (R1)(ft) for finding the correct percentage increase from their two volumes.
\nOR
\n1.4 × 79587.0… = 111421.81… (R1)(ft)
\nNote: Award (R1)(ft) for finding the capacity of a trash can 40% larger than the original.
\nClaim is correct (A1)(ft)
\nNote: Follow through from parts (b), (e) and within part (f). The final (R1)(A1)(ft) can be awarded for their correct reason and conclusion. Do not award (R0)(A1)(ft).
\n[4 marks]
\nChicken eggs are classified by grade (, , , or ), based on weight. A mixed carton contains eggs and could include eggs from any grade. As part of the science project, Rocky buys mixed cartons and sorts the eggs according to their weight.
\nState whether the weight of the eggs is a continuous or discrete variable.
\nWrite down the modal grade of the eggs.
\nUse your graphic display calculator to find an estimate for the standard deviation of the weight of the eggs.
\nThe mean weight of these eggs is 64.9 grams, correct to three significant figures.
\nUse the table and your answer to part (c) to find the smallest possible number of eggs that could be within one standard deviation of the mean.
\ncontinuous (A1) (C1)
\n[1 mark]
\n(A1) (C1)
\nNote: Award (A0) for an answer of .
\n[1 mark]
\n(g) (A2) (C2)
\n[2 marks]
\nOR (M1)
\nNote: Award (M1) for correct endpoints seen. If the answer to part (c) is , award (M1) for endpoints of .
\n(A1)(ft) (C2)
\nNote: Follow through from their part (c). For a standard deviation between and inclusive, the FT answer is .
\n[2 marks]
\nIron in the asteroid 16 Psyche is said to be valued at quadrillion euros , where one quadrillion .
\nJames believes the asteroid is approximately spherical with radius . He uses this information to estimate its volume.
\nWrite down the value of the iron in the form where .
\nCalculate James’s estimate of its volume, in .
\nThe actual volume of the asteroid is found to be .
\nFind the percentage error in James’s estimate of the volume.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1) (C2)
\n\n
Note: Award (A1) for , (A1) for . Award (A1)(A0) for .
Award (A0)(A0) for answers of the type .
\n
[2 marks]
\n(M1)
\n
Note: Award (M1) for correct substitution in volume of sphere formula.
(A1) (C2)
[2 marks]
\n(M1)
\n
Note: Award (M1) for their correct substitution into the percentage error formula (accept a consistent absence of “” from all terms).
(A1)(ft) (C2)
Note: Follow through from their answer to part (b). If the final answer is negative, award at most (M1)(A0).
[2 marks]
On journeys to his office, Isaac noted whether or not it rained. He also recorded his journey time to the office, and classified each journey as short, medium or long.
\nOf the journeys to the office, there were short journeys when it rained, medium journeys when it rained, and long journeys when it rained. There were also short journeys when it did not rain.
\nIsaac carried out a test at the level of significance on these data, looking at the weather and the types of journeys.
\nWrite down , the null hypothesis for this test.
\nFind the expected number of short trips when it rained.
\nThe -value for this test is .
\nState the conclusion to Isaac’s test. Justify your reasoning.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
type of journey and whether it rained are independent (A1) (C1)
\nNote: Accept “there is no association” or “not dependent”. Do not accept “not related” or “not correlated”. Accept equivalent terms for ‘type of journey’.
\n
[1 mark]
OR (A1)(M1)
\n
Note: Award (A1) for or seen. Award (M1) for OR seen.
(A1) (C3)
[3 marks]
reject (do not accept) (A1)
\nOR
\ntype of journey and whether it rained are not independent (A1)
\n
Note: Follow through from part (a) for their phrasing of the null hypothesis.
(R1) (C2)
Note: A comparison must be seen, either numerically or in words (e.g. -value < significance level). Do not award (R0)(A1).
[2 marks]
The diagram shows the graph of the quadratic function , with vertex .
\nThe equation has two solutions. One of these solutions is .
\nWrite down the other solution of .
\nComplete the table below placing a tick (✔) to show whether the unknown parameters and are positive, zero or negative. The row for has been completed as an example.
\nState the values of for which is decreasing.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
OR (M1)
\n
Note: Award (M1) for correct calculation of the left symmetrical point.
(A1) (C2)
[2 marks]
(A1)(A1) (C2)
Note: Award (A1) for each correct row.
[2 marks]
OR (A1)(A1) (C2)
\n
Note: Award (A1) for seen as part of an inequality, (A1) for completely correct notation. Award (A1)(A1) for correct equivalent statement in words, for example “decreasing when is greater than negative ”.
[2 marks]
Casanova restaurant offers a set menu where a customer chooses one of the following meals: pasta, fish or shrimp.
\nThe manager surveyed customers and recorded the customer’s age and chosen meal. The data is shown in the following table.
\nA test was performed at the significance level. The critical value for this test is .
\nWrite down
\nA customer is selected at random.
\nState , the null hypothesis for this test.
\nWrite down the number of degrees of freedom.
\nShow that the expected number of children who chose shrimp is , correct to two significant figures.
\nthe statistic.
\nthe -value.
\nState the conclusion for this test. Give a reason for your answer.
\nCalculate the probability that the customer is an adult.
\nCalculate the probability that the customer is an adult or that the customer chose shrimp.
\nGiven that the customer is a child, calculate the probability that they chose pasta or fish.
\nchoice of meal is independent of age (or equivalent) (A1)
\nNote: Accept \"not associated\" or \"not dependent\" instead of independent. In lieu of \"age\", accept an equivalent alternative such as \"being a child or adult\".
\n[1 mark]
\n2 (A1)
\n[1 mark]
\nOR (M1)
\nNote: Award (M1) for correct substitution into expected frequency formula.
\n(A1)
\n(AG)
\nNote: Both an unrounded answer that rounds to the given answer and rounded answer must be seen for the (A1) to be awarded.
\n[2 marks]
\n(G2)
\n[2 marks]
\n(-value ) (G1)
\nNote: Award (G0)(G2) if the statistic is missing or incorrect and the -value is correct.
\n[1 mark]
\nOR (R1)(ft)
\nthe null hypothesis is not rejected (A1)(ft)
\nOR
\nthe choice of meal is independent of age (or equivalent) (A1)(ft)
\nNote: Award (R1)(ft)) for a correct comparison of either their statistic to the critical value or their -value to the significance level.
Condone “accept” in place of “not reject”.
Follow through from parts (a) and (d).
Do not award (A1)(ft)(R0).
\n[2 marks]
\n(A1)(A1)(G2)
\nNote: Award (A1) for numerator, (A1) for denominator.
\n[2 marks]
\n(A1)(A1)(G2)
\nNote: Award (A1) for numerator, (A1) for denominator.
\n[2 marks]
\n(A1)(A1)(G2)
\nNote: Award (A1) for numerator, (A1) for denominator.
\n[2 marks]
\nThe diagram shows the straight line . Points , and are points on .
\nis the midpoint of .
\nLine is perpendicular to and passes through point .
\nThe point is on .
\nFind the gradient of .
\nFind the coordinates of point .
\nFind the equation of . Give your answer in the form , where .
\nFind the value of .
\nFind the distance between points and .
\nGiven that the length of is , find the area of triangle .
\n(M1)
\nNote: Award (M1) for correct substitution into the gradient formula.
\n(A1)(G2)
\n[2 marks]
\nand (M1)
\nNote: Award (M1) for correct substitution into the midpoint formula for both coordinates.
\nOR
\n (M1)
Note: Award (M1) for a sketch showing the horizontal displacement from to is and the vertical displacement is and the coordinates at .
\nOR
\nand (M1)
\nNote: Award (M1) for correct equations seen.
\n(A1)(G1)(G1)
\nNote: Accept . Award at most (M1)(A0) or (G1)(G0) if parentheses are missing.
\n[2 marks]
\ngradient of the normal (A1)(ft)
\nNote: Follow through from their gradient from part (a).
\nOR (M1)
\nNote: Award (M1) for correct substitution of and their gradient of normal into straight line formula.
\n(accept integer multiples) (A1)(ft)(G3)
\n[3 marks]
\n(M1)
\nNote: Award (M1) for substitution of into their equation of normal line or substitution of and into equation of gradient of normal.
\n(A1)(ft)(G2)
\nNote: Follow through from part (c).
\n[2 marks]
\n(M1)
\nNote: Award (M1) for correctly substituting point and their into distance formula.
\n(A1)(ft)
\nNote: Follow through from part (d).
\n[2 marks]
\n(M1)
\nNote: Award (M1) for their correct substitution into area of a triangle formula. Award (M0) for their without any evidence of multiplication by to find length . Accept any other correct method to find the area.
\n(A1)(ft)(G2)
\nNote: Accept from use of a sf value for . Follow through from part (e).
\n[2 marks]
\nThe diagram shows the curve .
\nThe equation of the vertical asymptote of the curve is .
\nWrite down the value of .
\nFind .
\nAt the point where , the gradient of the tangent to the curve is .
\nFind the value of .
\n(A1) (C1)
\nNote: Award (A1) for an answer of \"\".
\n[1 mark]
\n(A1)(A1)(A1) (C3)
\nNote: Award (A1) for , (A1) for , (A1) for or . Award at most (A1)(A1)(A0) if extra terms are seen.
\n[3 marks]
\n(M1)
\nNote: Award (M1) for their correctly substituted derivative equated to .
\n(A1)(ft) (C2)
\nNote: Follow through from part (b) providing their answer is not as this value contradicts the graph.
\n[2 marks]
\nJean-Pierre jumps out of an airplane that is flying at constant altitude. Before opening his parachute, he goes through a period of freefall.
\nJean-Pierre’s vertical speed during the time of freefall, , in , is modelled by the following function.
\n\nwhere , is the number of seconds after he jumps out of the airplane, and is a constant. A sketch of Jean-Pierre’s vertical speed against time is shown below.
\nJean-Pierre’s initial vertical speed is .
\nFind the value of .
\nIn the context of the model, state what the horizontal asymptote represents.
\nFind Jean-Pierre’s vertical speed after seconds. Give your answer in .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\n
Note: Award (M1) for correctly substituted function equated to zero.
(A1) (C2)
[2 marks]
the (vertical) speed that Jean-Pierre is approaching (as increases) (A1) (C1)
OR
the limit of the (vertical) speed of Jean-Pierre (A1) (C1)
Note: Accept “maximum speed” or “terminal speed”.
[1 mark]
(M1)
Note: Award (M1) for correctly substituted function.
(A1)(ft)
Note: Follow through from part (a).
(A1)(ft) (C3)
Note: Award the final (A1)(ft) for correct conversion of their speed to .
[3 marks]
Maegan designs a decorative glass face for a new Fine Arts Centre. The glass face is made up of small triangular panes. The first three levels of the glass face are illustrated in the following diagram.
\nThe level, at the bottom of the glass face, has triangular panes. The level has triangular panes, and the level has triangular panes. Each additional level has more triangular panes than the level below it.
\nMaegan has triangular panes to build the decorative glass face and does not want it to have any incomplete levels.
\nFind the number of triangular panes in the level.
\nShow that the total number of triangular panes, , in the first levels is given by:
\n.
\nHence, find the total number of panes in a glass face with levels.
\nFind the maximum number of complete levels that Maegan can build.
\nEach triangular pane has an area of .
\nFind the total area of the decorative glass face, if the maximum number of complete levels were built. Express your area to the nearest .
\n(M1)(A1)
\nNote: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions.
\n(A1)(G3)
\n[3 marks]
\n(M1)(A1)
\nNote: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions.
\nOR (M1)
\nNote: Award (M1) for evidence of expansion and simplification, or division by leading to the final answer.
\n(AG)
\nNote: The final line must be seen, with no incorrect working, for the final (M1) to be awarded.
\n[3 marks]
\n(M1)
\nNote: Award (M1) for correctly substituted formula for .
\n(A1)
\nNote: The use of “hence” in the question paper means that the formula (from part (b)) must be used.
\n[2 marks]
\nOR (or equivalent) (M1)
\nNote: Award (M1) for equating to or for equating the correctly substituted sum of arithmetic sequence formula to .
\nOR
\na sketch of the graphs and intersecting (M1)
\nNote: Award (M1) for a sketch of a quadratic and a horizontal line with at least one point of intersection.
\nOR
\na sketch of intersecting the -axis (M1)
\nNote: Award (M1) for a sketch of with at least one -intercept.
\nOR (A1)
\nNote: Award (A1) for or seen. Can be implied by a correct final answer.
\n(A1)(ft)(G2)
\nNote: Do not accept . Award a maximum of (M1)(A1)(A0) if two final answers are given. Follow though from their unrounded answer.
\nOR
\nand (A2)
\nNote: Award (A2) for both “crossover” values seen. Do not split this (A2) mark.
\n(A1)(G2)
\n\n
[3 marks]
\n(M1)(M1)
\nNote: Award (M1) for their correct substitution to find the total number of triangular panes. Award (M1) for multiplying their number of panes by .
\nOR
\n(A1)(ft)(M1)
\nNote: Award (A1)(ft) for their seen. Award (M1) for multiplying their number of panes by . Follow through from part (d).
\n(A1)(ft)(G2)
\n(A1)(ft)(G3)
\n[4 marks]
\nHafizah harvested mangoes from her farm. The weights of the mangoes, , in grams, are shown in the following grouped frequency table.
\nWrite down the modal group for these data.
\nUse your graphic display calculator to find an estimate of the standard deviation of the weights of mangoes from this harvest.
\nOn the grid below, draw a histogram for the data in the table.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1) (C1)
\n
Note: Accept alternative notation or
Do not accept \"\".
[1 mark]
(A2) (C2)
\n
Note: Award (A1)(A0) for an answer of .
[2 marks]
(A2)(A1) (C3)
Note: Award (A2) for all correct heights of bars or (A1) for three or four correct heights of bars.
Award (A1) for rectangular bars all with correct left and right end points ( and ) and for no gaps; the bars do not have to be shaded.
Award at most (A2)(A0) if a ruler is not used for all lines.
[3 marks]
is a simple, connected, planar graph with vertices and edges.
\nThe complement of has edges.
\nFind the maximum possible value of .
\nFind an expression for in terms of .
\nGiven that the complement of is also planar and connected, find the possible values of .
\nis a simple graph with vertices and edges.
\nGiven that both and its complement are planar and connected, find the maximum possible value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
substitutes into either or (M1)
\nthe maximum number of edges is A1
\n
[2 marks]
has (A1)
\nso A1
\n
[2 marks]
(M1)
\n(the possible values are and ) A1
\n
[2 marks]
recognises that (or equivalent) (A1)
\nuses and M1
\nto form A1
\n
Note: Award A1 for .
attempts to solve their quadratic inequality (equality) (M1)
the maximum possible value of is A1
\n
[5 marks]
Stephen was invited to perform a piano recital. In preparation for the event, Stephen recorded the amount of time, in minutes, that he rehearsed each day for the piano recital.
\nStephen rehearsed for days and data for all these days is displayed in the following box-and-whisker diagram.
\nStephen states that he rehearsed on each of the days.
\nWrite down the median rehearsal time.
\nState whether Stephen is correct. Give a reason for your answer.
\nOn days, Stephen practiced exactly minutes.
\nFind the possible values of .
\n(minutes) (A1) (C1)
\n[1 mark]
\nStephen is correct. (A1)
\n the minimum rehearsal time is greater than zero (R1)
OR
he rehearsed at least minutes every day (R1) (C2)
Note: Do not award (A1)(R0). Accept equivalent reasoning based on the box-and-whisker diagram.
\n[2 marks]
\n(A1)(A1)(A1) (C3)
\nNote: Award (A1)(A1) for each correct endpoint of the interval, (A1) for indication of integer values, except 1, between their endpoints.
\n[3 marks]
\nConsider the graph of the function .
\nThe equation of the tangent to the graph of at is .
\nWrite down .
\nWrite down the gradient of this tangent.
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)(A1) (C3)
Note: Award (A1) for , (A1) for , and (A1) for or .
Award at most (A1)(A1)(A0) if additional terms are seen.
\n
[3 marks]
\n (A1) (C1)
[1 mark]
\n (M1)
Note: Award (M1) for equating their gradient from part (b) to their substituted derivative from part (a).
(A1)(ft) (C2)
Note: Follow through from parts (a) and (b).
[2 marks]
A shop sells carrots and broccoli. The weights of carrots can be modelled by a normal distribution with variance and the weights of broccoli can be modelled by a normal distribution with variance . The shopkeeper claims that the mean weight of carrots is and the mean weight of broccoli is .
\nDong Wook decides to investigate the shopkeeper’s claim that the mean weight of carrots is . He plans to take a random sample of carrots in order to calculate a confidence interval for the population mean weight.
\nAnjali thinks the mean weight, , of the broccoli is less than . She decides to perform a hypothesis test, using a random sample of size . Her hypotheses are
\n.
\nShe decides to reject if the sample mean is less than .
\nAssuming that the shopkeeper’s claim is correct, find the probability that the weight of six randomly chosen carrots is more than two times the weight of one randomly chosen broccoli.
\nFind the least value of required to ensure that the width of the confidence interval is less than .
\nFind the significance level for this test.
\nGiven that the weights of the broccoli actually follow a normal distribution with mean and variance , find the probability of Anjali making a Type II error.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
Let M1
\n(M1)(A1)
\n(M1)(A1)
\nA1
\n
Note: Condone the notation only if the (M1) is awarded for the variance.
[6 marks]
(A1)
\nM1
\n\n\nA1
\n
Note: Condone the use of equal signs.
[3 marks]
variance (A1)
\nunder
\nsignificance level (M1)
\nor A1
\n
Note: Accept any answer that rounds to or .
[3 marks]
Type II error probability (M1)
\n(A1)
\nA1
\nNote: Accept any answer that rounds to .
\n
[3 marks]
Consider the graph of the function .
\nWrite down the zero of .
\nWrite down the coordinates of the local minimum point.
\nConsider the function .
\nSolve .
\n(M1)
\n
Note: Award (M1) for equating the function to zero.
(A1) (C2)
Note: Award (C1) for a correct -value given as part of a coordinate pair or alongside an explicitly stated -value.
[2 marks]
(A1)(A1) (C2)
\n
Note: Accept .
[2 marks]
(or equivalent) (M1)
\n
Note: Award (M1) for equating the functions or for a sketch of the two functions.
(A1) (C2)
Note: Do not award the final (A1) if the answer is seen as part of a coordinate pair or a -value is explicitly stated, unless already penalized in part (a).
[2 marks]
M-Line is a company that prints and sells custom designs on T-shirts. For each order, they charge an initial design fee and then an additional fee for each printed T-shirt.
\nM-Line charges euros per order. This charge is modelled by the linear function , where is the number of T-shirts in the order.
\nEnYear is another company that prints and sells T-shirts. The price, euros, that they charge for an order can be modelled by the linear function , where is the number of T-shirts in the order.
\nWrite down the initial design fee charged for each order.
\nFind the total amount charged for an order of T-shirts.
\nWrite down the number of T-shirts in an order for which EnYear charged euros.
\nAn order of T-shirts will be charged the same price by both M-Line and EnYear.
\nFind the value of .
\n(euros) (A1) (C1)
\n[1 mark]
\n(M1)
\nNote: Award (M1) for correct substitution of into given function.
\n(euros) (A1) (C2)
\n[2 marks]
\n(T-shirts) (A1) (C1)
\n[1 mark]
\n(M1)
\nNote: Award (M1) for equating the given functions. Accept a sketch showing both functions.
\n(T-shirts) (A1) (C2)
\n[2 marks]
\nThe graph of the quadratic function intersects the -axis at .
\nThe vertex of the function is .
\nThe equation has two solutions. The first solution is .
\nLet be the tangent at .
\nFind the value of .
\nWrite down the equation for the axis of symmetry of the graph.
\nUse the symmetry of the graph to show that the second solution is .
\nWrite down the -intercepts of the graph.
\nOn graph paper, draw the graph of for and . Use a scale of to represent unit on the -axis and to represent units on the -axis.
\nWrite down the equation of .
\nDraw the tangent on your graph.
\nGiven and , state whether the function, , is increasing or decreasing at . Give a reason for your answer.
\nOR (or equivalent) (M1)
\nNote: Award (M1) for evaluating .
\n(A1)(G2)
\nNote: Award (G2) if or seen.
\n[2 marks]
\n(A1)(A1)
\nNote: Award (A1) for “ constant”, (A1) for the constant being . The answer must be an equation.
\n[2 marks]
\n(M1)
\nOR
\n(M1)
\nOR
\n(M1)
\nOR
\ndiagram showing axis of symmetry and given points (-values labels, , and , are sufficient) and an indication that the horizontal distances between the axis of symmetry and the given points are . (M1)
\nNote: Award (M1) for correct working using the symmetry between and . Award (M0) if candidate has used and to show the axis of symmetry is . Award (M0) if candidate solved or evaluated and .
\n(AG)
\n[1 mark]
\nand (A1)(A1)
\nNote: Accept and or and , award at most (A0)(A1) if parentheses are omitted.
\n[2 marks]
\n (A1)(A1)(A1)(A1)(ft)
Note: Award (A1) for labelled axes with correct scale, correct window. Award (A1) for the vertex, , in correct location.
Award (A1) for a smooth continuous curve symmetric about their vertex. Award (A1)(ft) for the curve passing through their and intercepts in correct location. Follow through from their parts (a) and (d).
If graph paper is not used:
Award at most (A0)(A0)(A1)(A1)(ft). Their graph should go through their and for the last (A1)(ft) to be awarded.
\n
[4 marks]
\nOR (A1)(A1)
\nNote: Award (A1) for \" constant\", (A1) for the constant being . The answer must be an equation.
\n[2 marks]
\ntangent to the graph drawn at (A1)(ft)
\nNote: Award (A1) for a horizontal straight-line tangent to curve at approximately . Award (A0) if a ruler is not used. Follow through from their part (e).
\n[1 mark]
\ndecreasing (A1)
\ngradient (of tangent line) is negative (at ) OR (R1)
\nNote: Do not accept \"gradient (of tangent line) is \". Do not award (A1)(R0).
\n[2 marks]
\nIn a class of students, play tennis, play both tennis and volleyball, and do not play either sport.
\nThe following Venn diagram shows the events “plays tennis” and “plays volleyball”. The values and represent numbers of students.
\nFind the value of .
\nFind the value of .
\nFind the probability that a randomly selected student from the class plays tennis or volleyball, but not both.
\nvalid approach to find (M1)
\neg
\n(may be seen on Venn diagram) A1 N2
\n[2 marks]
\nvalid approach to find (M1)
\neg
\n(may be seen on Venn diagram) A1 N2
\n[2 marks]
\nvalid approach (M1)
\neg , students, ,
A1 N2
\n[2 marks]
\nAndre will play in the semi-final of a tennis tournament.
\nIf Andre wins the semi-final he will progress to the final. If Andre loses the semi-final, he will not progress to the final.
\nIf Andre wins the final, he will be the champion.
\nThe probability that Andre will win the semi-final is . If Andre wins the semi-final, then the probability he will be the champion is .
\nThe probability that Andre will not be the champion is .
\nComplete the values in the tree diagram.
\nFind the value of .
\nGiven that Andre did not become the champion, find the probability that he lost in the semi-final.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1) (C1)
Note: Award (A1) for the correct pair of probabilities.
\n
[1 mark]
\n (M1)
Note: Award (M1) for multiplying and adding correct probabilities for losing equated to .
OR
(M1)
Note: Award (M1) for multiplying correct probabilities for winning equated to or .
(A1)(ft) (C2)
Note: Follow through from their part (a). Award the final (A1)(ft) only if their is within the range .
[2 marks]
(A1)(ft)(A1)
Note: Award (A1)(ft) for their correct numerator. Follow through from part (b). Award (A1) for the correct denominator.
OR
(A1)(ft)(A1)(ft)
Note: Award (A1)(ft) for their correct numerator. Follow through from part (b). Award (A1)(ft) for their correct calculation of Andre losing the semi-final or winning the semi-final and then losing in the final. Follow through from their parts (a) and (b).
(A1)(ft) (C3)
Note: Follow through from parts (a) and (b).
[3 marks]
In a school, all Mathematical Studies SL students were given a test. The test contained four questions, each one on a different topic from the syllabus. The quality of each response was classified as satisfactory or not satisfactory. Each student answered only three of the four questions, each on a separate answer sheet.
\nThe table below shows the number of satisfactory and not satisfactory responses for each question.
\n![\"M17/5/MATSD/SP2/ENG/TZ2/01\"](\"../assets/uploads/tinymce_asset/asset/3605/Schermafbeelding_2017-08-16_om_17.16.22.png\"/)
A test is carried out at the 5% significance level for the data in the table.
\nThe critical value for this test is 7.815.
\nIf the teacher chooses a response at random, find the probability that it is a response to the Calculus question;
\nIf the teacher chooses a response at random, find the probability that it is a satisfactory response to the Calculus question;
\nIf the teacher chooses a response at random, find the probability that it is a satisfactory response, given that it is a response to the Calculus question.
\nThe teacher groups the responses by topic, and chooses two responses to the Logic question. Find the probability that both are not satisfactory.
\nState the null hypothesis for this test.
\nShow that the expected frequency of satisfactory Calculus responses is 12.
\nWrite down the number of degrees of freedom for this test.
\nUse your graphic display calculator to find the statistic for this data.
\nState the conclusion of this test. Give a reason for your answer.
\n(A1)(A1)(G2)
\n\n
Note: Award (A1) for correct numerator, (A1) for correct denominator.
\n\n
[2 marks]
\n(A1)(A1)(G2)
\n\n
Note: Award (A1) for correct numerator, (A1) for correct denominator.
\n\n
[2 marks]
\n(A1)(A1)(G2)
\n\n
Note: Award (A1) for correct numerator, (A1) for correct denominator.
\n\n
[2 marks]
\n(A1)(M1)
\n\n
Note: Award (A1) for two correct fractions seen, (M1) for multiplying their two fractions.
\n\n
(A1)(G2)
\n[3 marks]
\n: quality (of response) and topic (from the syllabus) are independent (A1)
\n\n
Note: Accept there is no association between quality (of response) and topic (from the syllabus). Do not accept “not related” or “not correlated” or “influenced”.
\n\n
[1 mark]
\nOR (M1)
\n\n
Note: Award (M1) for correct substitution in expected value formula.
\n\n
(AG)
\n\n
Note: The conclusion, , must be seen for the (A1) to be awarded.
\n\n
[1 mark]
\n3 (A1)
\n[1 mark]
\n(G2)
\n[2 marks]
\nOR (R1)
\nthe null hypothesis is not rejected (A1)(ft)
\nOR
\nthe quality of the response and the topic are independent (A1)(ft)
\n\n
Note: Award (R1) for a correct comparison of either their statistic to the critical value or the correct -value 0.690688… to the test level, award (A1)(ft) for the correct result from that comparison. Accept “” for the comparison, but only if their value is explicitly seen in part (f). Follow through from their answers to part (f) and part (c). Do not award (R0)(A1).
\n\n
[2 marks]
\nSrinivasa places the nine labelled balls shown below into a box.
\nSrinivasa then chooses two balls at random, one at a time, from the box. The first ball is not replaced before he chooses the second.
\nFind the probability that the first ball chosen is labelled .
\nFind the probability that the first ball chosen is labelled or labelled .
\nFind the probability that the second ball chosen is labelled , given that the first ball chosen was labelled .
\nFind the probability that both balls chosen are labelled .
\n(A1) (C1)
\n
[1 mark]
(A1) (C1)
\n
[1 mark]
(A1)(A1) (C2)
\n
Note: Award (A1) for correct numerator, (A1) for correct denominator.
[2 marks]
(M1)
\n
Note: Award (M1) for a correct compound probability calculation seen.
(A1) (C2)
[2 marks]
Haraya owns two triangular plots of land, and . The length of is , is and is . The size of is and is .
\nThe following diagram shows this information.
\nHaraya attaches a long rope to a vertical pole at point .
\nFind the length of .
\nFind the size of .
\nCalculate the area of the triangular plot of land .
\nDetermine whether the rope can extend into the triangular plot of land, . Justify your answer.
\n(or equivalent) (A1)
\nNote: Award (A1) for (or equivalent) seen.
\n(M1)(A1)
\nNote: Award (M1) for substitution into sine rule formula, (A1) for correct substitution.
\nOR
\n(A1)(M1)(A1)
\nNote: Award (A1) for or seen, (M1) for substitution into cosine rule formula, (A1) for correct substitution.
\n(A1)(G3)
\n[4 marks]
\n(M1)(A1)
\nNote: Award (M1) for substitution into cosine rule formula, (A1) for correct substitution.
\n(A1)(G2)
\n[3 marks]
\nUnits are required in part (c)
\n\n
(M1)(A1)(ft)
\nNote: Award (M1) for substitution into the area formula, (A1)(ft) for correct substitution. Award (M0)(A0)(A0) for .
\n(A1)(ft)(G2)
\nNote: Follow through from part (b).
\n[3 marks]
\nMETHOD 1 (equating part (c) to expression for area of triangle ABC)
\n(M1)(A1)(ft)
\nNote: Award (M1) for correctly substituted area of triangle formula. Award (A1)(ft) for equating the area formula to their area found in part (c).
\n(A1)(ft)
\nNote: Follow through from their part (c).
\n(R1)(ft)
\nNote: Accept “the length of the rope is greater than the altitude of triangle ”.
\nthe rope passes inside the triangular plot of land (A1)(ft)
\nNote: Follow through from their altitude. The final (A1) is contingent on (R1) being awarded.
\n\n
METHOD 2 (finding or with sine rule and then trig ratio)
\n(M1)
\nNote: Award (M1) for their correct substitution into sine rule formula to find or . Follow through from their part (b).
\n(M1)
\nNote: Award (M1) for correct substitution of their or into trig formula.
\n(A1)(ft)
\nNote: Follow through from their part (b).
\n(R1)(ft)
\nNote: Accept “the length of the rope is greater than the altitude of triangle ”.
\nthe rope passes inside the triangular plot of land (A1)(ft)
\nNote: Follow through from their altitude. The final (A1) is contingent on (R1) being awarded.
\n\n
METHOD 3 (finding or with with cosine rule and then trig ratio)
\n(M1)
\nNote: Award (M1) for for their correct substitution into cosine rule formula to find or .
\n(M1)
\nNote: Award (M1) for correct substitution of their or into trig formula.
\n(A1)(ft)
\n(R1)(ft)
\nNote: Accept “the length of the rope is greater than the altitude of triangle ”.
\nthe rope passes inside the triangular plot of land (A1)(ft)
\nNote: Follow through from their altitude. The final (A1) is contingent on (R1) being awarded.
\n\n
METHOD 4 (finding area of triangle with height , justifying the contradiction)
\n(M1)(A1)
\nNote: Award (M1) for correct substitution into area of a triangle formula for a triangle with height and base . Award (A1) for . Award (M0)(A0) for unsupported unless subsequent reasoning explains how the was found.
\n(R1)
\nif rope exactly touches the then this triangle has an area greater than
and as the distance is fixed the altitude must be less than (R1)
OR
\n(height perpendicular to ) and therefore height perpendicular to (R1)(ft)
\nNote: Award (R1) for an explanation that recognizes the actual triangle and this new triangle have the same base and hence the height of triangle is less than .
\ntherefore, the rope passes inside the triangular plot of land (A1)(ft)
\n\n
Note: Other methods, besides those listed here, may be possible. These methods can be summarized in two broad groups: the first is to find the altitude of the triangle, and compare it to , and the second is to create an artificial triangle with an altitude of and explain why this triangle is not by relating to area and the given lengths of the sides.
\n\n
[5 marks]
\nThe following diagram shows part of the graph of , for .
\nLet be any point on the graph of . Line is the tangent to the graph of at .
\nLine intersects the -axis at point and the -axis at point B.
\nFind in terms of and .
\nShow that the equation of is .
\nFind the area of triangle in terms of .
\nThe graph of is translated by to give the graph of .
In the following diagram:
Line is the tangent to the graph of at , and passes through and .
\nGiven that triangle and rectangle have equal areas, find the gradient of in terms of .
\n(A1)
\nA1 N2
\n[2 marks]
\nattempt to use point and gradient to find equation of M1
\neg
\ncorrect working leading to answer A1
\neg
\nAG N0
\n[2 marks]
\nMETHOD 1 – area of a triangle
\nrecognizing at (M1)
\ncorrect working to find -coordinate of null (A1)
\neg
\n-coordinate of null at (may be seen in area formula) A1
\ncorrect substitution to find area of triangle (A1)
\neg
\narea of triangle A1 N3
\n\n
METHOD 2 – integration
\nrecognizing to integrate between and (M1)
\neg
\ncorrect integration of both terms A1
\neg
\nsubstituting limits into their integrated function and subtracting (in either order) (M1)
\neg
\ncorrect working (A1)
\neg
\narea of triangle A1 N3
\n\n
[5 marks]
\nNote: In this question, the second M mark may be awarded independently of the other marks, so it is possible to award (M0)(A0)M1(A0)(A0)A0.
\n\n
recognizing use of transformation (M1)
\neg area of triangle = area of triangle , gradient of gradient of , one correct shift
\ncorrect working (A1)
\neg area of triangle
\ngradient of area of rectangle
\nvalid approach (M1)
\neg
\ncorrect working (A1)
\neg
\ncorrect expression for gradient (in terms of ) (A1)
\neg
\ngradient of is A1 N3
\n[6 marks]
\nLet the universal set, , be the set of all integers such that ≤ < .
, and are subsets of .
Write down .
\nComplete the following Venn diagram using all elements of .
\nWrite down an element that belongs to .
\n(A1) (C1)
\n[1 mark]
\n (A1)(A1)(A1)(A1) (C4)
Note: Award (A1) for in the correct place. Award (A1) for , , and in the correct places. Award (A1) for , , , in the correct places. Award (A1) for outside of the three circles and not shown in the diagram.
\nIf any entry is duplicated within its region, award at most (A3).
\n[4 marks]
\n9 (A1)(ft) (C1)
\nNote: Award (A1) for the correct element. Follow through from their Venn diagram in part (b). Award (A0) if additional incorrect elements are included in their answer.
\n[1 mark]
\nMia baked a very large apple pie that she cuts into slices to share with her friends. The smallest slice is cut first. The volume of each successive slice of pie forms a geometric sequence.
\nThe second smallest slice has a volume of . The fifth smallest slice has a volume of .
\nFind the common ratio of the sequence.
\nFind the volume of the smallest slice of pie.
\nThe apple pie has a volume of .
\nFind the total number of slices Mia can cut from this pie.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
and (M1)
Note: Award (M1) for both the given terms expressed in the formula for .
OR
(M1)
Note: Award (M1) for a correct equation seen.
(A1) (C2)
[2 marks]
OR (M1)
Note: Award (M1) for their correct substitution in geometric sequence formula.
(A1)(ft) (C2)
Note: Follow through from part (a).
[2 marks]
(M1)
Note: Award (M1) for correctly substituted geometric series formula equated to .
(slices) (A1)(ft) (C2)
Note: Follow through from parts (a) and (b).
[2 marks]
The base of an electric iron can be modelled as a pentagon ABCDE, where:
\n\n
![\"M17/5/MATSD/SP2/ENG/TZ1/02\"](\"../assets/uploads/tinymce_asset/asset/3599/Schermafbeelding_2017-08-16_om_11.05.55.png\"/)
Insulation tape is wrapped around the perimeter of the base of the iron, ABCDE.
\nF is the point on AB such that . A heating element in the iron runs in a straight line, from C to F.
\nWrite down an equation for the area of ABCDE using the above information.
\nShow that the equation in part (a)(i) simplifies to .
\nFind the length of CD.
\nShow that angle , correct to one decimal place.
\nFind the length of the perimeter of ABCDE.
\nCalculate the length of CF.
\n(M1)(M1)(A1)
\n\n
Note: Award (M1) for correct area of triangle, (M1) for correct area of rectangle, (A1) for equating the sum to 222.
\n\n
OR
\n(M1)(M1)(A1)
\n\n
Note: Award (M1) for area of bounding rectangle, (M1) for area of triangle, (A1) for equating the difference to 222.
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for complete expansion of the brackets, leading to the final answer, with no incorrect working seen. The final answer must be seen to award (M1).
\n\n
(AG)
\n[2 marks]
\n(A1)
\n(A1)(G2)
\n[2 marks]
\n(A1)(ft)
\n\n
Note: Follow through from part (b).
\n\n
(M1)
\n\n
Note: Award (M1) for their correct substitutions in tangent ratio.
\n\n
(A1)
\n(AG)
\n\n
Note: Do not award the final (A1) unless both the correct unrounded and rounded answers are seen.
\n\n
OR
\n(A1)(ft)
\n(M1)
\n\n
Note: Award (M1) for their correct substitutions in tangent ratio.
\n\n
\n
(A1)
\n(AG)
\n\n
Note: Do not award the final (A1) unless both the correct unrounded and rounded answers are seen.
\n\n
[3 marks]
\n(M1)(M1)
\n\n
Note: Award (M1) for correct substitution into Pythagoras. Award (M1) for the addition of 5 sides of the pentagon, consistent with their .
\n\n
(A1)(ft)(G3)
\n\n
Note: Follow through from part (b).
\n\n
[3 marks]
\n(M1)
\nOR
\n(M1)
\n(M1)(A1)(ft)
\n\n
Note: Award (M1) for substituted cosine rule formula and (A1) for correct substitutions. Follow through from part (b).
\n\n
(A1)(ft)(G3)
\nOR
\n(A1)
\n\n
Note: Award (A1) for angle , where G is the point such that CG is a projection/extension of CB and triangles BGF and CGF are right-angled triangles. The candidate may use another variable.
\n![\"M17/5/MATSD/SP2/ENG/TZ1/02.e/M\"](\"../assets/uploads/tinymce_asset/asset/3602/Schermafbeelding_2017-08-16_om_13.44.50.png\"/)
\n
AND (M1)
\n\n
Note: Award (M1) for correct substitution into trig formulas to find both GF and BG.
\n\n
(M1)
\n\n
Note: Award (M1) for correct substitution into Pythagoras formula to find CF.
\n\n
(A1)(ft)(G3)
\n[4 marks]
\nSungwon plays a game where she rolls a fair -sided die and spins a fair spinner with equal sectors. During each turn in the game, the die is rolled once and the spinner is spun once. The score for each turn is the sum of the two results. For example, on the die and on the spinner would receive a score of .
\nThe following diagram represents the sample space.
\nSungwon takes a second turn.
\nFind the probability that Sungwon’s score on her first turn is greater than .
\nFind the probability that Sungwon scores greater than on both of her first two turns.
\nSungwon will play the game for turns.
\nFind the expected number of times the score on a turn is greater than .
\n (A1)(A1) (C2)
Note: Award (A1) for correct numerator, (A1) for correct denominator.
[2 marks]
\n (M1)
Note: Award (M1) for the square of their probability in part (a).
(A1)(ft) (C2)
Note: Follow through from part (a), provided their answer is less than or equal to .
\n[2 marks]
\n (M1)
Note: Award (M1) for multiplying their part (a) by .
(A1)(ft) (C2)
Note: Follow through from part (a), provided their answer is less than or equal to .
\n[2 marks]
\nThe following diagram shows a triangle .
\n, and .
\nLet .
\nGiven that is acute, find .
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1 – (sine rule)
\nevidence of choosing sine rule (M1)
\neg
\ncorrect substitution (A1)
\neg
\nA1 N2
\n\n
METHOD 2 – (perpendicular from vertex )
\nvalid approach to find perpendicular length (may be seen on diagram) (M1)
\neg ,
correct perpendicular length (A1)
\neg
\nA1 N2
\n\n
Note: Do not award the final A mark if candidate goes on to state , as this demonstrates a lack of understanding.
\n\n
[3 marks]
\nattempt to substitute into double-angle formula for cosine (M1)
\n\ncorrect working (A1)
\neg
\nA1 N2
\n[3 marks]
\nAnne-Marie planted four sunflowers in order of height, from shortest to tallest.
\nFlower is tall.
\nThe median height of the flowers is .
\nThe range of the heights is . The height of Flower is and the height of Flower is .
\nThe mean height of the flowers is .
\nFind the height of Flower .
\nUsing this information, write down an equation in and .
\nWrite down a second equation in and .
\nUsing your answers to parts (b) and (c), find the height of Flower .
\nUsing your answers to parts (b) and (c), find the height of Flower .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nOR OR (M1)
\n
Note: Award (M1) for subtracting from the median, or equivalent.
(A1) (C2)
[2 marks]
(or equivalent) (A1) (C1)
\n
[1 mark]
OR (or equvalent) (A1)(ft) (C1)
\n
Note: Follow through from part (a).
[1 mark]
(A1)(ft) (C1)
\n
Note: Follow through from parts (b) and (c).
[1 mark]
(A1)(ft) (C1)
\n
Note: Follow through from parts (b) and (c).
[1 mark]
Let . The following diagram shows part of the graph of .
\nThe shaded region is enclosed by the graph of , the -axis and the -axis.
\nThe graph of intersects the -axis at the point .
\nFind the value of .
\nFind the volume of the solid formed when the shaded region is revolved about the -axis.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognize (M1)
\neg
\n(accept , ) A1 N2
\n[2 marks]
\nattempt to substitute either their limits or the function into volume formula (must involve ) (M1)
\neg
\ncorrect integration of each term A1 A1
\neg
\nsubstituting limits into their integrated function and subtracting (in any order) (M1)
\neg
\n\n
Note: Award M0 if candidate has substituted into or .
\n\n
volume A1 N2
\n[5 marks]
\nOlava’s Pizza Company supplies and delivers large cheese pizzas.
\nThe total cost to the customer, , in Papua New Guinean Kina (), is modelled by the function
\n\nwhere , is the number of large cheese pizzas ordered. This total cost includes a fixed cost for delivery.
\nState, in the context of the question, what the value of represents.
\nState, in the context of the question, what the value of represents.
\nWrite down the minimum number of pizzas that can be ordered.
\nKaelani has .
\nFind the maximum number of large cheese pizzas that Kaelani can order from Olava’s Pizza Company.
\nthe cost of each (large cheese) pizza / a pizza / one pizza / per pizza (A1) (C1)
Note: Award (A0) for “the cost of (large cheese) pizzas”. Do not accept “the minimum cost of a pizza”.
[1 mark]
the (fixed) delivery cost (A1) (C1)
[1 mark]
(A1) (C1)
[1 mark]
(M1)
\nNote: Award (M1) for equating the cost equation to (may be stated as an inequality).
\n
(A1)
(A1)(ft) (C3)
\n
Note: The final answer must be an integer.
The final (A1)(ft) is awarded for rounding their answer down to a whole number, provided their unrounded answer is seen.
[3 marks]
Yao drains the oil from his motorbike into two identical cuboids with rectangular bases of width cm and length cm. The height of each cuboid is cm.
\nThe oil from the motorbike fills the first cuboid completely and the second cuboid to a height of cm. The information is shown in the following diagram.
\nCalculate the volume of oil drained from Yao’s motorbike.
\nYao then pours all the oil from the cuboids into an empty cylindrical container. The height of the oil in the container is cm.
\nFind the internal radius, , of the container.
\nunits are required in both parts
\n\n
(M1)(M1)
\nNote: Award (M1) for correct substitutions in volume formula for both cuboids. Award (M1) for adding the volumes of both cuboids.
\ncm3 (A1) (C3)
\n[3 marks]
\nunits are required in both parts
\n\n
(M1)(M1)
\nNote: Award (M1) for correct substitution in volume of cylinder formula. Award (M1) for equating their expression (must include and ) to their .
\ncm (… cm) (A1)(ft) (C3)
\nNote: Follow through from their part (a).
\n[3 marks]
\nThe Maxwell Ohm Company is designing a portable Bluetooth speaker. The speaker is in the shape of a cylinder with a hemisphere at each end of the cylinder.
\nThe dimensions of the speaker, in centimetres, are illustrated in the following diagram where is the radius of the hemisphere, and is the length of the cylinder, with and .
\nThe Maxwell Ohm Company has decided that the speaker will have a surface area of .
\nThe quality of sound from the speaker will improve as increases.
\nWrite down an expression for , the volume (cm3) of the speaker, in terms of , and .
\nWrite down an equation for the surface area of the speaker in terms of , and .
\nGiven the design constraint that , show that .
\nFind .
\nUsing your answer to part (d), show that is a maximum when is equal to .
\nFind the length of the cylinder for which is a maximum.
\nCalculate the maximum value of .
\nUse your answer to part (f) to identify the shape of the speaker with the best quality of sound.
\n(or equivalent) (A1)(A1)
\nNote: Award (A1) for either the volume of a hemisphere formula multiplied by or the volume of a cylinder formula, and (A1) for completely correct expression. Accept equivalent expressions. Award at most (A1)(A0) if is used instead of .
\n[2 marks]
\n(A1)(A1)(A1)
\nNote: Award (A1) for the surface area of a hemisphere multiplied by . Award (A1) for the surface area of a cylinder. Award (A1) for the addition of their formulas equated to . Award at most (A1)(A1)(A0) if is used instead of , unless already penalized in part (a).
\n[3 marks]
\n(M1)
\nNote: Award (M1) for their correctly substituted formula for .
\n(M1)
\nNote: Award (M1) for correct expansion of brackets and simplification of the cylinder expression in leading to the final answer.
\n(AG)
\nNote: The final line must be seen, with no incorrect working, for the second (M1) to be awarded.
\n[2 marks]
\n(A1)(A1)
\nNote: Award (A1) for . Award (A1) for . Award maximum (A1)(A0) if extra terms seen.
\n[2 marks]
\nOR OR sketch of with -intercept indicated (M1)
\nNote: Award (M1) for equating their derivative to zero or a sketch of their derivative with -intercept indicated.
\nOR (A1)
\n(AG)
\nNote: The (AG) line must be seen for the preceding (A1) to be awarded.
\n[2 marks]
\n(M1)
\nNote: Award (M1) for correct substitution in the given formula for the length of the cylinder.
\n(A1)(G2)
\nNote: Award (M1)(A1) for correct substitution of the sf approximation leading to a correct answer of zero.
\n[2 marks]
\nOR (M1)
\nNote: Award (M1) for correct substitution in the formula for the volume of the speaker or the volume of a sphere.
\n(A1)(G2)
\nNote: Accept from use of sf value of . Award (M1)(A1)(ft) for correct substitution in their volume of speaker. Follow through from parts (a) and (f).
\n[2 marks]
\nsphere (spherical) (A1)(ft)
\nNote: Question requires the use of part (f) so if there is no answer to part (f), part (h) is awarded (A0). Follow through from their .
\n[1 mark]
\nGalois Airways has flights from Hong Kong International Airport to different destinations. The following table shows the distance, kilometres, between Hong Kong and the different destinations and the corresponding airfare, , in Hong Kong dollars (HKD).
\nThe Pearson’s product–moment correlation coefficient for this data is , correct to three significant figures.
\nThe distance from Hong Kong to Tokyo is km.
\nUse your graphic display calculator to find the equation of the regression line on .
\nUse your regression equation to estimate the cost of a flight from Hong Kong to Tokyo with Galois Airways.
\nExplain why it is valid to use the regression equation to estimate the airfare between Hong Kong and Tokyo.
\n\n
(A1)(A1) (C2)
\nNote: Award (A1) for , (A1) for . If the answer is not given as an equation, award a maximum of (A1)(A0).
\n[2 marks]
\n(M1)
\nNote: Award (M1) for substitution into their regression equation.
\n(, ) (HKD) (A1)(ft) (C2)
\nNote: Follow through from part (a).
\n[2 marks]
\nthe correlation is (very) strong (R1)
\n(km) is within the given data range (interpolation) (R1) (C2)
\nNote: Two correct reasons are required for the awarding of (C2).
\n[2 marks]
\nThe graph of a function passes through the point .
\nGiven that , find .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of integration (M1)
\neg
\ncorrect integration (accept missing ) (A1)
\neg
\nsubstituting initial condition into their integrated expression (must have ) M1
\neg
\n\n
Note: Award M0 if candidate has substituted into or .
\n\n
correct application of rule (seen anywhere) (A1)
\neg
\ncorrect application of rule (seen anywhere) (A1)
\neg
\ncorrect working (A1)
\neg
\nA1 N4
\n\n
[7 marks]
\nThere are four stations used by the fire wardens in a national forest.
\nOn the following Voronoi diagram, the coordinates of the stations are and where distances are measured in kilometres.
\nThe dotted lines represent the boundaries of the regions patrolled by the fire warden at each station. The boundaries meet at and .
\nTo reduce the areas of the regions that the fire wardens patrol, a new station is to be built within the quadrilateral . The new station will be located so that it is as far as possible from the nearest existing station.
\nThe Voronoi diagram is to be updated to include the region around the new station at . The edges defined by the perpendicular bisectors of and have been added to the following diagram.
\nShow that the new station should be built at .
\nWrite down the equation of the perpendicular bisector of .
\nHence draw the missing boundaries of the region around on the following diagram.
\n(the best placement is either point or point )
attempt at using the distance formula (M1)
OR
\nOR
\nOR
\nOR
\nOR
\n\n
( or or ) AND
\n( or or ) A1
\nOR (or or ) is greater than (or or ) A1
\npoint is the furthest away AG
\n
Note: Follow through from their values provided their (or or ) is greater than their (or or ).
\n
[3 marks]
\nA1
\n\n
[1 mark]
\n A1A1
Note: Award A1 for each correct straight line. Do not FT from their part (b)(i).
[1 mark]
\nIn part (a) many candidates realized that distances were required. Many candidates seemed to have an idea about Voronoi diagrams. However, several candidates did not realize that they had to consider point as well in their comparison. Hence, several candidates only calculated distances from . The numerical comparison of the distance from and from need to be clearly shown. It was a pity to see that some candidates lost marks due to incorrect rounding of the values to three significant figures. The most common error being . In part (b)(i) not many candidates seemed to understand what was required. A significant number of candidates wrote down the equation of the line through , , rather than the required line. In part (b)(ii), it seemed that much time was lost as many candidates attempted to find the equation of the perpendicular bisector of to draw the boundaries.
\nIn part (a) many candidates realized that distances were required. Many candidates seemed to have an idea about Voronoi diagrams. However, several candidates did not realize that they had to consider point as well in their comparison. Hence, several candidates only calculated distances from . The numerical comparison of the distance from and from need to be clearly shown. It was a pity to see that some candidates lost marks due to incorrect rounding of the values to three significant figures. The most common error being . In part (b)(i) not many candidates seemed to understand what was required. A significant number of candidates wrote down the equation of the line through , , rather than the required line. In part (b)(ii), it seemed that much time was lost as many candidates attempted to find the equation of the perpendicular bisector of to draw the boundaries.
\nIn part (a) many candidates realized that distances were required. Many candidates seemed to have an idea about Voronoi diagrams. However, several candidates did not realize that they had to consider point as well in their comparison. Hence, several candidates only calculated distances from . The numerical comparison of the distance from and from need to be clearly shown. It was a pity to see that some candidates lost marks due to incorrect rounding of the values to three significant figures. The most common error being . In part (b)(i) not many candidates seemed to understand what was required. A significant number of candidates wrote down the equation of the line through , , rather than the required line. In part (b)(ii), it seemed that much time was lost as many candidates attempted to find the equation of the perpendicular bisector of to draw the boundaries.
\nIn this question, all lengths are in metres and time is in seconds.
\nConsider two particles, and , which start to move at the same time.
\nParticle moves in a straight line such that its displacement from a fixed-point is given by , for .
\nFind an expression for the velocity of at time .
\nParticle also moves in a straight line. The position of is given by .
\nThe speed of is greater than the speed of when .
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognizing velocity is derivative of displacement (M1)
\neg
\nvelocity A1 N2
\n[2 marks]
\nvalid approach to find speed of (M1)
\neg , velocity
\ncorrect speed (A1)
\neg
\nrecognizing relationship between speed and velocity (may be seen in inequality/equation) R1
\neg , speed = | velocity | , graph of speed , speed velocity
correct inequality or equation that compares speed or velocity (accept any variable for ) A1
\neg
\n(seconds) (accept , do not accept ) A1 N2
\n\n
Note: Do not award the last two A1 marks without the R1.
\n[5 marks]
\nA geometric sequence has a first term of and a fourth term of .
\nFind the common ratio.
\nWrite down the second term of this sequence.
\nThe sum of the first terms is greater than .
\nFind the smallest possible value of .
\n(M1)
\nNote: Award (M1) for correctly substituted geometric sequence formula equated to .
\n(A1) (C2)
\n[2 marks]
\n(A1)(ft) (C1)
\nNote: Follow through from part (a).
\n[1 mark]
\n(M1)
\nNote: Award (M1) for their correctly substituted geometric series formula compared to .
\n(A1)(ft)
\n(A1)(ft) (C3)
\nNote: Answer must be an integer for the final (A1)(ft) to be awarded.
Follow through from part (a).
[3 marks]
\nA buoy is floating in the sea and can be seen from the top of a vertical cliff. A boat is travelling from the base of the cliff directly towards the buoy.
\nThe top of the cliff is 142 m above sea level. Currently the boat is 100 metres from the buoy and the angle of depression from the top of the cliff to the boat is 64°.
\nDraw and label the angle of depression on the diagram.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n (A1) (C1)
Note: The horizontal line must be shown and the angle of depression must be labelled. Accept a numerical or descriptive label.
\n[1 mark]
\nJoey is making a party hat in the form of a cone. The hat is made from a sector, , of a circular piece of paper with a radius of and as shown in the diagram.
\nTo make the hat, sides and are joined together. The hat has a base radius of .
\nWrite down the perimeter of the base of the hat in terms of .
\nFind the value of .
\nFind the surface area of the outside of the hat.
\nA1
\n
Note: Answer must be in terms of .
[1 mark]
\nMETHOD 1
\nOR (M1)
\n
Note: Award (M1) for correct substitution into length of an arc formula.
A1
\n
METHOD 2
\n(M1)
\nA1
\n\n
[2 marks]
\nEITHER
\n(M1)
\n
Note: Award (M1) for correct substitution into area of a sector formula.
\n
OR
\n(M1)
\n
Note: Award (M1) for correct substitution into curved area of a cone formula.
\n
THEN
\n(Area) A1
\n
Note: Allow FT from their part (a)(ii) even if their angle is not obtuse.
\n
[2 marks]
\nAlthough most candidates understood what to do in part (a), many of them wrote a decimal approximation instead and did not give their answer in terms of as required in this part. Many candidates were able to use the length of arc formula in (a)(ii). Some candidates went wrong with this question by confusing between the two values: and for the radius. In part (b), although candidates were able to find the surface area of the outside of the hat, several added the surface area of the base to their calculation. A few candidates used the cosine rule to find chord which was then used as the circumference of the base of the cone.
\nAlthough most candidates understood what to do in part (a), many of them wrote a decimal approximation instead and did not give their answer in terms of as required in this part. Many candidates were able to use the length of arc formula in (a)(ii). Some candidates went wrong with this question by confusing between the two values: and for the radius. In part (b), although candidates were able to find the surface area of the outside of the hat, several added the surface area of the base to their calculation. A few candidates used the cosine rule to find chord which was then used as the circumference of the base of the cone.
\nAlthough most candidates understood what to do in part (a), many of them wrote a decimal approximation instead and did not give their answer in terms of as required in this part. Many candidates were able to use the length of arc formula in (a)(ii). Some candidates went wrong with this question by confusing between the two values: and for the radius. In part (b), although candidates were able to find the surface area of the outside of the hat, several added the surface area of the base to their calculation. A few candidates used the cosine rule to find chord which was then used as the circumference of the base of the cone.
\nA hospital collected data from 1000 patients in four hospital wards to review the quality of its healthcare. The data, showing the number of patients who became infected during their stay in hospital, was recorded in the following table.
\nA -test was performed at the 5% significance level.
\nThe critical value for this test is 7.815.
\nThe null hypothesis for the test is
\n: Becoming infected during a stay in the hospital is independent of the ward.
\nFind the expected frequency of the patients who became infected whilst in Nightingale ward.
\nFor this test, write down the statistic.
\nState, giving a reason, whether the null hypothesis should be rejected.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
OR (M1)
\n(A1) (C2)
\n[2 marks]
\n8.21 (8.21497…) (A2) (C2)
\n[2 marks]
\nshould be rejected (A1)(ft)
\nOR (the -value) (R1) (C2)
\n\n
Note: Follow through from part (b). Do not award (A1)(R0).
\nAward (A1)(ft) for “ should be rejected” OR “Becoming infected during a stay in hospital is not independent of (is dependent on OR associated with) the ward”. Accept “Do not accept ” OR “YES”. Do not accept “Becoming infected during a stay in hospital is correlated (related OR linked) with the ward.”
\nAward (R1) for comparison of their statistic value from part (b) with the critical value OR a comparison of -value with 0.05.
\n\n
[2 marks]
\nTwo fixed points, A and B, are 40 m apart on horizontal ground. Two straight ropes, AP and BP, are attached to the same point, P, on the base of a hot air balloon which is vertically above the line AB. The length of BP is 30 m and angle BAP is 48°.
\nOn the diagram, draw and label with an x the angle of depression of B from P.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(A1) (C1)
[1 mark]
\nGive your answers in this question correct to the nearest whole number.
\nImon invested Singapore dollars () in a fixed deposit account with a nominal annual interest rate of , compounded monthly.
\nCalculate the value of Imon’s investment after years.
\nAt the end of the years, Imon withdrew from the fixed deposit account and reinvested this into a super-savings account with a nominal annual interest rate of , compounded half-yearly.
\nThe value of the super-savings account increased to after months.
\nFind the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)(A1)
\nNote: Award (M1) for substituted compound interest formula, (A1) for correct substitutions.
OR
(A1)(M1)
Note: Award (A1) for seen, (M1) for all other correct entries.
OR
(A1)(M1)
Note: Award (A1) for seen, (M1) for all other correct entries.
(A1) (C3)
Note: Do not award the final (A1) if answer is not given correct to the nearest integer.
[3 marks]
(M1)(A1)
\nNote: Award (M1) for substituted compound interest equated to . Award (A1) for correct substitutions.
OR
(A1)(M1)
Note: Award (A1) for seen, (M1) for all other correct entries.
OR
(A1)(M1)
Note: Award (A1) for seen, (M1) for all other correct entries.
(A1) (C3)
Note: Do not award the final (A1) if answer is not given correct to the nearest integer (unless already penalized in part(a)).
[3 marks]
Each athlete on a running team recorded the distance ( miles) they ran in minutes.
\nThe median distance is miles and the interquartile range is miles.
\nThis information is shown in the following box-and-whisker plot.
\nThe distance in miles, , can be converted to the distance in kilometres, , using the formula .
\nThe variance of the distances run by the athletes is .
\nThe standard deviation of the distances is miles.
\nA total of athletes from different teams compete in a race. The times the athletes took to run the race are shown in the following cumulative frequency graph.
\nThere were athletes who took between and minutes to complete the race.
\nFind the value of .
\nWrite down the value of the median distance in kilometres (km).
\nFind the value of .
\nFind .
\nThe first athletes that completed the race won a prize.
\nGiven that an athlete took between and minutes to complete the race, calculate the probability that they won a prize.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg
\nA1 N2
\n[2 marks]
\n(km) A1 N1
\n[1 mark]
\nMETHOD 1 (standard deviation first)
\nvalid approach (M1)
\neg
\nstandard deviation (km) (A1)
\nvalid approach to convert their standard deviation (M1)
\neg
\n(miles) A1 N3
\n\n
Note: If no working shown, award M1A1M0A0 for the value .
If working shown, and candidate’s final answer is , award M1A1M0A0.
\n
METHOD 2 (variance first)
\nvalid approach to convert variance (M1)
\neg
\nvariance (A1)
\nvalid approach (M1)
\neg
\n(miles) A1 N3
\n[4 marks]
\ncorrect frequency for minutes (A1)
\neg
\nadding their frequency (do not accept ) (M1)
\neg athletes
\n(minutes) A1 N3
\n[3 marks]
\n(minutes) (A1)
\ncorrect working (A1)
\neg athletes between and minutes,
\nevidence of conditional probability or reduced sample space (M1)
\neg
\ncorrect working (A1)
\neg
\nA1 N5
\n\n
Note: If no other working is shown, award A0A0M1A0A0 for answer of .
Award N0 for answer of with no other working shown.
\n
[5 marks]
\nSiân invests Australian dollars (AUD) into a savings account which pays a nominal annual interest rate of % compounded monthly.
\nCalculate the value of Siân’s investment after four years. Give your answer correct to two decimal places.
\nAfter the four-year period, Siân withdraws AUD from her savings account and uses this money to buy a car. It is known that the car will depreciate at a rate of % per year.
\nThe value of the car will be AUD after years.
\nFind the value of .
\n(M1)(A1)
\nNote: Award (M1) for substitution into compound interest formula, (A1) for correct substitution.
\n\n
OR
\n\n
\n
\n
\n
(A1)(M1)
\nNote: Award (A1) for seen, (M1) for all other correct entries.
\n\n
OR
\n\n
\n
\n
\n
(A1)(M1)
\nNote: Award (A1) for seen, (M1) for all other correct entries.
\n\n
(AUD) (A1) (C3)
\nNote: The final (A1) is not awarded if the answer is not correct to 2 decimal places.
\n[3 marks]
\nOR (M1)(A1)
\nNote: Award (M1) for substitution into compound interest formula or term of a geometric sequence formula and equating to , (A1) for correct substitution.
\n\n
OR
\n\n
\n
\n
\n
(A1)(M1)
\nNote: Award (A1) for seen, (M1) for all other correct entries. and must have opposite signs.
\n\n
OR
\n(M1)
\nand (A1)
\nNote: Award (M1) for a list of at least correct terms beginning with , (A1) for identifying and .
\n(A1) (C3)
\n\n
[3 marks]
\nIn a school, students in grades 9 to 12 were asked to select their preferred drink. The choices were milk, juice and water. The data obtained are organized in the following table.
\n![\"M17/5/MATSD/SP1/ENG/TZ1/06\"](\"../assets/uploads/tinymce_asset/asset/3567/Schermafbeelding_2017-08-15_om_11.06.16.png\"/)
A test is carried out at the 5% significance level with hypotheses:
\n\n
The critical value for this test is 12.6.
\nWrite down the value of .
\nWrite down the number of degrees of freedom for this test.
\nUse your graphic display calculator to find the statistic for this test.
\nState the conclusion for this test. Give a reason for your answer.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
30 (A1) (C1)
\n[1 mark]
\n6 (A1) (C1)
\n[1 mark]
\n(A2)(ft) (C2)
\n\n
Note: Follow through from part (a).
\nAward (A1) for truncation to 18.9.
\n\n
[2 marks]
\nreject (do not accept) ORaccept (A1)(ft)
\n\n
Note: Can be written in words.
\n\n
(R1) (C2)
\n\n
Note: Accept “” for the (R1) provided their value is explicitly seen in their part (c).
\n\n
OR
\n(R1) (C2)
\n\n
Note: Do not award (R0)(A1)(ft). Follow through from part (c). Numerical comparison must be seen to award the (R1).
\n\n
[2 marks]
\nA group of students sat a history exam. The cumulative frequency graph shows the scores obtained by the students.
\nThe students were awarded a grade from to , depending on the score obtained in the exam. The number of students receiving each grade is shown in the following table.
\nThe mean grade for these students is .
\nFind the median of the scores obtained.
\nFind an expression for in terms of .
\nFind the number of students who obtained a grade .
\nFind the minimum score needed to obtain a grade .
\nA1
\n\n
[1 mark]
\nrecognition that all entries add up to (M1)
\nOR A1
\n\n
[2 marks]
\n(M1)(A1)
\n
Note: Award (M1) for attempt to substitute into mean formula, LHS expression is sufficient for the M mark. Award (A1) for correct substitutions in one variable OR in two variables, followed by evidence of solving simultaneously with .
A1
\n
[3 marks]
\ntheir part (c)(i) seen (e.g. indicated on graph) (M1)
\nA1
\n\n
[2 marks]
\nMost candidates were able to identify the median from the cumulative frequency graph in part (a). In part (b), attention to the demand of this part of the question would have turned one mark into two for a significant number of candidates. Candidates realized that the sum of the frequencies in the table should add up to , however, some neglected to make the subject of the formula.
\nIn part (c)(i), there were mixed results when calculating the mean from the frequency table, with some candidates using only the frequencies and not including the grades in their calculation. Division by was often seen. More able candidates, however, did manage to arrive at the required value of . In part (c)(ii) some candidates scored follow-through marks from their incorrect value for but it seemed that many candidates did not realize that they had to use the cumulative frequency graph to find the answer.
\nMost candidates were able to identify the median from the cumulative frequency graph in part (a). In part (b), attention to the demand of this part of the question would have turned one mark into two for a significant number of candidates. Candidates realized that the sum of the frequencies in the table should add up to , however, some neglected to make the subject of the formula.
\nIn part (c)(i), there were mixed results when calculating the mean from the frequency table, with some candidates using only the frequencies and not including the grades in their calculation. Division by was often seen. More able candidates, however, did manage to arrive at the required value of . In part (c)(ii) some candidates scored follow-through marks from their incorrect value for but it seemed that many candidates did not realize that they had to use the cumulative frequency graph to find the answer.
\nMost candidates were able to identify the median from the cumulative frequency graph in part (a). In part (b), attention to the demand of this part of the question would have turned one mark into two for a significant number of candidates. Candidates realized that the sum of the frequencies in the table should add up to , however, some neglected to make the subject of the formula.
\nIn part (c)(i), there were mixed results when calculating the mean from the frequency table, with some candidates using only the frequencies and not including the grades in their calculation. Division by was often seen. More able candidates, however, did manage to arrive at the required value of . In part (c)(ii) some candidates scored follow-through marks from their incorrect value for but it seemed that many candidates did not realize that they had to use the cumulative frequency graph to find the answer.
\nMost candidates were able to identify the median from the cumulative frequency graph in part (a). In part (b), attention to the demand of this part of the question would have turned one mark into two for a significant number of candidates. Candidates realized that the sum of the frequencies in the table should add up to , however, some neglected to make the subject of the formula.
\nIn part (c)(i), there were mixed results when calculating the mean from the frequency table, with some candidates using only the frequencies and not including the grades in their calculation. Division by was often seen. More able candidates, however, did manage to arrive at the required value of . In part (c)(ii) some candidates scored follow-through marks from their incorrect value for but it seemed that many candidates did not realize that they had to use the cumulative frequency graph to find the answer.
\nDon took part in a project investigating wind speed, , and the time, minutes, to fully charge a solar powered robot.
\nThe investigation was carried out six times. The results are recorded in the table.
\nis the point with coordinates .
\nOn graph paper, draw a scatter diagram to show the results of Don’s investigation. Use a scale of to represent units on the -axis, and to represent units on the -axis.
\nCalculate , the mean wind speed.
\nCalculate , the mean time to fully charge the robot.
\nPlot and label the point on your scatter diagram.
\nCalculate , Pearson’s product–moment correlation coefficient.
\nDescribe the correlation between the wind speed and the time to fully charge the robot.
\nWrite down the equation of the regression line on , in the form .
\nDraw this regression line on your scatter diagram.
\nHence or otherwise estimate the charging time when the wind speed is .
\nDon concluded from his investigation: “There is no causation between wind speed and the time to fully charge the robot”.
\nIn the context of the question, briefly explain the meaning of “no causation”.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A4)
Note: Award (A1) for correct scales and labels.
Award (A3) for all six points correctly plotted.
Award (A2) for four or five points correctly plotted.
Award (A1) for two or three points correctly plotted.
Award at most (A0)(A3) if axes reversed.
If graph paper is not used, award at most (A1)(A0)(A0)(A0).
[4 marks]
(A1)
\n
[1 mark]
(minutes) (A1)
\n
[1 mark]
point in correct position, labelled (A1)(ft)(A1)
Note: Award (A1)(ft) for point plotted in correct position, (A1) for point labelled Follow through from their part (b).
[2 marks]
(G2)
Note: Award (G1) for (incorrect rounding).
[2 marks]
(very) strong positive correlation (A1)(ft)(A1)(ft)
Note: Award (A1)(ft) for (very) strong. Award (A1)(ft) for positive. Follow though from their part (d)(i). If there is no answer to part (d)(i), award at most (A0)(A1) for a correct direction.
[2 marks]
(A1)(A1)(G2)
Note: Award (A1) for . Award (A1) for . If the answer is not an equation, award at most (A1)(A0).
[2 marks]
regression line through their (A1)(ft)
regression line through their (A1)(ft)
Note: Award a maximum of (A1)(A0) if the line is not straight/ruler not used. Award (A0)(A0) if the points are connected.
Follow through from their point in part (b) and their -intercept in part (e)(i).
If is not plotted or labelled, then follow through from part (b).
[2 marks]
(M1)
Note: Award (M1) for correct substitution into their regression equation.
(minutes) (A1)(ft)(G2)
Note: Follow through from their equation in part (e)(i).
OR
an attempt to use their regression line to find the value at
\n
Note: Award (M1) for an indication of using their regression line. This must be illustrated by vertical and horizontal lines or marks at the correct place(s) on their scatter diagram.
(minutes) (A1)(ft)
Note: Follow through from part (e)(ii).
[2 marks]
wind speed does not cause a change in the time to charge (the robot) (A1)
\n
Note: Award (A1) for a statement that communicates the meaning of a non-causal relationship between the two variables.
[1 mark]
The diagram below shows part of the screen from a weather forecasting website showing the data for town . The percentages on the bottom row represent the likelihood of some rain during the hour leading up to the time given. For example there is a chance (a probability of ) of rain falling on any point in town between and .
\nPaula works at a building site in the area covered by this page of the website from to . She has lunch from to .
\nIn the following parts you may assume all probabilities are independent.
\nPaula needs to work outside between and and will also spend her lunchtime outside.
\nWrite down the probability it rains during Paula’s lunch break.
\nFind the probability it will not rain while Paula is outside.
\nFind the probability it will rain at least once while Paula is outside.
\nGiven it rains at least once while Paula is outside find the probability that it rains during her lunch hour.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nNote: Accept probabilities written as percentages throughout.
\n\n
A1
\n\n
[1 mark]
\nNote: Accept probabilities written as percentages throughout.
\n\n
(M1)
\nA1
\n\n
[2 marks]
\nNote: Accept probabilities written as percentages throughout.
\n\n
(M1)
\nA1
\n\n
[2 marks]
\nNote: Accept probabilities written as percentages throughout.
\n\n
M1A1
\nA1
\n\n
[3 marks]
\nOn her first day in a hospital, Kiri receives milligrams (mg) of a therapeutic drug. The amount of the drug Kiri receives increases by the same amount, , each day. On the seventh day, she receives 21 mg of the drug, and on the eleventh day she receives 29 mg.
\nKiri receives the drug for 30 days.
\nTed is also in a hospital and on his first day he receives a 20 mg antibiotic injection. The amount of the antibiotic Ted receives decreases by 50 % each day. On the second day, Ted receives a 10 mg antibiotic injection, on the third day he receives 5 mg, and so on.
\nWrite down an equation, in terms of and , for the amount of the drug that she receives on the seventh day.
\nWrite down an equation, in terms of and , for the amount of the drug that she receives on the eleventh day.
\nWrite down the value of and the value of .
\nCalculate the total amount of the drug, in mg, that she receives.
\nFind the amount of antibiotic, in mg, that Ted receives on the fifth day.
\nThe daily amount of antibiotic Ted receives will first be less than 0.06 mg on the th day. Find the value of .
\nHence find the total amount of antibiotic, in mg, that Ted receives during the first days.
\n(amount taken in the 7th day): (A1)
\nNote: Accept . The equations do not need to be simplified. They should be given in terms of and for the marks to be awarded.
\n[1 mark]
\n(amount taken in the 11th day): (A1)
\nNote: Accept . The equations do not need to be simplified. They should be given in terms of and for the marks to be awarded.
\n[1 mark]
\n( =) 9 (A1)(ft)
\n( =) 2 (A1)(ft)
\nNote: Follow through from part (a), but only if values are positive and < 21.
\n[2 marks]
\n(M1)(A1)(ft)
\nNote: Award (M1) for substitution in the sum of an arithmetic sequence formula; (A1)(ft) for their correct substitution.
\n1140 (mg) (A1)(ft)(G3)
\nNote: Follow through from their and from part (b).
\n[3 marks]
\n20 × (0.5)4 (M1)(A1)
\nNote: Award (M1) for substitution into the geometric sequence formula, (A1) for correct substitution.
\n1.25 (mg) (A1)(G3)
\n[3 marks]
\n(M1)(M1)
\nNote: Award (M1) for correct substitution into the geometric sequence formula; (M1) for comparing their expression to 0.06. Accept an equation instead of inequality.
\n( =) 10 (10th day) (A1)(ft)(G3)
\nNote: Follow through from part (d)(i), if 0 < < 1. Follow through answers must be rounded up for final mark.
\n[3 marks]
\n(M1)(A1)(ft)
\nNote: Award (M1) for substitution into sum of a geometric sequence formula, (A1)(ft) for correct substitution.
Follow through from their and in part (d)(i), if 0 < < 1. Follow through from their in part (d)(ii) but only if is a positive integer.
40.0 (39.9609…) (mg) (A1)(ft)(G2)
\n[3 marks]
\nA large underground tank is constructed at Mills Airport to store fuel. The tank is in the shape of an isosceles trapezoidal prism, .
\n, , , and . Angle and angle . The tank is illustrated below.
\nOnce construction was complete, a fuel pump was used to pump fuel into the empty tank. The amount of fuel pumped into the tank by this pump each hour decreases as an arithmetic sequence with terms .
\nPart of this sequence is shown in the table.
\nAt the end of the hour, the total volume of fuel in the tank was .
\nFind , the height of the tank.
\nShow that the volume of the tank is , correct to three significant figures.
\nWrite down the common difference, .
\nFind the amount of fuel pumped into the tank in the hour.
\nFind the value of such that .
\nWrite down the number of hours that the pump was pumping fuel into the tank.
\nFind the total amount of fuel pumped into the tank in the first hours.
\nShow that the tank will never be completely filled using this pump.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
OR (M1)
\n
Note: Award (M1) for correct substitutions in trig ratio.
OR
(M1)
\n
Note: Award (M1) for correct substitutions in Pythagoras’ theorem.
(A1)(G2)
[2 marks]
(M1)(M1)
\n
Note: Award (M1) for their correctly substituted area of trapezium formula, provided all substitutions are positive. Award (M1) for multiplying by . Follow through from part (a).
OR
(M1)(M1)
Note: Award (M1) for the addition of correct areas for two triangles and one rectangle. Award (M1) for multiplying by . Follow through from part (a).
OR
(M1)(M1)
Note: Award (M1) for their correct substitution in volume of cuboid formula. Award (M1) for correctly substituted volume of triangular prism(s). Follow through from part (a).
(A1)
(AG)
\n
Note: Both an unrounded answer that rounds to the given answer and the rounded value must be seen for the (A1) to be awarded.
[3 marks]
(A1)
\n
[1 mark]
(M1)
\n
Note: Award (M1) for correct substitutions in arithmetic sequence formula.
OR
Award (M1) for a correct term seen as part of list.
(A1)(ft)(G2)
Note: Follow through from part (c) for their value of .
[2 marks]
(M1)
\n
Note: Award (M1) for their correct substitution into arithmetic sequence formula, equated to zero.
(A1)(ft)(G2)
Note: Follow through from part (c). Award at most (M1)(A0) if their is not a positive integer.
[2 marks]
(A1)(ft)
\n
Note: Follow through from part (e)(i), but only if their final answer in (e)(i) is positive. If their in part (e)(i) is not an integer, award (A1)(ft) for the nearest lower integer.
[1 mark]
(M1)
\n
Note: Award (M1) for their correct substitutions in arithmetic series formula. If a list method is used, award (M1) for the addition of their correct terms.
(A1)(ft)(G2)
Note: Follow through from part (c). Award at most (M1)(A0) if their final answer is greater than .
[2 marks]
(M1)
\n
Note: Award (M1) for their correct substitutions into arithmetic series formula.
(A1)(ft)(G1)
Note: Award (M1)(A1) for correctly finding , provided working is shown e.g. , . Follow through from part (c) and either their (e)(i) or (e)(ii). If and their final answer is greater than , award at most (M1)(A1)(ft)(R0). If , there is no maximum, award at most (M1)(A0)(R0). Award no marks if their number of terms is not a positive integer.
(R1)
Hence it will never be filled (AG)
\n
Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.
For unsupported seen, award at most (G1)(R1)(AG). Working must be seen to follow through from parts (c) and (e)(i) or (e)(ii).
OR
(M1)
Note: Award (M1) for their correct substitution into arithmetic series formula, with .
Maximum of this function (A1)
Note: Follow through from part (c). Award at most (M1)(A1)(ft)(R0) if their final answer is greater than . Award at most (M1)(A0)(R0) if their common difference is not . Award at most (M1)(A0)(R0) if is not explicitly identified as the maximum of the function.
(R1)
Hence it will never be filled (AG)
Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.
OR
sketch with concave down curve and labelled horizontal line (M1)
Note: Accept a label of “tank volume” instead of a numerical value. Award (M0) if the line and the curve intersect.
curve explicitly labelled as or equivalent (A1)
Note: Award (A1) for a written explanation interpreting the sketch. Accept a comparison of values, e.g , where is the graphical maximum. Award at most (M1)(A0)(R0) if their common difference is not .
the line and the curve do not intersect (R1)
hence it will never be filled (AG)
\n
Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.
OR
(M1)
Note: Award (M1) for their correctly substituted arithmetic series formula equated to .
Demonstrates there is no solution (A1)
Note: Award (A1) for a correct working that the discriminant is less than zero OR correct working indicating there is no real solution in the quadratic formula.
There is no (real) solution (to this equation) (R1)
hence it will never be filled (AG)
\n
Note: At most (M1)(A0)(R0) for their correctly substituted arithmetic series formula or with a statement \"no solution\". Follow through from their part (b).
[3 marks]
As part of his mathematics exploration about classic books, Jason investigated the time taken by students in his school to read the book The Old Man and the Sea. He collected his data by stopping and asking students in the school corridor, until he reached his target of students from each of the literature classes in his school.
\nJason constructed the following box and whisker diagram to show the number of hours students in the sample took to read this book.
\n\n
Mackenzie, a member of the sample, took hours to read the novel. Jason believes Mackenzie’s time is not an outlier.
\nFor each student interviewed, Jason recorded the time taken to read The Old Man and the Sea , measured in hours, and paired this with their percentage score on the final exam . These data are represented on the scatter diagram.
\nJason correctly calculates the equation of the regression line on for these students to be
\n.
\nHe uses the equation to estimate the percentage score on the final exam for a student who read the book in hours.
\nJason found a website that rated the ‘top ’ classic books. He randomly chose eight of these classic books and recorded the number of pages. For example, Book is rated and has pages. These data are shown in the table.
\nJason intends to analyse the data using Spearman’s rank correlation coefficient, .
\nState which of the two sampling methods, systematic or quota, Jason has used.
\nWrite down the median time to read the book.
\nCalculate the interquartile range.
\nDetermine whether Jason is correct. Support your reasoning.
\nDescribe the correlation.
\nFind the percentage score calculated by Jason.
\nState whether it is valid to use the regression line on for Jason’s estimate. Give a reason for your answer.
\nCopy and complete the information in the following table.
\nCalculate the value of .
\nInterpret your result.
\nQuota sampling A1
\n\n
[1 mark]
\n(hours) A1
\n\n
[1 mark]
\n(M1)
\n
Note: Award M1 for and seen.
A1
[2 marks]
indication of a valid attempt to find the upper fence (M1)
\n\nA1
\n(accept equivalent answer in words) R1
\nJason is correct A1
\n
Note: Do not award R0A1. Follow through within this part from their , but only if their value is supported by a valid attempt or clearly and correctly explains what their value represents.
[4 marks]
“negative” seen A1
\n
Note: Strength cannot be inferred visually; ignore “strong” or “weak”.
[1 mark]
correct substitution (M1)
\n\nA1
\n\n
[2 marks]
\nnot reliable A1
\nextrapolation OR outside the given range of the data R1
\n\n
Note: Do not award A1R0. Only accept reasoning that includes reference to the range of the data. Do not accept a contextual reason such as hours is too short to read the book.
\n\n
[2 marks]
\n A1A1
Note: Do not award A1 for correct ranks for ‘number of pages’. Award A1 for correct ranks for ‘top 50 rating’.
\n
[2 marks]
\nA2
\n
Note: FT from their table.
\n
[2 marks]
\nEITHER
\nthere is a (strong/moderate) positive association between the number of pages and the top rating. A1
\n
OR
there is a (strong/moderate) agreement between the rank order of number of pages and the rank order top rating. A1
\n
OR
there is a (strong/moderate) positive (linear) correlation between the rank order of number of pages and the rank order top rating. A1
\n
Note: Follow through from their value of .
[1 mark]
Observations on pairs of values of the random variables yielded the following results.
\n\nCalculate the value of , the product moment correlation coefficient of the sample.
\nAssuming that the distribution of is bivariate normal with product moment correlation coefficient , calculate the -value of your result when testing the hypotheses .
\nState whether your -value suggests that and are independent.
\nGiven a further value from the distribution of , , predict the corresponding value of . Give your answer to one decimal place.
\nuse of
\nM1
\nA1
\nA1
\n
Note: Accept any answer that rounds to .
[3 marks]
(M1)
\nA1
\n-value A1
\n
Note: Accept any answer that rounds to or .
Note: Follow through their -value
\n
[3 marks]
this value indicates that are not independent A1
\n
[1 mark]
use of
\nM1
\nA1
\nputting gives A1
\n
[3 marks]
Abhinav carries out a χ2 test at the 1 % significance level to determine whether a person’s gender impacts their chosen professional field: engineering, medicine or law.
\nHe surveyed 220 people and the results are shown in the table.
\nState the null hypothesis, H0, for this test.
\nCalculate the expected number of male engineers.
\nFind the p-value for this test.
\nAbhinav rejects H0.
\nState a reason why Abhinav is incorrect in doing so.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
gender and chosen profession are independent (A1) (C1)
\nNote: Accept there is no association between chosen profession and gender. Accept “not dependent”. Do not accept “not related” or “not correlated” or “not influenced”, or “does not impact”.
\n\n
[1 mark]
\n(M1)
\nNote: Award (M1) for correct substitution into expectation formula.
\n= 45 (A1) (C2)
\n\n
[2 marks]
\n0.0193 (0.0192644…) (A2) (C2)
\n\n
[2 marks]
\n0.0193 > 0.01 (1%) (A1)(ft)
\nOR
\nthe p-value is greater than the significance level (1%) (A1)(ft) (C1)
\n\n
Note: A numerical value in (c) must be seen to award the (A1)(ft). Follow through from part (c), only if it is > 0.01.
\nAccept a correct answer from comparing both the numerical value of the Χ2 statistic and the numerical value of the critical value: 7.89898… < 9.21.
\n\n
[1 mark]
\nEmlyn plays many games of basketball for his school team. The number of minutes he plays in each game follows a normal distribution with mean minutes.
\nIn any game there is a chance he will play less than .
\nIn any game there is a chance he will play less than .
\nThe standard deviation of the number of minutes Emlyn plays in any game is .
\nThere is a chance Emlyn plays less than minutes in a game.
\nEmlyn will play in two basketball games today.
\nEmlyn and his teammate Johan each practise shooting the basketball multiple times from a point . A record of their performance over the weekend is shown in the table below.
\nOn Monday, Emlyn and Johan will practise and each will shoot times from point .
\nSketch a diagram to represent this information.
\nShow that .
\nFind the probability that Emlyn plays between and in a game.
\nFind the probability that Emlyn plays more than in a game.
\nFind the value of .
\nFind the probability he plays between and in one game and more than in the other game.
\nFind the expected number of successful shots Emlyn will make on Monday, based on the results from Saturday and Sunday.
\nEmlyn claims the results from Saturday and Sunday show that his expected number of successful shots will be more than Johan’s.
\nDetermine if Emlyn’s claim is correct. Justify your reasoning.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1)
Note: Award (A1) for bell shaped curve with mean or indicated. Award (A1) for approximately correct shaded region.
[2 marks]
(M1)
\n
OR
(M1)
Note: Award (M1) for correct probability equation using OR correctly shaded diagram indicating . Strict or weak inequalities are accepted in parts (b), (c) and (d).
OR (M1)
Note: Award (M0)(M1) for unsupported OR OR OR the midpoint of and is .
Award at most (M1)(M0) if the final answer is not seen. Award (M0)(M0) for using known values and to validate or .
(AG)
[2 marks]
(M1)
\n
OR
(M1)
Note: Award (M1) for correct probability equation OR correctly shaded diagram indicating and .
(A1)(G2)
[2 marks]
(M1)
\n
OR
(M1)
Note: Award (M1) for correct probability equation OR correctly shaded diagram indicating .
(A1)(G2)
[2 marks]
(M1)
\n
OR
(M1)
Note: Award (M1) for correct probability equation OR for a correctly shaded region with indicated to the right-hand side of the mean.
(A1)(G2)
[2 marks]
(M1)(M1)
\n
OR
(M1)(M1)
Note: Award (M1) for the multiplication of their parts (c)(i) and (c)(ii), (M1) for multiplying their product by or for adding their products twice. Follow through from part (c).
(A1)(ft)(G2)
Note: Award (G0) for an unsupported final answer of
[3 marks]
(M1)
\n
Note: Award (M1) for correct probability multiplied by .
(A1)(G2)
[2 marks]
(A1)
\n
Note: Award (M1) for or seen.
Emlyn is incorrect, (R1)
Note: To award the final (R1), both the conclusion and the comparison must be seen. Award at most (A0)(R1)(ft) for consistent incorrect methods in parts (f) and (g).
OR
(A1)
Note: Award (A1) for both correct probabilities seen.
Emlyn is incorrect, (R1)
Note: To award the final (R1), both the conclusion and the comparison must be seen. Award at most (A0)(R1)(ft) for consistent incorrect methods in parts (f) and (g).
[2 marks]
Points and have coordinates and respectively.
\nThe line , which passes through , has equation .
\nExpress in terms of .
\nFind the value of .
\nConsider a unit vector , such that , where .
\nPoint is such that .
\nFind the coordinates of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach to find (M1)
\neg
\nA1 N2
\n[2 marks]
\nvalid approach (M1)
\neg
\none correct equation (A1)
\neg
\ncorrect value for A1
\neg
\nsubstituting their value into their expression/equation to find (M1)
\neg
\nA1 N3
\n[5 marks]
\nvalid approach (M1)
\neg
\ncorrect working to find (A1)
\neg and
\ncorrect approach to find (seen anywhere) A1
\neg
\nrecognizing unit vector has magnitude of (M1)
\neg
\ncorrect working (A1)
\neg
\nA1
\nsubstituting their value of (M1)
\neg
\n(accept ) A1 N4
\n\n
Note: The marks for finding are independent of the first two marks.
For example, it is possible to award marks such as (M0)(A0)A1(M1)(A1)A1 (M0)A0 or (M0)(A0)A1(M1)(A0)A0 (M1)A0.
\n
[8 marks]
\nA food scientist measures the weights of potatoes taken from a single field and the distribution of the weights is shown by the cumulative frequency curve below.
\nFind the number of potatoes in the sample with a weight of more than grams.
\nFind the median weight.
\nFind the lower quartile.
\nFind the upper quartile.
\nThe weight of the smallest potato in the sample is grams and the weight of the largest is grams.
\nUse the scale shown below to draw a box and whisker diagram showing the distribution of the weights of the potatoes. You may assume there are no outliers.
\n\n
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\n(M1)A1
\n\n
[2 marks]
\nMedian A1
\n\n
[1 mark]
\nLower quartile A1
\n\n
[1 mark]
\nUpper quartile A1
\n\n
[1 mark]
\n M1A1
\n
Note: The M1 is for a box and whisker plot and the A1 for all statistics in the right places.
\n\n
[2 marks]
\nConsider the function
\nFind .
\nSolve .
\nThe graph of has a local minimum at point .
\nFind the coordinates of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to substitute (M1)
\neg
\n(exact) A1 N2
\n[2 marks]
\nA2 N2
\n[2 marks]
\nA1A1 N2
\n[2 marks]
\nThe following diagram shows the graph .
\nVerify that satisfies the handshaking lemma.
\nShow that cannot be redrawn as a planar graph.
\nState, giving a reason, whether contains an Eulerian circuit.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nsum of degrees of vertices A1
\nnumber of edges A1
\nthe sum is equal to twice the number of edges which
\nverifies the handshaking lemma R1
\n
METHOD 2
degrees of vertices A1
\nthere are vertices of odd order A1
\nthere is an even number of vertices of odd order
\nwhich verifies the handshaking lemma R1
\n
[3 marks]
if planar then M1
\nA1
\ninequality not satisfied R1
\ntherefore is not planar AG
\n
Note: method explaining that the graph contains is acceptable.
[3 marks]
there are vertices of odd degree R1
\nhence it does not contain an Eulerian circuit A1
\n
Note: Do not award R0A1.
[2 marks]
The coordinates of point A are and the coordinates of point B are . Point M is the midpoint of AB.
\nis the line through A and B.
\nThe line is perpendicular to and passes through M.
\nWrite down, in the form , the equation of .
\n(A1)(ft) (C1)
\n\n
Note: Follow through from parts (c)(i) and (a). Award (A0) if final answer is not written in the form .
\n[1 mark]
\nConsider a function , for . The derivative of is given by .
\nThe graph of is concave-down when .
\nShow that .
\nFind the least value of .
\nFind .
\nLet be the region enclosed by the graph of , the -axis and the lines and . The area of is , correct to three significant figures.
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nevidence of choosing the quotient rule (M1)
\neg
\nderivative of is (must be seen in rule) (A1)
\nderivative of is (must be seen in rule) (A1)
\ncorrect substitution into the quotient rule A1
\neg
\nAG N0
\n\n
METHOD 2
\nevidence of choosing the product rule (M1)
\neg
\nderivative of is (must be seen in rule) (A1)
\nderivative of is (must be seen in rule) (A1)
\ncorrect substitution into the product rule A1
\neg
\nAG N0
\n\n
[4 marks]
\nMETHOD 1 (2nd derivative) (M1)
\nvalid approach
\neg
\n(exact) A1 N2
\n\n
METHOD 2 (1st derivative)
\nvalid attempt to find local maximum on (M1)
\neg sketch with max indicated,
\n(exact) A1 N2
\n\n
[2 marks]
\nevidence of valid approach using substitution or inspection (M1)
\neg
\nA2 N3
\n[3 marks]
\nrecognizing that area (seen anywhere) (M1)
\nrecognizing that their answer to (c) is their (accept absence of ) (M1)
\neg
\ncorrect value for (seen anywhere) (A1)
\neg
\ncorrect integration for (seen anywhere) (A1)
\n\nadding their integrated expressions and equating to (do not accept an expression which involves an integral) (M1)
\neg
\n(A1)
\nA1 N4
\n[7 marks]
\nThe matrix is given by .
\nBy considering the determinant of a relevant matrix, show that the eigenvalues, , of satisfy the equation
\n,
\nwhere and are functions of to be determined.
\nVerify that
\n0.
\nAssuming that is non-singular, use the result in part (b)(i) to show that
\n.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1
\nM1A1
\nA1
\n\n
[4 marks]
(M1)A1
\n
M1
A2
0 AG
Note: Award A1A0 for a single error.
[5 marks]
multiply throughout by giving M1
\n0 A1
\nAG
\n
[2 marks]
The diagram below shows a circular clockface with centre . The clock’s minute hand has a length of . The clock’s hour hand has a length of .
\nAt pm the endpoint of the minute hand is at point and the endpoint of the hour hand is at point .
\n\n
Between pm and pm, the endpoint of the minute hand rotates through an angle, , from point to point . This is illustrated in the diagram.
\nA second clock is illustrated in the diagram below. The clock face has radius with minute and hour hands both of length . The time shown is am. The bottom of the clock face is located above a horizontal bookshelf.
\nThe height, centimetres, of the endpoint of the minute hand above the bookshelf is modelled by the function
\n\nwhere is the angle rotated by the minute hand from am.
\nThe height, centimetres, of the endpoint of the hour hand above the bookshelf is modelled by the function
\n\nwhere is the angle in degrees rotated by the minute hand from am.
\nFind the size of angle in degrees.
\nFind the distance between points and .
\nFind the size of angle in degrees.
\nCalculate the length of arc .
\nCalculate the area of the shaded sector, .
\nWrite down the height of the endpoint of the minute hand above the bookshelf at am.
\nFind the value of when .
\nWrite down the amplitude of .
\nThe endpoints of the minute hand and hour hand meet when .
\nFind the smallest possible value of .
\nOR (M1)
\nA1
\n
[2 marks]
substitution in cosine rule (M1)
\n(A1)
\nA1
\n
Note: Follow through marks in part (b) are contingent on working seen.
[3 marks]
(M1)
\nA1
\n
[2 marks]
substitution into the formula for arc length (M1)
\nOR
\nA1
\n
[2 marks]
substitution into the area of a sector (M1)
\nOR
\nA1
\n
[2 marks]
A1
\n
[1 mark]
correct substitution (M1)
\n\nA1
\n
[2 marks]
A1
\n
[1 mark]
EITHER
\n(M1)
\n
OR
(M1)
Note: Award M1 for equating the functions. Accept a sketch of and with point(s) of intersection marked.
\n
THEN
\nA1
\n
Note: The answer is incorrect but the correct method is implicit. Award (M1)A0.
[2 marks]
Line intersects the -axis at point A and the -axis at point B, as shown on the diagram.
\n![\"M17/5/MATSD/SP1/ENG/TZ2/04\"](\"../assets/uploads/tinymce_asset/asset/3581/Schermafbeelding_2017-08-15_om_17.18.01.png\"/)
The length of line segment OB is three times the length of line segment OA, where O is the origin.
\nPoint lies on .
\nFind the gradient of .
\nFind the equation of in the form .
\nFind the -coordinate of point A.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)(A1) (C2)
\n\n
Notes: Award (A1) for 3 and (A1) for a negative value.
\nAward (A1)(A0) for either or .
\n\n
[2 marks]
\nOR (M1)
\n\n
Note: Award (M1) for substitution of their gradient from part (a) into a correct equation with the coordinates correctly substituted.
\n(A1)(ft) (C2)
\n\n
Notes: Award (A1)(ft) for their correct equation. Follow through from part (a).
\nIf no method seen, award (A1)(A0) for .
\nAward (A1)(A0) for .
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for substitution of in their equation from part (b).
\n\n
(A1)(ft) (C2)
\n\n
Notes: Follow through from their equation from part (b). Do not follow through if no method seen. Do not award the final (A1) if the value of is negative or zero.
\n\n
[2 marks]
\nPoints in the plane are subjected to a transformation in which the point is transformed to the point where
\n.
\nDescribe, in words, the effect of the transformation .
\nShow that the points form a square.
\nDetermine the area of this square.
\nFind the coordinates of , the points to which are transformed under .
\nShow that is a parallelogram.
\nDetermine the area of this parallelogram.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
a stretch of scale factor in the direction
\nand a stretch of scale factor in the direction A1
\n
[1 mark]
the four sides are equal in length A1
\nA1
\nso product of gradients , therefore is perpendicular to A1
\ntherefore is a square AG
\n
[3 marks]
area of square A1
\n
[1 mark]
the transformed points are
\n\n\n\nA2
\n
Note: Award A1 if one point is incorrect.
[2 marks]
A1
\ntherefore is parallel to R1
\nA1
\ntherefore is parallel to
\ntherefore is a parallelogram AG
\n
[3 marks]
(M1)
\nA1
\n
[2 marks]
Lucy sells hot chocolate drinks at her snack bar and has noticed that she sells more hot chocolates on cooler days. On six different days, she records the maximum daily temperature, , measured in degrees centigrade, and the number of hot chocolates sold, . The results are shown in the following table.
\nThe relationship between and can be modelled by the regression line with equation .
\nFind the value of and of .
\nWrite down the correlation coefficient.
\nUsing the regression equation, estimate the number of hot chocolates that Lucy will sell on a day when the maximum temperature is .
\nvalid approach (M1)
\neg correct value for or (or for or seen in (ii))
\n\nA1A1 N3
\n[3 marks]
\nA1 N1
\n[1 mark]
\ncorrect substitution into their equation (A1)
\neg
\n( from )
\n(hot chocolates) A1 N2
\n[2 marks]
\nA school consists of students divided into grade levels. The numbers of students in each grade are shown in the table below.
\nThe Principal of the school wishes to select a sample of students. She wishes to ensure that, as closely as possible, the proportion of the students from each grade in the sample is the same as the proportions in the school.
\nCalculate the number of grade students who should be in the sample.
\nThe Principal selects the students for the sample by asking those who took part in a previous survey if they would like to take part in another. She takes the first of those who reply positively, up to the maximum needed for the sample.
\nState which two of the sampling methods listed below best describe the method used.
\nStratified Quota Convenience Systematic Simple random
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nM1(A1)
\n(students) A1
\n\n
[3 marks]
\nQuota and convenience A1A1
\n\n
Note: Award A1A0 for one correct and one incorrect answer.
\n\n
[2 marks]
\nEmily’s kite ABCD is hanging in a tree. The plane ABCDE is vertical.
\nEmily stands at point E at some distance from the tree, such that EAD is a straight line and angle BED = 7°. Emily knows BD = 1.2 metres and angle BDA = 53°, as shown in the diagram
\n![\"N17/5/MATSD/SP1/ENG/TZ0/10\"](\"../assets/uploads/tinymce_asset/asset/4172/Schermafbeelding_2018-02-12_om_18.18.28.png\"/)
T is a point at the base of the tree. ET is a horizontal line. The angle of elevation of A from E is 41°.
\nFind the length of EB.
\nWrite down the angle of elevation of B from E.
\nFind the vertical height of B above the ground.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nUnits are required in parts (a) and (c).
\n(M1)(A1)
\n\n
Note: Award (M1) for substitution into sine formula, (A1) for correct substitution.
\n\n
OROR (A1) (C3)
\n[3 marks]
\n34° (A1) (C1)
\n[1 mark]
\nUnits are required in parts (a) and (c).
\n(M1)
\n\n
Note: Award (M1) for correct substitution into a trigonometric ratio.
\n\n
OROR (A1)(ft) (C2)
\n\n
Note: Accept “BT” used for height. Follow through from parts (a) and (b). Use of 7.86 gives an answer of 4.39525….
\n\n
[2 marks]
\nA factory produces engraved gold disks. The cost of the disks is directly proportional to the cube of the radius of the disk.
\nA disk with a radius of cm costs US dollars (USD).
\nFind an equation which links and .
\nFind, to the nearest USD, the cost of disk that has a radius of cm.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\n(M1)
\n(M1)
\nA1
\n\n
[3 marks]
\n(M1)
\nA1
\n\n
Note: accept from use of .
\n\n
[2 marks]
\nThe of a solution is given by the formula where is the hydrogen ion concentration in a solution, measured in moles per litre ().
\nFind the value for a solution in which the hydrogen ion concentration is .
\nWrite an expression for in terms of .
\nFind the hydrogen ion concentration in a solution with . Give your answer in the form where and is an integer.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\n(M1)A1
\n\n
[2 marks]
\n(M1)A1
\n\n
Note: Award M1 for an exponential equation with as the base.
\n\n
[2 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nA discrete random variable has the following probability distribution.
\nFind an expression for in terms of .
\nFind the value of which gives the largest value of .
\nHence, find the largest value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of summing probabilities to (M1)
\neg
\nA1 N2
\n[2 marks]
\ncorrect substitution into formula (A1)
\neg
\nvalid approach to find when is a maximum (M1)
\neg max on sketch of , ,
\n(exact) (accept ) A1 N3
\n[3 marks]
\n(exact), A1 N1
\n[1 mark]
\nThe size of the population of migrating birds in a particular town can be approximately modelled by the equation , where is measured in months from the time of the initial measurements.
\nIn a month period the maximum population is and occurs when and the minimum population is and occurs when .
\nThis information is shown on the graph below.
\nFind the value of .
\nFind the value of .
\nFind the value of .
\nFind the value of at which the population first reaches .
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\n(M1)A1
\n\n
[2 marks]
\n(M1)A1
\n\n
Note: Accept .
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nSolve (M1)
\nA1
\n\n
[2 marks]
\nThe weights of apples on a tree can be modelled by a normal distribution with a mean of grams and a standard deviation of grams.
\nA sample of apples are taken from trees, and , in different parts of the orchard.
\nThe data is shown in the table below.
\nThe owner of the orchard wants to know whether the mean weight of the apples from tree is greater than the mean weight of the apples from tree so sets up the following test:
\nand
\nFind the probability that an apple from the tree has a weight greater than grams.
\nFind the -value for the owner’s test.
\nThe test is performed at the significance level.
\nState the conclusion of the test, giving a reason for your answer.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nLet the weight of an apple be
\n(M1)A1
\n\n
[2 marks]
\n-value (M1)A1
\n\n
[2 marks]
\nR1
\nSufficient evidence to reject the null hypothesis (that the weights of apples from the two trees are equal) A1
\n\n
[2 marks]
\nThe position of a helicopter relative to a communications tower at the top of a mountain at time (hours) can be described by the vector equation below.
\n\nThe entries in the column vector give the displacements east and north from the communications tower and above/below the top of the mountain respectively, all measured in kilometres.
\nFind the speed of the helicopter.
\nFind the distance of the helicopter from the communications tower at .
\nFind the bearing on which the helicopter is travelling.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\n(M1)
\nA1
\n\n
[2 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nBearing is or (M1)
\nA1
\n\n
[2 marks]
\nConsider the curve .
\nFind an expression for .
\nShow that the normal to the curve at the point where is .
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nA1
\n\n
[1 mark]
\nGradient at is M1
\nGradient of normal is A1
\nWhen (M1)A1
\n\n
EITHER
\nM1
\nor A1
\n\n
OR
\nM1
\nA1
\n\n
THEN
\nAG
\n\n
[6 marks]
\nThe diagram below shows part of the screen from a weather forecasting website showing the data for town . The percentages on the bottom row represent the likelihood of some rain during the hour leading up to the time given. For example there is a chance (a probability of ) of rain falling on any point in town between and .
\nPaula works at a building site in the area covered by this page of the website from to . She has lunch from to .
\nIn the following parts you may assume all probabilities are independent.
\nPaula needs to work outside between and .
\nPaula will also spend her lunchtime outside.
\nWrite down the probability it rains during Paula’s lunch break.
\nFind the probability it will rain in each of the three hours Paula is working outside.
\nFind the probability it will not rain while Paula is outside.
\nFind the probability it will rain at least once while Paula is outside.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nNote: Accept probabilities written as percentages throughout.
\n\n
A1
\n\n
[1 mark]
\nNote: Accept probabilities written as percentages throughout.
\n\n
(M1)
\nA1
\n\n
[2 marks]
\nNote: Accept probabilities written as percentages throughout.
\n\n
(M1)
\nA1
\n\n
[2 marks]
\nNote: Accept probabilities written as percentages throughout.
\n\n
(M1)
\nA1
\n\n
[2 marks]
\nLet and , for .
\nFind .
\nSolve the equation .
\nHence or otherwise, given that , find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to form composite (in any order) (M1)
\neg
\nA1 N2
\n[2 marks]
\nvalid approach using GDC (M1)
\neg ,
A1 N2
\n[2 marks]
\nMETHOD 1 – (using properties of functions)
\nrecognizing inverse relationship (M1)
\neg
\nequating to their from (i) (A1)
\neg
\n\nA1 N2
\n\n
METHOD 2 – (finding inverse)
\ninterchanging and (seen anywhere) (M1)
\neg
\ncorrect working (A1)
\neg , sketch showing intersection of and
\n\nA1 N2
\n\n
[3 marks]
\nJenna is a keen book reader. The number of books she reads during one week can be modelled by a Poisson distribution with mean .
\nDetermine the expected number of weeks in one year, of weeks, during which Jenna reads at least four books.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
let be the random variable “number of books Jenna reads per week.”
\nthen
\n(M1)(A1)
\n(M1)
\nA1
\n
Note: Accept weeks.
[4 marks]
Let .
\nPart of the graph of is shown below. Point is a local maximum and has coordinates and point is a local minimum with coordinates .
\nWrite down a sequence of transformations that will transform the graph of onto the graph of .
\nVertical stretch, scale factor A1
\nHorizontal stretch, scale factor A1
\nHorizontal translation of unit to the right A1
\n\n
Note: The vertical stretch can be at any position in the order of transformations. If the order of the final two transformations are reversed the horizontal translation is units to the right.
\n\n
[3 marks]
\nAn estate manager is responsible for stocking a small lake with fish. He begins by introducing fish into the lake and monitors their population growth to determine the likely carrying capacity of the lake.
\nAfter one year an accurate assessment of the number of fish in the lake is taken and it is found to be .
\nLet be the number of fish years after the fish have been introduced to the lake.
\nInitially it is assumed that the rate of increase of will be constant.
\nWhen the estate manager again decides to estimate the number of fish in the lake. To do this he first catches fish and marks them, so they can be recognized if caught again. These fish are then released back into the lake. A few days later he catches another fish, releasing each fish after it has been checked, and finds of them are marked.
\nLet be the number of marked fish caught in the second sample, where is considered to be distributed as . Assume the number of fish in the lake is .
\nThe estate manager decides that he needs bounds for the total number of fish in the lake.
\nThe estate manager feels confident that the proportion of marked fish in the lake will be within standard deviations of the proportion of marked fish in the sample and decides these will form the upper and lower bounds of his estimate.
\nThe estate manager now believes the population of fish will follow the logistic model where is the carrying capacity and .
\nThe estate manager would like to know if the population of fish in the lake will eventually reach .
\nUse this model to predict the number of fish in the lake when .
\nAssuming the proportion of marked fish in the second sample is equal to the proportion of marked fish in the lake, show that the estate manager will estimate there are now fish in the lake.
\nWrite down the value of and the value of .
\nState an assumption that is being made for to be considered as following a binomial distribution.
\nShow that an estimate for is .
\nHence show that the variance of the proportion of marked fish in the sample, , is .
\nTaking the value for the variance given in (d) (ii) as a good approximation for the true variance, find the upper and lower bounds for the proportion of marked fish in the lake.
\nHence find upper and lower bounds for the number of fish in the lake when .
\nGiven this result, comment on the validity of the linear model used in part (a).
\nAssuming a carrying capacity of use the given values of and to calculate the parameters and .
\nUse these parameters to calculate the value of predicted by this model.
\nComment on the likelihood of the fish population reaching .
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nM1
\nA1
\n\n
[2 marks]
\nM1A1
\nAG
\n\n
[2 marks]
\nA1A1
\n\n
[2 marks]
\nAny valid reason for example: R1
\nMarked fish are randomly distributed, so constant.
\nEach fish caught is independent of previous fish caught
\n\n
[1 mark]
\nM1
\nA1
\nAG
\n\n
[2 marks]
\nM1A1
\nAG
\n\n
[2 marks]
\n(M1)
\nand A1
\n\n
[2 marks]
\nM1
\nLower bound upper bound A1
\n\n
[2 marks]
\nLinear model prediction falls outside this range so unlikely to be a good model R1A1
\n\n
[2 marks]
\nM1
\nA1
\nM1
\n(M1)
\nA1
\n\n
[5 marks]
\nM1A1
\n\n
Note: Accept any answer that rounds to .
\n\n
[2 marks]
\nThis is much higher than the calculated upper bound for so the rate of growth of the fish is unlikely to be sufficient to reach a carrying capacity of . M1R1
\n\n
[2 marks]
\nA survey of British holidaymakers found that of those surveyed took a holiday in the Lake District in .
\nA random sample of British holidaymakers was taken. The number of people in the sample who took a holiday in the Lake District in can be modelled by a binomial distribution.
\nState two assumptions made in order for this model to be valid.
\nFind the probability that at least three people from the sample took a holiday in the Lake District in .
\nFrom a random sample of holidaymakers, the probability that at least one of them took a holiday in the Lake District in is greater than .
\nDetermine the least possible value of .
\npeople’s holidays are independent of each other R1
\nthe proportion is constant (at ) R1
\n
[2 marks]
(M1)A1
\n
[2 marks]
probability of at least one probability of none
\nOR (A1)
\nattempt to solve inequality (M1)
\n\nso least possible A1
\n
[3 marks]
Give your answers in parts (a), (d)(i), (e) and (f) to the nearest dollar.
\nDaisy invested Australian dollars () in a fixed deposit account with an annual interest rate of compounded quarterly.
\nAfter months, the amount of money in the fixed deposit account has appreciated to more than .
\nDaisy is saving to purchase a new apartment. The price of the apartment is .
\nDaisy makes an initial payment of and takes out a loan to pay the rest.
\nThe loan is for years, compounded monthly, with equal monthly payments of made by Daisy at the end of each month.
\nFor this loan, find
\nAfter years of paying off this loan, Daisy decides to pay the remainder in one final payment.
\nCalculate the value of Daisy’s investment after years.
\nFind the minimum value of , where .
\nWrite down the amount of the loan.
\nthe amount of interest paid by Daisy.
\nthe annual interest rate of the loan.
\nFind the amount of Daisy’s final payment.
\nFind how much money Daisy saved by making one final payment after years.
\nEITHER
\n
(M1)(A1)
Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct.
OR
(M1)(A1)
\n
Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct.
\n
OR
(M1)(A1)
\n
Note: Award M1 for substitution into compound interest formula, (A1) for correct substitution.
\n
A1
\n
Note: Award (M1)(A1)A0 for unsupported .
[3 marks]
EITHER
\n
(M1)(A1)
Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct. The final mark can still be awarded for the correct number of months (multiple of ).
OR
(M1)(A1)
\n
Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct.
\n
OR
OR (M1)(A1)
\n
Note: Award M1 for the correct inequality, and substituted compound interest formula. Allow an equation. Award A1 for correct substitution.
\n
THEN
\nOR (A1)
\nmonths A1
\n
Note: Award A1 for rounding their to the correct number of months. The final answer must be a multiple of . Follow through within this part.
[4 marks]
A1
\n\n
[1 mark]
\n(M1)
\nA1
\n\n
[2 marks]
\n
(M1)(A1)
Note: Award M1 for an attempt to use a financial app in their technology or an attempt to use an annuity formula or seen. If a compound interest formula is equated to zero, award M1, otherwise award M0 for a substituted compound interest formula.
Award A1 for all entries correct in financial app or correct substitution in annuity formula, but award A0 for a substituted compound interest formula. Follow through marks in part (d)(ii) are contingent on working seen.
\n
A1
\n\n
[3 marks]
\n
(M1)(A1)
Note: Award M1 for an attempt to use a financial app in their technology or an attempt to use an annuity formula. Award (M0) for a substituted compound interest formula. Award A1 for all entries correct. Follow through marks in part (e) are contingent on working seen.
\n
A1
\n\n
[3 marks]
\nOR (M1)(A1)
\n
Note: Award M1 for . Award M1 for subtracting their from their . Award at most M1M0 for their or M0M0 for their . Follow through from parts (d)(i) and (e). Follow through marks in part (f) are contingent on working seen.
\n
A1
\n\n
[3 marks]
\nThe water temperature in Lake Windermere is measured on the first day of eight consecutive months from January to August (months to ) and the results are shown below. The value for May (month ) has been accidently deleted.
\nAssuming the data follows a linear model for this period, find the regression line of on for the remaining data.
\nUse your line to find an estimate for the water temperature on the first day of May.
\nExplain why your line should not be used to estimate the value of at which the temperature is .
\nExplain in context why your line should not be used to predict the value for December (month ).
\nState a more appropriate model for the water temperature in the lake over an extended period of time. You are not expected to calculate any parameters.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nA1A1
\n\n
[2 marks]
\n(M1)A1
\n\n
[2 marks]
\nBecause the line should only be used to estimate from and not from . R1
\n\n
[1 mark]
\nBecause the temperatures are no longer going up at a steady rate, or because we know that winter is approaching so the temperature will go down, or temperatures are not likely to continue increasing. R1
\n\n
[1 mark]
\nTrigonometric or sinusoidal A1
\n\n
[1 mark]
\nThe stopping distances for bicycles travelling at are assumed to follow a normal distribution with mean and standard deviation .
\nUnder this assumption, find, correct to four decimal places, the probability that a bicycle chosen at random travelling at manages to stop
\nrandomly selected bicycles are tested and their stopping distances when travelling at are measured.
\nFind, correct to four significant figures, the expected number of bicycles tested that stop between
\nThe measured stopping distances of the bicycles are given in the table.
\nIt is decided to perform a goodness of fit test at the level of significance to decide whether the stopping distances of bicycles travelling at can be modelled by a normal distribution with mean and standard deviation .
\nin less than .
\nin more than .
\nand .
\nand .
\nState the null and alternative hypotheses.
\nFind the -value for the test.
\nState the conclusion of the test. Give a reason for your answer.
\nevidence of correct probability (M1)
\ne.g sketch OR correct probability statement,
\nA1
\n\n
[2 marks]
\nA1
\n
Note: Answers should be given to decimal place.
[1 mark]
multiplying their probability by (M1)
\nA1
\n
[2 marks]
A1
\n
Note: Answers should be given to 4 sf.
[1 mark]
stopping distances can be modelled by
stopping distances cannot be modelled by A1A1
Note: Award A1 for correct , including reference to the mean and standard deviation. Award A1 for the negation of their .
\n
[2 marks]
or seen (M1)
\nA2
\n
[3 marks]
R1
\nthere is insufficient evidence to reject (or “accept ”) A1
\n
Note: Do not award R0A1.
[2 marks]
An infinite geometric series has first term and second term , where .
\nFind the common ratio in terms of .
\nFind the values of for which the sum to infinity of the series exists.
\nFind the value of when .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of dividing terms (in any order) (M1)
\neg
\nA1 N2
\n[2 marks]
\nrecognizing (must be in terms of ) (M1)
\neg
\nA2 N3
\n[3 marks]
\ncorrect equation (A1)
\neg
\n(exact) A2 N3
\n[3 marks]
\nThe production of oil , in barrels per day, from an oil field satisfies the differential equation where is measured in days from the start of production.
\nThe production of oil at is barrels per day.
\nFind .
\nState in context what this value represents.
\nFind an expression for in terms of .
\nDetermine and state what it represents.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\n(barrels per day) A1
\n\n
[1 mark]
\nThis is the increase (change) in (production per day) between and (or during the first days) A1
\n\n
[1 mark]
\nMETHOD 1
\n(M1)A1
\n(M1)A1
\n\n\n
METHOD 2
\n(M1)
\nA1
\n\n
Note: A1 is for the correct integral, with the correct limits.
\n\n
(M1)A1
\n\n\n
[4 marks]
\n(barrels) A1
\nTotal production of oil in barrels in the first year (or first days) A1
\n\n
Note: For the final A1 “barrels”’ must be present either in the statement or as the units.
\n\n
Accept any value which rounds correctly to
\n\n
[2 marks]
\nConsider the function .
\nFor the curve has a single local maximum.
\nFind .
\nFind in terms of the value of at which the maximum occurs.
\nHence find the value of for which has the smallest possible maximum value.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\n\n\n
Note: M1 is for use of the chain rule.
\n\n
M1A1
\n\n
[2 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nValue of local maximum M1A1
\n\nThis has a minimum value when (M1)A1
\n\n
[4 marks]
\nThe following diagram shows a circle with centre and radius . Points and lie on the circumference of the circle and , where .
\nThe tangents to the circle at and intersect at point .
\n\n
Show that .
\nFind the value of when the area of the shaded region is equal to the area of sector .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct working for (seen anywhere) A1
\neg
\nAG N0
\n[1 mark]
\nMETHOD 1 (working with half the areas)
\narea of triangle or triangle (A1)
\neg
\ncorrect sector area (A1)
\neg
\ncorrect approach using their areas to find the shaded area (seen anywhere) (A1)
\neg
\ncorrect equation A1
\neg
\n\nA2 N4
\n\n
METHOD 2 (working with entire kite and entire sector)
\narea of kite (A1)
\neg
\ncorrect sector area (A1)
\neg
\ncorrect approach using their areas to find the shaded area (seen anywhere) (A1)
\neg
\ncorrect equation A1
\neg
\n\nA2 N4
\n\n
[6 marks]
\nThe cross-sectional view of a tunnel is shown on the axes below. The line represents a vertical wall located at the left side of the tunnel. The height, in metres, of the tunnel above the horizontal ground is modelled by , relative to an origin .
\nPoint has coordinates , point has coordinates , and point has coordinates .
\nWhen the height of the tunnel is and when the height of the tunnel is . These points are shown as and on the diagram, respectively.
\nFind .
\nHence find the maximum height of the tunnel.
\nUse the trapezoidal rule, with three intervals, to estimate the cross-sectional area of the tunnel.
\nWrite down the integral which can be used to find the cross-sectional area of the tunnel.
\nHence find the cross-sectional area of the tunnel.
\nevidence of power rule (at least one correct term seen) (M1)
\nA1
\n
[2 marks]
M1
\nA1
\n(M1)
\n\n
Note: Award M1 for substituting their zero for into .
\n
A1
Note: Award M0A0M0A0 for an unsupported .
Award at most M0A0M1A0 if only the last two lines in the solution are seen.
Award at most M1A0M1A1 if their is not seen.
[6 marks]
(A1)(M1)
\n\n
Note: Award A1 for seen. Award M1 for correct substitution into the trapezoidal rule (the zero can be omitted in working).
\n
A1
[3 marks]
OR A1A1
\n\n
Note: Award A1 for a correct integral, A1 for correct limits in the correct location. Award at most A0A1 if is omitted.
\n
[2 marks]
A2
\n
Note: As per the marking instructions, FT from their integral in part (c)(i). Award at most A1FTA0 if their area is , this is outside the constraints of the question (a rectangle).
[2 marks]
A hollow chocolate box is manufactured in the form of a right prism with a regular hexagonal base. The height of the prism is , and the top and base of the prism have sides of length .
\nGiven that , show that the area of the base of the box is equal to .
\nGiven that the total external surface area of the box is , show that the volume of the box may be expressed as .
\nSketch the graph of , for .
\nFind an expression for .
\nFind the value of which maximizes the volume of the box.
\nHence, or otherwise, find the maximum possible volume of the box.
\nThe box will contain spherical chocolates. The production manager assumes that they can calculate the exact number of chocolates in each box by dividing the volume of the box by the volume of a single chocolate and then rounding down to the nearest integer.
\nExplain why the production manager is incorrect.
\nevidence of splitting diagram into equilateral triangles M1
\narea A1
\nAG
\n
Note: The AG line must be seen for the final A1 to be awarded.
[2 marks]
total surface area of prism M1A1
\n
Note: Award M1 for expressing total surface areas as a sum of areas of rectangles and hexagons, and A1 for a correctly substituted formula, equated to .
A1
volume of prism (M1)
\nA1
\nAG
\n
Note: The AG line must be seen for the final A1 to be awarded.
[5 marks]
A1A1
Note: Award A1 for correct shape, A1 for roots in correct place with some indication of scale (indicated by a labelled point).
\n
[2 marks]
A1A1
\n
Note: Award A1 for a correct term.
[2 marks]
from the graph of or OR solving (M1)
\nA1
\n
[2 marks]
from the graph of OR substituting their value for into (M1)
\nA1
\n
[2 marks]
EITHER
wasted space / spheres do not pack densely (tesselate) A1
OR
the model uses exterior values / assumes infinite thinness of materials and hence the modelled volume is not the true volume A1
[1 mark]
The cars for a fairground ride hold four people. They arrive at the platform for loading and unloading every seconds.
\nDuring the hour from am the arrival of people at the ride in any interval of minutes can be modelled by a Poisson distribution with a mean of .
\nWhen the am car leaves there is no one in the queue to get on the ride.
\nShunsuke arrives at am.
\nFind the probability that more than people arrive at the ride before Shunsuke.
\nFind the probability there will be space for him on the car.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nLet be the number of people who arrive between am and am
\n\n(M1)
\nA1
\n\n
[2 marks]
\nMean number of people arriving each seconds is (M1)
\nLet be the number who arrive in the first seconds and the number who arrive in the second seconds.
\n(Shunsuke will be able to get on the ride)
\nM1M1
\n\n
Note: M1 for first term, M1 for any of the other terms.
\n\n
null (A1)(A1)
\n\n
Note: (A1) for one correct value, (A1)(A1) for four correct values.
\n\n
A1
\n\n
[6 marks]
\nSophia pays into a bank account at the end of each month. The annual interest paid on money in the account is which is compounded monthly.
\nThe average rate of inflation per year over the years was .
\nFind the value of her investment after a period of years.
\nFind an approximation for the real interest rate for the money invested in the account.
\nHence find the real value of Sophia’s investment at the end of years.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nNumber of time periods (A1)
\nN =
I% =
PV =
PMT =
P/Y =
C/Y =
Value (M1)A1
\n
[3 marks]
\nMETHOD 1
\nReal interest rate (M1)A1
\n\n
METHOD 2
\n(M1)
\n(accept ) A1
\n\n
[2 marks]
\nN =
I% =
PV =
PMT =
P/Y =
C/Y =
(M1)A1
\n
Note: Award A1 for only.
\n\n
[2 marks]
\nKatya approximates , correct to four decimal places, by using the following expression.
\n\nCalculate Katya’s approximation of , correct to four decimal places.
\nCalculate the percentage error in using Katya’s four decimal place approximation of , compared to the exact value of in your calculator.
\n(A1)
\nA1
\n\n
Note: Award A1 for correct rounding to 4 decimal places. Follow through within this part.
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award M1 for substitution of their final answer in part (a) into the percentage error formula. Candidates should use the exact value of from their GDC.
\n\n
A1
\n\n
[2 marks]
\nThe following diagram shows a water wheel with centre and radius metres. Water flows into buckets, turning the wheel clockwise at a constant speed.
\n
The height, metres, of the top of a bucket above the ground seconds after it passes through point is modelled by the function
, for .
\nA bucket moves around to point which is at a height of metres above the ground. It takes seconds for the top of this bucket to go from point to point .
\nThe chord is metres, correct to three significant figures.
\nFind the height of point above the ground.
\nCalculate the number of seconds it takes for the water wheel to complete one rotation.
\nHence find the number of rotations the water wheel makes in one hour.
\nFind .
\nFind .
\nDetermine the rate of change of when the top of the bucket is at .
\nvalid approach (M1)
\neg
\n(metres) A1 N2
\n[2 marks]
\nvalid approach to find the period (seen anywhere) (M1)
\neg , attempt to find two consecutive max/min,
\n\n
(seconds) (exact) A1 N2
\n[2 marks]
\ncorrect approach (A1)
\neg rotations per minute
\n(rotations) A1 N2
\n[2 marks]
\ncorrect substitution into equation (accept the use of ) (A1)
\neg
\nvalid attempt to solve their equation (M1)
\neg
A1 N3
\n[3 marks]
\nMETHOD 1
\nevidence of choosing the cosine rule or sine rule (M1)
\neg
\ncorrect working (A1)
\neg
\n\n\n
A1 N3
\n\n
METHOD 2
\nattempt to find the half central angle (M1)
\neg
\ncorrect working (A1)
\neg
\n\nA1 N3
\n\n
METHOD 3
\nvalid approach to find fraction of period (M1)
\neg
\ncorrect approach to find angle (A1)
\neg
\n( using )
\nA1 N3
\n\n
[3 marks]
\nrecognizing rate of change is (M1)
\neg
\n( from )
\nrate of change is A1 N2
\n( from )
\n[2 marks]
\nDeb used a thermometer to record the maximum daily temperature over ten consecutive days. Her results, in degrees Celsius (), are shown below.
\n\nFor this data set, find the value of
\nthe mode.
\nthe mean.
\nthe standard deviation.
\nA1
\n\n
[1 mark]
\n(M1)
\nA1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nIt is known that the weights of male Persian cats are normally distributed with mean and variance .
\nA group of male Persian cats are drawn from this population.
\nThe male cats are now joined by female Persian cats. The female cats are drawn from a population whose weights are normally distributed with mean and standard deviation .
\nTen female cats are chosen at random.
\nSketch a diagram showing the above information.
\nFind the proportion of male Persian cats weighing between and .
\nDetermine the expected number of cats in this group that have a weight of less than .
\nFind the probability that exactly one of them weighs over .
\nLet be the number of cats weighing over .
\nFind the variance of .
\nA cat is selected at random from all cats.
\nFind the probability that the cat was female, given that its weight was over .
\n A1A1
Note: Award A1 for a normal curve with mean labelled or , A1 for indication of SD marks on horizontal axis at and/or OR and/or on the correct side and approximately correct position.
[2 marks]
\nOR labelled sketch of region (M1)
\nA1
\n
[2 marks]
(A1)
\n(M1)
\nA1
\n
[3 marks]
,
\n(A1)
\nuse of binomial seen or implied (M1)
\nusing (M1)
\nA1
\n
[4 marks]
A1
\n
[1 mark]
(A1)
\nattempt use of tree diagram OR use of (M1)
\n(A1)
\nA1
\n
[4 marks]
Nymphenburg Palace in Munich has extensive grounds with points of interest (stations) within them.
\nThese nine points, along with the palace, are shown as the vertices in the graph below. The weights on the edges are the walking times in minutes between each of the stations and the total of all the weights is minutes.
\nAnders decides he would like to walk along all the paths shown beginning and ending at the Palace (vertex A).
\nUse the Chinese Postman algorithm, clearly showing all the stages, to find the shortest time to walk along all the paths.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nOdd vertices are B, F, H and I (M1)A1
\nPairing the vertices M1
\nBF and HI
BH and FI
BI and FH A2
Note: award A1 for two correct totals.
\n
Shortest time is (minutes) M1A1
\n\n
[7 marks]
\nA teacher is concerned about the amount of lesson time lost by students through arriving late at school. Over a period of weeks he records the total number of minutes they are late. He also asks them how far they live from school. The results are shown in the table below.
\nWhich of the correlation coefficients would you recommend is used to assess whether or not there is an association between total number of minutes late and distance from school? Fully justify your answer.
\nSpearman’s rank correlation should be used A1
\nBecause the product moment correlation coefficient is distorted by an outlier. R1
\n\n
Note: Do not award A1R0
\n\n
[2 marks]
\nIt is believed that the power of a signal at a point km from an antenna is inversely proportional to where .
\nThe value of is recorded at distances of to and the values of and are plotted on the graph below.
\nThe values of and are shown in the table below.
\nExplain why this graph indicates that is inversely proportional to .
\n\n
Find the equation of the least squares regression line of against .
\n\n
Use your answer to part (b) to write down the value of to the nearest integer.
\nFind an expression for in terms of .
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\na straight line with a negative gradient A1A1
\n\n
[2 marks]
\nA1A1
\n\n
Note: A1 for each correct term.
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\n(M1)
\nA1
\n\n
[2 marks]
\nA box of chocolates is to have a ribbon tied around it as shown in the diagram below.
\nThe box is in the shape of a cuboid with a height of cm. The length and width of the box are and cm.
\nAfter going around the box an extra cm of ribbon is needed to form the bow.
\nThe volume of the box is .
\nFind an expression for the total length of the ribbon in terms of and .
\nShow that
\nFind
\nSolve
\nHence or otherwise find the minimum length of ribbon required.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nA1A1
\n\n
Note: A1 for and A1 for or .
\n\n
[2 marks]
\nA1
\nA1
\nM1
\nAG
\n\n
[3 marks]
\n(M1)
\nA1A1
\n\n
Note: A1 for (and ), A1 for .
\n\n
[3 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\n(M1)A1
\n\n
[2 marks]
\nA graphic designer, Ben, wants to create an animation in which a sequence of squares is created by a composition of successive enlargements and translations and then rotated about the origin and reduced in size.
\nBen outlines his plan with the following storyboards.
\nThe first four frames of the animation are shown below in greater detail.
\nThe sides of each successive square are one half the size of the adjacent larger square. Let the sequence of squares be
\nThe first square, , has sides of length .
\nBen decides the animation will continue as long as the width of the square is greater than the width of one pixel.
\nBen decides to generate the squares using the transformation
\n\nwhere is a matrix that represents an enlargement, is a column vector that represents a translation, is a point in and is its image in .
\nBy considering the case where is ,
\nOnce the image of squares has been produced, Ben wants to continue the animation by rotating the image counter clockwise about the origin and having it reduce in size during the rotation.
\nLet be the enlargement matrix used when the original sequence of squares has been rotated through degrees.
\nBen decides the enlargement scale factor, , should be a linear function of the angle, , and after a rotation of the sequence of squares should be half of its original length.
\nFind an expression for the width of in centimetres.
\nGiven the width of a pixel is approximately , find the number of squares in the final image.
\nWrite down .
\nWrite down , in terms of .
\nstate the coordinates, , of its image in .
\nhence find .
\nshow that .
\nHence or otherwise, find the coordinates of the top left-hand corner in .
\nFind, , in the form .
\nWrite down .
\nHence find the image of after it is rotated and enlarged.
\nFind the value of at which the enlargement scale factor equals zero.
\nAfter the enlargement scale factor equals zero, Ben continues to rotate the image for another two revolutions.
\nDescribe the animation for these two revolutions, stating the final position of the sequence of squares.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nM1A1
\n\n
[2 marks]
\n(A1)
\n\n(A1)
\n\n
Note: Accept equations in place of inequalities.
\n\n
Hence there are squares A1
\n\n
[3 marks]
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\n(M1)
\nA1
\n\n
[2 marks]
\nRecognise the geometric series M1
\nEach component is equal to M1A1
\nAG
\n\n
[3 marks]
\nM1A1
\nA1
\n\n
[3 marks]
\nM1A1
\nA1
\nA1
\n\n\n
[4 marks]
\nA1
\n\n
[1 mark]
\nM1A1A1
\nA1
\n\n
[4 marks]
\nA1
\n\n
[1 mark]
\nThe image will expand from zero (accept equivalent answers)
\nIt will rotate counter clockwise
\nThe design will (re)appear in the opposite (third) quadrant A1A1
\n\n
Note: Accept any two of the above
\n\n
Its final position will be in the opposite (third) quadrant or from its original position or equivalent statement. A1
\n\n
[3 marks]
\nA piece of candy is made in the shape of a solid hemisphere. The radius of the hemisphere is .
\nCalculate the total surface area of one piece of candy.
\nThe total surface of the candy is coated in chocolate. It is known that gram of the chocolate covers an area of .
\nCalculate the weight of chocolate required to coat one piece of candy.
\nOR (M1)(A1)(M1)
\n
Note: Award M1 for use of surface area of a sphere formula (or curved surface area of a hemisphere), A1 for substituting correct values into hemisphere formula, M1 for adding the area of the circle.
\n
A1
\n\n
[4 marks]
\n(M1)
\nA1
\n
[2 marks]
The cross-sectional view of a tunnel is shown on the axes below. The line represents a vertical wall located at the left side of the tunnel. The height, in metres, of the tunnel above the horizontal ground is modelled by , relative to an origin .
\nPoint has coordinates , point has coordinates , and point has coordinates .
\nFind the height of the tunnel when
\nFind .
\nHence find the maximum height of the tunnel.
\n.
\n.
\nUse the trapezoidal rule, with three intervals, to estimate the cross-sectional area of the tunnel.
\nWrite down the integral which can be used to find the cross-sectional area of the tunnel.
\nHence find the cross-sectional area of the tunnel.
\nevidence of power rule (at least one correct term seen) (M1)
\nA1
\n
[2 marks]
M1
\nA1
\n(M1)
\n\n
Note: Award M1 for substituting their zero for into .
\n
A1
Note: Award M0A0M0A0 for an unsupported .
Award at most M0A0M1A0 if only the last two lines in the solution are seen.
Award at most M1A0M1A1 if their is not seen.
[6 marks]
One correct substitution seen (M1)
\nA1
\n
[2 marks]
A1
\n
[1 mark]
(A1)(M1)
\n\n
Note: Award A1 for seen. Award M1 for correct substitution into the trapezoidal rule (the zero can be omitted in working).
\n
A1
[3 marks]
OR A1A1
\n\n
Note: Award A1 for a correct integral, A1 for correct limits in the correct location. Award at most A0A1 if is omitted.
\n
[2 marks]
A2
\n
Note: As per the marking instructions, FT from their integral in part (d)(i). Award at most A1FTA0 if their area is , this is outside the constraints of the question (a rectangle).
[2 marks]
Consider the second order differential equation
\n\nwhere is the displacement of a particle for .
\nWrite the differential equation as a system of coupled first order differential equations.
\nWhen ,
\nUse Euler’s method with a step length of to find an estimate for the value of the displacement and velocity of the particle when .
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nM1
\nA1
\n\n
[2 marks]
\n(M1)(A1)
\n\n
Note: Award M1 for a correct attempt to substitute the functions in part (a) into the formula for Euler’s method for coupled systems.
\n\n
When
\nA1
\nA1
\n\n
Note: Accept .
\n\n
[4 marks]
\nIn this question, give all answers correct to 2 decimal places.
\nRaul and Rosy want to buy a new house and they need a loan of Australian dollars () from a bank. The loan is for years and the annual interest rate for the loan is , compounded monthly. They will pay the loan in fixed monthly instalments at the end of each month.
\nFind the amount they will pay the bank each month.
\nFind the amount Raul and Rosy will still owe the bank at the end of the first years.
\nUsing your answers to parts (a) and (b)(i), calculate how much interest they will have paid in total during the first years.
\n
(M1)(A1)
Note: Award (M1) for an attempt to use a financial app in their technology with at least two entries seen, award A1 for all entries correct. Accept a positive or negative value for .
A1
Note: Accept an answer of . Do not award final A1 if answer is not given correct to dp.
\n
[3 marks]
\n
(M1)(A1)
Note: Award (M1) for an attempt to use a financial app in their technology with at least two entries seen, award A1 for all entries correct. and must have opposite signs.
A1
Note: Do not award final A1 if answer is not given correct to dp, unless already penalized in part (a). Accept from use of exact value for .
\n
[3 marks]
\namount of money paid: (M1)
\nloan paid off: (M1)
\ninterest paid: A1
\n
Note: Allow or from use of some exact values from parts (a) and (b)(i). If their answer to part (b)(i) is greater than then award at most (M1)(M1)(A0) for follow through in part (b)(ii).
\n
[3 marks]
\nThis question was quite challenging for many candidates. In part (a), it was commendable that a significant minority of candidates arrived at the required answer, but others simply used or attempted to use the compound interest formula. Candidates were not using their GDC efficiently. Those candidates who showed their calculator entries could score at least one mark out of three for communicating their method. For those candidates who were comfortable with the financial application in their GDC in parts (a) and (b)(i), five out of six marks was a popular score. Some neglected to round their answers to two decimal places as required. Part (c) proved to be particularly difficult for candidates to decipher which values were necessary to find the amount of interest paid. Most candidates scored only one mark for the amount of money paid over the ten years. Many incorrect or unnecessary calculations were shown. Some candidates attempted to find an interest rate.
\nThis question was quite challenging for many candidates. In part (a), it was commendable that a significant minority of candidates arrived at the required answer, but others simply used or attempted to use the compound interest formula. Candidates were not using their GDC efficiently. Those candidates who showed their calculator entries could score at least one mark out of three for communicating their method. For those candidates who were comfortable with the financial application in their GDC in parts (a) and (b)(i), five out of six marks was a popular score. Some neglected to round their answers to two decimal places as required. Part (c) proved to be particularly difficult for candidates to decipher which values were necessary to find the amount of interest paid. Most candidates scored only one mark for the amount of money paid over the ten years. Many incorrect or unnecessary calculations were shown. Some candidates attempted to find an interest rate.
\nThis question was quite challenging for many candidates. In part (a), it was commendable that a significant minority of candidates arrived at the required answer, but others simply used or attempted to use the compound interest formula. Candidates were not using their GDC efficiently. Those candidates who showed their calculator entries could score at least one mark out of three for communicating their method. For those candidates who were comfortable with the financial application in their GDC in parts (a) and (b)(i), five out of six marks was a popular score. Some neglected to round their answers to two decimal places as required. Part (c) proved to be particularly difficult for candidates to decipher which values were necessary to find the amount of interest paid. Most candidates scored only one mark for the amount of money paid over the ten years. Many incorrect or unnecessary calculations were shown. Some candidates attempted to find an interest rate.
\nThe rate of change of the height of a ball above horizontal ground, measured in metres, seconds after it has been thrown and until it hits the ground, can be modelled by the equation
\n\nThe height of the ball when is .
\nFind an expression for the height of the ball at time .
\nFind the value of at which the ball hits the ground.
\nHence write down the domain of .
\nFind the range of .
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nM1
\nA1A1
\nWhen (M1)
\n(A1)
\nA1
\n\n
[6 marks]
\nseconds (M1)A1
\n\n
[2 marks]
\nA1
\n\n
Note: Accept .
\n\n
[1 mark]
\nMaximum value is (M1)
\nRange is A1A1
\n\n
Note: Accept .
\n\n
[3 marks]
\nIn a small village there are two doctors’ clinics, one owned by Doctor Black and the other owned by Doctor Green. It was noted after each year that of Doctor Black’s patients moved to Doctor Green’s clinic and of Doctor Green’s patients moved to Doctor Black’s clinic. All additional losses and gains of patients by the clinics may be ignored.
\nAt the start of a particular year, it was noted that Doctor Black had patients on their register, compared to Doctor Green’s patients.
\nWrite down a transition matrix indicating the annual population movement between clinics.
\nFind a prediction for the ratio of the number of patients Doctor Black will have, compared to Doctor Green, after two years.
\nFind a matrix , with integer elements, such that , where is a diagonal matrix.
\nHence, show that the long-term transition matrix is given by .
\nHence, or otherwise, determine the expected ratio of the number of patients Doctor Black would have compared to Doctor Green in the long term.
\nM1A1
\n
Note: Award M1A1 for .
Award the A1 for a transposed if used correctly in part (b) i.e. preceded by matrix rather than followed by a matrix.
[2 marks]
(M1)
\n\nso ratio is A1
\n
[2 marks]
to solve
\n(M1)
\n\nOR (A1)
\nattempt to find eigenvectors for at least one eigenvalue (M1)
\nwhen (or any real multiple) (A1)
\nwhen (or any real multiple) (A1)
\ntherefore (accept integer valued multiples of their eigenvectors and columns in either order) A1
\n
[6 marks]
(A1)
\n
Note: This mark is independent, and may be seen anywhere in part (d).
(A1)
(M1)A1
\n
Note: Award (M1)A0 for finding correctly.
as R1
so A1
\nAG
\n
Note: The AG line must be seen for the final A1 to be awarded.
[6 marks]
METHOD ONE
\n(M1)
\nso ratio is A1
\n\n
METHOD TWO
\nlong term ratio is the eigenvector associated with the largest eigenvalue (M1)
\nA1
\n
[2 marks]
Fiona walks from her house to a bus stop where she gets a bus to school. Her time, minutes, to walk to the bus stop is normally distributed with .
\nFiona always leaves her house at 07:15. The first bus that she can get departs at 07:30.
\nThe length of time, minutes, of the bus journey to Fiona’s school is normally distributed with . The probability that the bus journey takes less than minutes is .
\nIf Fiona misses the first bus, there is a second bus which departs at 07:45. She must arrive at school by 08:30 to be on time. Fiona will not arrive on time if she misses both buses. The variables and are independent.
\nFind the probability that it will take Fiona between minutes and minutes to walk to the bus stop.
\nFind .
\nFind the probability that the bus journey takes less than minutes.
\nFind the probability that Fiona will arrive on time.
\nThis year, Fiona will go to school on days.
\nCalculate the number of days Fiona is expected to arrive on time.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nA2 N2
\n[2 marks]
\nfinding standardized value for (A1)
\neg
\ncorrect substitution using their -value (A1)
\neg
\n\nA1 N3
\n[3 marks]
\nA2 N2
\n[2 marks]
\nvalid attempt to find one possible way of being on time (do not penalize incorrect use of strict inequality signs) (M1)
\neg and , and
\ncorrect calculation for (seen anywhere) (A1)
\neg
\ncorrect calculation for (seen anywhere) (A1)
\neg
\ncorrect working (A1)
\neg
\n\n(on time) A1 N2
\n[5 marks]
\nrecognizing binomial with (M1)
\neg
\n( from )
\nA1 N2
\n[2 marks]
\nWrite down in exponential form.
\nAn equilateral triangle is to be drawn on the Argand plane with one of the vertices at the point corresponding to and all the vertices equidistant from .
\nFind the points that correspond to the other two vertices. Give your answers in Cartesian form.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nA1A1
\n\n
Note: Accept equivalent answers:
\n\n
[2 marks]
\nmultiply by (M1)
\nA1A1
\n\n
[3 marks]
\nThe price of gas at Leon’s gas station is per litre. If a customer buys a minimum of litres, a discount of is applied.
\nThis can be modelled by the following function, , which gives the total cost when buying a minimum of litres at Leon’s gas station.
\n\nwhere is the number of litres of gas that a customer buys.
\nFind the total cost of buying litres of gas at Leon’s gas station.
\nFind .
\nThe price of gas at Erica’s gas station is per litre. A customer must buy a minimum of litres of gas. The total cost at Erica’s gas station is cheaper than Leon’s gas station when .
\nFind the minimum value of .
\n(M1)
\nA1
\n
[2 marks]
(M1)
\nlitres A1
\n
[2 marks]
(A1)
\n(M1)
\n
Note: Award M1 for a graph showing two intersecting linear functions, provided one function has a -intercept of and the other function has a negative -intercept.
(minimum value of ) A1
Note: Accept .
[3 marks]
A dice manufacturer claims that for a novelty die he produces the probability of scoring the numbers to are all equal, and the probability of a is two times the probability of scoring any of the other numbers.
\nTo test the manufacture’s claim one of the novelty dice is rolled times and the numbers scored on the die are shown in the table below.
\nA goodness of fit test is to be used with a significance level.
\nFind the probability of scoring a six when rolling the novelty die.
\nFind the probability of scoring more than sixes when this die is rolled times.
\nFind the expected frequency for each of the numbers if the manufacturer’s claim is true.
\nWrite down the null and alternative hypotheses.
\nState the degrees of freedom for the test.
\nDetermine the conclusion of the test, clearly justifying your answer.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nLet the probability of scoring be
\n(M1)(A1)
\nProbability of A1
\n\n
[3 marks]
\nLet the number of sixes be
\n(M1)
\nor (M1)
\n(M1)A1
\n\n
[4 marks]
\nExpected frequency is or (M1)
\n A1
\n
[2 marks]
\n The manufacture’s claim is correct A1
The manufacturer’s claim is not correct A1
\n
[2 marks]
\nDegrees of freedom A1
\n\n
[1 mark]
\np-value (M1)A1
\nR1
\nHence insufficient evidence to reject the manufacture’s claim. A1
\n\n
[4 marks]
\nThe Voronoi diagram below shows three identical cellular phone towers, and . A fourth identical cellular phone tower, is located in the shaded region. The dashed lines in the diagram below represent the edges in the Voronoi diagram.
\nHorizontal scale: unit represents .
Vertical scale: unit represents .
Tim stands inside the shaded region.
\nTower has coordinates and the edge connecting vertices and has equation .
\nExplain why Tim will receive the strongest signal from tower .
\nWrite down the coordinates of tower .
\nTower has coordinates .
\nFind the gradient of the edge of the Voronoi diagram between towers and .
\nevery point in the shaded region is closer to tower R1
\n
Note: Specific reference must be made to the closeness of tower .
[1 mark]
A1A1
\n
Note: Award A1 for each correct coordinate. Award at most A0A1 if parentheses are missing.
[2 marks]
correct use of gradient formula (M1)
\ne.g.
\ntaking negative reciprocal of their (at any point) (M1)
\nedge gradient A1
\n
[3 marks]
In a city, of people have blue eyes. If someone has blue eyes, the probability that they also have fair hair is . This information is represented in the following tree diagram.
\nIt is known that of people in this city have fair hair.
\nCalculate the value of
\nWrite down the value of .
\nFind an expression, in terms of , for the probability of a person not having blue eyes and having fair hair.
\n.
\n.
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\n(M1)
\n
Note: Award (M1) for setting up equation for fair-haired or equivalent.
\n
A1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nDespite its position on the paper, this probability question was reasonably well attempted by most candidates. Many candidates were able to score at least the first mark and the final FT mark. Even the weaker candidates recognized that probabilities added to in part (c)(ii). Parts (b) and (c)(i) seemed to be the most difficult part for the less able candidates.
\nDespite its position on the paper, this probability question was reasonably well attempted by most candidates. Many candidates were able to score at least the first mark and the final FT mark. Even the weaker candidates recognized that probabilities added to in part (c)(ii). Parts (b) and (c)(i) seemed to be the most difficult part for the less able candidates.
\nDespite its position on the paper, this probability question was reasonably well attempted by most candidates. Many candidates were able to score at least the first mark and the final FT mark. Even the weaker candidates recognized that probabilities added to in part (c)(ii). Parts (b) and (c)(i) seemed to be the most difficult part for the less able candidates.
\nDespite its position on the paper, this probability question was reasonably well attempted by most candidates. Many candidates were able to score at least the first mark and the final FT mark. Even the weaker candidates recognized that probabilities added to in part (c)(ii). Parts (b) and (c)(i) seemed to be the most difficult part for the less able candidates.
\nCharlotte decides to model the shape of a cupcake to calculate its volume.
\nFrom rotating a photograph of her cupcake she estimates that its cross-section passes through the points and , where all units are in centimetres. The cross-section is symmetrical in the -axis, as shown below:
\nShe models the section from to as a straight line.
\nCharlotte models the section of the cupcake that passes through the points and with a quadratic curve.
\nCharlotte thinks that a quadratic with a maximum point at and that passes through the point would be a better fit.
\nBelieving this to be a better model for her cupcake, Charlotte finds the volume of revolution about the -axis to estimate the volume of the cupcake.
\nFind the equation of the line passing through these two points.
\nFind the equation of the least squares regression quadratic curve for these four points.
\nBy considering the gradient of this curve when , explain why it may not be a good model.
\nFind the equation of the new model.
\nWrite down an expression for her estimate of the volume as a sum of two integrals.
\nFind the value of Charlotte’s estimate.
\nA1A1
\n
Note: Award A1 for , A1 for .
Award a maximum of A0A1 if not part of an equation.
[2 marks]
(M1)A1
\n\n
[2 marks]
gradient of curve is positive at R1
\n
Note: Accept a sensible rationale that refers to the gradient.
[1 mark]
METHOD 1
\nlet
\ndifferentiating or using (M1)
\n\nsubstituting in the coordinates
(A1)
(A1)
solve to get
OR A1
Note: Use of quadratic regression with points using the symmetry of the graph is a valid method.
\n
METHOD 2
\n(M1)
\n(M1)
\n(A1)
\nOR A1
\n\n
[4 marks]
\n(M1)(M1) (M1)A1
\n
Note: Award (M1)(M1)(M1)A0 if is omitted but response is otherwise correct. Award (M1) for an integral that indicates volume, (M1) for their part (a) within their volume integral, (M1) for their part (b)(i) within their volume integral, A1 for their correct two integrals with all correct limits.
\n
[4 marks]
\nA1
\n\n
[1 mark]
\nOn a school excursion, students visited an amusement park. The amusement park’s main attractions are rollercoasters (), water slides (), and virtual reality rides ().
\nThe students were asked which main attractions they visited. The results are shown in the Venn diagram.
\nA total of students visited the rollercoasters or the water slides.
\nFind the value of .
\nFind the value of .
\nFind the number of students who visited at least two types of main attraction.
\nWrite down the value of .
\nFind the probability that a randomly selected student visited the rollercoasters.
\nFind the probability that a randomly selected student visited the virtual reality rides.
\nHence determine whether the events in parts (d)(i) and (d)(ii) are independent. Justify your reasoning.
\nOR (M1)
\n
Note: Award (M1) for setting up a correct expression.
(A1)(G2)
[2 marks]
(M1)
\nOR
\n(M1)
\nOR
\n(M1)
\n
Note: Award (M1) for setting up a correct expression. Follow through from part (a)(i) but only for .
(A1)(ft)(G2)
Note: Follow through from part(a)(i). The value of must be greater or equal to zero for the (A1)(ft) to be awarded.
[2 marks]
(M1)
\n
Note: Award (M1) for adding and .
(A1)(G2)
[2 marks]
(A1)
\n[1 mark]
\n(A1)(A1)(G2)
\n
Note: Award (A1) for correct numerator. Award(A1) for the correct denominator. Award (A0) for only.
[2 marks]
(A1)(ft)
\n
Note: Follow through from their denominator from part (d)(i).
[1 mark]
they are not independent (A1)(ft)
\nOR (R1)
\n
Note: Comparison of numerical values must be seen for (R1) to be awarded.
Do not award (A1)(R0). Follow through from parts (d)(i) and (d)(ii).
[2 marks]
Arriane has geese on her farm. She claims the mean weight of eggs from her black geese is less than the mean weight of eggs from her white geese.
\nShe recorded the weights of eggs, in grams, from a random selection of geese. The data is shown in the table.
\nIn order to test her claim, Arriane performs a -test at a level of significance. It is assumed that the weights of eggs are normally distributed and the samples have equal variances.
\nState, in words, the null hypothesis.
\nCalculate the -value for this test.
\nState whether the result of the test supports Arriane’s claim. Justify your reasoning.
\nEITHER
\nThe population mean weight of eggs from (her/the) black geese is equal to/the same as the population mean weight of eggs from (her/the) white geese.
\nOR
\nThe population mean weight of eggs from (her/the) black geese is not less than the population mean weight of eggs from (her/the) white geese. A1
\n\n
Note: Reference to the \"population mean weight\" must be explicit for the A1 to be awarded. The term “population” can be implied by use of “all” or “on average” or “generally” when relating to the weight of eggs e.g. “the mean weight of eggs for all (her/the) black geese”.
Award A0 if reference is made to the mean weights from the sample or the table.
Award A0 for a null hypothesis written in symbolic form.
[1 mark]
-value A2
\n
Note: Award A1 for an answer of , from “unpooled” settings on GDC.
[2 marks]
R1
\n(insufficient evidence to reject )
\nArriane’s claim is not supported by the evidence A1
\n\n
Note: Accept or significance level provided is explicitly seen in part (b). Award A1 only if reference is specifically made to Arriane's claim.
Do not award R0A1.
[2 marks]
Hank sets up a bird table in his garden to provide the local birds with some food. Hank notices that a specific bird, a large magpie, visits several times per month and he names him Bill. Hank models the number of times per month that Bill visits his garden as a Poisson distribution with mean .
\nOver the course of consecutive months, find the probability that Bill visits the garden:
\nUsing Hank’s model, find the probability that Bill visits the garden on exactly four occasions during one particular month.
\non exactly occasions.
\nduring the first and third month only.
\nFind the probability that over a -month period, there will be exactly months when Bill does not visit the garden.
\nAfter the first year, a number of baby magpies start to visit Hank’s garden. It may be assumed that each of these baby magpies visits the garden randomly and independently, and that the number of times each baby magpie visits the garden per month is modelled by a Poisson distribution with mean .
\nDetermine the least number of magpies required, including Bill, in order that the probability of Hank’s garden having at least magpie visits per month is greater than .
\nA1
\n
[1 mark]
(M1)
\nA1
\n
[2 marks]
(M1)
\n(A1)
\nA1
\n
[3 marks]
(A1)
\n(M1)(A1)
\n
Note: Award M1 for recognizing binomial probability, and A1 for correct parameters.
A1
[4 marks]
METHOD ONE
\n (M1)(A1)(A1)
Note: Award M1 for evidence of a cumulative Poisson with , A1 for and A1 for .
so require magpies (including Bill) A1
\n
METHOD TWO
\nevidence of a cumulative Poisson with (M1)
\nsketch of curve and (A1)
\n(intersect at) (A1)
\nrounding up gives
\nso require magpies (including Bill) A1
\n\n
[4 marks]
\nThe surface area of an open box with a volume of and a square base with sides of length is given by where .
\nFind .
\nSolve .
\nInterpret your answer to (b)(i) in context.
\n(M1)
\n
Note: Award (M1) for expressing second term with a negative power. This may be implied by seen as part of their answer.
OR A1A1
Note: Award A1 for and A1 for . The first A1 is for differentiated correctly and is independent of the (M1).
\n
[3 marks]
\nEITHER
\nany correct manipulation of e.g. (M1)
\n
OR
sketch of graph of with root indicated (M1)
\n\n
OR
\nsketch of graph of with minimum indicated (M1)
\n
THEN
A1
\n
Note: Value must be positive. Follow through from their part (a) irrespective of working.
\n
[2 marks]
\nthe value of that will minimize surface area of the box A1
\n
Note: Accept ‘optimize’ in place of minimize.
\n
[1 mark]
\nIn part (a), many candidates scored at least the mark for correctly differentiating although differentiating proved to be more problematic, not realizing that the term could be written as . Some who did realize it, made a mistake while differentiating the negative index. In part (b)(i), the manipulation of the equation was frequently incorrect; those that used their GDC got the correct answer with no working. Many candidates could follow the instruction but where errors were made in part (a), valid solutions for part (b) proved tricky with some negative values seen. In part (b)(ii), a significant number of candidates did not appreciate what is meant by a gradient function equal to zero. Of those who had some idea, the words minimize and maximize were seen but not always in terms of the surface area. Many incorrect answers referred to the volume. Many candidates had difficulty communicating an interpretation of their answer in context. This resulted in several negative answers found for part (b)(i) being left as is, when contextually, negative answers would not make sense.
\nIn part (a), many candidates scored at least the mark for correctly differentiating although differentiating proved to be more problematic, not realizing that the term could be written as . Some who did realize it, made a mistake while differentiating the negative index. In part (b)(i), the manipulation of the equation was frequently incorrect; those that used their GDC got the correct answer with no working. Many candidates could follow the instruction but where errors were made in part (a), valid solutions for part (b) proved tricky with some negative values seen. In part (b)(ii), a significant number of candidates did not appreciate what is meant by a gradient function equal to zero. Of those who had some idea, the words minimize and maximize were seen but not always in terms of the surface area. Many incorrect answers referred to the volume. Many candidates had difficulty communicating an interpretation of their answer in context. This resulted in several negative answers found for part (b)(i) being left as is, when contextually, negative answers would not make sense.
\nIn part (a), many candidates scored at least the mark for correctly differentiating although differentiating proved to be more problematic, not realizing that the term could be written as . Some who did realize it, made a mistake while differentiating the negative index. In part (b)(i), the manipulation of the equation was frequently incorrect; those that used their GDC got the correct answer with no working. Many candidates could follow the instruction but where errors were made in part (a), valid solutions for part (b) proved tricky with some negative values seen. In part (b)(ii), a significant number of candidates did not appreciate what is meant by a gradient function equal to zero. Of those who had some idea, the words minimize and maximize were seen but not always in terms of the surface area. Many incorrect answers referred to the volume. Many candidates had difficulty communicating an interpretation of their answer in context. This resulted in several negative answers found for part (b)(i) being left as is, when contextually, negative answers would not make sense.
\nProfessor Wei observed that students have difficulty remembering the information presented in his lectures.
\nHe modelled the percentage of information retained, , by the function , , where is the number of days after the lecture.
\nHe found that day after a lecture, students had forgotten of the information presented.
\nBased on his model, Professor Wei believes that his students will always retain some information from his lecture.
\nFind the value of .
\nUse this model to find the percentage of information retained by his students hours after Professor Wei’s lecture.
\nState a mathematical reason why Professor Wei might believe this.
\nWrite down one possible limitation of the domain of the model.
\nEITHER
\nOR (M1)
\n
OR
\n
THEN
\nA1
\n
[2 marks]
(M1)
\nA1
\n
[2 marks]
OR has a horizontal asymptote R1
\n
[1 mark]
Award A1 for one reasonable limitation of the domain: A1
\nsmall values of produce unrealistic results
\n\nlarge values of are not possible
\npeople do not live forever
\nmodel is not valid at small or large values of
\n
The reason should focus on the domain . Do not accept answers such as:
recollection varies for different people
\nmemories are discrete not continuous
\nthe nature of the information will change how easily it is recalled
\nemotional/physical stress can affect recollection/concentration
\n
Note: Do not accept as this is a limitation that has been given in the question.
[1 mark]
Charlie and Daniella each began a fitness programme. On day one, they both ran . On each subsequent day, Charlie ran more than the previous day whereas Daniella increased her distance by of the distance ran on the previous day.
\nCalculate how far
\nCharlie ran on day of his fitness programme.
\nDaniella ran on day of her fitness programme.
\nOn day of the fitness programmes Daniella runs more than Charlie for the first time.
\nFind the value of .
\nattempt to find using an arithmetic sequence (M1)
\ne.g. and OR OR
\n(Charlie ran) A1
\n
[2 marks]
(A1)
\nattempt to find using a geometric sequence (M1)
\ne.g. and a value for OR OR
\n(Daniella ran) A1
\n
[3 marks]
(M1)
\nattempt to solve inequality (M1)
\n\nA1
\n
[3 marks]
The living accommodation on a university campus is in the shape of a rectangle with sides of length and .
\nThere are three offices for the management of the accommodation set at the points , and . These offices are responsible for all the students in the areas closest to the office. These areas are shown on the Voronoi diagram below. On this coordinate system the positions of , and are , and respectively.
\nThe equation of the perpendicular bisector of is .
\nThe manager of office believes that he has more than one third of the area of the campus to manage.
\nFind the area of campus managed by office .
\nHence or otherwise find the areas managed by offices and .
\nState a further assumption that must be made in order to use area covered as a measure of whether or not the manager of office is responsible for more students than the managers of offices and .
\n\n
A new office is to be built within the triangle formed by , and , at a point as far as possible from the other three offices.
\nFind the distance of this office from each of the other offices.
\nDivides area into two appropriate shapes
\nFor example,
\nArea of triangle (A1)
\nArea of rectangle (A1)
\nA1
\n\n
Note: The area can be found using different divisions. Award A1 for any two correct areas found and A1 for the final answer.
\n\n
[3 marks]
\nEITHER
\n(M1)A1
\n\n
OR
\n(M1)A1
\n\n
THEN
\nArea managed by both offices and is A1
\n\n
[3 marks]
\nDensity of accommodation/students is uniform R1
\n\n
[1 mark]
\n(M1)A1
\n\n
Note: M1 is for an attempt to find the distance from the intersection point to one of the offices.
\n\n
[2 marks]
\nIrina uses a set of coordinate axes to draw her design of a window. The base of the window is on the -axis, the upper part of the window is in the form of a quadratic curve and the sides are vertical lines, as shown on the diagram. The curve has end points and and its vertex is . Distances are measured in centimetres.
\nThe quadratic curve can be expressed in the form for .
\nWrite down the value of .
\nHence form two equations in terms of and .
\nHence find the equation of the quadratic curve.
\nFind the area of the shaded region in Irina’s design.
\nA1
\n\n
[1 mark]
\nA1
\nA1
\n
Note: Award A1 for each equivalent expression or A1 for the use of the axis of symmetry formula to find or from use of derivative. Award A0A1 for and .
[2 marks]
\nA1A1
\n
Note: Award A1A0 if one term is incorrect, A0A0 if two or more terms are incorrect. Award at most A1A0 if correct and values are seen but answer not expressed as an equation.
[2 marks]
\nrecognizing the need to integrate their expression (M1)
\n(A1)
\n
Note: Award (A1) for correct integral, including limits. Condone absence of .
\n
A1
\n
[3 marks]
Generally, the responses were good for this last question on the paper. The main issue here was to not give the two equations in part (a)(ii) with simplified coefficients of and . Several candidates understood what was required but left their answers with and un-simplified and lost marks. Some candidates used the coordinates to substitute in the equation with an incorrect equation of . Candidates were successful at writing the equations in part (a)(iii). In part (b), most candidates realized that they had to use integration to find the area of the shaded region and, for the most part, were able to find a correct value for the area using either the correct equation or their obtained equation from the previous part. A common error was to integrate between and instead of and .
\nGenerally, the responses were good for this last question on the paper. The main issue here was to not give the two equations in part (a)(ii) with simplified coefficients of and . Several candidates understood what was required but left their answers with and un-simplified and lost marks. Some candidates used the coordinates to substitute in the equation with an incorrect equation of . Candidates were successful at writing the equations in part (a)(iii). In part (b), most candidates realized that they had to use integration to find the area of the shaded region and, for the most part, were able to find a correct value for the area using either the correct equation or their obtained equation from the previous part. A common error was to integrate between and instead of and .
\nGenerally, the responses were good for this last question on the paper. The main issue here was to not give the two equations in part (a)(ii) with simplified coefficients of and . Several candidates understood what was required but left their answers with and un-simplified and lost marks. Some candidates used the coordinates to substitute in the equation with an incorrect equation of . Candidates were successful at writing the equations in part (a)(iii). In part (b), most candidates realized that they had to use integration to find the area of the shaded region and, for the most part, were able to find a correct value for the area using either the correct equation or their obtained equation from the previous part. A common error was to integrate between and instead of and .
\nGenerally, the responses were good for this last question on the paper. The main issue here was to not give the two equations in part (a)(ii) with simplified coefficients of and . Several candidates understood what was required but left their answers with and un-simplified and lost marks. Some candidates used the coordinates to substitute in the equation with an incorrect equation of . Candidates were successful at writing the equations in part (a)(iii). In part (b), most candidates realized that they had to use integration to find the area of the shaded region and, for the most part, were able to find a correct value for the area using either the correct equation or their obtained equation from the previous part. A common error was to integrate between and instead of and .
\nA triangular field is such that and , each measured correct to the nearest metre, and the angle at is equal to , measured correct to the nearest .
\nCalculate the maximum possible area of the field.
\nattempt to find any relevant maximum value (M1)
\nlargest sides are and (A1)
\nsmallest possible angle is (A1)
\nattempt to substitute into area of a triangle formula (M1)
\n\nA1
\n
[5 marks]
Long term experience shows that if it is sunny on a particular day in Vokram, then the probability that it will be sunny the following day is . If it is not sunny, then the probability that it will be sunny the following day is .
\nThe transition matrix is used to model this information, where .
\nThe matrix can be written as a product of three matrices, , where is a diagonal matrix.
\nIt is sunny today. Find the probability that it will be sunny in three days’ time.
\nFind the eigenvalues and eigenvectors of .
\nWrite down the matrix .
\nWrite down the matrix .
\nHence find the long-term percentage of sunny days in Vokram.
\nfinding OR use of tree diagram (M1)
\n\nthe probability of sunny in three days’ time is A1
\n\n
[2 marks]
\nattempt to find eigenvalues (M1)
\n\n
Note: Any indication that has been used is sufficient for the (M1).
A1
\nattempt to find either eigenvector (M1)
\nso an eigenvector is A1
\nso an eigenvector is A1
\n
Note: Accept multiples of the stated eigenvectors.
\n
[5 marks]
\nOR A1
\n
Note: Examiners should be aware that different, correct, matrices may be seen.
\n
[1 mark]
\nOR A1
\n
Note: and must be consistent with each other.
\n
[1 mark]
\n(M1)
\nOR (A1)
\n
Note: Award A1 only if their corresponds to their
(M1)
A1
\n\n
[4 marks]
\nA game is played where two unbiased dice are rolled and the score in the game is the greater of the two numbers shown. If the two numbers are the same, then the score in the game is the number shown on one of the dice. A diagram showing the possible outcomes is given below.
\nLet be the random variable “the score in a game”.
\nFind the probability that
\nComplete the table to show the probability distribution of .
\na player scores at least in a game.
\na player scores , given that they scored at least .
\nFind the expected score of a game.
\n A2
Note: Award A1 if three to five probabilities are correct.
[2 marks]
(A1)
\n
[1 mark]
use of conditional probability (M1)
\ne.g. denominator of OR denominator of , etc.
\nA1
\n
[2 marks]
(M1)
\nA1
\n
[2 marks]
A particle moves along the -axis. The velocity of is at time seconds, where for .
\nFind the times when is at instantaneous rest.
\nFind the magnitude of the particle’s acceleration at seconds.
\nFind the greatest speed of in the interval .
\nThe particle starts from the origin . Find an expression for the displacement of from at time seconds.
\nFind the total distance travelled by in the interval .
\nsolving M1
\nA1
\n\n
[2 marks]
\nuse of power rule (M1)
\n(A1)
\n\n(A1)
\nmagnitude A1
\n\n
[4 marks]
\nusing a sketch graph of (M1)
\nA1
\n\n
[2 marks]
\nMETHOD ONE
\n\nattempt at integration of (M1)
\nA1
\nattempt to find (use of ) (M1)
\nA1
\n\n\n
METHOD TWO
\n\nattempt at integration of (M1)
\nA1
\nattempt to substituted limits into their integral (M1)
\nA1
\n\n
[4 marks]
\n(M1)(A1)
\n
Note: Award M1 for using the absolute value of , or separating into two integrals, A1 for the correct expression.
A1
\n
[3 marks]
\nA theatre set designer is designing a piece of flat scenery in the shape of a hill. The scenery is formed by a curve between two vertical edges of unequal height. One edge is metres high and the other is metre high. The width of the scenery is metres.
\nA coordinate system is formed with the origin at the foot of the metres high edge. In this coordinate system the highest point of the cross‐section is at .
\nA set designer wishes to work out an approximate value for the area of the scenery .
\nIn order to obtain a more accurate measure for the area the designer decides to model the curved edge with the polynomial where metres is the height of the curved edge a horizontal distance from the origin.
\nExplain why .
\nBy dividing the area between the curve and the ‐axis into two trapezoids of unequal width show that , justifying the direction of the inequality.
\nWrite down the value of .
\nUse differentiation to show that .
\nDetermine two other linear equations in , and .
\nHence find an expression for .
\nUse the expression found in (f) to calculate a value for .
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nThe area is less than the rectangle containing the cross-section which is equal to R1
\n\n
Note: is not sufficient for R1.
\n\n
[1 mark]
\n(M1)(A1)
\nA1
\nThis is an underestimate as the trapezoids are enclosed by (are under) the curve. R1
\n\n
Note: This can be shown in a diagram.
\n\n
[4 marks]
\nA1
\n\n
[1 mark]
\nA1
\nM1
\nhence AG
\n\n
[2 marks]
\nSubstitute the points and (M1)
\n\nand
\nA1A1
\n\n
[3 marks]
\nSolve on a GDC (M1)
\nA2
\n\n\n
[3 marks]
\n(M1)A1
\n\n
Note: Accept from the three significant figure answer to part (g).
\n\n
[2 marks]
\nAn infinite geometric sequence, with terms , is such that and .
\nFind the common ratio, , for the sequence.
\nFind the least value of such that .
\n(M1)
\nA1
\n
[2 marks]
OR (M1)
\n(A1)
\nA1
\n
Note: If is seen, with or without seeing the value then award M1A1A0.
[3 marks]
A number of candidates did not attempt what should have been a straightforward question. Perhaps because it relied on a part of the syllabus that is restricted to HL and is not in common with SL. Some attempted it but were unaware of the formula for the sum of an infinite geometric sequence, although this is in the formula booklet. By far the biggest error was to fail to recognize that was the smallest integer value greater than that found from solving the equation. There were disappointingly few candidates who adopted a tabular or graphical approach to this question using technology. Some relied on trial-and-error.
\nA number of candidates did not attempt what should have been a straightforward question. Perhaps because it relied on a part of the syllabus that is restricted to HL and is not in common with SL. Some attempted it but were unaware of the formula for the sum of an infinite geometric sequence, although this is in the formula booklet. By far the biggest error was to fail to recognize that was the smallest integer value greater than that found from solving the equation. There were disappointingly few candidates who adopted a tabular or graphical approach to this question using technology. Some relied on trial-and-error.
\nA farmer owns a triangular field . The length of side is and side is . The angle between these two sides is .
\nFind the area of the field.
\nThe farmer would like to divide the field into two equal parts by constructing a straight fence from to a point on .
\nFind . Fully justify any assumptions you make.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nArea (M1)(A1)
\nA1
\n\n
Note: units must be given for the final A1 to be awarded.
\n\n
[3 marks]
\n(M1)A1
\nA1
\n\n
METHOD 1
\nBecause the height and area of each triangle are equal they must have the same length base R1
\nmust be placed half-way along A1
\nA1
\n\n
Note: the final two marks are dependent on the R1 being awarded.
\n\n
METHOD 2
\nLet
\nM1
\n\nUse of area formula
\nA1
\nA1
\n\n
[6 marks]
\nIf a shark is spotted near to Brighton beach, a lifeguard will activate a siren to warn swimmers.
\nThe sound intensity, , of the siren varies inversely with the square of the distance, , from the siren, where .
\nIt is known that at a distance of metres from the siren, the sound intensity is watts per square metre ().
\nShow that .
\nSketch the curve of on the axes below showing clearly the point .
\nWhilst swimming, Scarlett can hear the siren only if the sound intensity at her location is greater than .
\nFind the values of where Scarlett cannot hear the siren.
\n(M1)
\nM1
\nAG
\n
Note: The AG line must be seen for the second M1 to be awarded.
Award no marks for substituting and into (i.e., working backwards).
[2 marks]
A1A1
\n
Note: Award A1 for correct general shape (concave up) with no -intercept, passing through the marked point ; the point must be labelled with either the coordinates or the values and on the and axes. Award A1 for the curve showing asymptotic behavior (i.e. tends to , as tends to infinity), extending to at least ; the curve must not cross nor veer away from the horizontal asymptote.
\n
[2 marks]
(M1)
\nNote: Award (M1) for a correct inequality.
\n
A1
Note: Award A0 for .
[2 marks]
An ice-skater is skating such that her position vector when viewed from above at time seconds can be modelled by
\n\nwith respect to a rectangular coordinate system from a point , where the non-zero constants and can be determined. All distances are in metres.
\nAt time , the displacement of the ice-skater is given by and the velocity of the ice‑skater is given by .
\nFind the velocity vector at time .
\nShow that the magnitude of the velocity of the ice-skater at time is given by
\n.
\nFind the value of and the value of .
\nFind the magnitude of the velocity of the ice-skater when .
\nAt a point , the ice-skater is skating parallel to the -axis for the first time.
\nFind .
\nuse of product rule (M1)
\nA1A1
\n\n
[3 marks]
\nM1
\n
Note: It is more likely that an expression for is seen.
is not sufficient to award the M1, their part (a) must be substituted.
A1
use of within a factorized expression that leads to the final answer M1
\nA1
\nmagnitude of velocity is AG
\n
[4 marks]
when
\nA1
\n(M1)
\nA1
\n
Note: Use of result from part (b) is an alternative approach.
[3 marks]
(M1)
\nA1
\n
[2 marks]
(M1)
\n\n\n(A1)
\ncorrect substitution of their to find or (M1)
and (A1)
use of Pythagoras / distance formula (M1)
A1
[6 marks]
Juri skis from the top of a hill to a finishing point at the bottom of the hill. She takes the shortest route, heading directly to the finishing point .
\nLet define the height of the hill above at a horizontal distance from the starting point at the top of the hill.
\nThe graph of the derivative of is shown below. The graph of has local minima and maxima when is equal to and . The graph of intersects the -axis when is equal to , and .
\nIdentify the value of the point where has its maximum value.
\nInterpret this point in the given context.
\nJuri starts at a height of metres and finishes at , where .
\nSketch a possible diagram of the hill on the following pair of coordinate axes.
\nA1
\n
[1 mark]
the hill is at its steepest / largest slope of hill A1
\n
[1 mark]
A1A1A1
Note: Award (A1) for decreasing function from to and to and increasing from to ; (A1) for minimum at and max at ; (A1) for starting at height of and finishing at a height of at . If reasonable curvature not evident on graph (i.e. only straight lines used) award A1A0A1.
[3 marks]
This was one of the weakest questions on the paper. Many candidates failed to appreciate the significance of the absolute value and gave as the maximum value rather than . Another common error was to interpret the maximum value as greatest velocity or highest point rather than the point where the hill was steepest. A few candidates drew a graph that went from the starting point to the finishing point. What happened in between, often, showed little understanding of the relationship between the graphs of a function and its derivative. The section of the syllabus that mentions understanding derivatives through graphical methods needs more support from teachers.
\nThis was one of the weakest questions on the paper. Many candidates failed to appreciate the significance of the absolute value and gave as the maximum value rather than . Another common error was to interpret the maximum value as greatest velocity or highest point rather than the point where the hill was steepest. A few candidates drew a graph that went from the starting point to the finishing point. What happened in between, often, showed little understanding of the relationship between the graphs of a function and its derivative. The section of the syllabus that mentions understanding derivatives through graphical methods needs more support from teachers.
\nA disc is divided into sectors, number to . The angles at the centre of each of the sectors form an arithmetic sequence, with being the largest angle.
\nIt is given that .
\nWrite down the value of .
\nFind the value of .
\nA game is played in which the arrow attached to the centre of the disc is spun and the sector in which the arrow stops is noted. If the arrow stops in sector the player wins points, otherwise they lose points.
\nLet be the number of points won
\nFind .
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\n\n
A1
\n\n
[1 mark]
\nEITHER
\nM1
\nM1A1
\n\n
OR
\nM1
\nM1
\nSubstitute this value A1
\n\n
THEN
\nA1
\n\n
[4 marks]
\nM1A1
\n\n
[2 marks]
\nDana has collected some data regarding the heights (metres) of waves against a pier at randomly chosen times in a single day. This data is shown in the table below.
\nShe wishes to perform a -test at the significance level to see if the height of waves could be modelled by a normal distribution. Her null hypothesis is
\nThe data can be modelled by a normal distribution.
\nFrom the table she calculates the mean of the heights in her sample to be and the standard deviation of the heights to be .
\nShe calculates the expected values for each interval under this null hypothesis, and some of these values are shown in the table below.
\nUse the given value of to find the value of .
\nFind the value of and the value of , giving your answers correct to one decimal place.
\nFind the value of the test statistic for this test.
\nDetermine the degrees of freedom for Dana’s test.
\nIt is given that the critical value for this test is .
\nState the conclusion of the test in context. Use your answer to part (c) to justify your conclusion.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\n(M1)
\n\n
Note: M1 is for the use of the correct formula
\n\n
A1
\n\n
[2 marks]
\nUsing and (M1)
\nA1A1
\n\n
[3 marks]
\n(M1)A1
\n\n
[2 marks]
\nCombining columns with expected values less than leaves cells (M1)
\nA1
\n\n
[2 marks]
\nR1
\nhence insufficient evidence to reject that the heights of the waves are normally distributed. A1
\n\n
Note: The A1 can be awarded independently of the R1.
\n\n
[2 marks]
\nConsider the following system of coupled differential equations.
\n\n\nFind the value of
\nFind the eigenvalues and corresponding eigenvectors of the matrix .
\nHence, write down the general solution of the system.
\nDetermine, with justification, whether the equilibrium point is stable or unstable.
\n(i) at .
\n(ii) at .
\nSketch a phase portrait for the general solution to the system of coupled differential equations for , .
\n(M1)
\n(A1)
\nOR A1
\n\n(M1)
\n
Note: This M1 can be awarded for attempting to find either eigenvector.
possible eigenvector is (or any real multiple) A1
\n\n\n\npossible eigenvector is (or any real multiple) A1
\n
[6 marks]
(M1)A1
\n
Note: Award M1A1 for , M1A0 if LHS is missing or incorrect.
\n
[2 marks]
\ntwo (distinct) real negative eigenvalues R1
\n(or equivalent (eg both as ))
\n⇒ stable equilibrium point A1
\n
Note: Do not award R0A1.
\n
[2 marks]
\n(M1)
\n(i) A1
\n(ii) A1
\n\n
[3 marks]
\n A1A1A1A1
\n
Note: Award A1 for a phase plane, with correct axes (condone omission of labels) and at least three non-overlapping trajectories. Award A1 for all trajectories leading to a stable node at . Award A1 for showing gradient is negative at and . Award A1 for both eigenvectors on diagram.
\n\n
[4 marks]
\nKatie likes to cycle to work as much as possible. If Katie cycles to work one day then she has a probability of of not cycling to work on the next work day. If she does not cycle to work one day then she has a probability of of not cycling to work on the next work day.
\nComplete the following transition diagram to represent this information.
\nKatie works for days in a year.
\nFind the probability that Katie cycles to work on her final working day of the year.
\n A1A1
[2 marks]
(A1)
\n(M1)
\nA1
\n
[3 marks]
This question was often done well. Many textbooks teach the method of multiplying the transition matrix by an initial state vector. This was often seen in candidates’ responses. eg . Errors were often due to the figures being incorrectly placed in the transition matrix; not just the transpose, but other combinations of the four values as well.
\nThis question was often done well. Many textbooks teach the method of multiplying the transition matrix by an initial state vector. This was often seen in candidates’ responses. eg . Errors were often due to the figures being incorrectly placed in the transition matrix; not just the transpose, but other combinations of the four values as well.
\nEllis designs a gift box. The top of the gift box is in the shape of a right-angled triangle .
\nA rectangular section is inscribed inside this triangle. The lengths of , and are and respectively.
\nThe area of the top of the gift box is .
\nEllis wishes to find the value of that will minimize the area of the top of the gift box.
\nFind in terms of and .
\nShow that .
\nFind .
\nWrite down an equation Ellis could solve to find this value of .
\nHence, or otherwise, find this value of .
\nOR OR A1
\n\n
[1 mark]
\nvalid attempt to link and , using tangents, similar triangles or other method (M1)
\neg. and OR and OR
\ncorrect equation linking and A1
\neg. OR OR
\nsubstitute into a correct area expression M1
\neg. OR
\nAG
\n\n
Note: The AG line must be seen with no incorrect, intermediate working, for the final M1 to be awarded.
\n\n
[3 marks]
\nA1A1
\n
Note: Award A1 for , A1 for . Award A1A0 if extra terms are seen.
[2 marks]
A1
\n
[1 mark]
A1
\n
[1 mark]
A robot moves around the maze shown below.
\nWhenever it leaves a room it is equally likely to take any of the exits.
\nThe time interval between the robot entering and leaving a room is the same for all transitions.
\nFind the transition matrix for the maze.
\nA scientist sets up the robot and then leaves it moving around the maze for a long period of time.
\nFind the probability that the robot is in room when the scientist returns.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\n (M1)A1A1
\n
Note: Award A1A0 if there is one error in the matrix. A0A0 for more than one error.
\n\n
[3 marks]
\nSteady state column matrix is (M1)
\nProbability it is in room is A1
\n\n
[2 marks]
\nThe graph of is given on the following set of axes. The graph passes through the points and , and has a horizontal asymptote at .
\nLet .
\nFind .
\nOn the same set of axes draw the graph of , showing any intercepts and asymptotes.
\nM1A1
\n
[2 marks]
-asymptote A1
\nconcave up decreasing curve and passing through A1
\n
[2 marks]
This question was not particularly well done. Candidates often failed to apply the transformation of the function correctly and did not understand how to use the algebra of graphical transformations. Others applied the geometry of stretches and translations, often incorrectly. Even if the graph was drawn correctly, some candidates failed to follow the instruction to show the asymptote.
\nThis question was not particularly well done. Candidates often failed to apply the transformation of the function correctly and did not understand how to use the algebra of graphical transformations. Others applied the geometry of stretches and translations, often incorrectly. Even if the graph was drawn correctly, some candidates failed to follow the instruction to show the asymptote.
\nA biologist introduces rabbits to an island and records the size of their population over a period of time. The population growth of the rabbits can be approximately modelled by the following differential equation, where is time measured in years.
\n\nA population of foxes is introduced to the island when the population of rabbits has reached . The subsequent population growth of rabbits and foxes, where is the population of foxes at time , can be approximately modelled by the coupled equations:
\n\n\nUse Euler’s method with a step size of , to find
\nThe graph of the population sizes, according to this model, for the first years after the foxes were introduced is shown below.
\nDescribe the changes in the populations of rabbits and foxes for these years at
\nFind the population of rabbits year after they were introduced.
\n(i) the population of rabbits 1 year after the foxes were introduced.
\n(ii) the population of foxes 1 year after the foxes were introduced.
\npoint .
\npoint .
\nFind the non-zero equilibrium point for the populations of rabbits and foxes.
\n(M1)
\n\n(A1)
\n(M1)
\n(A1)
\nA1
\n
Note: Accept for the final A1.
[5 marks]
(A1)
\n
Note: This may be inferred from a correct column, where this is seen.
(A1)
(A1)
\n (A1)
\n
Note: Award A1 for whole line correct when or . The column may be omitted and implied by the correct and values. The formulas are implied by the correct and columns.
\n
(i) ( OR ) A1
(ii) OR A1
\n
[6 marks]
both populations are increasing A1
\n
[1 mark]
rabbits are decreasing and foxes are increasing A1A1
\n
[2 marks]
setting at least one to zero (M1)
\nA1A1
\n
[3 marks]
The following diagram shows a corner of a field bounded by two walls defined by lines and . The walls meet at a point , making an angle of .
\nFarmer Nate has of fencing to make a triangular enclosure for his sheep. One end of the fence is positioned at a point on , from . The other end of the fence will be positioned at some point on , as shown on the diagram.
\nHe wants the enclosure to take up as little of the current field as possible.
\nFind the minimum possible area of the triangular enclosure .
\nMETHOD 1
\nattempt to find using cosine rule M1
\n(A1)
\nattempt to solve a quadratic equation (M1)
\nAND (A1)
\n
Note: At least must be seen, or implied by subsequent working.
minimum area M1
Note: Do not award M1 if incorrect value for minimizing the area has been chosen.
A1
\n
METHOD 2
\nattempt to find using the sine Rule M1
\n(A1)
\nOR (A1)
\n
EITHER
(A1)
\narea M1
\n
OR
sine rule or cosine rule to find (A1)
\nminimum area M1
\n
THEN
A1
\n
Note: Award A0M1A0 if the wrong length or the wrong angle selected but used correctly finding a value of for the area.
\n
[6 marks]
\nAs has often been the case in the past, trigonometry is a topic that is poorly understood and candidates are poorly prepared for. Approaches to this question required the use of the cosine or sine rules. Some candidates tried to use right-angled trigonometry instead. A minority of candidates used the cosine rule approach and were more likely to be successful, navigating the roots of the quadratic equation formed. When using the sine rule the method involved the ambiguous case as the required angle was obtuse. Few candidates realized this and this was the most common mistake. In a few instances, the word “minimum” led candidates to attempt an approach using calculus.
\nJuliet is a sociologist who wants to investigate if income affects happiness amongst doctors. This question asks you to review Juliet’s methods and conclusions.
\nJuliet obtained a list of email addresses of doctors who work in her city. She contacted them and asked them to fill in an anonymous questionnaire. Participants were asked to state their annual income and to respond to a set of questions. The responses were used to determine a happiness score out of . Of the doctors on the list, replied.
\nJuliet’s results are summarized in the following table.
\nFor the remaining ten responses in the table, Juliet calculates the mean happiness score to be .
\nJuliet decides to carry out a hypothesis test on the correlation coefficient to investigate whether increased annual income is associated with greater happiness.
\nJuliet wants to create a model to predict how changing annual income might affect happiness scores. To do this, she assumes that annual income in dollars, , is the independent variable and the happiness score, , is the dependent variable.
\nShe first considers a linear model of the form
\n.
\nJuliet then considers a quadratic model of the form
\n.
\nAfter presenting the results of her investigation, a colleague questions whether Juliet’s sample is representative of all doctors in the city.
\nA report states that the mean annual income of doctors in the city is . Juliet decides to carry out a test to determine whether her sample could realistically be taken from a population with a mean of .
\nDescribe one way in which Juliet could improve the reliability of her investigation.
\nDescribe one criticism that can be made about the validity of Juliet’s investigation.
\nJuliet classifies response as an outlier and removes it from the data. Suggest one possible justification for her decision to remove it.
\nCalculate the mean annual income for these remaining responses.
\nDetermine the value of , Pearson’s product-moment correlation coefficient, for these remaining responses.
\nState why the hypothesis test should be one-tailed.
\nState the null and alternative hypotheses for this test.
\nThe critical value for this test, at the significance level, is . Juliet assumes that the population is bivariate normal.
\nDetermine whether there is significant evidence of a positive correlation between annual income and happiness. Justify your answer.
\nUse Juliet’s data to find the value of and of .
\nInterpret, referring to income and happiness, what the value of represents.
\nFind the value of , of and of .
\nFind the coefficient of determination for each of the two models she considers.
\nHence compare the two models.
\nJuliet decides to use the coefficient of determination to choose between these two models.
\nComment on the validity of her decision.
\nState the name of the test which Juliet should use.
\nState the null and alternative hypotheses for this test.
\nPerform the test, using a significance level, and state your conclusion in context.
\nAny one from: R1
\nincrease sample size / increase response rate / repeat process
check whether sample is representative
test-retest participants or do a parallel test
use a stratified sample
use a random sample
Note: Do not condone:
Ask different types of doctor
Ask for proof of income
Ask for proof of being a doctor
Remove anonymity
Remove response .
\n
[1 mark]
\nAny one from: R1
\nnon-random sampling means a subset of population might be responding
self-reported happiness is not the same as happiness
happiness is not a constant / cannot be quantified / is difficult to measure
income might include external sources
Juliet is only sampling doctors in her city
correlation does not imply causation
sample might be biased
Note: Do not condone the following common but vague responses unless they make a clear link to validity:
Sample size is too small
Result is not generalizable
There may be other variables Juliet is ignoring
Sample might not be representative
\n
[1 mark]
\nbecause the income is very different / implausible / clearly contrived R1
\n
Note: Answers must explicitly reference \"income\" to get credit.
\n
[1 mark]
\n(M1)A1
\n
[2 marks]
A2
\n
[2 marks]
EITHER
only looking for change in one direction R1
OR
only looking for greater happiness with greater income R1
OR
only looking for evidence of positive correlation R1
[1 mark]
A1A1
\n
Note: Award A1 for seen (do not accept ), A1 for both correct hypotheses, using their or . Accept an equivalent statement in words, however reference to “correlation for the population” or “association for the population” must be explicit for the first A1 to be awarded.
Watch out for a null hypothesis in words similar to “Annual income is not associated with greater happiness”. This is effectively saying and should not be condoned.
\n
[2 marks]
METHOD 1 – using critical value of
\nR1
\n(therefore significant evidence of) a positive correlation A1
\n
Note: Do not award R0A1.
\n
METHOD 2 – using -value
\nA1
\n
Note: Follow through from their -value from part (c)(ii).
(therefore significant evidence of) a positive correlation A1
Note: Do not award A0A1.
[2 marks]
A1
\n
[1 mark]
EITHER
the amount the happiness score increases for every increase in (annual) income A1
OR
rate of change of happiness with respect to (annual) income A1
Note: Accept equivalent responses e.g. an increase of in happiness for every increase in salary.
[1 mark]
,
\n,
\nA1
\n
[1 mark]
for quadratic model: A1
\nfor linear model: A1
\n
Note: Follow through from their value from part (c)(ii).
[2 marks]
EITHER
quadratic model is a better fit to the data / more accurate A1
OR
quadratic model explains a higher proportion of the variance A1
[1 mark]
EITHER
not valid, not a useful measure to compare models with different numbers of parameters A1
OR
not valid, quadratic model will always have a better fit than a linear model A1
Note: Accept any other sensible critique of the validity of the method. Do not accept any answers which focus on the conclusion rather than the method of model selection.
[1 mark]
\n(single sample) -test A1
\n
[1 mark]
EITHER
\nA1
\nOR
\n (sample is drawn from a population where) the population mean is
the population mean is not A1
Note: Do not allow FT from an incorrect test in part (f)(i) other than a -test.
[1 mark]
A1
\n
Note: For a -test follow through from part (f)(i), either (from biased estimate of variance) or (from unbiased estimate of variance).
R1
EITHER
no (significant) evidence that mean differs from A1
\n
OR
the sample could plausibly have been drawn from the quoted population A1
Note: Allow R1FTA1FT from an incorrect -value, but the final A1 must still be in the context of the original research question.
[3 marks]
The following table shows the time, in days, from December and the percentage of Christmas trees in stock at a shop on the beginning of that day.
\nThe following table shows the natural logarithm of both and on these days to decimal places.
\nUse the data in the second table to find the value of and the value of for the regression line, .
\n\n
Assuming that the model found in part (a) remains valid, estimate the percentage of trees in stock when .
\nA1A1
\n\n
[2 marks]
\nM1
\n(A1)
\nA1
\n\n
[3 marks]
\nThose candidates who did this question were often successful. There were a number, however, who found an equation of a line through two of the points instead of using their technology to find the equation of the regression line. A common problem was to introduce rounding errors at various stages throughout the problem. Some candidates failed to find the value of from that of .
\nThose candidates who did this question were often successful. There were a number, however, who found an equation of a line through two of the points instead of using their technology to find the equation of the regression line. A common problem was to introduce rounding errors at various stages throughout the problem. Some candidates failed to find the value of from that of .
\nSieun hits golf balls into the air. Each time she hits a ball she records , the angle at which the ball is launched into the air, and , the horizontal distance, in metres, which the ball travels from the point of contact to the first time it lands. The diagram below represents this information.
\nSieun analyses her results and concludes:
\n.
\nDetermine whether the graph of against is increasing or decreasing at .
\nSieun observes that when the angle is , the ball will travel a horizontal distance of .
\nFind an expression for the function .
\n(M1)
\nA1
\nthe curve is decreasing at A1
\n
Note: For the final A1, follow through within this question part for their value. Award A0 for an answer of \"decreasing\" with no work shown.
[3 marks]
recognition of need to integrate (e.g. reverse power rule or integral symbol or integrating at least one term correctly) (M1)
\nA1A1
\n(M1)
\n
Note: Award M1 for correct substitution of and . A constant of integration must be seen (can be implied by a correct answer).
A1
\n
Note: Accept any variable in the working, but for the final A1, the variable must be used in the expression.
[5 marks]
Jorge is carefully observing the rise in sales of a new app he has created.
\nThe number of sales in the first four months is shown in the table below.
\nJorge believes that the increase is exponential and proposes to model the number of sales in month with the equation
\n\nJorge plans to adapt Euler’s method to find an approximate value for .
\nWith a step length of one month the solution to the differential equation can be approximated using Euler’s method where
\n\nJorge decides to take the mean of these values as the approximation of for his model. He also decides the graph of the model should pass through the point .
\nThe sum of the square residuals for these points for the least squares regression model is approximately .
\nShow that Jorge’s model satisfies the differential equation
\n\nShow that
\nHence find three approximations for the value of .
\nFind the equation for Jorge’s model.
\nFind the sum of the square residuals for Jorge’s model using the values .
\nComment how well Jorge’s model fits the data.
\nGive two possible sources of error in the construction of his model.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\n(M1)A1
\n\n
Note: M1 is for an attempt to find
\n\n
AG
\n\n
Note: Accept solution of the differential equation by separating variables
\n\n
[2 marks]
\nM1
\nM1A1
\nAG
\n\n
Note: Do not penalize the use of the sign.
\n\n
[3 marks]
\nCorrect method (M1)
\n\n\nA2
\n\n
Note: A1 for a single error A0 for two or more errors.
\n\n
[3 marks]
\nor A1
\n(M1)
\n\nA1
\n\n
[3 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nThe sum of the square residuals is approximately times as large as the minimum possible, so Jorge’s model is unlikely to fit the data exactly R1
\n\n
[1 mark]
\nFor example
\nSelecting a single point for the curve to pass through
\nApproximating the gradient of the curve by the gradient of a chord R1R1
\n\n
[2 marks]
\nThe slope field for the differential equation is shown in the following two graphs.
\nOn the second graph,
\nCalculate the value of at the point .
\nSketch, on the first graph, a curve that represents the points where .
\n(i) sketch the solution curve that passes through the point .
\n(ii) sketch the solution curve that passes through the point .
\nA1
\n\n
[1 mark]
\ngradient at A1
\ncorrect shape A1
\n
Note: Award second A1 for horizontal asymptote of , and general symmetry about the -axis.
\n
[2 marks]
\n(i) positive gradient at origin A1
\ncorrect shape A1
\n
Note: Award second A1 for a single maximum in 1st quadrant and tending toward an asymptote.
\n
(ii) positive gradient at A1
\ncorrect shape A1
\n
Note: Award second A1 for a single minimum in 2nd quadrant, single maximum in 1st quadrant and tending toward an asymptote.
\n
[4 marks]
\nThere were many good attempts at this question. Care needs to be taken over graph sketching, and the existence of asymptotes or the position of intersections needs to be shown clearly. Many candidates correctly found at in part (a). However, they were then misled into finding a solution curve through this point rather than graphing the points where as required in part (b). Part (c) was answered well with a number of correct answers. Often the curve through had a flat central section and did not show a clear maximum and minimum. The asymptotes were generally poorly drawn with the curves meeting the -axis and stopping or worse still crossing over it.
\nThere were many good attempts at this question. Care needs to be taken over graph sketching, and the existence of asymptotes or the position of intersections needs to be shown clearly. Many candidates correctly found at in part (a). However, they were then misled into finding a solution curve through this point rather than graphing the points where as required in part (b). Part (c) was answered well with a number of correct answers. Often the curve through had a flat central section and did not show a clear maximum and minimum. The asymptotes were generally poorly drawn with the curves meeting the -axis and stopping or worse still crossing over it.
\nThere were many good attempts at this question. Care needs to be taken over graph sketching, and the existence of asymptotes or the position of intersections needs to be shown clearly. Many candidates correctly found at in part (a). However, they were then misled into finding a solution curve through this point rather than graphing the points where as required in part (b). Part (c) was answered well with a number of correct answers. Often the curve through had a flat central section and did not show a clear maximum and minimum. The asymptotes were generally poorly drawn with the curves meeting the -axis and stopping or worse still crossing over it.
\nGeorge goes fishing. From experience he knows that the mean number of fish he catches per hour is . It is assumed that the number of fish he catches can be modelled by a Poisson distribution.
\nOn a day in which George spends hours fishing, find the probability that he will catch more than fish.
\n(M1)
\n
Note: Award (M1) for calculating the mean, , of the distribution
OR (M1)
(M1)A1
\n\n
Note: Award (M1)(M0)(M1)A0 for finding OR .
\n
[4 marks]
A garden has a triangular sunshade suspended from three points and , relative to an origin in the corner of the garden. All distances are measured in metres.
\nFind .
\nFind .
\nFind .
\nHence find the area of the triangle .
\nA1
\n
[1 mark]
A1
\n
[1 mark]
(M1)A1
\n
Note: Do not award (M1) if less than 2 entries are correct.
[2 marks]
area is (M1)A1
\n
[2 marks]
A suitable site for the landing of a spacecraft on the planet Mars is identified at a point, . The shortest time from sunrise to sunset at point must be found.
\nRadians should be used throughout this question. All values given in the question should be treated as exact.
\nMars completes a full orbit of the Sun in Martian days, which is one Martian year.
\nOn day , where , the length of time, in hours, from the start of the Martian day until sunrise at point can be modelled by a function, , where
\n.
\nThe graph of is shown for one Martian year.
\nMars completes a full rotation on its axis in hours and minutes.
\nThe time of sunrise on Mars depends on the angle, , at which it tilts towards the Sun. During a Martian year, varies from to radians.
\nThe angle, , through which Mars rotates on its axis from the start of a Martian day to the moment of sunrise, at point , is given by , .
\nUse your answers to parts (b) and (c) to find
\nLet be the length of time, in hours, from the start of the Martian day until sunset at point on day . can be modelled by the function
\n.
\nThe length of time between sunrise and sunset at point , , can be modelled by the function
\n.
\nLet and hence .
\ncan be written in the form , where and are complex functions of .
\nShow that .
\nFind the angle through which Mars rotates on its axis each hour.
\nShow that the maximum value of , correct to three significant figures.
\nFind the minimum value of .
\nthe maximum value of .
\nthe minimum value of .
\nHence show that , correct to two significant figures.
\nFind the value of .
\nFind the value of .
\nWrite down and in exponential form, with a constant modulus.
\nHence or otherwise find an equation for in the form , where .
\nFind, in hours, the shortest time from sunrise to sunset at point that is predicted by this model.
\nrecognition that period (M1)
\nOR A1
\n
Note: Award A1 for a correct expression leading to the given value or for a correct value of to 4 sf or greater accuracy.
AG
[2 marks]
length of day hours (A1)
\n
Note: Award A1 for or .
(M1)
Note: Accept .
radians A1
[3 marks]
\nsubstitution of either value of into equation (M1)
\ncorrect use of arccos to find a value for (M1)
\n
Note: Both (M1) lines may be seen in either part (c)(i) or part (c)(ii).
A1
AG
\n
Note: For substitution of award M0A0.
\n
[3 marks]
\nA1
\n\n
[1 mark]
\n(M1)
\nA1
\n
Note: Accept from use of .
[2 marks]
\nA1
\n
Note: Accept and from use of rounded values.
[1 mark]
\nM1
\nA1
\n
Note: Award M1 for substituting their values into a correct expression. Award A1 for a correct value of from their expression which has at least 3 significant figures and rounds correctly to .
(correct to sf) AG
\n
[2 marks]
\nEITHER
\n(M1)
\n
OR
or
\n
THEN
A1
\n
Note: Accept from use of rounded values. Follow through on their answers to part (d) and .
[2 marks]
(M1)
\nA1
\n
Note: Follow through for minus their answer to part (f).
[2 marks]
at least one expression in the form (M1)
\nA1A1
\n
[3 marks]
EITHER
\n\n(M1)
\n(A1)(A1)
\n
OR
graph of or
\n(A1)
\nOR (M1)(A1)
\n
Note: The and variables (or equivalent) must be seen.
THEN
A1
\n\n
Note: Accept equivalent forms, e.g. .
Follow through on their answer to part (g) replacing .
[4 marks]
shortest time between sunrise and sunset
\n(M1)
\nhours A1
\n
Note: Accept from use of sf values.
[2 marks]
\nAn engineer plans to visit six oil rigs in the Gulf of Mexico, starting and finishing at . The travelling time, in minutes, between each of the rigs is shown in the table.
\nThe data above can be represented by a graph .
\nUse Prim’s algorithm to find the weight of the minimum spanning tree of the subgraph of obtained by deleting and starting at . List the order in which the edges are selected.
\nHence find a lower bound for the travelling time needed to visit all the oil rigs.
\nDescribe how an improved lower bound might be found.
\nuse of Prim’s algorithm M1
\nA1
\nA1
\n\n\nTotal A1
\n
Note: Award M0A0A0A1 for without correct working e.g. use of Kruskal’s, or with no working.
Award M1A0A0A1 for by using Prim’s from an incorrect starting point.
[4 marks]
(M1)
\nminutes A1
\n
[2 marks]
delete a different vertex A1
\n
[1 mark]
Consider , where .
\nFind when
\nPoint on the Argand diagram can be transformed to point by two transformations.
\n.
\n.
\nDescribe these two transformations and give the order in which they are applied.
\nHence, or otherwise, find the value of when .
\n(M1)
\nA1
\n
[2 marks]
A1
\n
[1 mark]
EITHER
\nrotation of (anticlockwise, centre at the origin) A1A1
\n
Note: Award A1 for “rotation” and A1 for “”.
followed by a translation of A1
OR
translation of A1
followed by rotation of (anticlockwise, centre at the origin) A1A1
Note: Award A1 for “rotation” and A1 for “”.
\n
[3 marks]
\nEITHER
\nmove to left to (M1)
\nthen rotate by to
\nA1
\n\n
OR
\n\n\n(M1)
\nA1
\n\n
[2 marks]
\nA change in grazing habits has resulted in two species of herbivore, and , competing for food on the same grasslands. At time environmentalists begin to record the sizes of both populations. Let the size of the population of be , and the size of the population be . The following model is proposed for predicting the change in the sizes of the two populations:
\n\n\nfor
\nFor this system of coupled differential equations find
\nWhen has a population of .
\nIt is known that has an initial population of .
\nthe eigenvalues.
\nthe eigenvectors.
\nHence write down the general solution of the system of equations.
\nSketch the phase portrait for this system, for .
\nOn your sketch show
\nWrite down a condition on the size of the initial population of if it is to avoid its population reducing to zero.
\nFind the value of at which .
\nFind the population of at this value of . Give your answer to the nearest herbivores.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\n(M1)(A1)
\nand A1
\n\n
[3 marks]
\nAttempt to solve either
\nor
\nor equivalent (M1)
\nor A1A1
\n\n
Note: accept equivalent forms
\n\n
[3 marks]
\nA1
\n\n
[1 mark]
\n A1A1A1
\n
Note: A1 for correctly labelled, A1 for at least two trajectories above and A1 for at least two trajectories below , including arrows.
\n\n
[3 marks]
\nA1
\n\n
[1 mark]
\nAt M1A1
\n\n
Note: Award M1 for the substitution of and
\n\n
Hence A1A1
\nM1
\n(years) A1
\n\n
[6 marks]
\n(M1)
\n\n(to the nearest animals) A1
\n\n
[2 marks]
\nA factory, producing plastic gifts for a fast food restaurant’s Jolly meals, claims that just of the toys produced are faulty.
\nA restaurant manager wants to test this claim. A box of toys is delivered to the restaurant. The manager checks all the toys in this box and four toys are found to be faulty.
\nThe restaurant manager performs a one-tailed hypothesis test, at the significance level, to determine whether the factory’s claim is reasonable. It is known that faults in the toys occur independently.
\nIdentify the type of sampling used by the restaurant manager.
\nWrite down the null and alternative hypotheses.
\nFind the -value for the test.
\nState the conclusion of the test. Give a reason for your answer.
\nConvenience A1
\n
[1 mark]
of the toys produced are faulty A1
More than are faulty A1
[2 marks]
(M1)
\nA1
\n
Note: Any attempt using Normal approximation to find -value is awarded M0A0.
[2 marks]
R1
\nso there is insufficient evidence to reject . A1
\n
Note: Do not award R0A1. Accept “fail to reject ” or “accept ”.
[2 marks]
A submarine is located in a sea at coordinates relative to a ship positioned at the origin . The direction is due east, the direction is due north and the direction is vertically upwards.
\nAll distances are measured in kilometres.
\nThe submarine travels with direction vector .
\nThe submarine reaches the surface of the sea at the point .
\nAssuming the submarine travels in a straight line, write down an equation for the line along which it travels.
\nFind the coordinates of .
\nFind .
\nA1A1
\n\n
Note: Award A1 for each correct vector. Award A0A1 if their “” is omitted.
\n
[2 marks]
(M1)
\n\n
(M1)
\nhas coordinates A1
\n
Note: Accept the coordinates of in vector form.
[3 marks]
(M1)
\nA1
\n
[2 marks]
The weights of apples from Tony’s farm follow a normal distribution with mean and standard deviation . The apples are sold in bags that contain six apples.
\nFind the mean weight of a bag of apples.
\nFind the standard deviation of the weights of these bags of apples.
\nFind the probability that a bag selected at random weighs more than .
\n(M1)A1
\n
[2 marks]
variance (M1)
\nA1
\n
[2 marks]
(M1)A1
\n
Note: Accept if sf value is used.
Award (M1)A1FT if the answer is correct for their SD, even if no working is shown. e.g. If the SD is then accept .
[2 marks]
The masses in kilograms of melons produced by a farm can be modelled by a normal distribution with a mean of and a standard deviation of .
\nFind the probability that two melons picked at random and independently of each other will
\nOne year due to favourable weather conditions it is thought that the mean mass of the melons has increased.
\nThe owner of the farm decides to take a random sample of melons to test this hypothesis at the significance level, assuming the standard deviation of the masses of the melons has not changed.
\nUnknown to the farmer the favourable weather conditions have led to all the melons having greater mass than the model described above.
\nFind the probability that a melon selected at random will have a mass greater than .
\nboth have a mass greater than .
\nhave a total mass greater than .
\nWrite down the null and alternative hypotheses for the test.
\nFind the critical region for this test.
\nFind the mean and standard deviation of the mass of the melons for this year.
\nFind the probability of a Type II error in the owner’s test.
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nLet represent the mass of a melon
\n(M1)A1
\n\n
[2 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nLet represent the total mass
\nA1
\n(M1)A1
\n\nA1
\n\n
[4 marks]
\nLet be the mean mass of the melons produced by the farm.
\n, only A1
\n\n
Note: Accept The mean mass of melons produced by the farm is equal to
The mean mass of melons produced by the farm is greater than
Note: Award A0 if does not appear in the hypothesis.
\n\n
[1 mark]
\nUnder A1
\n(M1)
\n(A1)
\nCritical region is A1
\n\n
[4 marks]
\nLet represent the new mass of the melons
\nA1
\nStandard deviation of (M1)
\nA1
\n\n
Note: award M1A0 for
\n\n
[3 marks]
\n(M1)
\nA1
\n\n
Note: Accept from use of the three‐figure answer to part (d)
\n\n
[2 marks]
\nThe diagram shows the slope field for the differential equation
\n.
\nThe graphs of the two solutions to the differential equation that pass through points and are shown.
\nFor the two solutions given, the local minimum points lie on the straight line .
\nFind the equation of , giving your answer in the form .
\nFor the two solutions given, the local maximum points lie on the straight line .
\nFind the equation of .
\nA1
\n(M1)
\n(the equation of is) A1
\n\n
[3 marks]
\nOR (M1)A1
\n\n
[2 marks]
\nA tank of water initially contains litres. Water is leaking from the tank such that after minutes there are litres remaining in the tank.
\nThe volume of water, litres, remaining in the tank after minutes, can be modelled by the differential equation
\n, where is a constant.
\nShow that .
\nFind the time taken for the tank to empty.
\nuse of separation of variables (M1)
\nA1
\nA1
\nconsidering initial conditions A1
\n\nA1
\n\nA1
\n\n
Note: Award A1 for any correct intermediate step that leads to the AG.
\n
AG
Note: Do not award the final A1 if the AG line is not stated.
[6 marks]
minutes (M1)A1
\n
[2 marks]
An ant is walking along the edges of a wire frame in the shape of a triangular prism.
\nThe vertices and edges of this frame can be represented by the graph below.
\nWrite down the adjacency matrix, , for this graph.
\nFind the number of ways that the ant can start at the vertex , and walk along exactly edges to return to .
\nA1A1A1
\n\n
Note: Award A1 for each two correct rows.
\n\n
[3 marks]
\ncalculating (M1)
\nA1
\n\n
[2 marks]
\nAlessia is an ecologist working for Mediterranean fishing authorities. She is interested in whether the mackerel population density is likely to fall below mackerel per , as this is the minimum value required for sustainable fishing. She believes that the primary factor affecting the mackerel population is the interaction of mackerel with sharks, their main predator.
\nThe population densities of mackerel ( thousands per ) and sharks ( per ) in the Mediterranean Sea are modelled by the coupled differential equations:
\n\n\nwhere is measured in years, and and are parameters.
\nThis model assumes that no other factors affect the mackerel or shark population densities.
\nThe term models the population growth rate of the mackerel in the absence of sharks.
The term models the death rate of the mackerel due to being eaten by sharks.
Suggest similar interpretations for the following terms.
\nAn equilibrium point is a set of values of and , such that and .
\nGiven that both species are present at the equilibrium point,
\nThe equilibrium point found in part (b) gives the average values of and over time.
Use the model to predict how the following events would affect the average value of . Justify your answers.
To estimate the value of , Alessia considers a situation where there are no sharks and the initial mackerel population density is .
\nBased on additional observations, it is believed that
\n,
\n,
\n,
\n.
\nAlessia decides to use Euler’s method to estimate future mackerel and shark population densities. The initial population densities are estimated to be and . She uses a step length of years.
\nAlessia will use her model to estimate whether the mackerel population density is likely to fall below the minimum value required for sustainable fishing, per , during the first nine years.
\nshow that, at the equilibrium point, the value of the mackerel population density is ;
\nfind the value of the shark population density at the equilibrium point.
\nToxic sewage is added to the Mediterranean Sea. Alessia claims this reduces the shark population growth rate and hence the value of is halved. No other parameter changes.
\nGlobal warming increases the temperature of the Mediterranean Sea. Alessia claims that this promotes the mackerel population growth rate and hence the value of is doubled. No other parameter changes.
\nWrite down the differential equation for that models this situation.
\nShow that the expression for the mackerel population density after years is
\nAlessia estimates that the mackerel population density increases by a factor of three every two years. Show that to three significant figures.
\nWrite down expressions for and in terms of and .
\nUse Euler’s method to find an estimate for the mackerel population density after one year.
\nUse Euler’s method to sketch the trajectory of the phase portrait, for and , over the first nine years.
\nUsing your phase portrait, or otherwise, determine whether the mackerel population density would be sufficient to support sustainable fishing during the first nine years.
\nState two reasons why Alessia’s conclusion, found in part (f)(ii), might not be valid.
\npopulation growth rate / birth rate of sharks (due to eating mackerel) A1
\n
[1 mark]
(net) death rate of sharks A1
\n
[1 mark]
A1
\nsince R1
\n
Note: Accept .
getting to given answer without further error by either cancelling or factorizing A1
AG
\n
[3 marks]
(M1)
\n(since ) A1
\n
[2 marks]
M1
\nNote: Accept equivalent in words.
\n
Doubles A1
Note: Do not accept “increases”.
[2 marks]
is not dependent on R1
\n
Note: Award R0 for any contextual argument.
no change A1
Note: Do not award R0A1.
[2 marks]
A1
\n
[1 mark]
M1
\n
Note: Award M1 is for an attempt to separate variables. This means getting to the point where the integral can be seen or implied by further work.
A1
Note: Accept . Condone missing constant of integration for this mark.
when M1
\n
Note: Award M1 for a clear attempt at using initial conditions to find a constant of integration. Only possible if the constant of integration exists. or “initially” or similar must be seen. Substitution may appear earlier, following the integration.
initial conditions and all other manipulations correct and clearly communicated to get to the final answer A1
AG
\n
[4 marks]
seen anywhere (A1)
\nsubstituting into equation (M1)
\n\nOR A1
\n
Note: The A1 requires either the exact answer or an answer to at least sf.
AG
\n
[3 marks]
\nan attempt to set up one recursive equation (M1)
\n
Note: Must include two given parameters and and and or for the (M1) to be awarded.
A1
A1
\n\n
[3 marks]
\nEITHER
A2
OR
(mackerel per ) A2
\n
[2 marks]
\nspiral or closed loop shape A1
\napproximately rotations (can only be awarded if a spiral) A1
\ncorrect shape, in approximately correct position (centred at approx. ) A1
\n
Note: Award A0A0A0 for any plot of or against .
[3 marks]
\nEITHER
\napproximate minimum is (which is greater than ) A1
\n
OR
the line clearly labelled on their phase portrait A1
\n
THEN
(the density will not fall below ) hence sufficient for sustainable fishing A1
Note: Do not award A0A1. Only if the minimum point is labelled on the sketch then a statement here that “the mackerel population is always above ” would be sufficient. Accept the value seen within a table of values.
[2 marks]
\nAny two from: A1A1
\n• Current values / parameters are only an estimate,
\n• The Euler method is only an approximate method / choosing might be too large.
\n• There might be random variation / the model has no stochastic component
\n• Conditions / parameters might change over the nine years,
\n• A discrete system is being approximated by a continuous system,
\nAllow any other sensible critique.
\n
If a candidate identifies factors which the model ignores, award A1 per factor identified. These factors could include:
• Other predators
\n• Seasonality
\n• Temperature
\n• The effect of fishing
\n• Environmental catastrophe
\n• Migration
\n
Note: Do not allow:
“You cannot have mackerel”.
It is only a model (as this is too vague).
Some factors have been ignored (without specifically identifying the factors).
Values do not always follow the equation / model. (as this is too vague).
\n
[2 marks]
\nThe graph of the function is translated by so that it then passes through the points and .
\nFind the value of and the value of .
\nnew function is (M1)
\nA1
\nA1
\n(M1)
\n\n\n\nA1
\n(M1)A1
\n\n
[7 marks]
\nA ball is attached to the end of a string and spun horizontally. Its position relative to a given point, , at time seconds, , is given by the equation
\nwhere all displacements are in metres.
\nThe string breaks when the magnitude of the ball’s acceleration exceeds .
\nShow that the ball is moving in a circle with its centre at and state the radius of the circle.
\nFind an expression for the velocity of the ball at time .
\nHence show that the velocity of the ball is always perpendicular to the position vector of the ball.
\nFind an expression for the acceleration of the ball at time .
\nFind the value of at the instant the string breaks.
\nHow many complete revolutions has the ball completed from to the instant at which the string breaks?
\n* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
\nM1
\nas R1
\n\n
Note: use of the identity needs to be explicitly stated.
\n\n
Hence moves in a circle as displacement from a fixed point is constant. R1
\nRadius A1
\n\n
[4 marks]
\nM1A1
\n\n
Note: M1 is for an attempt to differentiate each term
\n\n
[2 marks]
\nM1
\n\n
Note: M1 is for an attempt to find
\n\n
A1
\nHence velocity and position vector are perpendicular. AG
\n\n
[2 marks]
\nM1A1A1
\n\n
[3 marks]
\n(M1)(A1)
\n\n
Note: M1 is for an attempt to equate the magnitude of the acceleration to .
\n\n
A1
\n\n
[3 marks]
\nAngle turned through is M1
\nA1
\nM1
\n\ncomplete revolutions A1
\n\n
[4 marks]
\nA manufacturer of chocolates produces them in individual packets, claiming to have an average of chocolates per packet.
\nTalha bought of these packets in order to check the manufacturer’s claim.
\nGiven that the number of individual chocolates is , Talha found that, from his packets, and .
\nFind an unbiased estimate for the mean number of chocolates per packet.
\nUse the formula to determine an unbiased estimate for the variance of the number of chocolates per packet.
\nFind a confidence interval for . You may assume that all conditions for a confidence interval have been met.
\nSuggest, with justification, a valid conclusion that Talha could make.
\nA1
\n
[1 mark]
(M1)
\nA1
\n
[2 marks]
A2
\n
[2 marks]
is outside the confidence interval and therefore Talha would suggest that the manufacturer’s claim is incorrect R1
Note: The conclusion must refer back to the original claim.
Allow use of a two sided -test giving a -value rounding to and therefore Talha would suggest that the manufacturer’s claims in incorrect.
\n
[1 mark]
On Paul’s farm, potatoes are packed in sacks labelled . The weights of the sacks of potatoes can be modelled by a normal distribution with mean weight and standard deviation .
\nFind the probability that a sack is under its labelled weight.
\nFind the lower quartile of the weights of the sacks of potatoes.
\nThe sacks of potatoes are transported in crates. There are sacks in each crate and the weights of the sacks of potatoes are independent of each other.
\nFind the probability that the total weight of the sacks of potatoes in a crate exceeds .
\nlet be the random variable “the weight of a sack of potatoes”
\n(M1)
\nA1
\n\n
[2 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nattempt to sum independent random variables (M1)
\n(A1)
\nA1
\n\n
[3 marks]
\nThe first part of the question was often answered well but there were a number of candidates who interpreted finding by finding or something similar. Not all candidates, however, understood that the lower quartile is given by . Part (c) was less well understood. Attempts to sum independent random variables correctly involved multiplication of the mean by but the standard deviation and not the variance was incorrectly multiplied by .
\nThe first part of the question was often answered well but there were a number of candidates who interpreted finding by finding or something similar. Not all candidates, however, understood that the lower quartile is given by . Part (c) was less well understood. Attempts to sum independent random variables correctly involved multiplication of the mean by but the standard deviation and not the variance was incorrectly multiplied by .
\nThe first part of the question was often answered well but there were a number of candidates who interpreted finding by finding or something similar. Not all candidates, however, understood that the lower quartile is given by . Part (c) was less well understood. Attempts to sum independent random variables correctly involved multiplication of the mean by but the standard deviation and not the variance was incorrectly multiplied by .
\nThe diagram below shows a network of roads in a small village with the weights indicating the distance of each road, in metres, and junctions indicated with letters.
\nMusab is required to deliver leaflets to every house on each road. He wishes to minimize his total distance.
\n\n
Musab starts and finishes from the village bus-stop at . Determine the total distance Musab will need to walk.
\nInstead of having to catch the bus to the village, Musab’s sister offers to drop him off at any junction and pick him up at any other junction of his choice.
\nExplain which junctions Musab should choose as his starting and finishing points.
\nOdd vertices are A1
\nConsider pairings: M1
\n
Note: Award (M1) if there are four vertices not necessarily all correct.
has shortest route and
so repeated edges
Note: Condone in place of giving .
has shortest route and
so repeated edges
has shortest route and ,
so repeated edges A2
Note: Award A1 if only one or two pairings are correctly considered.
so best pairing is
weight of route is therefore A1
[5 marks]
least value of the pairings is therefore repeat R1
\n and A1
Note: Do not award R0A1.
[2 marks]
A firm wishes to review its recruitment processes. This question considers the validity and reliability of the methods used.
\nEvery year an accountancy firm recruits new employees for a trial period of one year from a large group of applicants.
\nAt the start, all applicants are interviewed and given a rating. Those with a rating of either Excellent, Very good or Good are recruited for the trial period. At the end of this period, some of the new employees will stay with the firm.
\nIt is decided to test how valid the interview rating is as a way of predicting which of the new employees will stay with the firm.
\nData is collected and recorded in a contingency table.
\nThe next year’s group of applicants are asked to complete a written assessment which is then analysed. From those recruited as new employees, a random sample of size is selected.
\nThe sample is stratified by department. Of the new employees recruited that year, were placed in the national department and in the international department.
\nAt the end of their first year, the level of performance of each of the employees in the sample is assessed by their department manager. They are awarded a score between (low performance) and (high performance).
\nThe marks in the written assessment and the scores given by the managers are shown in both the table and the scatter diagram.
\nThe firm decides to find a Spearman’s rank correlation coefficient, , for this data.
\nThe same seven employees are given the written assessment a second time, at the end of the first year, to measure its reliability. Their marks are shown in the table below.
\nThe written assessment is in five sections, numbered to . At the end of the year, the employees are also given a score for each of five professional attributes: and .
\nThe firm decides to test the hypothesis that there is a correlation between the mark in a section and the score for an attribute.
\nThey compare marks in each of the sections with scores for each of the attributes.
\nUse an appropriate test, at the significance level, to determine whether a new employee staying with the firm is independent of their interview rating. State the null and alternative hypotheses, the -value and the conclusion of the test.
\nShow that employees are selected for the sample from the national department.
\nWithout calculation, explain why it might not be appropriate to calculate a correlation coefficient for the whole sample of employees.
\nFind for the seven employees working in the international department.
\nHence comment on the validity of the written assessment as a measure of the level of performance of employees in this department. Justify your answer.
\nState the name of this type of test for reliability.
\nFor the data in this table, test the null hypothesis, , against the alternative hypothesis, , at the significance level. You may assume that all the requirements for carrying out the test have been met.
\nHence comment on the reliability of the written assessment.
\nWrite down the number of tests they carry out.
\nThe tests are performed at the significance level.
Assuming that:
find the probability that at least one of the tests will be significant.
\nThe firm obtains a significant result when comparing section of the written assessment and attribute . Interpret this result.
\nUse of test for independence (M1)
\n Staying (or leaving) the firm and interview rating are independent.
Staying (or leaving) the firm and interview rating are not independent A1
Note: For accept ‘…are dependent’ in place of ‘…not independent’.
-value A2
Note: Award A1 for if -value is omitted or incorrect.
R1
(the result is not significant at the level)
\ninsufficient evidence to reject the (or “accept ”) A1
\n
Note: Do not award R0A1. The final R1A1 can follow through from their incorrect -value
\n
[6 marks]
\nM1A1
\n
Note: Award A1 for anything that rounds to .
AG
[2 marks]
there seems to be a difference between the two departments (A1)
\nthe international department manager seems to be less generous than the national department manager R1
\n
Note: The A1 is for commenting there is a difference between the two departments and the R1 is for correctly commenting on the direction of the difference
[2 marks]
(M1)(A1)
Note: Award (M1) for an attempt to rank the data, and (A1) for correct ranks for both variables. Accept either set of rankings in reverse.
(M1)(A1)
Note: The (M1) is for calculating the PMCC for their ranks.
Note: If a final answer of is seen, from use of , award (M1)(A1)A1.
Accept if one set of ranks has been ordered in reverse.
[4 marks]
EITHER
\nthere is a (strong) association between the written assessment mark and the manager scores. A1
OR
\nthere is a (strong) agreement in the rank order of the written assessment marks and the rank order of the manager scores. A1
OR
\nthere is a (strong linear) correlation between the rank order of the written assessment marks and the rank order of the manager scores. A1
\n
Note: Follow through on a value for their value of in c(ii).
THEN
\nthe written assessment is likely to be a valid measure (of the level of employee performance) R1
\n
[2 marks]
test-retest A1
\n
[1 mark]
-value A2
\nR1
\n(the result is significant at the level)
(there is sufficient evidence to) reject A1
Note: Do not award R0A1. Accept “accept ”. The final R1A1 can follow through from their incorrect -value.
[4 marks]
the test seems reliable A1
\n
Note: Follow through from their answer in part (d)(ii). Do not award if there is no conclusion in d(ii).
[1 mark]
A1
\n
[1 mark]
probability of significant result given no correlation is (M1)
\nprobability of at least one significant result in tests is
\n(M1)(A1)
\n
Note: Award (M1) for use of or the binomial distribution with any value of .
A1
[4 marks]
(though the result is significant) it is very likely that one significant result would be achieved by chance, so it should be disregarded or further evidence sought R1
\n
[1 mark]
It is given that and .
\nIn parts (a)(i) and (a)(ii), give your answers in the form .
\nFind the value of .
\nFind the value of for .
\nFind the least value of such that .
\n A1A1
Note: Award A1 for and A1 for the angle in the correct form.
[2 marks]
A1A2
Note: Award A1 for , A2 for the angle in the correct form and A1 for the angle in incorrect form e.g. and/or . Award A1 if is given in place of .
[3 marks]
(M1)
\n\n(M1)
\n A1
[3 marks]
The graph below shows a small maze, in the form of a network of directed routes. The vertices to show junctions in the maze and the edges show the possible paths available from one vertex to another.
\nA mouse is placed at vertex and left to wander the maze freely. The routes shown by dashed lines indicate paths sprinkled with sugar.
\nWhen the mouse reaches any junction, she rests for a constant time before continuing.
\nAt any junction, it may also be assumed that
\nDetermine the transition matrix for this graph.
\nIf the mouse was left to wander indefinitely, use your graphic display calculator to estimate the percentage of time that the mouse would spend at point .
\nComment on your answer to part (b), referring to at least one limitation of the model.
\ntransition matrix is M1A1A1
Note: Allow the transposed matrix.
Award M1 for a matrix with all values between and , and all columns (or rows if transposed) adding up to , award A1 for one correct row (or column if transposed) and A1 for all rows (or columns if transposed) correct.
[3 marks]
attempting to raise the transition matrix to a large power (M1)
\nsteady state vector is (A1)
\nso percentage of time spent at vertex is A1
\n
Note: Accept .
[3 marks]
the model assumes instantaneous travel from junction to junction, R1
and hence the answer obtained would be an overestimate R1
OR
the mouse may eat the sugar over time R1
and hence the probabilities would change R1
Note: Accept any other sensible answer.
[3 marks]
Eduardo believes that there is a linear relationship between the age of a male runner and the time it takes them to run metres.
\nTo test this, he recorded the age, years, and the time, minutes, for eight males in a single race. His results are presented in the following table and scatter diagram.
\nEduardo looked in a sports science text book. He found that the following information about was appropriate for athletic performance.
\nFor this data, find the value of the Pearson’s product-moment correlation coefficient, .
\nComment on your answer to part (a), using the information that Eduardo found.
\nWrite down the equation of the regression line of on , in the form .
\nA 57-year-old male also ran in the race.
\nUse the equation of the regression line to estimate the time he took to complete the race.
\nA2
\n\n
[2 marks]
\nstrong A1
\n
Note: Answer may include “positive”, however this is not necessary for the mark.
\n
[1 mark]
\nA1
\n
Note: Condone in place of . Answer must be an equation.
\n
[1 mark]
\n(M1)
\n
Note: Award (M1) for correct substitution into their regression line.
A1
Note: Accept and from use of sf and/or sf values.
[2 marks]
Generally, a good starting point for most candidates. Most of them could find the correlation coefficient with a correct interpretation. They could find the regression equation and use it appropriately to make a prediction. It was observed that candidates used a mixture of two and three significant figures to reach an answer. Marks were generally awarded for their correct three significant answers or answers given to a higher degree of accuracy. The weaker candidates substituted instead of years into their equation.
\nGenerally, a good starting point for most candidates. Most of them could find the correlation coefficient with a correct interpretation. They could find the regression equation and use it appropriately to make a prediction. It was observed that candidates used a mixture of two and three significant figures to reach an answer. Marks were generally awarded for their correct three significant answers or answers given to a higher degree of accuracy. The weaker candidates substituted instead of years into their equation.
\nGenerally, a good starting point for most candidates. Most of them could find the correlation coefficient with a correct interpretation. They could find the regression equation and use it appropriately to make a prediction. It was observed that candidates used a mixture of two and three significant figures to reach an answer. Marks were generally awarded for their correct three significant answers or answers given to a higher degree of accuracy. The weaker candidates substituted instead of years into their equation.
\nGenerally, a good starting point for most candidates. Most of them could find the correlation coefficient with a correct interpretation. They could find the regression equation and use it appropriately to make a prediction. It was observed that candidates used a mixture of two and three significant figures to reach an answer. Marks were generally awarded for their correct three significant answers or answers given to a higher degree of accuracy. The weaker candidates substituted instead of years into their equation.
\nThe following diagram shows a frame that is made from wire. The total length of wire is equal to . The frame is made up of two identical sectors of a circle that are parallel to each other. The sectors have angle radians and radius . They are connected by lengths of wire perpendicular to the sectors. This is shown in the diagram below.
\nThe faces of the frame are covered by paper to enclose a volume, .
\nShow that .
\nFind an expression for in terms of .
\nFind the expression .
\nSolve algebraically to find the value of that will maximize the volume, .
\nM1
\nA1
\n
Note: Award A1 for any reasonable working leading to expected result e,g, factorizing .
AG
\n
[2 marks]
\nattempt to use sector area to find volume (M1)
\nvolume
\nA1
\n\n
[2 marks]
\nM1A1A1
\n\n\n
[3 marks]
\nM1
\n
Note: Award this M1 for simplified version equated to zero. The simplified version may have been seen in part (b)(ii).
A1
\n
[2 marks]
\nSeveral candidates missed that the angle was in radians and used arc and sector formulas with degrees instead. This aside, part (a) was often well done. Part (b)(i) was also correctly answered by many candidates, but their failure to make any attempt to simplify their answer often led to difficulties in part (b)(ii). Again, failing to simplify the result in part (b)(ii) led to yet more difficulties in part (b)(iii). Some candidates used the product rule to differentiate as rather than the quotient rule. This was fine but made solving the equation in (b)(iii) less straightforward.
\nSeveral candidates missed that the angle was in radians and used arc and sector formulas with degrees instead. This aside, part (a) was often well done. Part (b)(i) was also correctly answered by many candidates, but their failure to make any attempt to simplify their answer often led to difficulties in part (b)(ii). Again, failing to simplify the result in part (b)(ii) led to yet more difficulties in part (b)(iii). Some candidates used the product rule to differentiate as rather than the quotient rule. This was fine but made solving the equation in (b)(iii) less straightforward.
\nSeveral candidates missed that the angle was in radians and used arc and sector formulas with degrees instead. This aside, part (a) was often well done. Part (b)(i) was also correctly answered by many candidates, but their failure to make any attempt to simplify their answer often led to difficulties in part (b)(ii). Again, failing to simplify the result in part (b)(ii) led to yet more difficulties in part (b)(iii). Some candidates used the product rule to differentiate as rather than the quotient rule. This was fine but made solving the equation in (b)(iii) less straightforward.
\nSeveral candidates missed that the angle was in radians and used arc and sector formulas with degrees instead. This aside, part (a) was often well done. Part (b)(i) was also correctly answered by many candidates, but their failure to make any attempt to simplify their answer often led to difficulties in part (b)(ii). Again, failing to simplify the result in part (b)(ii) led to yet more difficulties in part (b)(iii). Some candidates used the product rule to differentiate as rather than the quotient rule. This was fine but made solving the equation in (b)(iii) less straightforward.
\nA geometric transformation is defined by
\n.
\nFind the coordinates of the image of the point .
\nGiven that , find the value of and the value of .
\nA triangle with vertices lying on the plane is transformed by .
\nExplain why both and its image will have exactly the same area.
\n(M1)
\nOR A1
\n
[2 marks]
(M1)
\n\n(A1)
\nsolve simultaneously:
\nA1
\n\n
Note: Award A0 if and are not labelled or are labelled the other way around.
\n
[3 marks]
A1
\nscale factor of image area is therefore (and the translation does not affect the area) A1
\n
[2 marks]
A medical centre is testing patients for a certain disease. This disease occurs in of the population.
\nThey test every patient who comes to the centre on a particular day.
\nIt is intended that if a patient has the disease, they test “positive”, and if a patient does not have the disease, they test “negative”.
\nHowever, the tests are not perfect, and only of people who have the disease test positive. Also, of people who do not have the disease test positive.
\nThe tree diagram shows some of this information.
\nWrite down the value of
\nUse the tree diagram to find the probability that a patient selected at random
\nThe staff at the medical centre looked at the care received by all visiting patients on a randomly chosen day. All the patients received at least one of these services: they had medical tests (), were seen by a nurse (), or were seen by a doctor (). It was found that:
\nState the sampling method being used.
\n.
\n.
\n.
\n.
\nwill not have the disease and will test positive.
\nwill test negative.
\nhas the disease given that they tested negative.
\nThe medical centre finds the actual number of positive results in their sample is different than predicted by the tree diagram. Explain why this might be the case.
\nDraw a Venn diagram to illustrate this information, placing all relevant information on the diagram.
\nFind the total number of patients who visited the centre during this day.
\nconvenience sampling (A1)
\n
[1 mark]
A1
\n
[1 mark]
A1
\n
[1 mark]
A1
\n
[1 mark]
A1
\n
[1 mark]
(M1)
\nA1
\n
[2 marks]
(M1)(M1)
\n
Note: Award M1 for summing two products and M1 for correct products seen.
A1
[3 marks]
recognition of conditional probability (M1)
\nA1
\nA1
\n
Note: Accept if used.
[3 marks]
EITHER
sample may not be representative of population A1
OR
sample is not randomly selected A1
OR
unrealistic to think expected and observed values will be exactly equal A1
[1 mark]
A1A1A1
Note: Award A1 for rectangle and labelled circles and in centre region; A1 for ; A1 for and .
[3 marks]
(M1)
\n A1
Note: Follow through from the entries on their Venn diagram in part (e). Working required for FT.
[2 marks]
A modern art painting is contained in a square frame. The painting has a shaded region bounded by a smooth curve and a horizontal line.
\nWhen the painting is placed on a coordinate axes such that the bottom left corner of the painting has coordinates and the top right corner has coordinates , the curve can be modelled by and the horizontal line can be modelled by the -axis. Distances are measured in metres.
\nThe artist used the equation to draw the curve.
\nUse the trapezoidal rule, with the values given in the following table, to approximate the area of the shaded region.
\nFind the exact area of the shaded region in the painting.
\nFind the area of the unshaded region in the painting.
\n(A1)(M1)
\n
Note: Award A1 for evidence of , M1 for a correct substitution into trapezoidal rule (allow for an incorrect only). The zero can be omitted in the working.
A1
\n
[3 marks]
\nOR (M1)
\n
Note: Award M1 for using definite integration with correct limits.
A1
\n
Note: Question requires exact answer, do not award final A1 for .
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award M1 for seen as part of a subtraction.
\n
A1
\n
[2 marks]
\nThere seemed a better attempt at using the trapezium rule in this session compared to the two 2021 sessions. Despite many incorrect values for , candidates obtained the method mark for a correctly substituted formula (excluding ).
\n\n
The exact answer of 2.925 was asked for in the question, yet candidates frequently rounded to three significant figures and hence lost the final mark.
\n\n
Many candidates were able to correctly find the area of the unshaded region.
\nThe Bermuda Triangle is a region of the Atlantic Ocean with Miami , Bermuda , and San Juan as vertices, as shown on the diagram.
\nThe distances between , and are given in the following table, correct to three significant figures.
\nCalculate the value of , the measure of angle .
\nFind the area of the Bermuda Triangle.
\nattempt at substituting the cosine rule formula (M1)
\n(A1)
\n(accept rad ) A1
\n
[3 marks]
correctly substituted area of triangle formula (M1)
\n\nA1
\n
Note: Accept from use of . Other angles and their corresponding sides may be used.
[2 marks]
Most candidates were successful at selecting the cosine rule formula in part (a). In most cases, the cosine rule formula was correctly substituted. Some candidates found it hard to choose the correct side to obtain the required angle. Most of the candidates scored one or two marks out of three in this part. In part (b) some candidates assumed the triangle was right angled and used instead of . In part (b) many candidates who answered part (a) incorrectly were able to recover. Many candidates managed to score full marks in this part despite an incorrect answer in part (a). Some found angle or in part (a) but used the correct sides to obtain the correct area. Final answers given in calculator notations (such as ) scored at most one mark out of two. Calculator notation should generally be avoided; it is considered too informal to earn A marks, and although it can imply a method and earn M marks, we advise that candidates still provide the necessary commentary to support any GDC notation.
\nMost candidates were successful at selecting the cosine rule formula in part (a). In most cases, the cosine rule formula was correctly substituted. Some candidates found it hard to choose the correct side to obtain the required angle. Most of the candidates scored one or two marks out of three in this part. In part (b) some candidates assumed the triangle was right angled and used instead of . In part (b) many candidates who answered part (a) incorrectly were able to recover. Many candidates managed to score full marks in this part despite an incorrect answer in part (a). Some found angle or in part (a) but used the correct sides to obtain the correct area. Final answers given in calculator notations (such as ) scored at most one mark out of two. Calculator notation should generally be avoided; it is considered too informal to earn A marks, and although it can imply a method and earn M marks, we advise that candidates still provide the necessary commentary to support any GDC notation.
\nA group of students were asked which electronic device they preferred. The results per age group are given in the following table.
\nA student from the group is chosen at random. Calculate the probability that the student
\nA test for independence was performed on the collected data at the significance level. The critical value for the test is .
\nprefers a tablet.
\nis years old and prefers a mobile phone.
\nprefers a laptop given that they are years old.
\nprefers a tablet or is years old.
\nState the null and alternative hypotheses.
\nWrite down the number of degrees of freedom.
\nWrite down the test statistic.
\nWrite down the -value.
\nState the conclusion for the test in context. Give a reason for your answer.
\nA1A1
\n
Note: Award A1 for correct numerator, A1 for correct denominator.
\n
[2 marks]
\nA1A1
\n
Note: Award A1 for correct numerator, A1 for correct denominator.
\n
[2 marks]
\nA1A1
\n
Note: Award A1 for correct numerator, A1 for correct denominator.
\n
[2 marks]
\nOR (M1)
\nA1A1
\n
Note: Award A1 for correct denominator seen, (M1) for correct calculation of the numerator, A1 for the correct answer.
\n
[3 marks]
\nthe variables are independent
\nthe variables are dependent A1
\n
Note: Award A1 for for both hypotheses correct. Do not accept “not correlated” or “not related” in place of “independent”.
\n
[1 mark]
\nA1
\n\n
[1 mark]
\nA2
\n\n
[2 marks]
\nOR A1
\n\n
[1 mark]
\nEITHER
\nR1
\nOR
\nR1
\n
THEN
(there is sufficient evidence to accept that) preferred device and age group are not independent A1
\n
Note: For the final A1 the answer must be in context. Do not award A1R0.
\n
[2 marks]
\nA common error was to treat these as compound probability events. Candidates who attempted to use the conditional probability formula in part (a)(iii) often accrued at least one mark for a correct method. Some lost one mark for not reducing the total sample space. In part (a)(iv) many found , failing to subtract the intersection of the two events. When stating test hypotheses, terminology such as \"related\", \"correlated\", \"data is independent\" are not acceptable. Few candidates attempted to frame this as a Goodness of Fit test. Candidates correctly wrote down the degrees of freedom. It was pleasing to see most found the correct -test statistic and associated p-value. Relative to past examination sessions, fewer candidates appeared to compare the critical value with the p-value or the -test statistic with the level of significance. A conclusion consistent with their hypotheses was usually seen, though this was not always expressed in context.
\nA common error was to treat these as compound probability events. Candidates who attempted to use the conditional probability formula in part (a)(iii) often accrued at least one mark for a correct method. Some lost one mark for not reducing the total sample space. In part (a)(iv) many found , failing to subtract the intersection of the two events. When stating test hypotheses, terminology such as \"related\", \"correlated\", \"data is independent\" are not acceptable. Few candidates attempted to frame this as a Goodness of Fit test. Candidates correctly wrote down the degrees of freedom. It was pleasing to see most found the correct -test statistic and associated p-value. Relative to past examination sessions, fewer candidates appeared to compare the critical value with the p-value or the -test statistic with the level of significance. A conclusion consistent with their hypotheses was usually seen, though this was not always expressed in context.
\nA common error was to treat these as compound probability events. Candidates who attempted to use the conditional probability formula in part (a)(iii) often accrued at least one mark for a correct method. Some lost one mark for not reducing the total sample space. In part (a)(iv) many found , failing to subtract the intersection of the two events. When stating test hypotheses, terminology such as \"related\", \"correlated\", \"data is independent\" are not acceptable. Few candidates attempted to frame this as a Goodness of Fit test. Candidates correctly wrote down the degrees of freedom. It was pleasing to see most found the correct -test statistic and associated p-value. Relative to past examination sessions, fewer candidates appeared to compare the critical value with the p-value or the -test statistic with the level of significance. A conclusion consistent with their hypotheses was usually seen, though this was not always expressed in context.
\nA common error was to treat these as compound probability events. Candidates who attempted to use the conditional probability formula in part (a)(iii) often accrued at least one mark for a correct method. Some lost one mark for not reducing the total sample space. In part (a)(iv) many found , failing to subtract the intersection of the two events. When stating test hypotheses, terminology such as \"related\", \"correlated\", \"data is independent\" are not acceptable. Few candidates attempted to frame this as a Goodness of Fit test. Candidates correctly wrote down the degrees of freedom. It was pleasing to see most found the correct -test statistic and associated p-value. Relative to past examination sessions, fewer candidates appeared to compare the critical value with the p-value or the -test statistic with the level of significance. A conclusion consistent with their hypotheses was usually seen, though this was not always expressed in context.
\nA common error was to treat these as compound probability events. Candidates who attempted to use the conditional probability formula in part (a)(iii) often accrued at least one mark for a correct method. Some lost one mark for not reducing the total sample space. In part (a)(iv) many found , failing to subtract the intersection of the two events. When stating test hypotheses, terminology such as \"related\", \"correlated\", \"data is independent\" are not acceptable. Few candidates attempted to frame this as a Goodness of Fit test. Candidates correctly wrote down the degrees of freedom. It was pleasing to see most found the correct -test statistic and associated p-value. Relative to past examination sessions, fewer candidates appeared to compare the critical value with the p-value or the -test statistic with the level of significance. A conclusion consistent with their hypotheses was usually seen, though this was not always expressed in context.
\nA common error was to treat these as compound probability events. Candidates who attempted to use the conditional probability formula in part (a)(iii) often accrued at least one mark for a correct method. Some lost one mark for not reducing the total sample space. In part (a)(iv) many found , failing to subtract the intersection of the two events. When stating test hypotheses, terminology such as \"related\", \"correlated\", \"data is independent\" are not acceptable. Few candidates attempted to frame this as a Goodness of Fit test. Candidates correctly wrote down the degrees of freedom. It was pleasing to see most found the correct -test statistic and associated p-value. Relative to past examination sessions, fewer candidates appeared to compare the critical value with the p-value or the -test statistic with the level of significance. A conclusion consistent with their hypotheses was usually seen, though this was not always expressed in context.
\nA common error was to treat these as compound probability events. Candidates who attempted to use the conditional probability formula in part (a)(iii) often accrued at least one mark for a correct method. Some lost one mark for not reducing the total sample space. In part (a)(iv) many found , failing to subtract the intersection of the two events. When stating test hypotheses, terminology such as \"related\", \"correlated\", \"data is independent\" are not acceptable. Few candidates attempted to frame this as a Goodness of Fit test. Candidates correctly wrote down the degrees of freedom. It was pleasing to see most found the correct -test statistic and associated p-value. Relative to past examination sessions, fewer candidates appeared to compare the critical value with the p-value or the -test statistic with the level of significance. A conclusion consistent with their hypotheses was usually seen, though this was not always expressed in context.
\nA common error was to treat these as compound probability events. Candidates who attempted to use the conditional probability formula in part (a)(iii) often accrued at least one mark for a correct method. Some lost one mark for not reducing the total sample space. In part (a)(iv) many found , failing to subtract the intersection of the two events. When stating test hypotheses, terminology such as \"related\", \"correlated\", \"data is independent\" are not acceptable. Few candidates attempted to frame this as a Goodness of Fit test. Candidates correctly wrote down the degrees of freedom. It was pleasing to see most found the correct -test statistic and associated p-value. Relative to past examination sessions, fewer candidates appeared to compare the critical value with the p-value or the -test statistic with the level of significance. A conclusion consistent with their hypotheses was usually seen, though this was not always expressed in context.
\nA common error was to treat these as compound probability events. Candidates who attempted to use the conditional probability formula in part (a)(iii) often accrued at least one mark for a correct method. Some lost one mark for not reducing the total sample space. In part (a)(iv) many found , failing to subtract the intersection of the two events. When stating test hypotheses, terminology such as \"related\", \"correlated\", \"data is independent\" are not acceptable. Few candidates attempted to frame this as a Goodness of Fit test. Candidates correctly wrote down the degrees of freedom. It was pleasing to see most found the correct -test statistic and associated p-value. Relative to past examination sessions, fewer candidates appeared to compare the critical value with the p-value or the -test statistic with the level of significance. A conclusion consistent with their hypotheses was usually seen, though this was not always expressed in context.
\nLeo is investigating whether a six-sided die is fair. He rolls the die times and records the observed frequencies in the following table:
\nLeo carries out a goodness of fit test at a significance level.
\nWrite down the null and alternative hypotheses.
\nWrite down the degrees of freedom.
\nWrite down the expected frequency of rolling a .
\nFind the -value for the test.
\nState the conclusion of the test. Give a reason for your answer.
\nThe die is fair OR OR probabilities are equal
\nThe die is not fair OR OR probabilities are not equal A1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\n(-value ) A2
\n\n
[2 marks]
\nR1
\n
EITHER
Insufficient evidence to reject the null hypothesis A1
\n
OR
Insufficient evidence to reject that the die is fair A1
\n\n
Note: Do not award R0A1. Condone “accept the null hypothesis” or “the die is fair”. Their conclusion must be consistent with their -value and their hypothesis.
\n\n
[2 marks]
\nMany candidates confused this goodness of fit question with a χ2 test for independence and so incorrect statements for the hypotheses were seen frequently.
\n\n
Many candidates stated the degrees of freedom correctly.
\n\n
\n
Expected value was not well understood. The most popular but erroneous answer was where the candidate calculated .
\n\n
Some confusion between the required -value and the χ2 value. In the case of the latter, no further marks were available for this question. Too many candidates wrote down a value with .
\n\n
Comparing their -value with 0.05 or 5% correctly earned a mark for reasoning. Obtaining the reasoning mark enabled even those candidates with incorrect hypotheses in part (a), to be credited the final mark provided the conclusion was clear.
\nThe number of coffees sold per hour at an independent coffee shop is modelled by a Poisson distribution with a mean of coffees per hour.
\nSheila, the shop’s owner wants to increase the number of coffees sold in the shop. She decides to offer a discount to customers who buy more than one coffee.
\nTo test how successful this strategy is, Sheila records the number of coffees sold over a single -hour period. Sheila decides to use a level of significance in her test.
\nState the null and alternative hypotheses for the test.
\nFind the probability that Sheila will make a type I error in her test conclusion.
\nSheila finds coffees were sold during the -hour period.
\nState Sheila’s conclusion to the test. Justify your answer.
\nA1
\nNote: Accept other appropriate variables for the mean.
Accept in place of .
[1 mark]
(M1)(A1)
\n(M1)
\n(probability of making a type I error is) A1
\n
Note: If other probabilities are seen, the final A1 cannot be awarded unless is clearly identified as the final answer.
[4 marks]
OR recognizing or R1
\nso there is insufficient evidence to reject A1
\n(ie there is insufficient evidence to suggest that the number of coffees being sold has increased)
\n
Note: Accept ‘Accept ’.
Do not award R0A1.
[2 marks]
A ship is travelling with a constant velocity, , measured in kilometres per hour, where
\n.
\nAt time the ship is at a point relative to an origin , where distances are measured in kilometres.
\nA lighthouse is located at a point .
\nFind the position vector of the ship at time hours.
\nFind the value of when the ship will be closest to the lighthouse.
\nAn alarm will sound if the ship travels within kilometres of the lighthouse.
\nState whether the alarm will sound. Give a reason for your answer.
\nA1
\n\n
[1 mark]
\nattempt to find the vector from to (M1)
\nA1
\n
EITHER
(M1)(A1)
\nminimize to find on GDC (M1)
\n
OR
closest when (M1)
\n\n(M1)(A1)
\n
OR
closest when (M1)
\n\n(A1)
\n\nSolving simultaneously (M1)
\n
THEN
A1
\n\n
[6 marks]
\nthe alarm will sound A1
\nR1
\n
Note: Do not award A1R0.
\n
[2 marks]
\nPart (a) was generally well answered. Following that there were several possible approaches to this question. It was clear that many candidates understood that the closest approach could be found using the scalar product to find orthogonal vectors. However, typical efforts involved attempting to find the value of by assuming that and are perpendicular. Other methods incorrectly applied were equating and . Of course, this led to separate values of for each component. The method of minimizing was not commonly employed.
\nPart (a) was generally well answered. Following that there were several possible approaches to this question. It was clear that many candidates understood that the closest approach could be found using the scalar product to find orthogonal vectors. However, typical efforts involved attempting to find the value of by assuming that and are perpendicular. Other methods incorrectly applied were equating and . Of course, this led to separate values of for each component. The method of minimizing was not commonly employed.
\nPart (a) was generally well answered. Following that there were several possible approaches to this question. It was clear that many candidates understood that the closest approach could be found using the scalar product to find orthogonal vectors. However, typical efforts involved attempting to find the value of by assuming that and are perpendicular. Other methods incorrectly applied were equating and . Of course, this led to separate values of for each component. The method of minimizing was not commonly employed.
\nThe owner of a convenience store installs two security cameras, represented by points and . Both cameras point towards the centre of the store’s cash register, represented by the point .
\nThe following diagram shows this information on a cross-section of the store.
\nThe cameras are positioned at a height of , and the horizontal distance between the cameras is . The cash register is sitting on a counter so that its centre, , is above the floor.
\nThe distance from Camera to the centre of the cash register is .
\nDetermine the angle of depression from Camera to the centre of the cash register. Give your answer in degrees.
\nCalculate the distance from Camera to the centre of the cash register.
\nWithout further calculation, determine which camera has the largest angle of depression to the centre of the cash register. Justify your response.
\nOR (M1)
\nA1
\n\n
[2 marks]
\nMETHOD 1
\nOR OR (M1)
\n
Note: Award M1 for attempt to use Pythagorean Theorem with seen or for attempt to use cosine or tangent ratio.
(A1)
Note: Award the M1A1 if is seen in part (a).
(A1)
\n
Note: Award A1 for or equivalent seen, either as a separate calculation or in Pythagorean Theorem.
A1
\n\n
METHOD 2
\nattempt to use cosine rule (M1)
\n(A1)(A1)
\n
Note: Award A1 for substituted into cosine rule formula, A1 for correct substitution.
A1
\n
[4 marks]
\ncamera is closer to the cash register (than camera and both cameras are at the same height on the wall) R1
\nthe larger angle of depression is from camera A1
\n
Note: Do not award R0A1. Award R0A0 if additional calculations are completed and used in their justification, as per the question. Accept “” or “” as evidence for the R1.
\n
[2 marks]
\nMany candidates calculated the angle from vertical rather than the angle of depression.
\nCandidates could successfully use their vertical angle from (a) or other correct trigonometry, such as Pythagorean theorem or cosine rule, to find the distance from camera 2 to the cash register. This question is a good example of how premature rounding can affect a final answer, and some had an inaccurate final answer because they had rounded intermediate values.
\nMany provided reasonable justification for their response, even though they often followed correct reasoning with an incorrect conclusion about the larger angle of depression.
\nNatasha carries out an experiment on the growth of mould. She believes that the growth can be modelled by an exponential function
\n,
\nwhere is the area covered by mould in , is the time in days since the start of the experiment and and are constants.
\nThe area covered by mould is at the start of the experiment and after days.
\nWrite down the value of .
\nFind the value of .
\nA1
\n
[1 mark]
(M1)
\n
Note: Award (M1) for their correct equation.
EITHER
graph of and with indication of point of intersection (M1)
\n
OR
(M1)
\n
Note: Award (M1) for correct rearranging and use of .
\n
THEN
\nA1
\n
Note: Award (M1)(M1)(A0) for .
\n
[3 marks]
\nIn part (a), there were some problems for a few candidates to identify the value of . Many answers were left as and thus scored no marks. Those candidates who could identify the value of were generally able to find a correct solution. Most candidates were able to substitute into the given formula, and many were able to find a correct solution to the resulting equation. The exponential function did not seem to have put candidates off. In several responses the use of logs was seen or implied in the candidates' work; this topic is off syllabus and candidates are expected to use technology (and not logs) to solve such problems. However, very few candidates showed workings between the substitution and the final answer, which was to their detriment in the awarding of marks for their method whenever an incorrect answer was seen. A few candidates did not seem to understand the function notation . In part (b) a few candidates wrote for and multiplied the two values.
\nIn part (a), there were some problems for a few candidates to identify the value of . Many answers were left as and thus scored no marks. Those candidates who could identify the value of were generally able to find a correct solution. Most candidates were able to substitute into the given formula, and many were able to find a correct solution to the resulting equation. The exponential function did not seem to have put candidates off. In several responses the use of logs was seen or implied in the candidates' work; this topic is off syllabus and candidates are expected to use technology (and not logs) to solve such problems. However, very few candidates showed workings between the substitution and the final answer, which was to their detriment in the awarding of marks for their method whenever an incorrect answer was seen. A few candidates did not seem to understand the function notation . In part (b) a few candidates wrote for and multiplied the two values.
\nA farmer owns a field in the shape of a triangle such that and .
\nThe local town is planning to build a highway that will intersect the borders of the field at points and , where and , as shown in the diagram below.
\nThe town wishes to build a carpark here. They ask the farmer to exchange the part of the field represented by triangle . In return the farmer will get a triangle of equal area , where lies on the same line as and , as shown in the diagram above.
\nFind the size of .
\nFind .
\nFind the area of triangle .
\nEstimate . You may assume the highway has a width of zero.
\nuse of cosine rule (M1)
\n(A1)
\n A1
[3 marks]
use of sine rule (M1)
\n(A1)
\nA1
\n
[3 marks]
METHOD 1
\n(M1)
\nOR (A1)
\nsubstituted area of triangle formula (M1)
\n(A1)
\nA1
\n\n
METHOD 2
\n(M1)
\n(A1)
\nsubstituted area of triangle formula (M1)
\n
EITHER
(A1)
\n
OR
(A1)
\n
THEN
A1
\n\n
METHOD 3
\n(M1)
\n(A1)
\nsubstituted area of triangle formula (M1)
\n(A1)
\n A1
[5 marks]
\nOR (A1)
\nequating answer to part (c) to area of a triangle formula (M1)
\n(A1)
\nA1
\n\n
[4 marks]
\nThe admissions team at a new university are trying to predict the number of student applications they will receive each year.
\nLet be the number of years that the university has been open. The admissions team collect the following data for the first two years.
\nIt is assumed that the number of students that apply to the university each year will follow a geometric sequence, .
\nIn the first year there were places at the university available for applicants. The admissions team announce that the number of places available will increase by every year.
\nLet represent the number of places available at the university in year .
\nFor the first years that the university is open, all places are filled. Students who receive a place each pay an acceptance fee.
\nWhen , the number of places available will, for the first time, exceed the number of students applying.
\nCalculate the percentage increase in applications from the first year to the second year.
\nWrite down the common ratio of the sequence.
\nFind an expression for .
\nFind the number of student applications the university expects to receive when . Express your answer to the nearest integer.
\nWrite down an expression for .
\nCalculate the total amount of acceptance fees paid to the university in the first years.
\nFind .
\nState whether, for all , the university will have places available for all applicants. Justify your answer.
\n(M1)
\nA1
\n\n
[2 marks]
\nA1
\n
Note: Follow through from part (a).
\n
[1 mark]
\nA1
\n\n
[1 mark]
\n(M1)
\nA1
\n
Note: Answer must be to the nearest integer. Do not accept .
\n
[2 marks]
\nOR M1A1
\n
Note: Award M1 for substituting into arithmetic sequence formula, A1 for correct substitution.
\n
[2 marks]
\n(M1)(M1)
\n
Note: Award (M1) for multiplying by and (M1) for substitution into sum of arithmetic sequence formula.
A1
\n
[3 marks]
\nor equivalent (M1)
\n
Note: Award (M1) for equating their expressions from parts (b) and (c).
EITHER
graph showing and (M1)
\nOR
\ngraph showing (M1)
\nOR
\nlist of values including, and (M1)
\nOR
\nfrom graphical method or solving numerical equality (M1)
\n
Note: Award (M1) for a valid attempt to solve.
THEN
A1
\n\n
[3 marks]
\nthis will not guarantee enough places. A1
\nEITHER
\nA written statement that , with range of . R1
\nExample: “when (or greater), the number of applications will exceed the number of places again” (“”).
\n
OR
exponential growth will always exceed linear growth R1
\n
Note: Accept an equivalent sketch. Do not award A1R0.
\n
[2 marks]
\nThe percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by . In part (e), candidates were usually able to find the first point of intersection of and , but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where for . Though the answer \" is geometric, is arithmetic\", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of and .
\nThe percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by . In part (e), candidates were usually able to find the first point of intersection of and , but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where for . Though the answer \" is geometric, is arithmetic\", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of and .
\nThe percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by . In part (e), candidates were usually able to find the first point of intersection of and , but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where for . Though the answer \" is geometric, is arithmetic\", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of and .
\nThe percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by . In part (e), candidates were usually able to find the first point of intersection of and , but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where for . Though the answer \" is geometric, is arithmetic\", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of and .
\nThe percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by . In part (e), candidates were usually able to find the first point of intersection of and , but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where for . Though the answer \" is geometric, is arithmetic\", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of and .
\nThe percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by . In part (e), candidates were usually able to find the first point of intersection of and , but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where for . Though the answer \" is geometric, is arithmetic\", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of and .
\nThe percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by . In part (e), candidates were usually able to find the first point of intersection of and , but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where for . Though the answer \" is geometric, is arithmetic\", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of and .
\nThe percentage increase proved more difficult than anticipated, with many using an incorrect denominator. Part (b) was accessible with many candidates earning at least three marks. The expression for the general term of the arithmetic sequence was found in part (c). A surprising number of candidates found the acceptance fees paid in the tenth year, rather than the required total acceptance fees found in the first ten years. Most candidates were able to earn one mark for multiplying their value by . In part (e), candidates were usually able to find the first point of intersection of and , but did not always realize the answer must be an integer. Part (f) required candidates to support their answer through a comparison of growth rates, or by finding a range of values/single data point where for . Though the answer \" is geometric, is arithmetic\", inferred some understanding, this was insufficient justification and required a description that went a little bit further. It is recommended that teachers provide opportunities for candidates to explore the more advanced features of the GDC. An inappropriate choice of calculator window made it difficult for candidates to assess and appreciate the behaviour of and .
\nThe sides of a bowl are formed by rotating the curve , about the -axis, where and are measured in centimetres. The bowl contains water to a height of .
\nShow that the volume of water, , in terms of is .
\nHence find the maximum capacity of the bowl in .
\nattempt to use (M1)
\nor any reasonable attempt to find in terms of (M1)
\nA1
\n
Note: Correct limits must be seen for the A1 to be awarded.
(A1)
Note: Condone the absence of limits for this A1 mark.
A1
AG
\n
Note: If the variable used in the integral is instead of (i.e. ) and the candidate has not stated that they are interchanging and then award at most M1M1A0A1A1AG.
\n
[5 marks]
\nmaximum volume when (M1)
\nmax volume A1
\n\n
[2 marks]
\nA number of candidates switched variables so that and then used . Other candidates who correctly found in terms of failed to use the limits and , using and instead. As part (a) was to show that the volume was equal to the final expression it was necessary for examiners to see steps in obtaining the result. It was common to miss out any expression involving . Since the value could be written from the answer given, where this value came from needed to be shown. It was encouraging to see correct answers to (b), even when candidates had failed to gain marks for (a). Some candidates successfully used their GDC to calculate the value of the definite integral numerically.
\nA number of candidates switched variables so that and then used . Other candidates who correctly found in terms of failed to use the limits and , using and instead. As part (a) was to show that the volume was equal to the final expression it was necessary for examiners to see steps in obtaining the result. It was common to miss out any expression involving . Since the value could be written from the answer given, where this value came from needed to be shown. It was encouraging to see correct answers to (b), even when candidates had failed to gain marks for (a). Some candidates successfully used their GDC to calculate the value of the definite integral numerically.
\nDilara is designing a kite on a set of coordinate axes in which one unit represents .
\nThe coordinates of , and are and respectively. Point lies on the -axis. is perpendicular to . This information is shown in the following diagram.
\nFind the gradient of the line through and .
\nWrite down the gradient of the line through and .
\nFind the equation of the line through and . Give your answer in the form , where and are integers.
\nWrite down the -coordinate of point .
\n(M1)A1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nan equation of line with a correct intercept and either of their gradients from (a) or (b) (M1)
\ne.g. OR
\n
Note: Award (M1) for substituting either of their gradients from parts (a) or (b) and point or into equation of a line.
or any integer multiple A1
\n
[2 marks]
\nA1
\n\n
[1 mark]
\nThis question was overall well done by most candidates. In part (a) calculating the gradient was correctly done with few errors noted where candidates swapped the and coordinates in the gradient formula. Some candidates left their answer as which resulted in a loss of the final mark. There was some confusion with the gradient of a line and the gradient of the perpendicular line in part (a). Some candidates found the perpendicular gradient in part (a). Although many candidates were able to write an appropriate equation of the line through , several did not express their answer in the required form , where , and are integers. Many final answers were given as or . In part (d), writing the -coordinate of point was well done by most candidates. Some candidates wrote a coordinate pair rather than just the -coordinate as required.
\nThis question was overall well done by most candidates. In part (a) calculating the gradient was correctly done with few errors noted where candidates swapped the and coordinates in the gradient formula. Some candidates left their answer as which resulted in a loss of the final mark. There was some confusion with the gradient of a line and the gradient of the perpendicular line in part (a). Some candidates found the perpendicular gradient in part (a). Although many candidates were able to write an appropriate equation of the line through , several did not express their answer in the required form , where , and are integers. Many final answers were given as or . In part (d), writing the -coordinate of point was well done by most candidates. Some candidates wrote a coordinate pair rather than just the -coordinate as required.
\nThis question was overall well done by most candidates. In part (a) calculating the gradient was correctly done with few errors noted where candidates swapped the and coordinates in the gradient formula. Some candidates left their answer as which resulted in a loss of the final mark. There was some confusion with the gradient of a line and the gradient of the perpendicular line in part (a). Some candidates found the perpendicular gradient in part (a). Although many candidates were able to write an appropriate equation of the line through , several did not express their answer in the required form , where , and are integers. Many final answers were given as or . In part (d), writing the -coordinate of point was well done by most candidates. Some candidates wrote a coordinate pair rather than just the -coordinate as required.
\nThis question was overall well done by most candidates. In part (a) calculating the gradient was correctly done with few errors noted where candidates swapped the and coordinates in the gradient formula. Some candidates left their answer as which resulted in a loss of the final mark. There was some confusion with the gradient of a line and the gradient of the perpendicular line in part (a). Some candidates found the perpendicular gradient in part (a). Although many candidates were able to write an appropriate equation of the line through , several did not express their answer in the required form , where , and are integers. Many final answers were given as or . In part (d), writing the -coordinate of point was well done by most candidates. Some candidates wrote a coordinate pair rather than just the -coordinate as required.
\nThe pH of a solution measures its acidity and can be determined using the formula pH , where is the concentration of hydronium ions in the solution, measured in moles per litre. A lower pH indicates a more acidic solution.
\nThe concentration of hydronium ions in a particular type of coffee is moles per litre.
\nA different, unknown, liquid has times the concentration of hydronium ions of the coffee in part (a).
\nCalculate the pH of the coffee.
\nDetermine whether the unknown liquid is more or less acidic than the coffee. Justify your answer mathematically.
\n(pH =) (M1)
\nA1
\n\n
[2 marks]
\nEITHER
\ncalculating pH
\n(pH =) (M1)
\nA1
\n(, therefore) the unknown liquid is more acidic (than coffee). A1
\n
Note: Follow through within the part for the final A1. A correct conclusion must be supported by a mathematical justification linking the -value to the pH level to earn the final A1; a comparison of -values only earns M0A0A0.
\n
OR
\nreferencing the graph
\nThe graph of shows that as the value of increases, the value of decreases. M1
\nSince the -value (-value) of the unknown liquid is larger than that of the coffee, the pH level (-value) is lower. R1
\nThe unknown liquid is more acidic (than coffee). A1
\n
Note: Follow through within the part for the final A1. A correct conclusion must be supported by a mathematical justification linking the -value to the pH level to earn the final A1; a comparison of -values only earns M0R0A0.
\n
[3 marks]
\nEvaluation of logarithms was well done, although the notation when substituting into the logarithmic formula was not always correct, with several candidates including a multiplication sign between the base and the argument. Even when the substitution was done correctly, some candidates still used multiplication, so not fully understanding logarithmic notation.
\nSeveral candidates multiplied their answer to part (a) by 10 rather than multiplying the C-value by 10, and several attempted to compare the C-values rather than calculating the pH of the unknown liquid. Most were able to make a correct contextual interpretation of their result.
\nArianne plays a game of darts.
\nThe distance that her darts land from the centre, , of the board can be modelled by a normal distribution with mean and standard deviation .
\nFind the probability that
\nEach of Arianne’s throws is independent of her previous throws.
\nIn a competition a player has three darts to throw on each turn. A point is scored if a player throws all three darts to land within a central area around . When Arianne throws a dart the probability that it lands within this area is .
\nIn the competition Arianne has ten turns, each with three darts.
\na dart lands less than from .
\na dart lands more than from .
\nFind the probability that Arianne throws two consecutive darts that land more than from .
\nFind the probability that Arianne does not score a point on a turn of three darts.
\nFind the probability that Arianne scores at least points in the competition.
\nFind the probability that Arianne scores at least points and less than points.
\nGiven that Arianne scores at least points, find the probability that Arianne scores less than points.
\nLet be the random variable “distance from ”.
\n\n(M1)(A1)
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\n(M1)
\nA1
\n\n
[2 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nMETHOD 1
\nlet be the random variable “number of points scored”
\nevidence of use of binomial distribution (M1)
\n(A1)
\n. A1
\n\n
METHOD 2
\nlet be the random variable “number of times a point is not scored”
\nevidence of use of binomial distribution (M1)
\n(A1)
\nA1
\n\n
[3 marks]
\n(M1)
\nA1
\n
Note: Award M1 for a correct probability statement or indication of correct lower and upper bounds, and .
[2 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nCandidates appeared well prepared for straightforward questions using the normal distribution. Most were able to earn marks through recognition of the compound probability event in part (b). The wording of the information in part (c) required careful thought. This acted as a clear discriminator, causing difficulty for most candidates. One common error in this part was the calculation . Though at the end of the paper, it was pleasing to see many candidates identify the event in part (d) as binomial. A common error was the use of . It is recommended that candidates write down the distribution with associated parameters and support this with a probability statement. This will allow method and follow-through marks to be awarded in subsequent parts. Weaker candidates incorrectly used as the upper bound. Those who made it to the end of the paper were often rewarded for correct division of their probabilities found in parts (d)(i)&(ii).
\nCandidates appeared well prepared for straightforward questions using the normal distribution. Most were able to earn marks through recognition of the compound probability event in part (b). The wording of the information in part (c) required careful thought. This acted as a clear discriminator, causing difficulty for most candidates. One common error in this part was the calculation . Though at the end of the paper, it was pleasing to see many candidates identify the event in part (d) as binomial. A common error was the use of . It is recommended that candidates write down the distribution with associated parameters and support this with a probability statement. This will allow method and follow-through marks to be awarded in subsequent parts. Weaker candidates incorrectly used as the upper bound. Those who made it to the end of the paper were often rewarded for correct division of their probabilities found in parts (d)(i)&(ii).
\nCandidates appeared well prepared for straightforward questions using the normal distribution. Most were able to earn marks through recognition of the compound probability event in part (b). The wording of the information in part (c) required careful thought. This acted as a clear discriminator, causing difficulty for most candidates. One common error in this part was the calculation . Though at the end of the paper, it was pleasing to see many candidates identify the event in part (d) as binomial. A common error was the use of . It is recommended that candidates write down the distribution with associated parameters and support this with a probability statement. This will allow method and follow-through marks to be awarded in subsequent parts. Weaker candidates incorrectly used as the upper bound. Those who made it to the end of the paper were often rewarded for correct division of their probabilities found in parts (d)(i)&(ii).
\nCandidates appeared well prepared for straightforward questions using the normal distribution. Most were able to earn marks through recognition of the compound probability event in part (b). The wording of the information in part (c) required careful thought. This acted as a clear discriminator, causing difficulty for most candidates. One common error in this part was the calculation . Though at the end of the paper, it was pleasing to see many candidates identify the event in part (d) as binomial. A common error was the use of . It is recommended that candidates write down the distribution with associated parameters and support this with a probability statement. This will allow method and follow-through marks to be awarded in subsequent parts. Weaker candidates incorrectly used as the upper bound. Those who made it to the end of the paper were often rewarded for correct division of their probabilities found in parts (d)(i)&(ii).
\nCandidates appeared well prepared for straightforward questions using the normal distribution. Most were able to earn marks through recognition of the compound probability event in part (b). The wording of the information in part (c) required careful thought. This acted as a clear discriminator, causing difficulty for most candidates. One common error in this part was the calculation . Though at the end of the paper, it was pleasing to see many candidates identify the event in part (d) as binomial. A common error was the use of . It is recommended that candidates write down the distribution with associated parameters and support this with a probability statement. This will allow method and follow-through marks to be awarded in subsequent parts. Weaker candidates incorrectly used as the upper bound. Those who made it to the end of the paper were often rewarded for correct division of their probabilities found in parts (d)(i)&(ii).
\nCandidates appeared well prepared for straightforward questions using the normal distribution. Most were able to earn marks through recognition of the compound probability event in part (b). The wording of the information in part (c) required careful thought. This acted as a clear discriminator, causing difficulty for most candidates. One common error in this part was the calculation . Though at the end of the paper, it was pleasing to see many candidates identify the event in part (d) as binomial. A common error was the use of . It is recommended that candidates write down the distribution with associated parameters and support this with a probability statement. This will allow method and follow-through marks to be awarded in subsequent parts. Weaker candidates incorrectly used as the upper bound. Those who made it to the end of the paper were often rewarded for correct division of their probabilities found in parts (d)(i)&(ii).
\nCandidates appeared well prepared for straightforward questions using the normal distribution. Most were able to earn marks through recognition of the compound probability event in part (b). The wording of the information in part (c) required careful thought. This acted as a clear discriminator, causing difficulty for most candidates. One common error in this part was the calculation . Though at the end of the paper, it was pleasing to see many candidates identify the event in part (d) as binomial. A common error was the use of . It is recommended that candidates write down the distribution with associated parameters and support this with a probability statement. This will allow method and follow-through marks to be awarded in subsequent parts. Weaker candidates incorrectly used as the upper bound. Those who made it to the end of the paper were often rewarded for correct division of their probabilities found in parts (d)(i)&(ii).
\nA factory produces bags of sugar with a labelled weight of . The weights of the bags are normally distributed with a mean of and a standard deviation of .
\nA bag that weighs less than is rejected by the factory for being underweight.
\nWrite down the percentage of bags that weigh more than .
\nFind the probability that a randomly chosen bag is rejected for being underweight.
\nA bag that weighs more than grams is rejected by the factory for being overweight. The factory rejects of bags for being overweight.
\nFind the value of .
\nA1
\n
Note: Do not accept or .
\n
[1 mark]
\nA2
\n\n
[2 marks]
\nOR (M1)
\n
Note: Award (M1) for a sketch with correct region identified.
\n
A2
\n\n
[3 marks]
\nMany candidates made no attempt at this question, suggesting poor preparation, although it has appeared frequently in past sessions.
\nFew candidates recognized that a normal distribution is symmetric.
\n\n
Of the few candidates who were able to interpret the question correctly and use their calculator to identify the numerical digits satisfying the probability requirements, a number were out by a factor of 10.
\n\n
Few correct answers were seen. Of those that tried to show a diagram, there was often not enough detail to award a method mark (e.g. the shaded area was numerically identified as 2%).
\nA polygraph test is used to determine whether people are telling the truth or not, but it is not completely accurate. When a person tells the truth, they have a chance of failing the test. Each test outcome is independent of any previous test outcome.
\npeople take a polygraph test and all tell the truth.
\nCalculate the expected number of people who will pass this polygraph test.
\nCalculate the probability that exactly people will fail this polygraph test.
\nDetermine the probability that fewer than people will pass this polygraph test.
\n(M1)
\n(people) A1
\n\n
[2 marks]
\nrecognition of binomial probability (M1)
\nA1
\n\n
[2 marks]
\nand seen OR and seen (A1)
\nattempt to use binomial probability (M1)
\nA1
\n\n
[3 marks]
\nCalculating expected value was well done, with some finding the probability of passing first and then multiplying by 10, while others calculated the expected number who would fail and then subtracted from 10.
\nThere were some candidates who did not recognize binomial probability, and attempted to calculate probability using other methods. For the candidates who did recognize binomial probability, part (b) was well done with most selecting correct calculator entries for the probability. In part (c), there was some confusion as to what value to use in their binomial cumulative distribution function for “less than 7”, with the most common error being the use of 7 rather than 6 as the parameter in the calculation.
\nA new concert hall was built with seats in the first row. Each subsequent row of the hall has two more seats than the previous row. The hall has a total of rows.
\nFind:
\nThe concert hall opened in . The average number of visitors per concert during that year was . In , the average number of visitors per concert increased by .
\nThe concert organizers use this data to model future numbers of visitors. It is assumed that the average number of visitors per concert will continue to increase each year by .
\nthe number of seats in the last row.
\nthe total number of seats in the concert hall.
\nFind the average number of visitors per concert in .
\nDetermine the first year in which this model predicts the average number of visitors per concert will exceed the total seating capacity of the concert hall.
\nIt is assumed that the concert hall will host concerts each year.
\nUse the average number of visitors per concert per year to predict the total number of people expected to attend the concert hall from when it opens until the end of .
\nrecognition of arithmetic sequence with common difference (M1)
\nuse of arithmetic sequence formula (M1)
\n\nA1
\n\n
[3 marks]
\nuse of arithmetic series formula (M1)
\n\nA1
\n\n
[2 marks]
\nOR (M1)
\nA1
\n
Note: Award M0A0 if incorrect used in part (b), and FT with their in parts (c) and (d).
[2 marks]
\nrecognition of geometric sequence (M1)
\nequating their th geometric sequence term to their (M1)
\n
Note: Accept inequality.
METHOD 1
EITHER
\nA1
\n\nA1
\nA1
\n
OR
A1
\nA1
\nA1
\n\n
METHOD 2
\n11th term (M1)A1
\n12th term (M1)A1
\nA1
\n
Note: The last mark can be awarded if both their 11th and 12th correct terms are seen.
[5 marks]
\nseen (A1)
\n
EITHER
(M1)
\nmultiplying their sum by (M1)
\n
OR
sum of the number of visitors for their and their seven years (M1)
\nmultiplying their sum by (M1)
\n
OR
(M1)(M1)
\n
THEN
A1
\n
Note: Follow though from their from part (b).
[4 marks]
\nLet the function represent the height in centimetres of a cylindrical tin can with diameter .
\nfor .
\nThe function is the inverse function of .
\nFind the range of .
\nFind .
\nIn the context of the question, interpret your answer to part (b)(i).
\nWrite down the range of .
\nOR (M1)
\n
Note: Award (M1) for substituting or into . This can be implicit from seeing or .
A1A1
Note: Award A1 for both correct endpoints seen, A1 for the endpoints in a correct interval.
\n
[3 marks]
\nOR OR (M1)
\nA1
\n\n
[2 marks]
\na tin that is high will have a diameter of A1
\n
Note: Condone a correct answer expressed as the converse.
\n
[1 mark]
\nA1
\n
Note: Accept . Do not FT in this part.
\n
[1 mark]
\nPart (a) was reasonably well done. Many candidates were able to find the endpoints but there was some confusion about whether to use strict or weak inequalities. Some candidates wrote their answer as while some others wrote . A few candidates used integer values from to to find corresponding values for and gave the full list as their final answer. In part (b), the most popular incorrect answer seen was with weaker candidates simply finding . Several candidates equated to but missed out in their equation. Finding a value of the inverse of a function still proves to be difficult for candidates. There were many candidates who attempted to find an expression for the inverse before substituting by and this proved to be difficult for this function. Regardless of what answer candidates derived for part (b), very few of them could write an interpretation of their answer in context. There was significant confusion between the value for the height and value for the diameter. In part (d), there were very few candidates who realized the relationship between the domain of the function and the range of the inverse function. Many candidates simply reverted to their answer to part (a).
\nPart (a) was reasonably well done. Many candidates were able to find the endpoints but there was some confusion about whether to use strict or weak inequalities. Some candidates wrote their answer as while some others wrote . A few candidates used integer values from to to find corresponding values for and gave the full list as their final answer. In part (b), the most popular incorrect answer seen was with weaker candidates simply finding . Several candidates equated to but missed out in their equation. Finding a value of the inverse of a function still proves to be difficult for candidates. There were many candidates who attempted to find an expression for the inverse before substituting by and this proved to be difficult for this function. Regardless of what answer candidates derived for part (b), very few of them could write an interpretation of their answer in context. There was significant confusion between the value for the height and value for the diameter. In part (d), there were very few candidates who realized the relationship between the domain of the function and the range of the inverse function. Many candidates simply reverted to their answer to part (a).
\nPart (a) was reasonably well done. Many candidates were able to find the endpoints but there was some confusion about whether to use strict or weak inequalities. Some candidates wrote their answer as while some others wrote . A few candidates used integer values from to to find corresponding values for and gave the full list as their final answer. In part (b), the most popular incorrect answer seen was with weaker candidates simply finding . Several candidates equated to but missed out in their equation. Finding a value of the inverse of a function still proves to be difficult for candidates. There were many candidates who attempted to find an expression for the inverse before substituting by and this proved to be difficult for this function. Regardless of what answer candidates derived for part (b), very few of them could write an interpretation of their answer in context. There was significant confusion between the value for the height and value for the diameter. In part (d), there were very few candidates who realized the relationship between the domain of the function and the range of the inverse function. Many candidates simply reverted to their answer to part (a).
\nPart (a) was reasonably well done. Many candidates were able to find the endpoints but there was some confusion about whether to use strict or weak inequalities. Some candidates wrote their answer as while some others wrote . A few candidates used integer values from to to find corresponding values for and gave the full list as their final answer. In part (b), the most popular incorrect answer seen was with weaker candidates simply finding . Several candidates equated to but missed out in their equation. Finding a value of the inverse of a function still proves to be difficult for candidates. There were many candidates who attempted to find an expression for the inverse before substituting by and this proved to be difficult for this function. Regardless of what answer candidates derived for part (b), very few of them could write an interpretation of their answer in context. There was significant confusion between the value for the height and value for the diameter. In part (d), there were very few candidates who realized the relationship between the domain of the function and the range of the inverse function. Many candidates simply reverted to their answer to part (a).
\nThe amount, in milligrams, of a medicinal drug in the body hours after it was injected is given by . Before this injection, the amount of the drug in the body was zero.
\nWrite down
\nthe initial dose of the drug.
\nthe percentage of the drug that leaves the body each hour.
\nCalculate the amount of the drug remaining in the body hours after the injection.
\nA1
\n
[1 mark]
OR OR (M1)
\nA1
\n
[2 marks]
(M1)
\nA1
\n
[2 marks]
The function is defined by .
\nFind .
\nFind the equation of the normal to the curve at in the form , where .
\nOR A1(M1)A1
\n
Note: Award A1 for seen, and (M1) for expressing as (this can be implied from either or seen in their final answer), A1 for . Award at most A1(M1)A0 if any additional terms are seen.
\n
[3 marks]
\nfinding gradient at
\nA1
\nfinding the perpendicular gradient M1
\n\nOR M1
\n
Note: Award M1 for correctly substituting and and their .
A1
Note: Do not award the final A1 if the answer is not in the required form. Accept integer multiples of the equation.
\n
[4 marks]
\nDifferentiating the function was challenging for many candidates. The most frequently obtained mark was for the term . Handling the term was problematic and consequently the method mark and final accuracy mark were lost.
\n\n
Some good attempts at finding the equation of the normal were seen amongst the few that answered this part. Of those that found an equation in the form most included fractions thus hardly any fully correct answers were seen.
\nKarl has three brown socks and four black socks in his drawer. He takes two socks at random from the drawer.
\nComplete the tree diagram.
\nFind the probability that Karl takes two socks of the same colour.
\nGiven that Karl has two socks of the same colour find the probability that he has two brown socks.
\n A1
Note: Award A1 for both missing probabilities correct.
\n\n
[1 mark]
\nmultiplying along branches and then adding outcomes (M1)
\n\nA1
\n\n
[2 marks]
\nuse of conditional probability formula M1
\nA1
\nA1
\n\n
[3 marks]
\n(a) It was pleasing to see that, for those candidates who made a reasonable attempt at the paper, many were able to identify the correct values on the tree diagram.
\n(b) Many identified at least one correct branch. Some who identified both correct branches and the respective probabilities failed to add their results.
\n(c) Some candidates identified conditional probability evidenced by dividing a probability by their previous answer.
\nMany identified at least one correct branch. Some who identified both correct branches and the respective probabilities failed to add their results.
\nSome candidates identified conditional probability evidenced by dividing a probability by their previous answer.
\nA wind turbine is designed so that the rotation of the blades generates electricity. The turbine is built on horizontal ground and is made up of a vertical tower and three blades.
\nThe point is on the base of the tower directly below point at the top of the tower. The height of the tower, , is . The blades of the turbine are centred at and are each of length . This is shown in the following diagram.
\nThe end of one of the blades of the turbine is represented by point on the diagram. Let be the height of above the ground, measured in metres, where varies as the blade rotates.
\nFind the
\nThe blades of the turbine complete rotations per minute under normal conditions, moving at a constant rate.
\nThe height, , of point can be modelled by the following function. Time, , is measured from the instant when the blade first passes and is measured in seconds.
\n\nLooking through his window, Tim has a partial view of the rotating wind turbine. The position of his window means that he cannot see any part of the wind turbine that is more than above the ground. This is illustrated in the following diagram.
\nmaximum value of .
\nminimum value of .
\nFind the time, in seconds, it takes for the blade to make one complete rotation under these conditions.
\nCalculate the angle, in degrees, that the blade turns through in one second.
\nWrite down the amplitude of the function.
\nFind the period of the function.
\nSketch the function for , clearly labelling the coordinates of the maximum and minimum points.
\nFind the height of above the ground when .
\nFind the time, in seconds, that point is above a height of , during each complete rotation.
\nAt any given instant, find the probability that point is visible from Tim’s window.
\nThe wind speed increases. The blades rotate at twice the speed, but still at a constant rate.
\nAt any given instant, find the probability that Tim can see point from his window. Justify your answer.
\nmaximum metres A1
\n\n
[1 mark]
\nminimum metres A1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\n(M1)
\n
Note: Award (M1) for divided by their time for one revolution.
A1
\n
[2 marks]
\n(amplitude =) A1
\n\n
[1 mark]
\n(period ) A1
\n\n
[1 mark]
\nMaximum point labelled with correct coordinates. A1
\nAt least one minimum point labelled. Coordinates seen for any minimum points must be correct. A1
\nCorrect shape with an attempt at symmetry and “concave up\" evident as it approaches the minimum points. Graph must be drawn in the given domain. A1
\n\n
[3 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nevidence of on graph OR (M1)
\ncoordinates OR or equivalent (A1)
\n
Note: Award A1 for either -coordinate seen.
seconds A1
\n
[3 marks]
\n(M1)
\n(M1)
\nA1
\n\n
[3 marks]
\nMETHOD 1
\nchanging the frequency/dilation of the graph will not change the proportion of time that point is visible. A1
\nA1
\n\n
METHOD 2
\ncorrect calculation of relevant found values
\nA1
\nA1
\n
Note: Award A0A1 for an unsupported correct probability.
\n
[2 marks]
\nJudging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the \"trace\" feature in their GDC. Most were able to find the height of point when and make an attempt to find a time at which point is at a height of . It was pleasing to see a number of candidates draw on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.
\nJudging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the \"trace\" feature in their GDC. Most were able to find the height of point when and make an attempt to find a time at which point is at a height of . It was pleasing to see a number of candidates draw on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.
\nJudging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the \"trace\" feature in their GDC. Most were able to find the height of point when and make an attempt to find a time at which point is at a height of . It was pleasing to see a number of candidates draw on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.
\nJudging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the \"trace\" feature in their GDC. Most were able to find the height of point when and make an attempt to find a time at which point is at a height of . It was pleasing to see a number of candidates draw on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.
\nJudging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the \"trace\" feature in their GDC. Most were able to find the height of point when and make an attempt to find a time at which point is at a height of . It was pleasing to see a number of candidates draw on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.
\nJudging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the \"trace\" feature in their GDC. Most were able to find the height of point when and make an attempt to find a time at which point is at a height of . It was pleasing to see a number of candidates draw on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.
\nJudging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the \"trace\" feature in their GDC. Most were able to find the height of point when and make an attempt to find a time at which point is at a height of . It was pleasing to see a number of candidates draw on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.
\nJudging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the \"trace\" feature in their GDC. Most were able to find the height of point when and make an attempt to find a time at which point is at a height of . It was pleasing to see a number of candidates draw on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.
\nJudging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the \"trace\" feature in their GDC. Most were able to find the height of point when and make an attempt to find a time at which point is at a height of . It was pleasing to see a number of candidates draw on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.
\nJudging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the \"trace\" feature in their GDC. Most were able to find the height of point when and make an attempt to find a time at which point is at a height of . It was pleasing to see a number of candidates draw on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.
\nJudging by the responses in parts (a), (b) and (c), transferring and interpreting the information from a diagram is a skill that requires further nurturing. The amplitude should be expressed as a positive value. Overall, the sketch of reflected the correct general shape. Common flaws included a lack of symmetry about the mean, 'concave up' not evident as the curve approached the minimum points, and the curve being drawn beyond the given domain. At least one correct pair of coordinates was seen, though some gave their answers inaccurately, suggesting they found an approximate solution using the \"trace\" feature in their GDC. Most were able to find the height of point when and make an attempt to find a time at which point is at a height of . It was pleasing to see a number of candidates draw on their sketch, which would no doubt have assisted the candidates in visualizing the solution. Part (f) proved to be a high-grade discriminator, with few attaining full marks. Premature rounding in part (f)(i) resulted in an inaccurate final answer. It is recommended that candidates retrieve and use unrounded values from previous calculations in their GDC. Though many recognized the probability was independent of the speed of rotation, most were not able to support their answer through a correct calculation or written explanation.
\nThe graphs of and intersect at and , as shown in the following diagrams.
\nIn diagram 1, the region enclosed by the lines , , and the -axis has been shaded.
\nIn diagram 2, the region enclosed by the curve , and the lines , and the -axis has been shaded.
\nCalculate the area of the shaded region in diagram 1.
\nWrite down an integral for the area of the shaded region in diagram 2.
\nCalculate the area of this region.
\nHence, determine the area enclosed between and .
\nEITHER
\nattempt to substitute and into area of a trapezoid formula (M1)
\n\n
OR
given line expressed as an integral (M1)
\n\n
OR
attempt to sum area of rectangle and area of triangle (M1)
\n\n
THEN
(square units) A1
\n\n
[3 marks]
\nA1A1
\n
Note: Award A1 for the limits , in correct location. Award A1 for an integral of the quadratic function, must be included. Do not accept “” in place of the function, given that two equations are in the question.
\n
[2 marks]
\n(square units) A1
\n\n
[1 mark]
\n(M1)
\n(square units) A1
\n\n
[2 marks]
\nThere were a variety of methods used or attempted – area of trapezoid, integration, area of triangle plus area of rectangle, area of large rectangle minus area of top triangle, trapezoidal rule. All these methods, except for trapezoidal rule, proved successful for candidates, with the most common being the use of integration.
\nThis was reasonably well done except for a few notation issues such as not including dx with their integrand. Those who attempted integration manually were not successful.
\nRecognition that areas had to be subtracted was very evident.
\nAn inclined railway travels along a straight track on a steep hill, as shown in the diagram.
\nThe locations of the stations on the railway can be described by coordinates in reference to , and -axes, where the and axes are in the horizontal plane and the -axis is vertical.
\nThe ground level station has coordinates and station , located near the top of the hill, has coordinates . All coordinates are given in metres.
\nStation is to be built halfway between stations and .
\nFind the distance between stations and .
\nFind the coordinates of station .
\nWrite down the height of station , in metres, above the ground.
\nattempt at substitution into 3D distance formula (M1)
\n\nA1
\n
[2 marks]
attempt at substitution in the midpoint formula (M1)
\n\nA1
\n
[2 marks]
A1
\n
[1 mark]
Inspectors are investigating the carbon dioxide emissions of a power plant. Let be the rate, in tonnes per hour, at which carbon dioxide is being emitted and be the time in hours since the inspection began.
\nWhen is plotted against , the total amount of carbon dioxide produced is represented by the area between the graph and the horizontal -axis.
\nThe rate, , is measured over the course of two hours. The results are shown in the following table.
\n\n
Use the trapezoidal rule with an interval width of to estimate the total amount of carbon dioxide emitted during these two hours.
\nThe real amount of carbon dioxide emitted during these two hours was tonnes.
\nFind the percentage error of the estimate found in part (a).
\nattempt at using trapezoidal rule formula (M1)
\n(A1)
\n(total carbon =) tonnes A1
\n\n
[3 marks]
\n(M1)
\n
Note: Award (M1) for correct substitution of final answer in part (a) into percentage error formula.
A1
\n
[2 marks]
\nAlthough there were successful attempts at using the trapezoidal rule formula, there was quite a bit of confusion among candidates as to which values were to be substituted. It seemed that a significant number of candidates were approaching it with some confusion due to a lack of practice of using the trapezoidal rule formula. Calculation of percentage error in part (b) was generally well done by most candidates.
\nAlthough there were successful attempts at using the trapezoidal rule formula, there was quite a bit of confusion among candidates as to which values were to be substituted. It seemed that a significant number of candidates were approaching it with some confusion due to a lack of practice of using the trapezoidal rule formula. Calculation of percentage error in part (b) was generally well done by most candidates.
\nA newspaper vendor in Singapore is trying to predict how many copies of The Straits Times they will sell. The vendor forms a model to predict the number of copies sold each weekday. According to this model, they expect the same number of copies will be sold each day.
\nTo test the model, they record the number of copies sold each weekday during a particular week. This data is shown in the table.
\nA goodness of fit test at the significance level is used on this data to determine whether the vendor’s model is suitable. The critical value for the test is .
\nFind an estimate for how many copies the vendor expects to sell each day.
\nState the null and alternative hypotheses for this test.
\nWrite down the degrees of freedom for this test.
\nWrite down the conclusion to the test. Give a reason for your answer.
\nA1
\n
[1 mark]
The data satisfies the model A1
\nThe data does not satisfy the model A1
\n
Note: Do not accept “ The same number of copies will be sold each day” but accept a similar statement if the word ‘expect’ or ‘expected’ is included. Similarly for .
[2 marks]
A1
\n
[1 mark]
OR -value A2
\nOR R1
\ntherefore there is insufficient evidence to reject A1
\n(i.e. the data satisfies the model)
\n
Note: Do not award R0A1. Accept “accept” or “do not reject” in place of “insufficient evidence to reject”.
Award the R1 for comparing their -value with or their value with and then FT their final conclusion.
[4 marks]
A college runs a mathematics course in the morning. Scores for a test from this class are shown below.
\n\nFor these data, the lower quartile is and the upper quartile is .
\nThe box and whisker diagram showing these scores is given below.
\nTest scores
\nAnother mathematics class is run by the college during the evening. A box and whisker diagram showing the scores from this class for the same test is given below.
\nTest scores
\nA researcher reviews the box and whisker diagrams and believes that the evening class performed better than the morning class.
\nShow that the test score of would not be considered an outlier.
\nWith reference to the box and whisker diagrams, state one aspect that may support the researcher’s opinion and one aspect that may counter it.
\nOR seen anywhere OR seen anywhere (M1)
\n\nA1
\nR1
\nso is not an outlier AG
\n\n
[3 marks]
\nThe median score for the evening class is higher than the median score for the morning class. A1
\n
THEN
but the scores are more spread out in the evening class than in the morning class A1
OR
the scores are more inconsistent in the evening class A1
OR
the lowest scores are in the evening class A1
OR
the interquartile range is lower in the morning class A1
OR
the lower quartile is lower in the evening class A1
\n\n
Note: If an incorrect comparison is also made, award at most A1A0.
\nAward A0 for a comparison that references “the mean score” unless working is shown for the estimated means of the data sets, calculated from the mid-points of the 4 intervals. The estimated mean for the morning class is and the estimated mean for the evening class is .
\n\n
[2 marks]
\nThere were mixed results calculating the boundary value for outliers. Some determined the correct value of 23, but did not relate it back to 25. Some did not realize that a calculation had to be performed, and instead tried to present an argument referencing the box and whisker diagram.
\nThe majority of candidates were able to compare the medians as evidence supporting the researcher’s belief. However, some incorrectly referred to the median values as mean values. There were more counterarguments available to be presented, and again, candidates were generally able to communicate one of these. There were occasions where the candidate did not indicate which argument was in support of the researcher and which argument was the counterargument, which is an important element in the labelling/communication of their response.
\nThe strength of earthquakes is measured on the Richter magnitude scale, with values typically between and where is the most severe.
\nThe Gutenberg–Richter equation gives the average number of earthquakes per year, , which have a magnitude of at least . For a particular region the equation is
\n, for some .
\nThis region has an average of earthquakes per year with a magnitude of at least .
\nThe equation for this region can also be written as .
\nThe expected length of time, in years, between earthquakes with a magnitude of at least is .
\nWithin this region the most severe earthquake recorded had a magnitude of .
\nFind the value of .
\nFind the value of .
\nGiven , find the range for .
\nFind the expected length of time between this earthquake and the next earthquake of at least this magnitude. Give your answer to the nearest year.
\n(M1)
\nA1
\n\n
[2 marks]
\nEITHER
\n(M1)
\n\n
OR
(M1)
\n
THEN
A1
\n\n
[2 marks]
\nA1A1
\n
Note: Award A1 for correct endpoints and A1 for correct inequalities/interval notation.
\n
[2 marks]
\n(M1)
\nlength of time
\nyears A1
\n\n
[2 marks]
\nMany candidates did not attempt this question. Of those who did attempt the question, most of these candidates arrived at the correct answer to this part with the most common incorrect answer being 103.
\n\n
Those that were successful in part (a) answered this well.
\nThis was only answered correctly by the strongest candidates.
\nThis part of the question was a discriminator as correct responses were few and far between.
\nA wind turbine is designed so that the rotation of the blades generates electricity. The turbine is built on horizontal ground and is made up of a vertical tower and three blades.
\nThe point is on the base of the tower directly below point at the top of the tower. The height of the tower, , is . The blades of the turbine are centred at and are each of length . This is shown in the following diagram.
\nThe end of one of the blades of the turbine is represented by point on the diagram. Let be the height of above the ground, measured in metres, where varies as the blade rotates.
\nFind the
\nThe blades of the turbine complete rotations per minute under normal conditions, moving at a constant rate.
\nThe height, , of point can be modelled by the following function. Time, , is measured from the instant when the blade first passes and is measured in seconds.
\n\nmaximum value of .
\nminimum value of .
\nFind the time, in seconds, it takes for the blade to make one complete rotation under these conditions.
\nCalculate the angle, in degrees, that the blade turns through in one second.
\nWrite down the amplitude of the function.
\nFind the period of the function.
\nSketch the function for , clearly labelling the coordinates of the maximum and minimum points.
\nFind the height of above the ground when .
\nFind the time, in seconds, that point is above a height of , during each complete rotation.
\nThe wind speed increases and the blades rotate faster, but still at a constant rate.
\nGiven that point is now higher than for second during each complete rotation, find the time for one complete rotation.
\nmaximum metres A1
\n\n
[1 mark]
\nminimum metres A1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\n(M1)
\n
Note: Award (M1) for divided by their time for one revolution.
A1
\n
[2 marks]
\n(amplitude =) A1
\n\n
[1 mark]
\n(period ) A1
\n\n
[1 mark]
\nMaximum point labelled with correct coordinates. A1
\nAt least one minimum point labelled. Coordinates seen for any minimum points must be correct. A1
\nCorrect shape with an attempt at symmetry and “concave up\" evident as it approaches the minimum points. Graph must be drawn in the given domain. A1
\n\n
[3 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nevidence of on graph OR (M1)
\ncoordinates OR or equivalent (A1)
\n
Note: Award A1 for either -coordinate seen.
seconds A1
\n
[3 marks]
\nMETHOD 1
\n(M1)
\n\n(A1)
\n(M1)
\n(A1)
\nperiod seconds A1
\n\n
METHOD 2
\nattempt at diagram (M1)
\n(or recognizing special triangle) (M1)
\nangle made by , (A1)
\none third of a revolution in second (M1)
\nhence one revolution seconds A1
\n\n
METHOD 3
\nconsidering on original function (M1)
\nor (A1)
\n(A1)
\n
Note: Accept or equivalent.
so period is of original period (R1)
so new period is seconds A1
\n\n
[5 marks]
\nThis was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.
\nThis was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.
\nThis was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.
\nThis was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.
\nThis was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.
\nThis was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.
\nThis was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.
\nThis was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.
\nThis was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.
\nThis was perhaps the question with the best responses on the paper. Many candidates got close to full marks on this problem. The issues associated with the question were sometimes due to a lack of understanding of the definitions of amplitude and period. A good number of candidates solved both parts of part (e) suggesting that they had a good understanding of the concept of a function and how it can be applied to mathematical models. Part (f) was also well done by a surprisingly large number of candidates using a variety of approaches. This is evidence that candidates had good problem-solving skills.
\nThe diagram shows a sector, , of a circle with centre and radius , such that .
\nSam measured the value of to be and the value of to be .
\nIt is found that Sam’s measurements are accurate to only one significant figure.
\nUse Sam’s measurements to calculate the area of the sector. Give your answer to four significant figures.
\nFind the upper bound and lower bound of the area of the sector.
\nFind, with justification, the largest possible percentage error if the answer to part (a) is recorded as the area of the sector.
\n(M1)
\nA1
\n
Note: Do not award the final mark if the answer is not correct to 4 sf.
\n
[2 marks]
\nattempt to substitute any two values from or into area of sector formula (M1)
\nA1
\nA1
\n
Note: Given the nature of the question, accept correctly rounded OR correctly truncated significant figure answers.
\n
[3 marks]
\nA1
\nA1
\nso the largest percentage error is A1
\n
Note: Accept (), from use of full accuracy answers. Given the nature of the question, accept correctly rounded OR correctly truncated significant figure answers. Award A0A1A0 if is the only value found.
\n
[3 marks]
\nIn part (a), the area was almost always found correctly although some candidates gave the answer 1.0472 which is correct to 4 decimal places, not 4 significant figures as required. In part (b), many candidates failed to realize that the upper bounds for r and θ were 2.5 and 35° and lower bounds were 1.5 and 25°. Consequently, the bounds for the area were incorrect. In many cases, the incorrect values in part (b) were followed through into part (c) although in the percentage error calculations, many candidates had 1.047 in the denominator instead of the appropriate bound.
\nA storage container consists of a box of length , width and height , and a lid in the shape of a half-cylinder, as shown in the diagram. The lid fits the top of the box exactly. The total exterior surface of the storage container is to be painted.
\nFind the area to be painted.
\nAND (A1)
\n(A1)
\n(A1)
\n(M1)
\nuse of curved surface area formula (M1)
\n(A1)
\nA1
\n
[7 marks]
A study was conducted to investigate whether the mean reaction time of drivers who are talking on mobile phones is the same as the mean reaction time of drivers who are talking to passengers in the vehicle. Two independent groups were randomly selected for the study.
\nTo gather data, each driver was put in a car simulator and asked to either talk on a mobile phone or talk to a passenger. Each driver was instructed to apply the brakes as soon as they saw a red light appear in front of the car. The reaction times of the drivers, in seconds, were recorded, as shown in the following table.
\nAt the level of significance, a -test was used to compare the mean reaction times of the two groups. Each data set is assumed to be normally distributed, and the population variances are assumed to be the same.
\nLet and be the population means for the two groups. The null hypothesis for this test is .
\nState the alternative hypothesis.
\nCalculate the -value for this test.
\nState the conclusion of the test. Justify your answer.
\nState what your conclusion means in context.
\nA1
\n
Note: Accept an equivalent statement in words, however reference to “population mean” must be explicit for A1 to be awarded.
\n
[1 mark]
\nA2
\n
Note: Award A1 for an answer of from not using a pooled estimate of the variance.
\n
[2 marks]
\nR1
\nreject the null hypothesis A1
\n
Note: Do not award R0A1.
\n
[2 marks]
\nthere is (significant evidence of) a difference between the (population) mean reaction times A1
\n
Note: Their conclusion in (c)(ii) must match their conclusion in (c)(i) to earn A1. Award A0 if their conclusion refers to mean reaction times in the sample.
\n
[1 mark]
\nCandidates who attempted to write the alternative hypothesis symbolically were successful. Those who tried to write in words generally did not make it clear whether they were referring to “population mean” and hence, were unsuccessful.
\nSeveral candidates gave the p-value from not using a pooled estimate of the variance. As stated in the Mathematics: application & interpretation guide to the syllabus, for the t-test, candidates should assume that the variance of the two groups is equal and therefore, the pooled two-sample t-test should be used.
\nThe justification and generic conclusion were well done. However, candidates struggled when attempting to write their conclusion in context, generally referring to the reaction times rather than the mean reaction times. The conclusions were often vague as to whether the candidates were referring to the population means or the sample means; hypothesis testing is a good example on where candidates need to work to improve the clarity of their writing.
\nThe diagram below shows a helicopter hovering at point , vertically above a lake. Point is the point on the surface of the lake, directly below the helicopter.
\nMinta is swimming at a constant speed in the direction of point . Minta observes the helicopter from point as she looks upward at an angle of . After minutes, Minta is at point and she observes the same helicopter at an angle of .
\nWrite down the size of the angle of depression from to .
\nFind the distance from to .
\nFind the distance from to .
\nFind Minta’s speed, in metres per hour.
\nA1
\n
[1 mark]
OR OR (M1)
\nA1
\n
[2 marks]
METHOD 1
\nattempt to find (M1)
\n\n(A1)
\n\nA1
\n\n
METHOD 2
\nattempt to find (M1)
\n\n(A1)
\n\nA1
\n
[3 marks]
A1
\n
[1 mark]
A psychologist records the number of digits () of that a sample of IB Mathematics higher level candidates could recall.
\nThe psychologist has read that in the general population people can remember an average of digits of . The psychologist wants to perform a statistical test to see if IB Mathematics higher level candidates can remember more digits than the general population.
\nis the null hypothesis for this test.
\nFind an unbiased estimate of the population mean of .
\nFind an unbiased estimate of the population variance of .
\nState the alternative hypothesis.
\nGiven that all assumptions for this test are satisfied, carry out an appropriate hypothesis test. State and justify your conclusion. Use a significance level.
\nA1
\n\n
[1 mark]
\n(A1)
\nA1
\n
Note: Award A0A0 for an answer of from biased estimate.
\n
[2 marks]
\nA1
\n
[1 mark]
METHOD 1
\nusing a -test (M1)
\nA1
\nR1
\nreject null hypothesis A1
\n(therefore there is significant evidence that the IB HL math students know more digits of than the population in general)
\n
Note: Do not award R0A1. Allow R1A1 for consistent conclusion following on from their -value.
\n
METHOD 2
\nusing a -test (M1)
\nA1
\nR1
\nreject null hypothesis A1
\n(therefore there is significant evidence that the IB HL math students know more digits of than the population in general)
\n
Note: Do not award R0A1. Allow R1A1 for consistent conclusion following on from their -value.
[4 marks]
In parts (a) and (b), candidates used the 1-Var Stats facility to find the estimates of mean and variance although some forgot to include the frequency list so that they just found the mean and variance of the numbers 2, 3, …6, 7. Candidates who looked ahead realized that the answers to parts (a) and (b) would be included in the output from using their test. In part (c), the question was intended to use the t-test (as the population variance was unknown), however since the population could not be assumed to be normally distributed, the Principal Examiner condoned the use of the z-test (with the estimated variance from part (b)). As both methods could only produce an approximate p-value, either method (and the associated p-value) was awarded full marks.
\nA company’s profit per year was found to be changing at a rate of
\n\nwhere is the company’s profit in thousands of dollars and is the time since the company was founded, measured in years.
\nDetermine whether the profit is increasing or decreasing when .
\nOne year after the company was founded, the profit was thousand dollars.
\nFind an expression for , when .
\nMETHOD 1
\n(when )
\nOR (equivalent in words) OR M1
\ntherefore is decreasing A1
\n\n
METHOD 2
\nsketch with indicated in 4th quadrant OR -intercepts identified M1
\ntherefore is decreasing A1
\n\n
[2 marks]
\nA1A1
\n(M1)
\n
Note: Award M1 for substituting into their equation with seen.
A1
\n\n
[4 marks]
\nEven some weaker candidates were able to score in this part of the question as many candidates understood the process required to determine whether the profit is increasing or decreasing.
\nMany candidates failed to recognize that they needed to integrate the original function. Of the few that attempted to find the value of the constant the vast majority substituted 4000 rather than 4. So, a correct final expression for was rarely seen.
\nThe function is defined for .
\nFind an expression for . You are not required to state a domain.
\nSolve .
\nan attempt to isolate (or if switched) (M1)
\n\n\n\nswitching and (seen anywhere) M1
\nA1
\n
[3 marks]
sketch of and (M1)
\nA1
\n
[2 marks]
This question was well answered by most candidates.
\nThe sex of cuttlefish is difficult to determine visually, so it is often found by weighing the cuttlefish.
\nThe weights of adult male cuttlefish are known to be normally distributed with mean and standard deviation .
\nThe weights of adult female cuttlefish are known to be normally distributed with mean and standard deviation .
\nA zoologist uses the null hypothesis that in the absence of information a cuttlefish is male.
\nIf the weight is found to be above the cuttlefish is classified as female.
\nof adult cuttlefish are male.
\nFind the probability of making a Type I error when weighing a male cuttlefish.
\nFind the probability of making a Type II error when weighing a female cuttlefish.
\nFind the probability of making an error using the zoologist’s method.
\n(Type I error) (stating female when male)
\n(M1)
\nA1
\n\n
[2 marks]
\n(Type II error) (stating male when female)
\n(M1)
\nA1
\n\n
[2 marks]
\nattempt to use the total probability (M1)
\n(error)
\nA1
\n\n
[2 marks]
\nThis was a straightforward problem on Type I and Type II errors which some candidates answered successfully in a couple of lines but many candidates were unable to do the correct calculations.
\nA ball is dropped from a height of metres and bounces on the ground. The maximum height reached by the ball, after each bounce, is of the previous maximum height.
\nShow that the maximum height reached by the ball after it has bounced for the sixth time is , to the nearest .
\nFind the number of times, after the first bounce, that the maximum height reached is greater than .
\nFind the total vertical distance travelled by the ball from the point at which it is dropped until the fourth bounce.
\nuse of geometric sequence with M1
\n
EITHER
OR OR A1
\n\nAG
\n
OR
OR A1
\nAG
\n\n
[2 marks]
\nEITHER
\nOR (M1)
\n
Note: If (or ) is used then (M1) only awarded for use of in .
If (or ) is used then (M1) only awarded for use of in .
A1
OR
and (M1)
\nA1
\n
OR
solving to find (M1)
\nA1
\n
Note: Evidence of solving may be a graph OR the “solver” function OR use of logs to solve the equation. Working may use .
\n
[2 marks]
\nEITHER
\ndistance (in one direction) travelled between first and fourth bounce
\n(A1)
\nrecognizing distances are travelled twice except first distance (M1)
\n\nA1
\n\n
OR
\ndistance (in one direction) travelled between drop and fourth bounce
\n(A1)
\nrecognizing distances are travelled twice except first distance (M1)
\n\nA1
\n\n
OR
\ndistance (in one direction) travelled between first and fourth bounce
\n(A1)
\nrecognizing distances are travelled twice except first distance (M1)
\n\nA1
\n
Note: Answers may be given in .
\n
[3 marks]
\nMost of the candidates who tackled this question effectively realized that they were dealing with a geometric sequence and were able to correctly identify the common ratio and identify the sixth term.
\nMany candidates misunderstood the instruction: ‘Find the number of times, after the first bounce…’ So, the incorrect answers of 16 or 18 were seen frequently.
\nFew candidates saw that they needed to calculate the distances identified by the seven dotted lines on the given diagram. Those that attempted the question often scored just one mark for using a correctly substituted formula determining the distance travelled in one direction.
\nThe graph below shows the average savings, thousand dollars, of a group of university graduates as a function of , the number of years after graduating from university.
\nThe equation of the model can be expressed in the form , where and are real constants.
\nThe graph of the model must pass through the following four points.
\nA negative value of indicates that a graduate is expected to be in debt.
\nWrite down one feature of this graph which suggests a cubic function might be appropriate to model this scenario.
\nWrite down the value of .
\nWrite down three simultaneous equations for and .
\nHence, or otherwise, find the values of and .
\nUse the model to determine the total length of time, in years, for which a graduate is expected to be in debt after graduating from university.
\nAccept any one of the following (or equivalent):
\none minimum and one maximum point
three -intercepts or three roots (or zeroes)
one point of inflexion R1
Note: Do not accept “S shape” as a justification.
\n
[1 mark]
\nA1
\n\n
[1 mark]
\nA2
\n
Note: Award A2 if all three equations are correct.
Award A1 if at least one is correct. Award A1 for three correct equations that include the letter “”.
\n
[2 marks]
\nA1
\n\n
[1 mark]
\nequating found expression to zero (M1)
\n\n(A1)
\n(so total time in debt is )
\nyears A1
\n\n
[3 marks]
\nProved to be difficult with several referring to the shape of the graph, the graph increasing and decreasing, or positive and negative values fitting the context.
\nIt seemed easy to find the d-value in the function. Most candidates could derive at least one correct equation, but not always three. Many candidates did not write their equations in proper mathematical form, leaving exponents and like terms in their equations. Even those candidates who did not write correct equations in part (ii) were able to correctly find the values of a, b, and c in part (iii) using cubic regression (an off-syllabus method, but still valid and credited full marks). There were some candidates who attempted an analytic method to solve the system of equations which did not usually prove successful.
\nSome candidates realized they had to find the roots, but then did not know what to do with them. Several candidates selected one of the roots as the answer to the question, usually the largest root, clearly not understanding the relationship between the roots and the length of time in debt. Others found only one root and stated that as the answer.
\nA large water reservoir is built in the form of part of an upside-down right pyramid with a horizontal square base of length metres. The point is the centre of the square base and point is the vertex of the pyramid.
\nThe bottom of the reservoir is a square of length metres that is parallel to the base of the pyramid, such that the depth of the reservoir is metres as shown in the diagram.
\nThe second diagram shows a vertical cross section, , of the reservoir.
\nEvery day of water from the reservoir is used for irrigation.
\nJoshua states that, if no other water enters or leaves the reservoir, then when it is full there is enough irrigation water for at least one year.
\nFind the angle of depression from to .
\nFind .
\nHence or otherwise, show that the volume of the reservoir is .
\nBy finding an appropriate value, determine whether Joshua is correct.
\nTo avoid water leaking into the ground, the five interior sides of the reservoir have been painted with a watertight material.
\nFind the area that was painted.
\n(M1)
\nOR A1
\n\n
[2 marks]
\nOR (M1)
\n
Note: Award (M1) for an attempt at trigonometry or similar triangles (e.g. ratios).
A1
\n
[2 marks]
\nM1A1A1
\n
Note: Award M1 for finding the difference between the volumes of two pyramids, A1 for each correct volume expression. The final A1 is contingent on correct working leading to the given answer.
If the correct final answer is not seen, award at most M1A1A0. Award M0A0A0 for any height derived from , including or .
AG
\n
[3 marks]
\nMETHOD 1
\n(days) A1
\nJoshua is correct A1
\n
Note: Award A0A0 for unsupported answer of “Joshua is correct”. Accept for the first A1 mark.
\n
METHOD 2
\nOR A1
\nJoshua is correct A1
\n
Note: The second A1 can be awarded for an answer consistent with their result.
\n
[2 marks]
\nheight of trapezium is (M1)
\narea of trapezium is (M1)(A1)
\n(M1)
\n
Note: Award M1 for adding times their () trapezium area to the area of the () base.
A1
Note: No marks are awarded if the correct shape is not identified.
\n
[5 marks]
\nFinding the angle of depression from to proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle . Most candidates attempted to find through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid as .
\nFinding the angle of depression from to proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle . Most candidates attempted to find through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid as .
\nFinding the angle of depression from to proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle . Most candidates attempted to find through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid as .
\nFinding the angle of depression from to proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle . Most candidates attempted to find through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid as .
\nFinding the angle of depression from to proved problematic for many. Some chose the angle inclined to the vertical or attempted a less efficient method such as solving triangle . Most candidates attempted to find through trigonometry rather than similar triangles. Consistent with past examination sessions, it is clear that candidate do not always understand the demands of a 'show that' question. Working backwards to verify the given volume of accrued no marks. Various approaches were used, with most candidates earning at least one mark for one correct volume seen. Part (c) was accessible to all. It was surprising to see some candidates identify as the number of days in a calendar year. Part (d) was a true high-order discriminator, proving to be challenging for even very capable candidates. Some candidates found the surface area of a shape/object that is not part of the reservoir, while the majority incorrectly identified the height of trapezoid as .
\nAn electric circuit has two power sources. The voltage, , provided by the first power source, at time , is modelled by
\n.
\nThe voltage, , provided by the second power source is modelled by
\n.
\nThe total voltage in the circuit, , is given by
\n.
\nFind an expression for in the form , where and are real constants.
\nHence write down the maximum voltage in the circuit.
\nMETHOD 1
\nrecognizing that the real part is distributive (M1)
\n\n(A1)
\n(from the GDC) (A1)
\n
Note: Accept arguments differing by e.g. .
therefore A1
Note: Award the last A1 for the correct values of and seen either in the required form or not. If method used is unclear and answer is partially incorrect, assume Method 2 and award appropriate marks eg. (M1)A1A0A0 if only value is correct.
\n
METHOD 2
\nconverting given expressions to cos form (M1)
\n\n(from graph) A1
\n\neither by considering transformations or inserting points
\nA1
\nA1
\n
Note: Accept arguments differing by e.g. .
(so, )
Note: It is possible to have , with OR , with OR , with due to properties of the cosine curve.
\n
[4 marks]
\nmaximum voltage is (units) A1
\n\n
[1 mark]
\nThe crucial step in this question was to realize that . Candidates who failed to do this step were usually unable to obtain the required result.
\nThe number of sick days taken by each employee in a company during a year was recorded. The data was organized in a box and whisker diagram as shown below:
\nFor this data, write down
\nthe minimum number of sick days taken during the year.
\nthe lower quartile.
\nthe median.
\nPaul claims that this box and whisker diagram can be used to infer that the percentage of employees who took fewer than six sick days is smaller than the percentage of employees who took more than eleven sick days.
\nState whether Paul is correct. Justify your answer.
\nA1
\n
[1 mark]
A1
\n
[1 mark]
A1
\n
[1 mark]
EITHER
\nEach of these percentages represent approximately of the employees. R1
\n
OR
The diagram is not explicit enough to show what is happening at the quartiles regarding and / we do not have the data points R1
OR
Discrete data not clear how to interpret “fewer”. R1
THEN
Hence, Paul is not correct (OR no such inference can be made). A1
\n
Note: Do not award R0A1.
[2 marks]
The cross-section of a beach is modelled by the equation for where is the height of the beach (in metres) at a horizontal distance metres from an origin. is the time in hours after low tide.
\nAt the water is at the point . The height of the water rises at a rate of metres per hour. The point indicates where the water level meets the beach at time .
\n\n
A snail is modelled as a single point. At it is positioned at . The snail travels away from the incoming water at a speed of metre per hour in the direction along the curve of the cross-section of the beach. The following diagram shows this for a value of , such that .
\nWhen has an -coordinate equal to , find the horizontal component of the velocity of .
\nFind the time taken for the snail to reach the point .
\nHence show that the snail reaches the point before the water does.
\nuse of chain rule (M1)
\n\nattempt to find at (M1)
\n\nA1
\n\n
[3 marks]
\nif the position of the snail is
\nfrom part (a)
\nsince speed is :
\nfinding modulus of velocity vector and equating to (M1)
\nOR
\nOR
\nOR (A1)
\nOR (M1)
\nhours A1
\n\n
[4 marks]
\nEITHER
\ntime for water to reach top is hours (seen anywhere) A1
\n
OR
or at time , height of water is A1
\n
THEN
so the water will not reach the snail AG
\n\n
[1 mark]
\nIn part (a), a small minority of candidates found the horizontal component of velocity correctly. Few candidates made any significant progress in part (b).
\nThe shape of a vase is formed by rotating a curve about the -axis.
\nThe vase is high. The internal radius of the vase is measured at intervals along the height:
\nUse the trapezoidal rule to estimate the volume of water that the vase can hold.
\nOR (M1)
\n\nM1A1
\nA1
\n
Note: Do not award the second M1 If the terms are not squared.
\n
[4 marks]
\nThis was a straightforward question on the trapezoidal rule, presented in an unfamiliar way, but only a tiny minority answered it correctly. It may be that candidates were introduced to the trapezium rule as an approximation to the area under a curve and here they were being asked to find an approximation to a volume and they were unable to see how that could be done.
\nBoris recorded the number of daylight hours on the first day of each month in a northern hemisphere town.
\nThis data was plotted onto a scatter diagram. The points were then joined by a smooth curve, with minimum point and maximum point as shown in the following diagram.
\nLet the curve in the diagram be , where is the time, measured in months, since Boris first recorded these values.
\nBoris thinks that might be modelled by a quadratic function.
\nPaula thinks that a better model is , , for specific values of and .
\nFor Paula’s model, use the diagram to write down
\nThe true maximum number of daylight hours was hours and minutes.
\nWrite down one reason why a quadratic function would not be a good model for the number of hours of daylight per day, across a number of years.
\nthe amplitude.
\nthe period.
\nthe equation of the principal axis.
\nHence or otherwise find the equation of this model in the form:
\n\nFor the first year of the model, find the length of time when there are more than hours and minutes of daylight per day.
\nCalculate the percentage error in the maximum number of daylight hours Boris recorded in the diagram.
\nEITHER
annual cycle for daylight length R1
OR
there is a minimum length for daylight (cannot be negative) R1
OR
a quadratic could not have a maximum and a minimum or equivalent R1
Note: Do not accept “Paula's model is better”.
[1 mark]
A1
\n
[1 mark]
A1
\n
[1 mark]
A1A1
\n
Note: Award A1 “ (a constant)” and A1 for that constant being .
\n
[2 marks]
\nOR A1A1A1
\n
Note: Award A1 for (or ), A1 for , and A1 for . Award at most A1A1A0 if extra terms are seen or form is incorrect. Award at most A1A1A0 if is used instead of .
\n
[3 marks]
\n(M1)
\n
EITHER
(A1)(A1)
\n
OR
(A1)
\n(A1)
\n
THEN
months ( years) A1
\n\n
Note: Award M1A1A1A0 for an unsupported answer of . If there is only one intersection point, award M1A1A0A0.
\n\n
[4 marks]
\n(M1)(M1)
\n\n
Note: Award M1 for correct values and absolute value signs, M1 for .
\n\n
A1
\n\n
[3 marks]
\nPart (a) indicated a lack of understanding of quadratic functions and the cyclical nature of daylight hours. Some candidates seemed to understand the limitations of a quadratic model but were not always able to use appropriate mathematical language to explain the limitations clearly.
\nIn part (b), many candidates struggled to write down the amplitude, period, and equation of the principal axis.
\nIn part (c), very few candidates recognized that it would be a negative cosine graph here and most did not know how to find the “b” value even if they had originally found the period in part (b). Some candidates used the regression features in their GDC to find the equation of the model; this is outside the SL syllabus but is a valid approach and earned full credit.
\nIn part (d), very few candidates were awarded “follow through” marks in this part. Some substituted 10.5 into their equation rather that equate their equation to 10.5 and attempt to solve it using their GDC to graph the equations or using the “solver” function.
\nPart (e) was perhaps the best answered part in this question. However, due to premature rounding, many candidates did not gain full marks. A common error was to write the true number of daylight hours as 16.14.
\n\n
A vertical pole stands on a sloped platform. The bottom of the pole is used as the origin, , of a coordinate system in which the top, , of the pole has coordinates . All units are in metres.
\nThe pole is held in place by ropes attached at .
\nOne of these ropes is attached to the platform at point . The rope forms a straight line from to .
\nFind .
\nFind the length of the rope.
\nFind , the angle the rope makes with the platform.
\nA1
\n
[1 mark]
(M1)
\nA1
\n
[2 marks]
EITHER
\nA1
\n\n(A1)
\n(M1)
\n(A1)
\n\n
Note: If is used in place of then will be negative.
Award A1(A1)(M1)(A1) as above. In order to award the final A1, some justification for changing the resulting obtuse angle to its supplementary angle must be seen.
\n
OR
\n(A1)
\n(M1)(A1)
\n(A1)
\n
THEN
OR A1
\n\n
[5 marks]
\nMany candidates appeared unfamiliar with the notation for a vector and interpreted it to mean the magnitude of the vector. This notation is defined in the notation list in the subject guide. Candidates are recommended to be aware of this list and to use it throughout the course; this list indicates notation that will be used in the examination questions without introduction and thus it is important that candidates are familiar. It was also common to see in place of . Often, candidates were able to find the magnitude correctly even though they failed to write the vector. Correct answers to part (c) were more elusive. The use of scalar product or cosine rule was common although some incorrectly assumed that was a right angle. Other errors were to find one of the other angles of the triangle. Some candidates using the scalar product found the obtuse angle between and , but failed to find its supplementary angle. In a question like this, it is important that a suitable level of accuracy is maintained throughout so that early rounding will not affect subsequent parts of the question. Candidates must also learn to final round answers and not to truncate them.
\nEddie decides to construct a path across his rectangular grass lawn using pairs of tiles.
\nEach tile is wide and long. The following diagrams show the path after Eddie has laid one pair and three pairs of tiles. This pattern continues until Eddie reaches the other side of his lawn. When pairs of tiles are laid, the path has a width of centimetres and a length centimetres.
\nThe following diagrams show this pattern for one pair of tiles and for three pairs of tiles, where the white space around each diagram represents Eddie’s lawn.
\nThe following table shows the values of and for the first three values of .
\nFind the value of
\nWrite down an expression in terms of for
\nEddie’s lawn has a length .
\nThe tiles cost per square metre and are sold in packs of five tiles.
\nTo allow for breakages Eddie wants to have at least more tiles than he needs.
\nThere is a fixed delivery cost of .
\n.
\n.
\n.
\n.
\nShow that Eddie needs tiles.
\nFind the value of for this path.
\nFind the total area of the tiles in Eddie’s path. Give your answer in the form where and is an integer.
\nFind the cost of a single pack of five tiles.
\nFind the minimum number of packs of tiles Eddie will need to order.
\nFind the total cost for Eddie’s order.
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\narithmetic formula chosen (M1)
\nA1
\n\n
[2 marks]
\narithmetic formula chosen
\nA1
\n\n
[1 mark]
\nOR M1
\nA1
\ntiles AG
\n
Note: The AG line must be stated for the final A1 to be awarded.
\n
[2 marks]
\nA1
\n\n
[1 mark]
\n(M1)
\n(A1)
\nA1
\n
Note: Follow through within the question for correctly converting their intermediate value into standard form (but only if the pre-conversion value is seen).
\n
[3 marks]
\nEITHER
\nsquare metre (M1)
\n(so, tiles) and hence packs of tiles in a square metre (A1)
\n(so each pack is )
\n
OR
area covered by one pack of tiles is (A1)
\n(M1)
\n
THEN
per pack (of tiles) A1
\n\n
[3 marks]
\n(M1)(M1)
\n
Note: Award M1 for correct numerator, M1 for correct denominator.
(packs of tiles) A1
\n
[3 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nIn part (a), most candidates were able to find the correct values for and .
\nIn part (b), most candidates were able to write down the correct expressions for and .
\nIn part (c)(i), candidates continued to struggle with “show that” questions. Some substituted 144 for and worked backwards, however this is never the intention of the question; candidates should progress towards (not “away from”) the given result. Part (ii) was well answered by many candidates.
\nPart (d) was poorly answered with many candidates multiplying 720 with 730 instead of 10 with 20. However, the majority managed to convert their answer correctly to standard form which gained them a mark for that particular skill.
\nPart (e) saw very few candidates find the cost of one packet of tiles. The main reason was the failure to convert cm2 to m2.
\nIn part (f), about half of the candidates managed to find the correct number of packets. Some gained a mark for finding 8% or dividing by 5.
\nIn part (g), most candidates could use their answers to parts (e) and (f) to score “follow through” marks and find the total cost of their order.
\nThe equation of the line can be expressed in vector form .
\nThe matrix is defined by .
\nThe line (where ) is transformed into a new line using the transformation described by matrix .
\nFind the vectors and in terms of and/or .
\nFind the value of .
\nShow that the equation of the resulting line does not depend on or .
\n(one vector to the line is therefore) A1
\nthe line goes up for every across
\n(so the direction vector is) A1
\n
Note: Although these are the most likely answers, many others are possible.
\n
[2 marks]
\n(from GDC OR ) A1
\n\n
[1 mark]
\nMETHOD 1
\nM1A1
\nA1
\ntherefore the new line has equation A1
\nwhich is independent of or AG
\n
Note: The AG line (or equivalent) must be seen for the final A1 line to be awarded.
\n
METHOD 2
\ntake two points on the line, e.g and M1
\nthese map to
\nand A1
\ntherefore a direction vector is
\n(since ) a direction vector is
\nthe line passes through therefore it always has the origin as a jump-on vector A1
\nthe vector equation is therefore A1
\nwhich is independent of or AG
\n
Note: The AG line (or equivalent) must be seen for the final A1 line to be awarded.
\n
METHOD 3
\nM1A1
\nA1
\n\nwhere is an arbitrary parameter. A1
\nwhich is independent of or (as can take any value) AG
\n
Note: The AG line (or equivalent) must be seen for the final A1 line to be awarded.
\n
[4 marks]
\nIn part (a), most candidates were unable to convert the Cartesian equation of a line into its vector form. In part (b), almost every candidate showed that the value of the determinant was zero. In part (c), the great majority of candidates failed to come up with any sort of strategy to solve the problem.
\nThe scores of the eight highest scoring countries in the Eurovision song contest are shown in the following table.
\nFor this data, find
\nChester is investigating the relationship between the highest-scoring countries’ Eurovision score and their population size to determine whether population size can reasonably be used to predict a country’s score.
\nThe populations of the countries, to the nearest million, are shown in the table.
\nChester finds that, for this data, the Pearson’s product moment correlation coefficient is .
\nChester then decides to find the Spearman’s rank correlation coefficient for this data, and creates a table of ranks.
\nWrite down the value of:
\nthe upper quartile.
\nthe interquartile range.
\nDetermine if the Netherlands’ score is an outlier for this data. Justify your answer.
\nState whether it would be appropriate for Chester to use the equation of a regression line for on to predict a country’s Eurovision score. Justify your answer.
\n.
\n.
\n.
\nFind the value of the Spearman’s rank correlation coefficient .
\nInterpret the value obtained for .
\nWhen calculating the ranks, Chester incorrectly read the Netherlands’ score as . Explain why the value of the Spearman’s rank correlation does not change despite this error.
\n(M1)
\n
Note: This (M1) can also be awarded for either a correct or a correct in part (a)(ii).
A1
\n
[2 marks]
\ntheir part (a)(i) – their (clearly stated) (M1)
\nIQR A1
\n\n
[2 marks]
\n((IQR) ) (M1)
\n\nsince R1
\nNetherlands is not an outlier A1
\n
Note: The R1 is dependent on the (M1). Do not award R0A1.
\n
[3 marks]
\nnot appropriate (“no” is sufficient) A1
\nas is too close to zero / too weak a correlation R1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\nA2
\n\n
[2 marks]
\nEITHER
\nthere is a (positive) association between the population size and the score A1
\n
OR
there is a (positive) linear correlation between the ranks of the population size and the ranks of the scores (when compared with the PMCC of ). A1
\n\n
[1 mark]
\nlowering the top score by does not change its rank so is unchanged R1
\n
Note: Accept “this would not alter the rank” or “Netherlands still top rank” or similar. Condone any statement that clearly implies the ranks have not changed, for example: “The Netherlands still has the highest score.”
\n
[1 mark]
\nIn part (a), many candidates could use their GDC to find the upper quartile, but many forgot how to find the inter-quartile range.
\nIn part (b), very few candidates knew how to show if a score is an outlier. Many candidates did not know that there is a mathematical definition to “outlier” and simply wrote sentences explaining why or why not a value was an outlier.
\nIn part (c), candidates were able to assess the validity of a regression line. The justifications for their conclusion revealed a partial or imprecise understanding of the topic. Examples of this include “no correlation”, “weak value of ”, “low relationship”, “not close to 1”.
\nIn part (d), about half of the candidates managed to find the correct values missing from the table.
\nIn part (e), many candidates knew how to use their GDC to find Spearman’s rank correlation coefficient. Some mistakenly wrote down the value for instead of . Very few candidates could correctly interpret the value for as they became confused by the fact that linear correlation must go with the rank, otherwise it is about association. They could either have said “there is an association between population size and score” or “there is a linear correlation between the rank order of the population size and the ranks of the scores”.
\nIn part (f), most candidates were able to work out that, even if the score changed, the rank remained the same.
\nTwo schools are represented by points and on the graph below. A road, represented by the line with equation , passes near the schools. An architect is asked to determine the location of a new bus stop on the road such that it is the same distance from the two schools.
\nFind the equation of the perpendicular bisector of . Give your equation in the form .
\nDetermine the coordinates of the point on where the bus stop should be located.
\ngradient (A1)
\nmidpoint (A1)
\ngradient of bisector (M1)
\nperpendicular bisector: OR (M1)
\nperpendicular bisector: A1
\n
[5 marks]
attempt to solve simultaneous equations (M1)
\n\nA1
\n
[2 marks]
The masses of Fuji apples are normally distributed with a mean of and a standard deviation of .
\nWhen Fuji apples are picked, they are classified as small, medium, large or extra large depending on their mass. Large apples have a mass of between and .
\nApproximately of Fuji apples have a mass within the medium-sized category, which is between and .
\nDetermine the probability that a Fuji apple selected at random will be a large apple.
\nFind the value of .
\nsketch of normal curve with shaded region to the right of the mean and correct values (M1)
\nA1
\n\n
[2 marks]
\nEITHER
\n\n(A1)
\n(A1)
\n
OR
(A1)
\n
OR
OR (A1)
\n
OR
(A1)(A1)
\n
Note: Award A1 for a normal distribution curve with a vertical line on each side of the mean and a correct probability of either or or shown, A1 for a probability of seen.
\n
THEN
A1
\n\n
[3 marks]
\nA straightforward calculation of probability for a normal distribution was well done.
\nThe majority of candidates used 0.68 as the area to the left of k. Very few candidates knew how to approach the question when the probability given was not the complete area to the left or right of k.
\nA slope field for the differential equation is shown.
\nSome of the solutions to the differential equation have a local maximum point and a local minimum point.
\nWrite down the equation of the curve on which all these maximum and minimum points lie.
\nSketch this curve on the slope field.
\nThe solution to the differential equation that passes through the point has both a local maximum point and a local minimum point.
\nOn the slope field, sketch the solution to the differential equation that passes through .
\nA1
\n
[1 mark]
drawn on diagram (correct shape with a maximum at ) A1
\n
[1 mark]
correct shape with a local maximum and minimum, passing through A1
\nlocal maximum and minimum on the graph of A1
\n
[2 marks]
This question was very poorly done and frequently left blank. Few candidates understood the connection between the differential equation and maximum and minimum points. Even when the equation was correctly solved, it was rare to see the curve correctly drawn on the slope field. Some were able to draw a solution to the differential equation on the slope field though often not through the given initial condition.
\nA function is defined by for .
\nFind the range of .
\nFind the value of .
\nand (A1)
\nrange is A1A1
\n
Note: Award at most A1A1A0 if strict inequalities are used.
[3 marks]
EITHER
\nsketch of and or sketch of and (M1)
\n
OR
finding the correct expression of (M1)
\n
OR
(M1)
OR
(M1)
THEN
A1
[2 marks]
The position vector of a particle, , relative to a fixed origin at time is given by
\n.
\nFind the velocity vector of .
\nShow that the acceleration vector of is never parallel to the position vector of .
\nattempt at chain rule (M1)
\nA1
\n\n
[2 marks]
\nattempt at product rule (M1)
\nA1
\n
METHOD 1
let and
\nfinding using
\nM1
\n\n\n\nif is the angle between them, then
\nA1
\nso therefore the vectors are never parallel R1
\n\n
METHOD 2
\nsolve
\nM1
\nthen
\n\n
Note: Condone candidates not excluding the division by zero case here. Some might go straight to the next line.
A1
\nthis is never true so the two vectors are never parallel R1
\n\n
METHOD 3
\nembedding vectors in a 3d space and taking the cross product: M1
\n\nA1
\nsince the cross product is never zero, the two vectors are never parallel R1
\n\n
[5 marks]
\nIn part (a), many candidates found the velocity vector correctly. In part (b), however, many candidates failed to use the product rule correctly to find the acceleration vector. To show that the acceleration vector is never parallel to the position vector, a few candidates put presumably hoping to show that no value of the constant k existed for any t but this usually went nowhere.
\nAt Springfield University, the weights, in , of chinchilla rabbits and sable rabbits were recorded. The aim was to find out whether chinchilla rabbits are generally heavier than sable rabbits. The results obtained are summarized in the following table.
\nA -test is to be performed at the significance level.
\nWrite down the null and alternative hypotheses.
\nFind the -value for this test.
\nWrite down the conclusion to the test. Give a reason for your answer.
\n(let population mean for chinchilla rabbits, population mean for sable rabbits)
\nA1
\nA1
\n
Note: Accept an equivalent statement in words, must include mean and reference to “population mean” / “mean for all chinchilla rabbits” for the first A1 to be awarded.
Do not accept an imprecise “the means are equal”.
[2 marks]
-value A2
\n
Note: Award A1 for an answer of , from “unpooled” settings on GDC.
[2 marks]
. R1
\n(there is sufficient evidence to) reject (or not accept) A1
\n(there is sufficient evidence to suggest that chinchilla rabbits are heavier than sable rabbits)
\n
Note: Do not award R0A1. Accept ‘accept ’.
[2 marks]
Consider the function .
\nLine is a tangent to at the point .
\nFind .
\nUse your answer to part (a) to find the gradient of .
\nDetermine the number of lines parallel to that are tangent to . Justify your answer.
\nA1A1
\n
Note: Award A1 for , A1 for OR
[2 marks]
\nattempt to substitute into their part (a) (M1)
\n\n A1
[2 marks]
\nEITHER
\nM1
\nA1
\n
OR
sketch of with line M1
\nthree points of intersection marked on this graph A1
\n(and it can be assumed no further intersections occur outside of this window)
\n
THEN
there are two other tangent lines to that are parallel to A1
\n\n
Note: The final A1 can be awarded provided two solutions other than are shown OR three points of intersection are marked on the graph.
\nAward M1A1A1 for an answer of “3 lines” where is considered to be parallel with itself (given guide definition of parallel lines), but only if working is shown.
\n\n
[3 marks]
\nWas reasonably well done, with the stronger candidates able to handle a negative exponent appropriately when finding the derivative. There were a few who confused the notation for derivative with the notation for inverse.
\nMost knew to substitute into the derivative to find the gradient at that point, but some also tried to substitute the y-coordinate for .
\nThere was a lot of difficulty understanding what approach would help them determine the number of tangents to that are parallel to L. Several wrote just an answer, which is not adequate when justification is required.
\nArianne plays a game of darts.
\nThe distance that her darts land from the centre, , of the board can be modelled by a normal distribution with mean and standard deviation .
\nFind the probability that
\nEach of Arianne’s throws is independent of her previous throws.
\nIn a competition a player has three darts to throw on each turn. A point is scored if a player throws all three darts to land within a central area around . When Arianne throws a dart the probability that it lands within this area is .
\nIn the competition Arianne has ten turns, each with three darts.
\na dart lands less than from .
\na dart lands more than from .
\nFind the probability that Arianne throws two consecutive darts that land more than from .
\nFind the probability that Arianne does not score a point on a turn of three darts.
\nFind Arianne’s expected score in the competition.
\nFind the probability that Arianne scores at least points in the competition.
\nFind the probability that Arianne scores at least points and less than points.
\nGiven that Arianne scores at least points, find the probability that Arianne scores less than points.
\nLet be the random variable “distance from ”.
\n\n(M1)(A1)
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\n(M1)
\nA1
\n\n
[2 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nlet be the random variable “number of points scored”
\nevidence of use of binomial distribution (M1)
\n(A1)
\n(M1)
\nA1
\n\n
[4 marks]
\nA1
\n\n
[1 mark]
\n(M1)
\nA1
\n
Note: Award M1 for a correct probability statement or indication of correct lower and upper bounds, and .
[2 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nMost candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.
\nMost candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.
\nMost candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.
\nMost candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.
\nMost candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.
\nMost candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.
\nMost candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.
\nMost candidates started this question well and applied the normal distribution correctly in part (a). The same goes for part (b) where the candidates were able to combine probabilities correctly. Part (c) was not very well done, and there were a surprising number of incorrect approaches on a seemingly straightforward problem. This suggests that candidates were not interpreting the problem correctly and there was a lack of careful reading to be sure of the scenario being described. In part (d) most candidates recognized the need to model the situation with the binomial distribution. However, many candidates did not choose a correct probability for the Bernoulli trial in this question and oversimplified the problem. This again seems to be a problem of “interpretation” rather than “conceptual understanding”.
\nConsider the curve .
\nFind .
\nFind .
\nThe curve has a point of inflexion at .
\nFind the value of .
\nuse of product rule (M1)
\nA1
\n\n
[2 marks]
use of product rule (M1)
\nA1
\n\n\n
[2 marks]
OR sketch of with -intercept indicated OR finding the local maximum of at (M1)
\nA1
\n
[2 marks]
Some candidates attempted to apply the product rule in parts (a)(i) and (ii) but often incorrectly, particularly in part (ii) when finding . In part (b) there was little understanding shown of the point of inflexion. There were some attempts, some of which were correct, but many where either the function or the first derivative were set to zero rather than the second derivative.
\nIt is known that the weights of male Persian cats are normally distributed with mean and variance .
\nA group of male Persian cats are drawn from this population.
\nSketch a diagram showing the above information.
\nFind the proportion of male Persian cats weighing between and .
\nDetermine the expected number of cats in this group that have a weight of less than .
\nIt is found that of the cats weigh more than . Estimate the value of .
\nTen of the cats are chosen at random. Find the probability that exactly one of them weighs over .
\n A1A1
Note: Award A1 for a normal curve with mean labelled or , A1 for indication of SD marks on horizontal axis at and/or OR and/or on the correct side and approximately correct position.
[2 marks]
\nOR labelled sketch of region (M1)
\nA1
\n
[2 marks]
(A1)
\n(M1)
\nA1
\n
[3 marks]
OR (A1)
\nOR OR labelled sketch of region (M1)
\nA1
\n
[3 marks]
(A1)
\nrecognition of binomial (M1)
\ne.g.
\nA2
\n
[4 marks]
The cross-section of an arched entrance into the ballroom of a hotel is in the shape of a parabola. This cross-section can be modelled by part of the graph , where is the height of the archway, in metres, at a horizontal distance, metres, from the point , in the bottom corner of the archway.
\nTo prepare for an event, a square-based crate that is wide and high is to be moved through the archway into the ballroom. The crate must remain upright while it is being moved.
\nDetermine an equation for the axis of symmetry of the parabola that models the archway.
\nDetermine whether the crate will fit through the archway. Justify your answer.
\nOR coordinates of maximum point (M1)
\nA1
\n\n
[2 marks]
\nMETHOD 1
\nthe cart is centred in the archway when it is between
\nand , A1
\nwhere (which is greater than ) R1
\nthe archway is tall enough for the crate A1
\n
Note: Do not award R0A1.
\n
METHOD 2
\nthe height of the archway is greater or equal to between
\nand A1
\nwidth of this section of archway =
\n(which is greater than ) R1
\nthe archway is wide enough for the crate A1
\n\n
Note: Do not award R0A1.
\n\n
[3 marks]
\nMost candidates were able to substitute into the formula for axis of symmetry or find the vertex of the parabola correctly, both being appropriate methods, but neglected to write an equation from that, even though the question specifically asked for an equation.
\nDetermining a process to see if the crate would fit through the archway proved to be difficult for many candidates. It was common to see the maximum heights compared, the maximum widths compared, or the area of the front surface of the crate compared to the area of the archway opening. Other candidates merely calculated the height at , positioning the corner of the crate at O, and made their conclusion based on this value, without consideration of how the crate would be moving through the archway.
\nA garden includes a small lawn. The lawn is enclosed by an arc of a circle with centre and radius , such that . The straight border of the lawn is defined by chord .
\nThe lawn is shown as the shaded region in the following diagram.
\nA footpath is to be laid around the curved side of the lawn. Find the length of the footpath.
\nFind the area of the lawn.
\n(M1)(A1)
\nA1
\n
[3 marks]
evidence of splitting region into two areas (M1)
\n(M1)(M1)
\n
Note: Award M1 for correctly substituting into area of sector formula, M1 for evidence of substituting into area of triangle formula.
A1
\n
[4 marks]
The aircraft for a particular flight has seats. The airline’s records show that historically for this flight only of the people who purchase a ticket arrive to board the flight. They assume this trend will continue and decide to sell extra tickets and hope that no more than passengers will arrive.
\nThe number of passengers that arrive to board this flight is assumed to follow a binomial distribution with a probability of .
\nEach passenger pays for a ticket. If too many passengers arrive, then the airline will give in compensation to each passenger that cannot board.
\nThe airline sells tickets for this flight. Find the probability that more than passengers arrive to board the flight.
\nWrite down the expected number of passengers who will arrive to board the flight if tickets are sold.
\nFind the maximum number of tickets that could be sold if the expected number of passengers who arrive to board the flight must be less than or equal to .
\nFind, to the nearest integer, the expected increase or decrease in the money made by the airline if they decide to sell tickets rather than .
\n(let be the number of passengers who arrive)
\nOR (A1)
\nOR (M1)
\nA1
\n
Note: Using the distribution , to work with the that do not arrive for the flight, here and throughout this question, is a valid approach.
\n
[3 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\n(M1)
\n80 A1
\n\n
[2 marks]
\nMETHOD 1
\nEITHER
\nwhen selling tickets
\ntop row A1A1
\nbottom row A1A1
\n
Note: Award A1A1 for each row correct. Award A1 for one correct entry and A1 for the remaining entries correct.
(M1)A1
OR
income is (A1)
\nexpected compensation is
\n(M1)A1A1
\nexpected income when selling tickets is (M1)
\nA1
\n
THEN
income for tickets (A1)
\nso expected gain A1
\n\n
METHOD 2
\nfor tickets sold, let be the compensation paid out
\nA1A1
\n(M1)A1A1
\nextra expected revenue (A1)(M1)
\n
Note: Award A1 for the and M1 for the subtraction.
(to the nearest dollar) A1
\n
METHOD 3
\nlet be the change in income when selling tickets.
\n (A1)(A1)
Note: Award A1 for one error, however award A1A1 if there is no explicit mention that would result in and the other two are correct.
A1A1
(M1)A1A1
\nA1
\n\n
[8 marks]
\nIn part (a) Stronger candidates were able to recognize that they needed to use the binomial to find the probability. Some candidates confused binomialpdf and binomialcdf functions. Some did not understand that “more than 72” means “73 or 74” and how their GDC uses the lower boundary parameter.
\nIn part (b) many candidates could find the expected number of passengers and the maximum number of tickets. This part was well attempted.
\nPart (c) was expected to challenge the strongest candidates and had little scaffolding. However, this may have been too much for this cohort and resulted in few marks being awarded. A variety of methods were used but few progressed beyond finding values of income minus compensation. Only a few candidates took probabilities into consideration.
\nJuliana plans to invest money for years in an account paying interest, compounded annually. She expects the annual inflation rate to be per year throughout the -year period.
\nJuliana would like her investment to be worth a real value of , compared to current values, at the end of the -year period. She is considering two options.
\nOption 1: Make a one-time investment at the start of the -year period.
\nOption 2: Invest at the start of the -year period and then invest into the account
at the end of each year (including the first and last years).
For option 1, determine the minimum amount Juliana would need to invest. Give your answer to the nearest dollar.
\nFor option 2, find the minimum value of that Juliana would need to invest each year. Give your answer to the nearest dollar.
\nMETHOD 1 – (with )
\nEITHER
\n
(A1)(M1)
Note: Award A1 for seen and M1 for all other entries correct.
OR
(A1)(M1)
\n
Note: Award A1 for or seen, M1 for attempt to substitute into compound interest formula and equating to .
THEN
A1
\n
Note: Award A0 if not rounded to a whole number or a negative sign given.
\n
METHOD 2 – (With including inflation)
\ncalculate with inflation
\n(A1)
\n\n
EITHER
(M1)
\n
OR
(M1)
Note: Award M1 for their and all other entries correct.
THEN
A1
\n
Note: Award A0 if not rounded to a whole number or a negative sign given.
\n
METHOD 3 – (Using formula to calculate real rate of return)
\n(real rate of return =) (A1)
\n
EITHER
(A1)
\n
OR
(M1)
Note: Award M1 for all entries correct.
THEN
A1
\n\n
[3 marks]
\nMETHOD 1 – (Finding the future value of the investment using from part (a))
\n
(from Method 1) OR (from Methods 2, 3)
(M1)
Note: Award M1 for interest rate and answer to part (a) as .
OR (A1)
so payment required (from TVM) will be OR A1
\n
Note: Award A0 if a negative sign given, unless already penalized in part (a).
\n
METHOD 2 – (Using )
\n
(A1)(M1)
Note: Award A1 for and , M1 for all other entries correct and opposite and signs.
A1
Note: Correct 3sf answer is , however accept an answer of given that the context supports rounding up. Award A0 if a negative sign given, unless already penalized in part (a).
\n
[3 marks]
\nVery few candidates appeared to be familiar with real rate of return on an investment and attempted the question by ignoring the inflation rate. There was a mix of candidates who attempted to use the financial app on their graphic display calculator and those who attempted to use the compound interest formula, both of which are accepted methods. The preferred method of the IB for calculating the real rate of return is by simply subtracting the inflation rate from the nominal interest rate, however the exact formula is of course accepted too.
\nWas very poorly done, both in recognition of an appropriate future value with inflation and realizing that PV and FV must have opposite signs when using the financial app on their graphic display calculator. In both parts of the question, there seemed to be little consideration as to the appropriateness of the answers, and often unreasonable answers were presented.
\nA company produces bags of sugar with a labelled weight of . The weights of the bags are normally distributed with a mean of and a standard deviation of . In an inspection, if the weight of a randomly chosen bag is less than then the company fails the inspection.
\nFind the probability that the company fails the inspection.
\nA statistician in the company suggests it would be fairer if the company passes the inspection when the mean weight of five randomly chosen bags is greater than .
\nFind the probability of passing the inspection if the statistician’s suggestion is followed.
\nlet be the weight of sugar in the bag
\n(M1)A1
\n
[2 marks]
METHOD 1
\nlet be the mean weight of bags of sugar
\n(A1)
\nuse of (M1)
\n(A1)
\n\nA1
\n\n
METHOD 2
\nlet be the total weight of bags of sugar
\n(A1)
\nuse of for independent random variables (M1)
\n(A1)
\n\nA1
\n\n
[4 marks]
\nPart (a) was straightforward, and a good number of candidates showed their knowledge in achieving a correct answer. Candidates are advised to not use calculator notation, as examiners cannot be familiar with all variations of GDC syntax; instead, correct mathematical notation and/or a written commentary will ensure the method is communicated to the examiner. Rounding errors once again caused problems for some. Good answers to part (b) were much less common and this was a challenging question for many. A few understood how to use the central limit theorem to find the sampling distribution of the sample mean and a few used the mean and variance of the sum of independent random variables.
\nConsider the following directed network.
\nWrite down the adjacency matrix for this network.
\nDetermine the number of different walks of length that start and end at the same vertex.
\nA2
\n\n
Note: Award A2 for the transposed matrix. Presentation in markscheme assumes columns/rows ordered A-E; accept a matrix with rows and/or columns in a different order only if appropriately communicated. Do not FT from part (a) into part (b).
\n\n
[2 marks]
\nraising their matrix to a power of (M1)
\n\n
(A1)
\n\n
Note: The numbers along the diagonal are sufficient to award M1A1.
\n
(the required number is ) A1
\n
[3 marks]
\nThis was well answered by the majority of candidates with most writing down the correct adjacency matrix and then raising it to the power 5.
\nA sector of a circle, centre and radius , is shown in the following diagram.
\nA square field with side has a goat tied to a post in the centre by a rope such that the goat can reach all parts of the field up to from the post.
\n[Source: mynamepong, n.d. Goat [image online] Available at: https://thenounproject.com/term/goat/1761571/
This file is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)
https://creativecommons.org/licenses/by-sa/3.0/deed.en [Accessed 22 April 2010] Source adapted.]
Let be the volume of grass eaten by the goat, in cubic metres, and be the length of time, in hours, that the goat has been in the field.
\nThe goat eats grass at the rate of .
\nFind the angle .
\nFind the area of the shaded segment.
\nFind the area of the field that can be reached by the goat.
\nFind the value of at which the goat is eating grass at the greatest rate.
\nThe goat is tied in the field for hours.
\nFind the total volume of grass eaten by the goat during this time.
\n(M1)(A1)
\n( radians) A1
\n\n
Note: Other methods may be seen; award (M1)(A1) for use of a correct trigonometric method to find an appropriate angle and then A1 for the correct answer.
\n\n
[3 marks]
\nfinding area of triangle
\nEITHER
\narea of triangle (M1)
\n
Note: Award M1 for correct substitution into formula.
(A1)
OR
\n(M1)
\narea triangle
\n(A1)
\n\n
finding area of sector
\nEITHER
\narea of sector (M1)
\n(A1)
\nOR
\narea of sector (M1)
\n(A1)
\n\n
THEN
\narea of segment
\nA1
\n\n
[5 marks]
\nMETHOD 1
\n(A1)
\n(A1)
\nsubtraction of four segments from area of circle (M1)
\nA1
\n\n
METHOD 2
\nangle of sector (A1)
\narea of sector (A1)
\narea is made up of four triangles and four sectors (M1)
\ntotal area
\n\n
A1
\n\n
[4 marks]
\nsketch of OR OR attempt to find where (M1)
\nhour A1
\n\n
[2 marks]
\nrecognizing (M1)
\n(A1)
\nvolume eaten is A1
\n\n
[3 marks]
\nGenerally, this question was answered well but provided a good example of final marks being lost due to premature rounding. Some candidates gave a correct three significant figure intermediate answer of 27.3˚ for the angle in the right-angles triangle and then doubled it to get 54.6˚ as a final answer. This did not receive the final answer mark as the correct answer is 54.5˚ to three significant figures. Premature rounding needs to be avoided in all questions.
\nUnfortunately, many candidates failed to see the connection to part (a). Indeed, the most common answer was to assume the goat could eat all the grass in a circle of radius 4.5m.
\nMost candidates completed this question successfully by graphing the function. A few tried to differentiate the function again and, in some cases, also managed to obtain the correct answer.
\nThis was a question that was pleasingly answered correctly by many candidates who recognized that integration was needed to find the answer. As in part (c) a few tried to do the integration ‘by hand’, and were largely unsuccessful.
\nTommaso and Pietro have each been given euro to save for college.
\nPietro invests his money in an account that pays a nominal annual interest rate of , compounded half-yearly.
\nCalculate the amount Pietro will have in his account after years. Give your answer correct to decimal places.
\nTommaso wants to invest his money in an account such that his investment will increase to times the initial amount in years. Assume the account pays a nominal annual interest of compounded quarterly.
\nDetermine the value of .
\nMETHOD 1
\n OR
(M1)(A1)
Note: Award M1 for an attempt to use a financial app in their technology, A1 for all entries correct.
\n
METHOD 2
\n(M1)(A1)
\neuro A1
\n
[3 marks]
METHOD 1
\n OR
(M1)(A1)
Note: Award M1 for an attempt to use a financial app in their technology, A1 for all entries correct. and must have opposite signs.
\n
METHOD 2
\nOR (M1)(A1)
\n
Note: Award M1 for substitution in compound interest formula, A1 for correct substitution and for equating to (if using LHS equation) or to (if using RHS equation).
A1
Note: Accept .
Accept a trial and error method which leads to .
[3 marks]
The Voronoi diagram below shows four supermarkets represented by points with coordinates , , and . The vertices , , are also shown. All distances are measured in kilometres.
\nThe equation of is and the equation of is .
\nThe coordinates of are and the coordinates of are .
\nA town planner believes that the larger the area of the Voronoi cell , the more people will shop at supermarket .
\nFind the midpoint of .
\nFind the equation of .
\nFind the coordinates of .
\nDetermine the exact length of .
\nGiven that the exact length of is , find the size of in degrees.
\nHence find the area of triangle .
\nState one criticism of this interpretation.
\n(M1)
\nA1
\n\n
Note: Award A0 if parentheses are omitted in the final answer.
\n
[2 marks]
attempt to substitute values into gradient formula (M1)
\n(A1)
\ntherefore the gradient of perpendicular bisector is (M1)
\nso A1
\n\n
[4 marks]
\nidentifying the correct equations to use: (M1)
\nand
\nevidence of solving their correct equations or of finding intersection point graphically (M1)
\nA1
\n\n
Note: Accept an answer expressed as “”.
\n\n
[3 marks]
\nattempt to use distance formula (M1)
\n\nA1
\n\n
[2 marks]
\nMETHOD 1 (cosine rule)
\nlength of is (A1)
\n
Note: Accept and .
attempt to substitute into cosine rule (M1)
(A1)
\n
Note: Award A1 for correct substitution of , , values in the cos rule. Exact values do not need to be used in the substitution.
A1
Note: Last A1 mark may be lost if prematurely rounded values of , and/or are used.
\n
METHOD 2 (splitting isosceles triangle in half)
\nlength of is (A1)
\n
Note: Accept and .
required angle is (M1)(A1)
Note: Award A1 for correct substitution of (or ), values in the cos rule. Exact values do not need to be used in the substitution.
\n
A1
\n
Note: Last A1 mark may be lost if prematurely rounded values of , and/or are used.
\n
[4 marks]
\n(area =) OR (area =) (M1)
\nA1
\n\n
[2 marks]
\nAny sensible answer such as:
\nThere might be factors other than proximity which influence shopping choices.
\nA larger area does not necessarily result in an increase in population.
\nThe supermarkets might be specialized / have a particular clientele who visit even if other shops are closer.
\nTransport links might not be represented by Euclidean distances.
\netc. R1
\n\n
[1 mark]
\nPart (a) was answered very well and demonstrated that the candidates had good knowledge of the midpoint formula. Some candidates did not write the midpoint they found as a coordinate pair and lost a mark there. Part (b) was answered well. Most candidates were able to find the gradient of [BD] and then the gradient of the perpendicular [XZ] and its equation. Candidates who lost marks in (b) were able to collect follow through marks in parts (c), (d), and (e). In part (c), not all candidates were able to identify the correct equations that they needed to find the coordinates of point X. In part (d), many candidates overlooked the fact that the question called for the exact length of [YZ] – the majority of candidates gave the answer correct to three significant figures and hence lost a mark. In part (d) most candidates were able to correctly use the cosine rule. Marks were lost here if the candidates calculated the length of [XZ] incorrectly, or substituted the [XZ], [YZ] or [XY] values incorrectly. Often candidates used rounded [XZ], [YZ] or [XY] values prematurely, for which they lost the last mark. Part (f) was mostly well answered. If candidates found an angle in part (e), they were usually able to find the area in (f) correctly. Again, the contextual question in part (g) was challenging to many candidates. Some answers showed misconceptions about Voronoi diagrams, for example some candidates stated that “if the cell were larger, then some people living there will be much closer to supermarkets A, B, C than to supermarket D.” Many candidates offered explanations, which were long and unclear, which often did not address the issue they were asked to comment on, nor displayed a sound understanding of the topic.
\nA sector of a circle, centre and radius , is shown in the following diagram.
\nA square field with side has a goat tied to a post in the centre by a rope such that the goat can reach all parts of the field up to from the post.
\n[Source: mynamepong, n.d. Goat [image online] Available at: https://thenounproject.com/term/goat/1761571/
This file is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)
https://creativecommons.org/licenses/by-sa/3.0/deed.en [Accessed 22 April 2010] Source adapted.]
Let be the volume of grass eaten by the goat, in cubic metres, and be the length of time, in hours, that the goat has been in the field.
\nThe goat eats grass at the rate of .
\nFind the angle .
\nFind the area of the shaded segment.
\nFind the area of a circle with radius .
\nFind the area of the field that can be reached by the goat.
\nFind the value of at which the goat is eating grass at the greatest rate.
\n(M1)(A1)
\n( radians) A1
\n\n
Note: Other methods may be seen; award (M1)(A1) for use of a correct trigonometric method to find an appropriate angle and then A1 for the correct answer.
\n\n
[3 marks]
\nfinding area of triangle
\nEITHER
\narea of triangle (M1)
\n
Note: Award M1 for correct substitution into formula.
(A1)
OR
\n\narea triangle (M1)
\n(A1)
\n\n
finding area of sector
\nEITHER
\narea of sector (M1)
\n(A1)
\nOR
\narea of sector (M1)
\n(A1)
\n\n
THEN
\narea of segment
\nA1
\n\n
[5 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nMETHOD 1
\n(A1)
\nsubtraction of four segments from area of circle (M1)
\nA1
\n\n
METHOD 2
\n(M1)
\n(A1)
\nA1
\n\n
[3 marks]
\nsketch of OR OR attempt to find where (M1)
\nhour A1
\n\n
[2 marks]
\nPart (a)(i) proved to be difficult for many candidates. About half of the candidates managed to correctly find the angle . A variety of methods were used: cosine to find half of then double it; sine to find angle , then find half of A and double it; Pythagoras to find half of AB and then sine rule to find half of angle then double it; Pythagoras to find half of AB, then double it and use cosine rule to find angle . Many candidates lost a mark here due to premature rounding of an intermediate value and hence the final answer was not correct (to three significant figures).
\nIn part (a)(ii) very few candidates managed to find the correct area of the shaded segment and include the correct units. Some only found the area of the triangle or the area of the sector and then stopped.
\nIn part (b)(i), nearly all candidates managed to find the area of a circle.
\nIn part (b)(ii), finding the area of the field reached by the goat proved troublesome for most of the candidates. It appeared as if the candidates did not fully understand the problem. Very few candidates realized the connection to part (a)(ii).
\nPart (c) was accessed by only a handful of candidates. The candidates could simply have graphed the function on their GDC to find the greatest value, but most did not realize this.
\nThe following Argand diagram shows a circle centre with a radius of units.
\nA set of points, , on the Argand plane are defined by the equation
\n, where .
\nPlot on the Argand diagram the points corresponding to
\nConsider the case where .
\n.
\n.
\n.
\nFind this value of .
\nFor this value of , plot the approximate position of on the Argand diagram.
\n A1
\n
Note: Award A1 for correct modulus and A1 for correct argument for part (a)(i), and A1 for other two points correct.
The points may not be labelled, and they may be shown by line segments.
\n
[1 mark]
\n A1
\n
Note: Award A1 for correct modulus and A1 for correct argument for part (a)(i), and A1 for other two points correct.
The points may not be labelled, and they may be shown by line segments.
\n
[1 mark]
\n A1
\n
Note: Award A1 for correct modulus and A1 for correct argument for part (a)(i), and A1 for other two points correct.
The points may not be labelled, and they may be shown by line segments.
\n
[1 mark]
\n(M1)
\nA1
\n\n
[2 marks]
\n
is shown in the diagram above A1A1
\n
Note: Award A1 for a point plotted on the circle and A1 for a point plotted in the second quadrant.
\n
[2 marks]
\nThis question was challenging to many candidates, and some left the answer blank. Those who attempted it often failed to gain any marks. It would have helped examiners credit responses if points that were plotted on the Argand diagram were labelled. Certainly, there was some confusion caused by the appearance of θ both in the modulus and argument of the complex numbers in Euler form. Better use of technology to help visualize the complex numbers by simply getting decimal approximations of values in terms of π or by converting from Euler to Cartesian form would have helped in this question.
\nA newspaper vendor in Singapore is trying to predict how many copies of The Straits Times they will sell. The vendor forms a model to predict the number of copies sold each weekday. According to this model, they expect the same number of copies will be sold each day.
\nTo test the model, they record the number of copies sold each weekday during a particular week. This data is shown in the table.
\nA goodness of fit test at the significance level is used on this data to determine whether the vendor’s model is suitable.
\nThe critical value for the test is and the hypotheses are
\n The data satisfies the model.
The data does not satisfy the model.
Find an estimate for how many copies the vendor expects to sell each day.
\nWrite down the degrees of freedom for this test.
\nWrite down the conclusion to the test. Give a reason for your answer.
\nA1
\n
[1 mark]
A1
\n
[1 mark]
OR -value A2
\nOR R1
\ntherefore there is insufficient evidence to reject A1
\n(i.e. the data satisfies the model)
\n
Note: Do not award R0A1. Accept “accept” or “do not reject” in place of “insufficient evidence to reject”.
Award the R1 for comparing their -value with or their value with and then FT their final conclusion.
[4 marks]
A Principal would like to compare the students in his school with a national standard. He decides to give a test to eight students made up of four boys and four girls. One of the teachers offers to find the volunteers from his class.
\nThe marks out of , for the students who took the test, are:
\n\nFor the eight students find
\nThe national standard mark is out of .
\nTwo additional students take the test at a later date and the mean mark for all ten students is and the standard deviation is .
\nFor further analysis, a standardized score out of for the ten students is obtained by multiplying the scores by and adding .
\nFor the ten students, find
\nName the type of sampling that best describes the method used by the Principal.
\nthe mean mark.
\nthe standard deviation of the marks.
\nPerform an appropriate test at the significance level to see if the mean marks achieved by the students in the school are higher than the national standard. It can be assumed that the marks come from a normal population.
\nState one reason why the test might not be valid.
\ntheir mean standardized score.
\nthe standard deviation of their standardized score.
\nquota A1
\n\n
[1 mark]
\n(M1)A1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\n(let be the national mean)
\n\nA1
\n
Note: Accept hypotheses in words if they are clearly expressed and ‘population mean’ or ‘school mean’ is referred to. Do not accept unless is explicitly defined as “national standard mark” or given as .
recognizing -test (M1)
-value A1
\nR1
\n
Note: The R1 mark is for the comparison of their -value with .
insufficient evidence to reject the null hypothesis (that the mean for the school is ) A1
Note: Award the final A1 only if the null hypothesis is also correct (e.g. or (population) mean ) and the conclusion is consistent with both the direction of the inequality and the alternative hypothesis.
\n
[5 marks]
\nEITHER
\nthe sampling process is not random R1
\nFor example:
\nthe school asked for volunteers
\nthe students were selected from a single class
\n
OR
the quota might not be representative of the student population R1
\nFor example:
\nthe school may have only boys and girls.
Note: Do not accept ‘the sample is too small’.
\n
[1 mark]
\nA1
\n\n
[1 mark]
\n(A1)
\nA1
\n\n
[2 marks]
\nThe most common answer to this question was ‘convenience sampling’. Though it is a convenience sample because four boys and four girls were required the most appropriate response was ‘quota sampling’.
\nSome candidates still try to calculate a mean and standard deviation by hand. This is not expected.
\nThis was surprisingly poorly answered given that statistical testing forms a large part of the course. Candidates need to give the null and alternative hypotheses, find a p-value, compare this to the significance level and write their conclusion, in context of the question; examination questions may ask for each element individually or the question may say “Perform the test” wherein it is expected that each individual element will be clearly stated (as the test is incomplete if any are omitted). Many candidates had the null hypothesis as an inequality. The easiest way to write the null hypothesis is , but it could also be stated in words so long as it is clear that the population mean is being referred to rather than the sample mean. For example, H0: The mean score of the whole school is equal to 25.2.
\nThe answer that the sample was self-selecting or unrepresentative was the expected response. The sample being small was also accepted if the additional reason of therefore ‘not able to assume a normal population’ was also given. In general, a small sample can be valid (though will probably not be reliable).
\nSome candidates missed the point of this question, that it was concerned with transformations of the mean and standard deviation, and instead tried to work out the actual values for the extra two candidates.
\nThe sum of an infinite geometric sequence is .
\nThe first term is more than the second term.
\nFind the third term. Justify your answer.
\nMETHOD 1
\nA1
\ntherefore
\nA1
\nsubstitute or solve graphically: M1
\nOR
\n\nor
\nonly is possible as the sum to infinity exists R1
\nthen
\nA1
\n\n
METHOD 2
\nA1
\nA1
\nattempt to solve M1
\n\n\n\n\nattempting to solve both possible sequences
\nor
\nor
\nonly is possible as the sum to infinity exists R1
\nA1
\n\n
[5 marks]
\nMany candidates submitted quite poor attempts at this question. Many managed to state the equation obtained by considering the sum to infinity but few managed to find the second equation . Common errors in failing to obtain this equation were that “four more” meant multiplied by four or thinking that the second term was four more than the first term. Even those candidates who obtained both equations were often unable to solve them. Attempted solutions often filled the page with algebra going nowhere. Most of those candidates who actually found the third term correctly then failed to realize that there were two solutions to the equations, one of which had to be rejected. Consequently, the final “reasoning” mark was seldom awarded.
\nThis question is about modelling the spread of a computer virus to predict the number of computers in a city which will be infected by the virus.
\n
A systems analyst defines the following variables in a model:
The following data were collected:
\nA model for the early stage of the spread of the computer virus suggests that
\n\nwhere is the total number of computers in a city and is a measure of how easily the virus is spreading between computers. Both and are assumed to be constant.
\nThe data above are taken from city X which is estimated to have million computers.
The analyst looks at data for another city, Y. These data indicate a value of .
An estimate for , can be found by using the formula:
\n.
\nThe following table shows estimates of for city X at different values of .
\nAn improved model for , which is valid for large values of , is the logistic differential equation
\n\nwhere and are constants.
\nBased on this differential equation, the graph of against is predicted to be a straight line.
\nFind the equation of the regression line of on .
\nWrite down the value of , Pearson’s product-moment correlation coefficient.
\nExplain why it would not be appropriate to conduct a hypothesis test on the value of found in (a)(ii).
\nFind the general solution of the differential equation .
\nUsing the data in the table write down the equation for an appropriate non-linear regression model.
\nWrite down the value of for this model.
\nHence comment on the suitability of the model from (b)(ii) in comparison with the linear model found in part (a).
\nBy considering large values of write down one criticism of the model found in (b)(ii).
\nUse your answer from part (b)(ii) to estimate the time taken for the number of infected computers to double.
\nFind in which city, X or Y, the computer virus is spreading more easily. Justify your answer using your results from part (b).
\nDetermine the value of and of . Give your answers correct to one decimal place.
\nUse linear regression to estimate the value of and of .
\nThe solution to the differential equation is given by
\n\nwhere is a constant.
\nUsing your answer to part (f)(i), estimate the percentage of computers in city X that are expected to have been infected by the virus over a long period of time.
\nA1A1
\n
Note: Award at most A1A0 if answer is not an equation. Award A1A0 for an answer including either or .
\n
[2 marks]
\nA1
\n\n
[1 mark]
\nis not a random variable OR it is not a (bivariate) normal distribution
\nOR data is not a sample from a population
\nOR data appears nonlinear
\nOR only measures linear correlation R1
\n\n
Note: Do not accept “ is not large enough”.
\n\n
[1 mark]
\nattempt to separate variables (M1)
\n\nA1A1A1
\n\n
Note: Award A1 for LHS, A1 for , and A1 for .
\nAward full marks for OR .
\nAward M1A1A1A0 for
\n\n
[4 marks]
\nattempt at exponential regression (M1)
\nA1
\nOR
\nattempt at exponential regression (M1)
\nA1
\n\n
Note: Condone answers involving or . Condone absence of “” Award M1A0 for an incorrect answer in correct format.
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\ncomparing something to do with and something to do with M1
\n\n
Note: Examples of where the M1 should be awarded:
\n
The “correlation coefficient” in the exponential model is larger.
Model B has a larger
Examples of where the M1 should not be awarded:
\nThe exponential model shows better correlation (since not clear how it is being measured)
Model 2 has a better fit
Model 2 is more correlated
\n
an unambiguous comparison between and or and leading to the conclusion that the model in part (b) is more suitable / better A1
\n\n
Note: Condone candidates claiming that is the “correlation coefficient” for the non-linear model.
\n\n
[2 marks]
\nit suggests that there will be more infected computers than the entire population R1
\n\n
Note: Accept any response that recognizes unlimited growth.
\n\n
[1 mark]
\nOR OR OR using the model to find two specific times with values of which double M1
\n(days) A1
\n\n
Note: Do not FT from a model which is not exponential. Award M0A0 for an answer of which comes from using from the data or any other answer which finds a doubling time from figures given in the table.
\n\n
[2 marks]
\nan attempt to calculate for city X (M1)
\n
OR
A1
\nthis is larger than so the virus spreads more easily in city X R1
\n\n
Note: It is possible to award M1A0R1.
Condone “so the virus spreads faster in city X” for the final R1.
\n
[3 marks]
\nA1A1
\n\n
Note: Award A1A0 if values are correct but not to dp.
\n\n
[2 marks]
\n(A1)(A1)
\n
Note: Award A1 for each coefficient seen – not necessarily in the equation. Do not penalize seeing in the context of and .
identifying that the constant is OR that the gradient is (M1)
therefore A1
\n\nA1
\n
Note: Accept a value of of from use of sf value of , or any other value from plausible pre-rounding.
Allow follow-through within the question part, from the equation of their line to the final two A1 marks.
\n
[5 marks]
\nrecognizing that their is the eventual number of infected (M1)
\nA1
\n
Note: Accept any final answer consistent with their answer to part (f)(i) unless their is less than in which case award at most M1A0.
\n
[2 marks]
\nA significant minority were unable to attempt 1(a) which suggests poor preparation for the use of the GDC in this statistics-heavy course. Large numbers of candidates appeared to use and interchangeably. Accurate use of notation is an important skill which needs to be developed.
\n1(a)(iii) was a question at the heart of the Applications and interpretations course. In modern statistics many of the calculations are done by a computer so the skill of the modern statistician lies in knowing which tests are appropriate and how to interpret the results. Very few candidates seemed familiar with the assumptions required for the use of the standard test on the correlation coefficient. Indeed, many candidates answered this by claiming that the value was either too large or too small to do a hypothesis test, indicating a major misunderstanding of the purpose of hypothesis tests.
\n1(b)(i) was done very poorly. It seems that perhaps adding parameters to the equation confused many candidates – if the equation had been many more would have successfully attempted this. However, the presence of parameters is a fundamental part of mathematical modelling so candidates should practise working with expressions involving them.
\n1(b)(ii) and (iii) were done relatively well, with many candidates using the data to recognize an exponential model was a good idea. Part (iv) was often communicated poorly. Many candidates might have done the right thing in their heads but just writing that the correlation was better did not show which figures were being compared. Many candidates who did write down the numbers made it clear that they were comparing an value with an value.
\n1(c) was not meant to be such a hard question. There is a standard formula for half-life which candidates were expected to adapt. However, large numbers of candidates conflated the data and the model, finding the time for one of the data points (which did not lie on the model curve) to double. Candidates also thought that the value of t found was equivalent to the doubling time, often giving answers of around 40 days which should have been obviously wrong.
\n1(d) was quite tough. Several candidates realized that was the required quantity to be compared but very few could calculate for city X using the given information.
\n1(e) was meant to be relatively straightforward but many candidates were unable to interpret the notation given to do the quite straightforward calculation.
\n1(f) was meant to be a more unusual problem-solving question getting candidates to think about ways of linearizing a non-linear problem. This proved too much for nearly all candidates.
\nConsider the function . The graph of is shown in the diagram. The vertex of the graph has coordinates . The graph intersects the -axis at two points, and .
\nFind the value of .
\nFind the value of
\n(i) .
\n(ii) .
\n(iii) .
\nWrite down the equation of the axis of symmetry of the graph.
\nA1
\n
Note: Accept seen.
[1 mark]
METHOD 1
\n(M1)(A1)
\n(i) A1
\n(ii) A1
\n(iii) A1
\n
Note: Award the (M1)(A1) if at least one correct value is seen. Do not apply FT form part (a) if workings are not shown.
METHOD 2
(M1)
\n(i) A1
\n\n
(M1)
\n(ii) A1
\n(iii) A1
\n
[5 marks]
A1
\n
Note: Do not FT from their part (b), this is a contradiction with the diagram.
[1 mark]
The Texas Star is a Ferris wheel at the state fair in Dallas. The Ferris wheel has a diameter of . To begin the ride, a passenger gets into a chair at the lowest point on the wheel, which is above the ground, as shown in the following diagram. A ride consists of multiple revolutions, and the Ferris wheel makes revolutions per minute.
\nThe height of a chair above the ground, , measured in metres, during a ride on the Ferris wheel can be modelled by the function , where is the time, in seconds, since a passenger began their ride.
\nCalculate the value of
\nA ride on the Ferris wheel lasts for minutes in total.
\nFor exactly one ride on the Ferris wheel, suggest
\nBig Tex is a metre-tall cowboy statue that stands on the horizontal ground next to the Ferris wheel.
\n
[Source: Aline Escobar., n.d. Cowboy. [image online] Available at: https://thenounproject.com/search/?q=cowboy&i=1080130
This file is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0)
https://creativecommons.org/licenses/by-sa/3.0/deed.en [Accessed 13/05/2021]. Source adapted.]
There is a plan to relocate the Texas Star Ferris wheel onto a taller platform which will increase the maximum height of the Ferris wheel to . This will change the value of one parameter, , or , found in part (a).
\n.
\n.
\n.
\nCalculate the number of revolutions of the Ferris wheel per ride.
\nan appropriate domain for .
\nan appropriate range for .
\nBy considering the graph of , determine the length of time during one revolution of the Ferris wheel for which the chair is higher than the cowboy statue.
\nIdentify which parameter will change.
\nFind the new value of the parameter identified in part (e)(i).
\nan attempt to find the amplitude (M1)
\nOR
\nA1
\n
Note: Accept an answer of .
\n
[2 marks]
\n(period ) (A1)
\n()
\nA1
\n
Note: Accept an answer of .
\n
[2 marks]
\nattempt to find (M1)
\nOR
\nA1
\n\n
[2 marks]
\nOR (M1)
\n(revolutions per ride) A1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nA1A1
\n\n
Note: Award A1 for correct endpoints of domain and A1 for correct endpoints of range. Award A1 for correct direction of both inequalities.
\n\n
[2 marks]
\ngraph of and OR (M1)
\nand (A1)
\nA1
\n\n
[3 marks]
\nA1
\n\n
[1 mark]
\nEITHER
\n(A1)
\n
OR
(A1)
\n
OR
(new platform height) (A1)
\n
THEN
A1
\n\n
[2 marks]
\nOverall, this question was not well answered. Part (a) proved to be problematic for most candidates – hardly any candidates determined all three parameters correctly. The value of b was rarely found. Most candidates were able to find the number of revolutions in part (b). Only a small number of candidates were able to determine the domain and range in part (c) correctly. In part (d), a number of candidates understood that they needed to solve the equation , and gained the method mark, but very few candidates gained all three marks. In part (e), determining the parameter which would change proved challenging, and few were able to determine correctly how the parameter d would change.
\nThe matrix has eigenvalues and .
\nA switch has two states, and . Each second it either remains in the same state or moves according to the following rule: If it is in state it will move to state with a probability of and if it is in state it will move to state with a probability of .
\nFind an eigenvector corresponding to the eigenvalue of . Give your answer in the form , where .
\nUsing your answer to (a), or otherwise, find the long-term probability of the switch being in state . Give your answer in the form , where .
\nOR (M1)
\n(A1)
\nan eigenvector is (or equivalent with integer values) A1
\n\n
[3 marks]
\nEITHER
\n(the long-term probability matrix is given by the eigenvector corresponding to the eigenvalue equal to , scaled so that the sum of the entries is )
\n(M1)
\n
OR
(M1)
\n
OR
considering high powers of the matrix e.g. (M1)
\n\n
THEN
probability of being in state is A1
\n\n
[2 marks]
\nIn part (a), some candidates could correctly use either or to find an eigenvector but many did not pay attention to the fact that integer values of the eigenvector were required. Some candidates used the method of finding the steady state by finding for some high value of n in part (b) but ignored the fact that they needed to express their answer in rational form. Some did try to convert their calculated answer of 0.467 to but this could only receive partial credit as an exact answer was required.
\nThis question is about a metropolitan area council planning a new town and the location of a new toxic waste dump.
\n
A metropolitan area in a country is modelled as a square. The area has four towns, located at the corners of the square. All units are in kilometres with the -coordinate representing the distance east and the -coordinate representing the distance north from the origin at .
The metropolitan area council decides to build a new town called Isaacopolis located at .
\nA new Voronoi diagram is to be created to include Isaacopolis. The equation of the perpendicular bisector of is .
\nThe metropolitan area is divided into districts based on the Voronoi regions found in part (c).
\nA toxic waste dump needs to be located within the metropolitan area. The council wants to locate it as far as possible from the nearest town.
\nThe toxic waste dump, , is connected to the towns via a system of sewers.
\nThe connections are represented in the following matrix, , where the order of rows and columns is ().
\n\nA leak occurs from the toxic waste dump and travels through the sewers. The pollution takes one day to travel between locations that are directly connected.
\nThe digit in represents a direct connection. The values of in the leading diagonal of mean that once a location is polluted it will stay polluted.
\nThe model assumes that each town is positioned at a single point. Describe possible circumstances in which this modelling assumption is reasonable.
\nSketch a Voronoi diagram showing the regions within the metropolitan area that are closest to each town.
\nFind the equation of the perpendicular bisector of .
\nGiven that the coordinates of one vertex of the new Voronoi diagram are , find the coordinates of the other two vertices within the metropolitan area.
\nSketch this new Voronoi diagram showing the regions within the metropolitan area which are closest to each town.
\nA car departs from a point due north of Hamilton. It travels due east at constant speed to a destination point due North of Gaussville. It passes through the Edison, Isaacopolis and Fermitown districts. The car spends of the travel time in the Isaacopolis district.
\nFind the distance between Gaussville and the car’s destination point.
\nFind the location of the toxic waste dump, given that this location is not on the edge of the metropolitan area.
\nMake one possible criticism of the council’s choice of location.
\nFind which town is last to be polluted. Justify your answer.
\nWrite down the number of days it takes for the pollution to reach the last town.
\nA sewer inspector needs to plan the shortest possible route through each of the connections between different locations. Determine an appropriate start point and an appropriate end point of the inspection route.
\nNote that the fact that each location is connected to itself does not correspond to a sewer that needs to be inspected.
\nthe size of each town is small (in comparison with the distance between the towns)
OR
if towns have an identifiable centre
OR
the centre of the town is at that point R1
Note: Accept a geographical landmark in place of “centre”, e.g. “town hall” or “capitol”.
\n
[1 mark]
\n A1
Note: There is no need for a scale / coordinates here. Condone boundaries extending beyond the metropolitan area.
\n
[1 mark]
\nthe gradient of is (A1)
\nnegative reciprocal of any gradient (M1)
\ngradient of perpendicular bisector
\n
Note: Seeing (for example) used clearly as a gradient anywhere is evidence of the “negative reciprocal” method despite being applied to an inappropriate gradient.
\n
midpoint is (A1)
\nequation of perpendicular bisector is A1
\n
Note: Accept equivalent forms e.g. or .
Allow FT for the final A1 from their midpoint and gradient of perpendicular bisector, as long as the M1 has been awarded
\n
[4 marks]
\nthe perpendicular bisector of is (A1)
\n
Note: Award this A1 if seen in the -coordinate of any final answer or if is used as the -value in the equation of any other perpendicular bisector.
attempt to use symmetry OR intersecting two perpendicular bisectors (M1)
A1
\nA1
\n\n
[4 marks]
\n M1A1
\n
Note: Award M1 for exactly four perpendicular bisectors around (, , and ) seen, even if not in exactly the right place.
\nAward A1 for a completely correct diagram. Scale / coordinates are NOT necessary. Vertices should be in approximately the correct positions but only penalized if clearly wrong (condone northern and southern vertices appearing to be very close to the boundary).
\nCondone the Voronoi diagram extending outside of the square.
\nDo not award follow-though marks in this part.
\n\n
[2 marks]
\nof is (A1)
\nrecognizing line intersects bisectors at (or equivalent) but different -values (M1)
\nand
\nfinding an expression for the distance in Isaacopolis in terms of one variable (M1)
\n\nequating their expression to
\n\n\ndistance A1
\n\n
[4 marks]
\nmust be a vertex (award if vertex given as a final answer) (R1)
\nattempt to calculate the distance of at least one town from a vertex (M1)
\n
Note: This must be seen as a calculation or a value.
correct calculation of distances A1
OR AND OR
\nA1
\n
Note: Award R1M0A0A0 for a vertex written with no other supporting calculations.
Award R1M0A0A1 for correct vertex with no other supporting calculations.
The final A1 is not dependent on the previous A1. There is no follow-through for the final A1.
Do not accept an answer based on “uniqueness” in the question.
\n\n
[4 marks]
\nFor example, any one of the following:
\ndecision does not take into account the different population densities
\ncloser to a city will reduce travel time/help employees
\nit is closer to some cites than others R1
\n\n
Note: Accept any correct reason that engages with the scenario.
Do not accept any answer to do with ethical issues about whether toxic waste should ever be dumped, or dumped in a metropolitan area.
\n
[1 mark]
\nMETHOD 1
\nattempting M1
\nattempting M1
\ne.g.
\nlast row/column of
\nlast row/column of
\nhence Isaacopolis is the last city to be polluted A1
\n
Note: Do not award the A1 unless both and are considered.
Award M1M0A0 for a claim that the shortest distance is from to and that it is , without any support.
\n
METHOD 2
\nattempting to translate to a graph or a list of cities polluted on each day (M1)
\ncorrect graph or list A1
\nhence Isaacopolis is the last city to be polluted A1
\n
Note: Award M1A1A1 for a clear description of the graph in words leading to the correct answer.
\n
[3 marks]
\nit takes days A1
\n\n
[1 mark]
\nEITHER
\nthe orders of the different vertices are:
\n
(A1)
Note: Accept a list where each order is greater than listed above.
OR
a correct diagram/graph showing the connections between the locations (A1)
\n
Note: Accept a diagram with loops at each vertex.
This mark should be awarded if candidate is clearly using their correct diagram from the previous part.
THEN
“Start at F and end at I” OR “Start at I and end at F” A1
\n
Note: Award A1A0 for “it could start at either F or I”.
Award A1A1 for “IGEHTF” OR “FTHEGI”.
Award A1A1 for “F and I” OR “I and F”.
\n
[2 marks]
\nQuestion 2 was based on Voronoi diagrams, but a substantial number of candidates appeared to have not met this topic.
\n2(a) was generally well answered, although a strangely common answer was to claim that the towns must be 1km by 1km! Perhaps this came from thinking of coordinates as representing pixels rather than points.
\n2(b) was done well by most candidates who knew what a Voronoi diagram was.
\n2(c)(i) showed some poor planning skills by many candidates who found a line perpendicular to the given Voronoi edge rather than finding the gradient of IF.
\nIn 2(c)(ii) candidates would have benefited from taking a step back and thinking about the symmetry of the situation before unthinkingly intersecting a lot of lines.
\n2(c)(iii) was done well by some, but many did not seem to have any intuition for the effect of adding a point to a Voronoi diagram.
\n2(d) was probably the worst answered question. Candidates did not seem to have the problem-solving tools to deal with this slightly unfamiliar situation.
\nIn 2(e) many candidates clearly did not know that the solution to the toxic waste problem (as described in the syllabus) occurs at a vertex of the Voronoi diagram.
\nA pleasing number of candidates were able to approach 2(f) even if they were unable to do earlier parts of the question, and in these very long questions candidates should be advised that sometimes later parts are not necessarily harder or impossible to access. Very few who attempted the matrix power approach could interpret what zeroes meant in the matrices produced, but a good number successfully turned the matrix into a graph and proceeded well from there.
\nA particle moves such that its displacement, metres, from a point at time seconds is given by the differential equation
\n\nThe equation for the motion of the particle is amended to
\n.
\nWhen the particle is stationary at .
\nUse the substitution to show that this equation can be written as
\n.
\nFind the eigenvalues for the matrix .
\nHence state the long-term velocity of the particle.
\nUse the substitution to write the differential equation as a system of coupled, first order differential equations.
\nUse Euler’s method with a step length of to find the displacement of the particle when .
\nFind the long-term velocity of the particle.
\nOR M1
\n
Note: Award M1 for substituting for .
AG
\n
[1 mark]
\n(M1)
\n
Note: Award M1 for an attempt to find eigenvalues. Any indication that has been used is sufficient for the (M1).
OR (A1)
A1
\n\n
[3 marks]
\n(on a phase portrait the particle approaches as increases so long term velocity () is)
\nA1
\n
Note: Only award A1 for if both eigenvalues in part (a)(ii) are negative. If at least one is positive accept an answer of ‘no limit’ or ‘infinity’, or in the case of one positive and one negative also accept ‘no limit or (depending on initial conditions)’.
\n
[1 mark]
\n(A1)
\nA1
\n\n
[2 marks]
\nrecognition that in any recurrence formula (M1)
\n\n(A1)
\n(A1)
\n(when ,) A2
\n\n
[5 marks]
\nrecognizing that is the velocity
\nA1
\n\n
[2 marks]
\nIt was clear that second order differential equations had not been covered by many schools. Fortunately, many were able to successfully answer part (ii) as this was independent of the other two parts. For part (iii) it was expected that candidates would know that two negative eigenvalues mean the system tends to the origin and so the long-term velocity is 0. Some candidates tried to solve the system. It should be noted that when the command term is ‘state’ then no further working out is expected to be seen.
\nForming a coupled system from a second order differential equation and solving it using Euler’s method is a technique included in the course guide. Candidates who had learned this technique were successful in this question.
\nA company produces and sells electric cars. The company’s profit, , in thousands of dollars, changes based on the number of cars, , they produce per month.
\nThe rate of change of their profit from producing electric cars is modelled by
\n.
\nThe company makes a profit of (thousand dollars) when they produce electric cars.
\nFind an expression for in terms of .
\nThe company regularly increases the number of cars it produces.
\nDescribe how their profit changes if they increase production to over cars per month and up to cars per month. Justify your answer.
\nrecognition of need to integrate (eg reverse power rule or integral symbol) (M1)
\nA1A1
\n(M1)
\n
Note: Award M1 for correct substitution of and . A constant of integration must be seen (can be implied by a correct answer).
A1
\n
[5 marks]
profit will decrease (with each new car produced) A1
\nEITHER
\nbecause the profit function is decreasing / the gradient is negative / the rate of change of is negative R1
\n
OR
R1
\n
OR
evidence of finding and R1
\n
Note: Award at most R1A0 if or or both have incorrect values.
[2 marks]
The strength of earthquakes is measured on the Richter magnitude scale, with values typically between and where is the most severe.
\nThe Gutenberg–Richter equation gives the average number of earthquakes per year, , which have a magnitude of at least . For a particular region the equation is
\n, for some .
\nThis region has an average of earthquakes per year with a magnitude of at least .
\nThe equation for this region can also be written as .
\nWithin this region the most severe earthquake recorded had a magnitude of .
\nThe number of earthquakes in a given year with a magnitude of at least can be modelled by a Poisson distribution, with mean . The number of earthquakes in one year is independent of the number of earthquakes in any other year.
\nLet be the number of years between the earthquake of magnitude and the next earthquake of at least this magnitude.
\nFind the value of .
\nFind the value of .
\nFind the average number of earthquakes in a year with a magnitude of at least .
\nFind .
\n(M1)
\nA1
\n\n
[2 marks]
\nEITHER
\n(M1)
\n\n
OR
(M1)
\n
THEN
A1
\n\n
[2 marks]
\nA1
\n
Note: Do not accept an answer of .
\n
[1 mark]
\nMETHOD 1
\nno earthquakes in the first years (M1)
\n
EITHER
let be the number of earthquakes of at least magnitude in a year
\n\n(M1)
\n
OR
let be the number of earthquakes in years
\n(M1)
\n\n
THEN
A1
\n\n
METHOD 2
\nno earthquakes in the first years (M1)
\nlet be the number of earthquakes in years
\nsince is large and is small
\n(M1)
\n\nA1
\n\n
[3 marks]
\nParts (a), (b), and (c) were accessible to many candidates who earned full marks with the manipulation of logs and indices presenting no problems. Part (d), however, proved to be too difficult for most and very few correct attempts were seen. As in question 9, most candidates relied on calculator notation when using the Poisson distribution. The discipline of defining a random variable in terms of its distribution and parameters helps to conceptualize the problem in terms that aid a better understanding. Most candidates who attempted this question blindly entered values into the Poisson distribution calculator and were unable to earn any marks. There were a couple of correct solutions using a binomial distribution to approximate the given quantity.
\nA function is defined by for .
\nFind the range of .
\nFind an expression for the inverse function. The domain is not required.
\nWrite down the range of .
\nand (A1)
\nrange is A1A1
\n
Note: Award at most A1A1A0 if strict inequalities are used.
[3 marks]
interchanging at any stage (A1)
\n\n\n(A1)
\n\nA1
\n
[3 marks]
range is A1
\n
[1 mark]
This question uses statistical tests to investigate whether advertising leads to increased profits for a grocery store.
\n
Aimmika is the manager of a grocery store in Nong Khai. She is carrying out a statistical analysis on the number of bags of rice that are sold in the store each day. She collects the following sample data by recording how many bags of rice the store sells each day over a period of days.
She believes that her data follows a Poisson distribution.
\nAimmika knows from her historic sales records that the store sells an average of bags of rice each day. The following table shows the expected frequency of bags of rice sold each day during the day period, assuming a Poisson distribution with mean .
\nAimmika decides to carry out a goodness of fit test at the significance level to see whether the data follows a Poisson distribution with mean .
\nAimmika claims that advertising in a local newspaper for Thai Baht per day will increase the number of bags of rice sold. However, Nichakarn, the owner of the store, claims that the advertising will not increase the store’s overall profit.
\nNichakarn agrees to advertise in the newspaper for the next days. During that time, Aimmika records that the store sells bags of rice with a profit of on each bag sold.
\nAimmika wants to carry out an appropriate hypothesis test to determine whether the number of bags of rice sold during the days increased when compared with the historic sales records.
\nFind the mean and variance for the sample data given in the table.
\nHence state why Aimmika believes her data follows a Poisson distribution.
\nState one assumption that Aimmika needs to make about the sales of bags of rice to support her belief that it follows a Poisson distribution.
\nFind the value of , of , and of . Give your answers to decimal places.
\nWrite down the number of degrees of freedom for her test.
\nPerform the goodness of fit test and state, with reason, a conclusion.
\nBy finding a critical value, perform this test at a significance level.
\nHence state the probability of a Type I error for this test.
\nBy considering the claims of both Aimmika and Nichakarn, explain whether the advertising was beneficial to the store.
\nmean A1
\nvariance A1
\n\n
[2 marks]
\nmean is close to the variance A1
\n\n
[1 mark]
\nOne of the following:
\nthe number of bags sold each day is independent of any other day
\nthe sale of one bag is independent of any other bag sold
\nthe sales of bags of rice (each day) occur at a constant mean rate A1
\n\n
Note: Award A1 for a correct answer in context. Any statement referring to independence must refer to either the independence of each bag sold or the independence of the number of bags sold each day. If the third option is seen, the statement must refer to a “constant mean” or “constant average”. Do not accept “the number of bags sold each day is constant”.
\n\n
[1 mark]
\nattempt to find Poisson probabilities and multiply by (M1)
\nA1
\nA1
\n
EITHER
(M1)
\nA1
\n
OR
(M1)
\nA1
\n\n
Note: Do not penalize the omission of clear , and labelling as this will be penalized later if correct values are interchanged.
\n\n
[5 marks]
\nA1
\n\n
[1 mark]
\nThe number of bags of rice sold each day follows a Poisson distribution with mean . A1
\nThe number of bags of rice sold each day does not follow a Poisson distribution with mean . A1
\n
Note: Award A1A1 for both hypotheses correctly stated and in correct order. Award A1A0 if reference to the data and/or “mean ” is not included in the hypotheses, but otherwise correct.
\n
evidence of attempting to group data to obtain the observed frequencies for and (M1)
\n-value A2
\nR1
\nthe result is not significant so there is no reason to reject (the number of bags sold each day follows a Poisson distribution) A1
\n\n
Note: Do not award R0A1. The conclusion MUST follow through from their hypotheses. If no hypotheses are stated, the final A1 can still be awarded for a correct conclusion as long as it is in context (e.g. therefore the data follows a Poisson distribution).
\n\n
[7 marks]
\nMETHOD 1
\nevidence of multiplying (seen anywhere) M1
\n\nA1
\n
Note: Accept and for the A1.
evidence of finding probabilities around critical region (M1)
Note: Award (M1) for any of these values seen:
OR
\nOR
\nOR
\n
critical value A1
, R1
\nthe null hypothesis is rejected A1
\n(the advertising increased the number of bags sold during the days)
\n
Note: Do not award R0A1. Accept statements referring to the advertising being effective for A1 as long as the R mark is satisfied. For the R1A1, follow through within the part from their critical value.
\n
METHOD 2
\nevidence of dividing by (or seen anywhere) M1
\n\nA1
\nattempt to find critical value using central limit theorem (M1)
\n(e.g. sample standard deviation , etc.)
\n
Note: Award (M1) for a -value of seen.
critical value A1
R1
\nthe null hypothesis is rejected A1
\n(the advertising increased the number of bags sold during the days)
\n
Note: Do not award R0A1. Accept statements referring to the advertising being effective for A1 as long as the R mark is satisfied. For the R1A1, follow through within the part from their critical value.
\n
[6 marks]
\nA1
\n
Note: If a candidate uses METHOD 2 in part (e)(i), allow an FT answer of for this part but only if the candidate has attempted to find a -value.
\n
[1 mark]
\nattempt to compare profit difference with cost of advertising (M1)
\n
Note: Award (M1) for evidence of candidate mathematically comparing a profit difference with the cost of the advertising.
EITHER
(comparing profit from extra bags of rice with cost of advertising)
A1
OR
(comparing total profit with and without advertising)
A1
OR
(comparing increase of average daily profit with daily advertising cost)
A1
THEN
EITHER
\nEven though the number of bags of rice increased, the advertising is not worth it as the overall profit did not increase. R1
\n
OR
The advertising is worth it even though the cost is less than the increased profit, since the number of customers increased (possibly buying other products and/or returning in the future after advertising stops) R1
\n\n
Note: Follow through within the part for correct reasoning consistent with their comparison.
\n\n
[3 marks]
\nCandidates generally did well in finding the mean, although some wasted time by calculating it by hand rather than by using their GDC. Many candidates were able to find a correct variance. However, there were also many who gave the standard deviation as their variance or simply made the variance the same as their mean without performing a calculation, possibly looking ahead to part (a)(ii). Many candidates successfully used the clue given by the command term “hence” and correctly answered (a)(ii).
\nIt was clear candidates understood that independence was the key term needed in the response. However, a number of candidates struggled either by not being precise enough in their responses (e.g. simply stating “they are independent”) or by incorrectly stating that the bags of rice sold must be independent of the number of days. “Communication” is an assessment objective for the course, and candidates should aim for clarity in their responses thereby ensuring the examiner can be confident in awarding credit.
\nThis question was generally done well by the candidates. The two most common mistakes both stemmed from candidates not paying attention to the instructions given. Candidates either did not give their answer correct to 3 decimal places or they incorrectly attempted to use the normal distribution to find the expected frequencies. Another frequent mistake involved candidates multiplying their probabilities by 100 rather than by 90.
\nWhile most candidates were able to gain the mark in part (i), many candidates arrived at a correct answer but by using the incorrect 𝜒2 test for independence method of determining the degrees of freedom. In part (ii), several common mistakes led to very few candidates receiving the full seven marks for this question part. Candidates struggled to correctly write the hypotheses as they either wrote them in reverse order or they did not correctly reference the data and/or a Poisson distribution with mean 4.2. Unfortunately, some did not state the hypotheses at all. Another common mistake involved candidates incorrectly combining columns to create a column for days when at least 7 bags of rice were sold, possibly from incorrectly thinking that an observed value less than 5 is not allowed when carrying out goodness of fit test. Many candidates also missed out on possible follow through marks for their p-value by not fully writing out the observed and expected values they were inputting into their GDC. Although communication was not being assessed here, it highlights how it is easier to credit a correct method (even if leading to an incorrect answer) if there is appropriate working and/or a running commentary present.
\nIn contrast to part 1(d) where a method was given, many candidates struggled to know how to find a critical value in this question part. Although it was possible for candidates to find a critical value by either using Poisson probabilities or probabilities from a normal approximation, few knew how to begin. As a result, very few candidates scored full marks here.
\nVery few candidates managed to get full marks in this question part. Again, without being guided into a particular method, candidates struggled to understand how to begin the problem. Many candidates did not realize that some calculations were necessary to give a proper conclusion. A subset of these candidates thought it was a continuation of part (e) and made some statement related to their conclusion from the hypothesis test. For the candidates that did try to make some calculations, many simply calculated the profit from selling 282 bags of rice and compared that to the cost of the advertising. These candidates did not realize they needed to compare the difference in the expected profit without advertising and the actual profit with advertising.
\nAt a ship is east and north of a harbour. A coordinate system is defined with the harbour at the origin. The position vector of the ship at is given by .
\nThe ship has a constant velocity of kilometres per hour ().
\nWrite down an expression for the position vector of the ship, hours after .
\nFind the time at which the bearing of the ship from the harbour is .
\nA1
\n
Note: Do not condone the use of or any other variable apart from .
\n
[1 mark]
\nwhen the bearing from the port is , the distance east from the port is equal to the distance north from the port (M1)
\n(A1)
\n\n( hour minutes) (A1)
\ntime is A1
\n\n
[4 marks]
\nMost candidates were able to answer part (a) correctly but there were some very poor examples of vector notation. The question asked for an expression of in terms of and although a failure to write was condoned the use of or some other variable was penalized. In part (b) few candidates recognized that the eastern and northern distances would be equal with a bearing of 045°. Those who correctly obtained a value of often did not use this to find the time as required.
\nConsider the curve .
\nThe shape of a piece of metal can be modelled by the region bounded by the functions , , the -axis and the line segment , as shown in the following diagram. The units on the and axes are measured in metres.
\nThe piecewise function is defined by
\n\nThe graph of is obtained from the graph of by:
\nPoint lies on the graph of and has coordinates . Point is the image of under the given transformations and has coordinates .
\nThe piecewise function is given by
\n\nThe area enclosed by , the -axis and the line is correct to six significant figures.
\nFind .
\nHence show that the equation of the tangent to the curve at the point is .
\nFind the value of and the value of .
\nFind an expression for.
\nFind the value of .
\nFind the value of .
\nFind the area enclosed by , the -axis and the line .
\nFind the area of the shaded region on the diagram.
\n(M1)
\nA1
\n\n
[2 marks]
\ngradient at is M1
\n\n
EITHER
M1
\n
OR
M1
\n\n
Note: Do not allow working backwards from the given answer.
\n\n
THEN
\nhence AG
\n\n
[2 marks]
\n(or ) (accept \" \") A1A1
\n\n
[2 marks]
\nA2
\n
Note: Award A1 if only two correct transformations are seen.
\n
[2 marks]
\nA1
\n
[1 mark]
EITHER
\nCorrect substitution of their part (b) (or ) into the given expression (M1)
\n
OR
(M1)
\n
Note: Award M1 for transforming the equivalent expression for correctly.
THEN
A1
\n
[2 marks]
recognizing need to add two integrals (M1)
\n(A1)
\n
Note: The second integral could be replaced by the formula for the area of a trapezoid .
A1
[3 marks]
EITHER
\narea of a trapezoid (M1)(A1)
\n
OR
(M1)(A1)
\n
Note: If the rounded answer of from part (b) is used, the integral is which would be awarded (M1)(A1).
\n
THEN
\nshaded area (M1)
\n
Note: Award (M1) for the subtraction of both and their area for the trapezoid from their answer to (a)(i).
\n
A1
\n\n
[4 marks]
\nThe differentiation using the power rule was well done. In part (ii) some candidates felt it was sufficient to refer to the equation being the same as the one generated by their calculator. Generally, for ‘show that’ questions an algebraic derivation is expected.
\nThe candidates were successful at applying transformations to points but very few were able to apply these transformations to derive the correct function h. In most cases it was due to not appreciating the effect the horizontal transformations have on x.
\nThe candidates were successful at applying transformations to points but very few were able to apply these transformations to derive the correct function h. In most cases it was due to not appreciating the effect the horizontal transformations have on x.
\nPart (i) was frequently done well using the inbuilt functionality of the GDC. Part (ii) was less structured, and candidates needed to create a clear diagram so they could easily see which areas needed to be subtracted. Most of those who were successful used the formula for the trapezoid for the area they needed to find, though others were also successful through finding the equation of the line AB.
\nThe region bounded by and the -axis is rotated through about the -axis to form a solid.
\nExpand .
\nFind .
\nFind the volume of the solid formed. Give your answer in the form , where .
\nA1
\n\n
[1 mark]
\nOR (M1)
\nA1A1
\n
Note: Award A1 for first expression, A1 for second two expressions.
Award A1A0 for a final answer of .
[3 marks]
volume M1
\nA1
\n\nuse of log laws seen, for example M1
\nOR
\nOR and A1
\n
Note: Other correct integer solutions are possible and should be accepted for example .
\n
[4 marks]
\nSome candidates could answer part (a) (i). The link between parts (i) and (ii) was, however, lost to the majority. Those who did see the link were often able to give a reasonable answer to (ii). But some candidates lacked the skills to integrate without the use of technology, so an indefinite integral presented many problems. Even those who successfully navigated part (a) went on to fail to see the link to part (b). Either the integration was attempted as something totally new and unconnected or it was simply found with the GDC which did not lead to a final answer in the required form.
\nRoger buys a new laptop for himself at a cost of . At the same time, he buys his daughter Chloe a higher specification laptop at a cost of .
\nIt is anticipated that Roger’s laptop will depreciate at a rate of per year, whereas Chloe’s laptop will depreciate at a rate of per year.
\nRoger and Chloe’s laptops will have the same value years after they were purchased.
\nEstimate the value of Roger’s laptop after years.
\nFind the value of .
\nComment on the validity of your answer to part (b).
\n(M1)A1
\n
[2 marks]
(M1)
\nA1
\n
Note: Award M1A0 for in place of .
[2 marks]
\ndepreciation rates unlikely to be constant (especially over a long time period) R1
\n
Note: Accept reasonable answers based on the magnitude of or the fact that “value” depends on factors other than time.
[1 mark]
This question compares possible designs for a new computer network between multiple school buildings, and whether they meet specific requirements.
\n
A school’s administration team decides to install new fibre-optic internet cables underground. The school has eight buildings that need to be connected by these cables. A map of the school is shown below, with the internet access point of each building labelled .
Jonas is planning where to install the underground cables. He begins by determining the distances, in metres, between the underground access points in each of the buildings.
\nHe finds , and .
\nThe cost for installing the cable directly between and is .
\nJonas estimates that it will cost per metre to install the cables between all the other buildings.
\nJonas creates the following graph, , using the cost of installing the cables between two buildings as the weight of each edge.
\nThe computer network could be designed such that each building is directly connected to at least one other building and hence all buildings are indirectly connected.
\nThe computer network fails if any part of it becomes unreachable from any other part. To help protect the network from failing, every building could be connected to at least two other buildings. In this way if one connection breaks, the building is still part of the computer network. Jonas can achieve this by finding a Hamiltonian cycle within the graph.
\nAfter more research, Jonas decides to install the cables as shown in the diagram below.
\nEach individual cable is installed such that each end of the cable is connected to a building’s access point. The connection between each end of a cable and an access point has a probability of failing after a power surge.
\nFor the network to be successful, each building in the network must be able to communicate with every other building in the network. In other words, there must be a path that connects any two buildings in the network. Jonas would like the network to have less than a probability of failing to operate after a power surge.
\nFind .
\nFind the cost per metre of installing this cable.
\nState why the cost for installing the cable between and would be higher than between the other buildings.
\nBy using Kruskal’s algorithm, find the minimum spanning tree for , showing clearly the order in which edges are added.
\nHence find the minimum installation cost for the cables that would allow all the buildings to be part of the computer network.
\nState why a path that forms a Hamiltonian cycle does not always form an Eulerian circuit.
\nStarting at , use the nearest neighbour algorithm to find the upper bound for the installation cost of a computer network in the form of a Hamiltonian cycle.
\nNote: Although the graph is not complete, in this instance it is not necessary to form a table of least distances.
\nBy deleting , use the deleted vertex algorithm to find the lower bound for the installation cost of the cycle.
\nShow that Jonas’s network satisfies the requirement of there being less than a probability of the network failing after a power surge.
\n(M1)(A1)
\n
Note: Award (M1) for substitution into the cosine rule and (A1) for correct substitution.
A1
\n
[3 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nany reasonable statement referring to the lake R1
\n(eg. there is a lake between and , the cables would need to be installed under/over/around the lake, special waterproof cables are needed for lake, etc.)
\n\n
[1 mark]
\nedges (or weights) are chosen in the order
\n
A1A1A1
Note: Award A1 for the first two edges chosen in the correct order. Award A1A1 for the first six edges chosen in the correct order. Award A1A1A1 for all seven edges chosen in the correct order. Accept a diagram as an answer, provided the order of edges is communicated.
\n\n
[3 marks]
\nFinding the sum of the weights of their edges (M1)
total cost A1
\n\n
[2 marks]
\na Hamiltonian cycle is not always an Eulerian circuit as it does not have to include all edges of the graph (only all vertices) R1
\n\n
[1 mark]
\nedges (or weights) are chosen in the order
\n
A1A1A1
Note: Award A1 for the first two edges chosen in the correct order. Award A1A1 for the first five edges chosen in the correct order. Award A1A1A1 for all eight edges chosen in the correct order. Accept a diagram as an answer, provided the order of edges is communicated.
\n
finding the sum of the weights of their edges (M1)
upper bound A1
\n\n
[5 marks]
\nattempt to find MST after deleting vertex D (M1)
these edges (or weights) (in any order)
A1
Note: Prim’s or Kruskal’s algorithm could be used at this stage.
reconnect to MST with two different edges (M1)
A1
Note: This A1 is independent of the first A mark and can be awarded if both and are chosen to reconnect to the MST, even if the MST is incorrect.
finding the sum of the weights of their edges (M1)
Note: For candidates with an incorrect MST or no MST, the weights of at least seven of the edges being summed (two of which must connect to ) must be shown to award this (M1).
lower bound A1
\n
[6 marks]
\nMETHOD 1
\nrecognition of a binomial distribution (M1)
finding the probability that a cable fails (at least one of its connections fails)
OR A1
recognition that two cables must fail for the network to go offline M1
\nrecognition of binomial distribution for network, (M1)
\nOR A1
\ntherefore, the diagram satisfies the requirement since AG
\n
Note: Evidence of binomial distribution may be seen as combinations.
\n
METHOD 2
\nrecognition of a binomial distribution (M1)
finding the probability that at least two connections fail
OR A1
recognition that the previous answer is an overestimate M1
\nfinding probability of two ends of the same cable failing, ,
and the ends of the other cables not failing,
(A1)
\n\n
A1
\ntherefore, the diagram satisfies the requirement since AG
\n\n
METHOD 3
\nrecognition of a binomial distribution M1
finding the probability that the network remains secure if or connections fail or if connections fail provided that the second failed connection occurs at the other end of the cable with the first failure (M1)
\n(remains secure) A1
\nA1
\n(network fails) A1
\ntherefore, the diagram satisfies the requirement since AG
\n\n
METHOD 4
\n(network failing)
\nM1
\nA1A1A1
\n
Note: Award A1 for each of 2nd, 3rd and last terms.
A1
therefore, the diagram satisfies the requirement since AG
\n\n
[5 marks]
\nThis question part was intended to be an easy introduction to help candidates begin working with the larger story and most candidates handled it well. However, it was surprisingly common for a candidate to correctly choose the cosine rule and to make the correct substitutions into the formula but then arrive at an incorrect answer. The frequency of this mistake suggests that candidates were either making simple entry mistakes into their GDC or forgetting to ensure that their GDC was set to degrees rather than radians.
\n(b) and (c) Most candidates were able to gain the three marks available.
\n(d), (f) and (g) These three question parts required candidates to demonstrate their ability to carry out graph theory algorithms. Kruskal’s algorithm was split into two different question parts to guide candidates to show their work. As a result, many were able to score well in part (d)(ii) either from having the correct MST or from “follow through” marks from an incorrect MST. However, without this guidance in 2(f) and 2(g), many candidates did a poor job of showing the process they were using to apply the algorithms. The candidates that scored well were detailed in showing the order of how edges were selected and how they were being summed to arrive at the final answers. Although “follow through” within the problem was not available for the final answer in parts 2(f) and 2(g), many candidates missed the opportunity to gain the final method mark in both parts by not fully showing the process they used.
\nMany candidates were able to state the definitions of Hamiltonian cycles and Eulerian circuits. However, the question was not asking for definitions but rather a distinct conclusion of why a Hamiltonian cycle is not always an Eulerian circuit. Disappointingly, many incorrect answers contained five or more lines or writing that may have used up exam time that could have been devoted to other question parts. Another common mistake seen here was candidates incorrectly trying to state a reason based on the number of odd vertices.
\nThis question was very challenging for almost all the candidates. Although there were several different methods that candidates could have used to answer this question, most candidates were only able to gain one or two marks here. Many candidates did recognize that something binomial was needed, but few knew how to setup the correct parameters for the distribution.
\nThe number of cars arriving at a junction in a particular town in any given minute between and is historically known to follow a Poisson distribution with a mean of cars per minute.
\nA new road is built near the town. It is claimed that the new road has decreased the number of cars arriving at the junction.
\nTo test the claim, the number of cars, , arriving at the junction between and on a particular day will be recorded. The test will have the following hypotheses:
\n the mean number of cars arriving at the junction has not changed,
the mean number of cars arriving at the junction has decreased.
The alternative hypothesis will be accepted if .
\nAssuming the null hypothesis to be true, state the distribution of .
\nFind the probability of a Type I error.
\nFind the probability of a Type II error, if the number of cars now follows a Poisson distribution with a mean of cars per minute.
\nA1
\n
Note: Both distribution and mean must be seen for A1 to be awarded.
[1 mark]
(M1)
\nA1
\n
[2 marks]
(mean number of cars =) (A1)
\n(M1)
\n
Note: Award M1 for using to evaluate a probability.
OR (M1)
A1
\n\n
[4 marks]
\nPart (a) should have been routine as all the information needed to answer it was there in the question but here again a reliance of the use of a calculator’s probability distribution functions has meant that simply stating a distribution is too frequently neglected. Many candidates failed to progress beyond part (a). In parts (b) and (c), a lack of knowledge of Type I and Type II errors prevented candidates from tackling what was otherwise a relatively straightforward question to answer. Some had difficulty with the mechanics of using their own GDC model where must be interpreted as either or to be able to perform the calculation.
\nA meteorologist models the height of a hot air balloon launched from the ground. The model assumes the balloon travels vertically upwards and travels metres in the first minute.
\nDue to the decrease in temperature as the balloon rises, the balloon will continually slow down. The model suggests that each minute the balloon will travel only of the distance travelled in the previous minute.
\nFind how high the balloon will travel in the first minutes after it is launched.
\nThe balloon is required to reach a height of at least metres.
Determine whether it will reach this height.
Suggest a limitation of the given model.
\nrecognition of geometric sequence eg (M1)
\n(A1)
\nA1
\n
[3 marks]
(M1)
\nso the balloon will not reach the required height. A1
\n
[2 marks]
horizontal motion not taken into account,
\nrate of cooling will not likely be linear,
\nballoon is considered a point mass / size of balloon not considered,
\neffects of wind/weather unlikely to be consistent,
\na discrete model has been used, whereas a continuous one may offer greater accuracy R1
\n
Note: Accept any other sensible answer.
[1 mark]
The wind chill index is a measure of the temperature, in , felt when taking into account the effect of the wind.
\nWhen Frieda arrives at the top of a hill, the relationship between the wind chill index and the speed of the wind in kilometres per hour is given by the equation
\n\nFind an expression for .
\nWhen Frieda arrives at the top of a hill, the speed of the wind is kilometres per hour and increasing at a rate of .
\nFind the rate of change of at this time.
\nuse of power rule (M1)
\nOR A1
\n\n
[2 marks]
\n(A1)
\n(M1)
\n\nwhen
\n(M1)
\nA2
\n\n
Note: Accept a negative answer communicated in words, “decreasing at a rate of…”.
Accept a final answer of from use of .
Accept (or ).
\n
[5 marks]
\nThere was some success in using the power rule to differentiate the function in part (a). Many failed to recognize that part (b) was a related rates of change problem. There was also confusion about the term “rate of change” and with the units used in this question.
\nA transformation, , of a plane is represented by , where is a matrix, is a vector, is the position vector of a point in the plane and the position vector of its image under .
\nThe triangle has coordinates , and . Under T, these points are transformed to , and respectively.
\ncan be written as , where and are matrices.
\nrepresents an enlargement with scale factor , centre .
\nrepresents a rotation about .
\nThe transformation can also be described by an enlargement scale factor , centre , followed by a rotation about the same centre .
\nBy considering the image of , find .
\nBy considering the image of and , show that
\n.
\nWrite down the matrix .
\nUse to find the matrix .
\nHence find the angle and direction of the rotation represented by .
\nWrite down an equation satisfied by .
\nFind the value of and the value of .
\n(M1)
\nA1
\n\n
[2 marks]
\nEITHER
\nM1
\nhence A1
\nM1
\nhence A1
\n
OR
M1
\nhence A1
\n\nM1
\nA1
\n\n
THEN
AG
\n\n
[4 marks]
\nA1
\n\n
[1 mark]
\nEITHER
\n(A1)
\n(M1)
\n
Note: The M1 is for an attempt at rearranging the matrix equation. Award even if the order of the product is reversed.
(A1)
OR
let
\nattempt to solve a system of equations M1
\n\nA2
\n
Note: Award A1 for two correct equations, A2 for all four equations correct.
\n
THEN
\nOR OR A1
\n\n
Note: The correct answer can be obtained from reversing the matrices, so do not award if incorrect product seen. If the given answer is obtained from the product , award (A1)(M1)(A0)A0.
\n\n
[4 marks]
\nclockwise A1
\narccosine or arcsine of value in matrix seen (M1)
\nA1
\n
Note: Both A1 marks are dependent on the answer to part (c)(i) and should only be awarded for a valid rotation matrix.
\n
[3 marks]
\nMETHOD 1
\nA1
\n\n
METHOD 2
\nA1
\n\n
Note: Accept substitution of and (and and ) with particular points given in the question.
\n[1 mark]
\nMETHOD 1
\nsolving using simultaneous equations or (M1)
\nA1A1
\n\n\n
METHOD 2
\n(M1)
\n\n
Note: This line, with any of the points substituted, may be seen in part (d)(i) and if so the M1 can be awarded there.
\nA1A1
\n\n\n
[3 marks]
\nPart (i) proved to be straightforward for most candidates. A common error in part (ii) was for candidates to begin with the matrix P and to show it successfully transformed the points to their images. This received no marks. For a ‘show that’ question it is expected that the work moves to rather than from the given answer.
\n(b), (c) These two parts dealt generally with more familiar aspects of matrix transformations and were well done.
\n(b), (c) These two parts dealt generally with more familiar aspects of matrix transformations and were well done.
\nThe trick of recognizing that was invariant was generally not seen and as such the question could not be successfully answered.
\nA function is of the form . Part of the graph of is shown.
\nThe points and have coordinates and , and lie on .
\nThe point is a local maximum and the point is a local minimum.
\nFind the value of , of and of .
\nsubstitute coordinates of
\n\n(A1)
\n
substitute coordinates of
EITHER
(M1)
\nminimum occurs when
\n\n(A1)
\n
OR
minimum value occurs when (M1)
\n(A1)
\n
OR
period (A1)
\n(M1)
\n
THEN
A1
\n(A1)
\neliminate or (M1)
\nOR
\nA1
\nA1
\n\n
[8 marks]
\nThis was a challenging question and suitably positioned at the end of the examination. Candidates who attempted it were normally able to substitute points A and B into the given equation. Some were able to determine the first derivative. Only a few candidates were able to earn significant marks for this question.
\nTwo lines and are given by the following equations, where .
\n\n\nIt is known that and are perpendicular.
\nFind the possible value(s) for .
\nIn the case that , determine whether the lines intersect.
\nsetting a dot product of the direction vectors equal to zero (M1)
\n\n(A1)
\n\n\nA1
\n
[3 marks]
a common point would satisfy the equations
\n\n(M1)
\n\n
\n
METHOD 1
\nsolving the first two equations simultaneously
\nA1
\nsubstitute into the third equation: M1
\n\nso lines do not intersect. R1
\n
Note: Accept equivalent methods based on the order in which the equations are considered.
METHOD 2
attempting to solve the equations using a GDC M1
\nGDC indicates no solution A1
\nso lines do not intersect R1
\n
[4 marks]
The straight metal arm of a windscreen wiper on a car rotates in a circular motion from a pivot point, , through an angle of . The windscreen is cleared by a rubber blade of length that is attached to the metal arm between points and . The total length of the metal arm, , is .
\nThe part of the windscreen cleared by the rubber blade is shown unshaded in the following diagram.
\nCalculate the length of the arc made by , the end of the rubber blade.
\nDetermine the area of the windscreen that is cleared by the rubber blade.
\nattempt to substitute into length of arc formula (M1)
\n\nA1
\n\n
[2 marks]
\nsubtracting two substituted area of sectors formulae (M1)
\nOR (A1)
\nA1
\n\n
[3 marks]
\nThere was some difficulty determining the correct radius to substitute, with several candidates substituting a radius of 46.
\nIt was common to see candidates subtracting the radii before substituting into the area formula, rather than subtracting the sector areas after calculating each. Using the π key on the calculator rather than an approximated value was prevalent and pleasing to see.
\nThis question explores how graph algorithms can be applied to a graph with an unknown edge weight.
\n
Graph is shown in the following diagram. The vertices of represent tourist attractions in a city. The weight of each edge represents the travel time, to the nearest minute, between two attractions. The route between and is currently being resurfaced and this has led to a variable travel time. For this reason, has an unknown travel time minutes, where .
Daniel plans to visit all the attractions, starting and finishing at . He wants to minimize his travel time.
\nTo find a lower bound for Daniel’s travel time, vertex and its adjacent edges are first deleted.
\nDaniel makes a table to show the minimum travel time between each pair of attractions.
\nWrite down the value of
\nTo find an upper bound for Daniel’s travel time, the nearest neighbour algorithm is used, starting at vertex .
\nConsider the case where .
\nConsider the case where .
\nWrite down a Hamiltonian cycle in .
\nUse Prim’s algorithm, starting at vertex , to find the weight of the minimum spanning tree of the remaining graph. You should indicate clearly the order in which the algorithm selects each edge.
\nHence, for the case where , find a lower bound for Daniel’s travel time, in terms of .
\n.
\n.
\n.
\nUse the nearest neighbour algorithm to find two possible cycles.
\nFind the best upper bound for Daniel’s travel time.
\nFind the least value of for which the edge will definitely not be used by Daniel.
\nHence state the value of the upper bound for Daniel’s travel time for the value of found in part (e)(i).
\nThe tourist office in the city has received complaints about the lack of cleanliness of some routes between the attractions. Corinne, the office manager, decides to inspect all the routes between all the attractions, starting and finishing at . The sum of the weights of all the edges in graph is .
\nCorinne inspects all the routes as quickly as possible and takes hours.
\nFind the value of during Corinne’s inspection.
\ne.g. A1
\n
Note: Accept any other correct answers starting at any vertex.
\n
[1 mark]
\nvertices, so edges required for MST (M1)
\n
Note: To award (M1), their edges should not form a cycle.
M1A1A1
Note: Award M1 for the first three edges in correct order, A1 for in correct order and A1 for all of the edges correct.
weight of MST A1
Note: The final A1 can be awarded independently of previous marks.
\n
[5 marks]
\nlower bound (M1)
\nA1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\nattempt to use nearest neighbour algorithm (M1)
\nany two correct cycles from
A1A1
Note: Bracketed vertices may be omitted in candidate’s answer.
Award M1A0A1 for candidates who list two correct sequences of vertices, but omit the final vertex .
\n
[3 marks]
\nuse OR their shortest cycle from (d)(i) (M1)
\nupper bound A1
\n\n
[2 marks]
\ncycle starts:
\nreturn to has two options, or (M1)
\nhence least value of A1
\n\n
[2 marks]
\nupper bound A2
\n\n
[2 marks]
\nrecognition that edges will be repeated / there are odd vertices (M1)
\nOR A1
\nrecognizing and is lowest weight and is repeated (M1)
\nsolution to CPP A1
\nA1
\n
Note: Award M1A0M0A1A1 if only pairing and is considered, leading to a correct answer.
\n
[5 marks]
\nMostly well done, although some candidates wrote down a path instead of a cycle and some candidates wrote a cycle that did not include all the vertices.
\nMany candidates could apply the algorithm correctly to find the weight of the minimum spanning tree. A common misconception was selecting the shortest edge adjacent to the previous vertex, instead of selecting the shortest edge adjacent to the existing tree. This approach will not necessarily find the minimum spanning tree. A small number of candidates found the correct minimum spanning tree, but did not show evidence of using Prim’s algorithm, which only received partial credit; where a method/algorithm is explicit in the question, working must be seen to demonstrate that approach. Part (b)(ii) was also answered reasonably well, however several candidates did not read the instruction to give their answer in terms of , instead choosing a specific value.
\nMany candidates could apply the algorithm correctly to find the weight of the minimum spanning tree. A common misconception was selecting the shortest edge adjacent to the previous vertex, instead of selecting the shortest edge adjacent to the existing tree. This approach will not necessarily find the minimum spanning tree. A small number of candidates found the correct minimum spanning tree, but did not show evidence of using Prim’s algorithm, which only received partial credit; where a method/algorithm is explicit in the question, working must be seen to demonstrate that approach. Part (b)(ii) was also answered reasonably well, however several candidates did not read the instruction to give their answer in terms of , instead choosing a specific value.
\nMostly well-answered.
\nMostly well-answered.
\nMostly well-answered.
\nMany candidates could successfully apply the nearest neighbour algorithm to find a correct cycle, but some made errors finding a second one. Part (ii) was done reasonably well, with many candidates either giving the correct answer or gaining follow-through marks for selecting their shortest cycle from part (d)(i). A small number of candidates incorrectly chose their longest cycle.
\nMany candidates could successfully apply the nearest neighbour algorithm to find a correct cycle, but some made errors finding a second one. Part (ii) was done reasonably well, with many candidates either giving the correct answer or gaining follow-through marks for selecting their shortest cycle from part (d)(i). A small number of candidates incorrectly chose their longest cycle.
\nThis question had a mixed response. Some candidates used the table or the graph to realize that the two choices to get from to are either or . Some incorrectly stated . These candidates often achieved success in part (e)(ii), making the connection to their previous answer in part (d).
\nThis question had a mixed response. Some candidates used the table or the graph to realize that the two choices to get from to are either or . Some incorrectly stated . These candidates often achieved success in part (e)(ii), making the connection to their previous answer in part (d).
\nA surprising number of candidates did not seem to realize this was an application of the Chinese Postman Problem and simply equated the given total weight to minutes. Not only does this show a lack of understanding of the problem, but it also showed a lack of appreciation of the amount of work required to answer a 5 mark question. Out of the many candidates who recognized the need to repeat edges connecting the odd vertices, some did not show complete working to explain why they chose to connect and , only gaining partial credit.
\nThe front view of a doghouse is made up of a square with an isosceles triangle on top.
\nThe doghouse is high and wide, and sits on a square base.
\nThe top of the rectangular surfaces of the roof of the doghouse are to be painted.
\nFind the area to be painted.
\nheight of triangle at roof (A1)
\n
Note: Award A1 for (height of triangle) seen on the diagram.
OR (M1)
A1
\n
Note: If using then (A1) for angle of , (M1) for a correct trig statement.
area of one rectangle on roof M1
area painted
\nA1
\n\n
[5 marks]
\nAlthough the first question on the paper, with appropriate low-level mathematics, the interpretation required seems to have been quite high, and many candidates found this challenging. Many candidates scored only the one mark for the height of the triangle. The most common wrong method seen was calculation of the area of the triangle and adding their result to the calculation 2 × 0.9 × 0.9. Stronger candidates lost the final mark either through premature rounding or incorrect units; both are aspects that can and do occur throughout candidate responses and hence clearly require focus in the classroom.
\nA group of applicants applied for admission into either the Arts programme or the Sciences programme at a university. The outcomes of their applications are shown in the following table.
\nAn applicant is chosen at random from this group. It is found that they were accepted into the programme of their choice.
\nFind the probability that a randomly chosen applicant from this group was accepted by the university.
\nFind the probability that the applicant applied for the Arts programme.
\nTwo different applicants are chosen at random from the original group.
\nFind the probability that both applicants applied to the Arts programme.
\nA1
\n\n
[1 mark]
\nA1A1
\n\n
Note: Award A1 for correct numerator and A1 for correct denominator.
Award A1A0 for working of if followed by an incorrect answer.
\n
[2 marks]
\nA1M1
\n
Note: Award A1 for two correct fractions seen, M1 for multiplying their fractions.
\n
A1
\n\n
[3 marks]
\nCandidates are reasonably proficient at calculating simple probabilities from a table.
\nSeveral candidates did not consider the condition, with some merely finding the probability of an applicant applying for Arts programme with no condition, some considering those who were accepted into Arts with no condition, and some finding the probability of being accepted into Arts given the condition that they applied for Arts.
\nOnly a few candidates recognized dependent events, with most calculating as if the events were independent.
\nA vertical pole stands on horizontal ground. The bottom of the pole is taken as the origin, , of a coordinate system in which the top, , of the pole has coordinates . All units are in metres.
\n
The pole is held in place by ropes attached at .
One of the ropes is attached to the ground at a point with coordinates . The rope forms a straight line from to .
\nFind the length of the rope connecting to .
\nFind , the angle the rope makes with the ground.
\n(M1)
\nA1
\n\n
[2 marks]
\nOR OR (M1)
\nA1
\n\n
[2 marks]
\nThis question was the first of its type to be tested and the problem was reduced (erroneously) to 2D trigonometry. Whilst a minority of candidates did tackle this part correctly, many simply arrived at an incorrect length with metres.
\nThe incorrect interpretation of the diagram as being 2D, meant that there was an assumption that OA was of length 3.2 metres and so ° was seen on many scripts. If, using their incorrect answer for part (a), the candidate had used it was possible to award “follow through” marks.
\nThis question explores models for the height of water in a cylindrical container as water drains out.
\n
The diagram shows a cylindrical water container of height metres and base radius metre. At the base of the container is a small circular valve, which enables water to drain out.
Eva closes the valve and fills the container with water.
\nAt time , Eva opens the valve. She records the height, metres, of water remaining in the container every minutes.
\nEva first tries to model the height using a linear function, , where .
\nEva uses the equation of the regression line of on , to predict the time it will take for all the water to drain out of the container.
\nEva thinks she can improve her model by using a quadratic function, , where .
\nEva uses this equation to predict the time it will take for all the water to drain out of the container and obtains an answer of minutes.
\nLet be the volume, in cubic metres, of water in the container at time minutes.
Let be the radius, in metres, of the circular valve.
Eva does some research and discovers a formula for the rate of change of .
\n\nEva measures the radius of the valve to be metres. Let be the time, in minutes, it takes for all the water to drain out of the container.
\nEva wants to use the container as a timer. She adjusts the initial height of water in the container so that all the water will drain out of the container in minutes.
\nEva has another water container that is identical to the first one. She places one water container above the other one, so that all the water from the highest container will drain into the lowest container. Eva completely fills the highest container, but only fills the lowest container to a height of metre, as shown in the diagram.
\nAt time Eva opens both valves. Let be the height of water, in metres, in the lowest container at time .
\nFind the equation of the regression line of on .
\nInterpret the meaning of parameter in the context of the model.
\nSuggest why Eva’s use of the linear regression equation in this way could be unreliable.
\nFind the equation of the least squares quadratic regression curve.
\nFind the value of .
\nHence, write down a suitable domain for Eva’s function .
\nShow that .
\nBy solving the differential equation , show that the general solution is given by , where .
\nUse the general solution from part (d) and the initial condition to predict the value of .
\nFind this new height.
\nShow that , where .
\nUse Euler’s method with a step length of minutes to estimate the maximum value of .
\nA1A1
\n
Note: Award A1 for an equation in and and A1 for the coefficient and constant .
\n
[2 marks]
\nEITHER
\nthe rate of change of height (of water in metres per minute) A1
\n
Note: Accept “rate of decrease” or “rate of increase” in place of “rate of change”.
OR
the (average) amount that the height (of the water) decreases each minute A1
\n\n
[1 mark]
\nEITHER
\nunreliable to use on equation to estimate A1
\n
OR
unreliable to extrapolate from original data A1
\n
OR
rate of change (of height) might not remain constant (as the water drains out) A1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\n(M1)
\nA1
\n\n
[2 marks]
\nEITHER
\nA1
\n
OR
(due to range of original data / interpolation) A1
\n\n
[1 mark]
\n(A1)
\nEITHER
\nM1
\n
OR
attempt to use chain rule M1
\n\n
THEN
A1
\nAG
\n\n
[3 marks]
\nattempt to separate variables M1
\nA1
\nA1A1
\n
Note: Award A1 for each correct side of the equation.
A1
Note: Award the final A1 for any correct intermediate step that clearly leads to the given equation.
AG
\n
[5 marks]
\n(M1)
\n(A1)
\nsubstituting and their non-zero value of (M1)
\n\n(minutes) A1
\n\n
[4 marks]
\n(A1)
\n(M1)
\n(metres) A1
\n\n
[3 marks]
\nlet be the height of water in the highest container from parts (d) and (e) we get
\n(M1)(A1)
\nso M1A1
\n\n\nAG
\n\n
[4 marks]
\nevidence of using Euler’s method correctly
\ne.g. (A1)
\nmaximum value of (metres) (at minutes) A2
\n( metres)
\n\n
[3 marks]
\nAll parts were answered well. In part(a)(i) a few candidates lost a mark from either not writing an equation or not using the variables and . In part (a)(ii) some candidates incorrectly stated it was the rate of change of water, instead of the rate of change of the height of the water. A few weaker candidates simply stated it is the gradient of the line. In part (a)(iii) some candidates incorrectly criticized the linear model, instead of addressing the question about why it could be unreliable to use the model to make a prediction about the future.
\nAll parts were answered well. In part(a)(i) a few candidates lost a mark from either not writing an equation or not using the variables and . In part (a)(ii) some candidates incorrectly stated it was the rate of change of water, instead of the rate of change of the height of the water. A few weaker candidates simply stated it is the gradient of the line. In part (a)(iii) some candidates incorrectly criticized the linear model, instead of addressing the question about why it could be unreliable to use the model to make a prediction about the future.
\nAll parts were answered well. In part(a)(i) a few candidates lost a mark from either not writing an equation or not using the variables and . In part (a)(ii) some candidates incorrectly stated it was the rate of change of water, instead of the rate of change of the height of the water. A few weaker candidates simply stated it is the gradient of the line. In part (a)(iii) some candidates incorrectly criticized the linear model, instead of addressing the question about why it could be unreliable to use the model to make a prediction about the future.
\nThis question was answered well by many candidates. In part (b)(ii) a small number of candidates incorrectly
gave two answers for , showing a lack of understanding of the context of the model.
This question was answered well by many candidates. In part (b)(ii) a small number of candidates incorrectly
gave two answers for , showing a lack of understanding of the context of the model.
This question was answered well by many candidates. In part (b)(ii) a small number of candidates incorrectly
gave two answers for , showing a lack of understanding of the context of the model.
Many candidates recognized the need to use related rates of change, but could not present coherent working to reach the given answer. Often candidates either did not appreciate the need to use the equation for the volume of a cylinder or did not simplify their equation using . Many candidates wrote nonsense arguments trying to cancel the factor of . In these long paper 3 questions, the purpose of “show that” parts is often to enable candidates to re-enter a question if they are unable to do a previous part.
\nMany candidates were able to correctly separate the variables, but many found the integral of to be too difficult. A common incorrect approach was to use logarithms. A surprising number also incorrectly wrote , showing a lack of understanding of the difference between a parameter and a variable. Given that most questions in this course will be set in context, it is important that candidates learn to distinguish these differences.
\nGenerally done well.
\nMany candidates found this question too difficult.
\nPart (g)(i) was often left blank and was the worst answered question on the paper. Part (g)(ii) was answered correctly by a number of candidates, who made use of the given answer from part (g)(i).
\nPart (g)(i) was often left blank and was the worst answered question on the paper. Part (g)(ii) was answered correctly by a number of candidates, who made use of the given answer from part (g)(i).
\nThe height of a baseball after it is hit by a bat is modelled by the function
\n\nwhere is the height in metres above the ground and is the time in seconds after the ball was hit.
\nWrite down the height of the ball above the ground at the instant it is hit by the bat.
\nFind the value of when the ball hits the ground.
\nState an appropriate domain for in this model.
\nmetres A1
\n\n
[1 mark]
\n(M1)
\nA1
\n\n
Note: If both values for are seen do not award the A1 mark unless the negative is explicitly excluded.
[2 marks]
\nOR A1A1
\n
Note: Award A1 for correct endpoints and A1 for expressing answer with correct notation. Award at most A1A0 for use of instead of .
[2 marks]
\nProbably the best answered question on the paper with many correct answers seen.
\n\n
Many candidates correctly solved the quadratic equation.
\n\n
Some cases, the lower bound was given as 1.2 from confusing height with time. Often the variable was used in the interval notation which lost a mark.
\nUsing geometry software, Pedro draws a quadrilateral . and . Angle and angle . This information is shown in the diagram.
\n, where point is the midpoint of .
\nCalculate the length of .
\nShow that angle , correct to three significant figures.
\nCalculate the area of triangle .
\nPedro draws a circle, with centre at point , passing through point . Part of the circle is shown in the diagram.
\nShow that point lies outside this circle. Justify your reasoning.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)(A1)
\n
Note: Award (M1) for substituted sine rule, (A1) for correct substitution.
(A1)(G2)
Note: If radians are used the answer is award at most (M1)(A1)(A0).
[3 marks]
(A1)(ft)(M1)(A1)(ft)
\n
Note: Award (A1) for or seen, (M1) for substituted cosine rule, (A1)(ft) for correct substitutions.
(A1)
( sig figures) (AG)
\n
Note: Both an unrounded answer that rounds to the given answer and the rounded value must be seen for the final (M1) to be awarded.
Award at most (A1)(ft)(M1)(A1)(ft)(A0) if the known angle is used to validate the result. Follow through from their in part (a).
[4 marks]
Units are required in this question.
\n
(M1)(A1)(ft)
Note: Award (M1) for substituted area formula. Award (A1) for correct substitution.
(A1)(ft)(G3)
Note: Follow through from part (a).
[3 marks]
(A1)(M1)(A1)(ft)
\n
Note: Award (A1) for seen. Award (M1) for substituted cosine rule to find , (A1)(ft) for correct substitutions.
(A1)(ft)(G3)
Note: Follow through from part (a).
OR
(A1)(M1)(A1)(ft)
Note: Award (A1) for or seen. Award (M1) for substituted cosine rule to find (do not award (M1) for cosine or sine rule to find ), (A1)(ft) for correct substitutions.
\n
(A1)(ft)(G3)
\n
Note: Follow through from part (a).
. (A1)(ft)
point is outside the circle. (AG)
\n
Note: Award (A1) for a numerical comparison of and . Follow through for the final (A1)(ft) within the part for their . The final (A1)(ft) is contingent on a valid method to find the value of .
Do not award the final (A1)(ft) if the (AG) line is not stated.
Do not award the final (A1)(ft) if their point is inside the circle.
[5 marks]
A flying drone is programmed to complete a series of movements in a horizontal plane relative to an origin and a set of --axes.
\nIn each case, the drone moves to a new position represented by the following transformations:
\nAll the movements are performed in the listed order.
\nWrite down each of the transformations in matrix form, clearly stating which matrix represents each transformation.
\nFind a single matrix that defines a transformation that represents the overall change in position.
\nFind .
\nHence state what the value of indicates for the possible movement of the drone.
\nThree drones are initially positioned at the points , and . After performing the movements listed above, the drones are positioned at points , and respectively.
\nShow that the area of triangle is equal to the area of triangle .
\nFind a single transformation that is equivalent to the three transformations represented by matrix .
\nNote: For clarity, exact answers are used throughout this markscheme. However it is perfectly acceptable for candidates to write decimal values .
\n\n
rotation anticlockwise is OR (M1)A1
\nreflection in
\n(M1)
\n(A1)
\nmatrix is OR A1
\nrotation clockwise is OR A1
\n\n
[6 marks]
\nNote: For clarity, exact answers are used throughout this markscheme. However it is perfectly acceptable for candidates to write decimal values .
\n\n
an attempt to multiply three matrices (M1)
\n(A1)
\nOR A1
\n\n
[3 marks]
\nNote: For clarity, exact answers are used throughout this markscheme. However it is perfectly acceptable for candidates to write decimal values .
\n\n
A1
\n
Note: Do not award A1 if final answer not resolved into the identity matrix .
\n
[1 mark]
\nNote: For clarity, exact answers are used throughout this markscheme. However it is perfectly acceptable for candidates to write decimal values .
\n\n
if the overall movement of the drone is repeated A1
\nthe drone would return to its original position A1
\n\n
[2 marks]
\nNote: For clarity, exact answers are used throughout this markscheme. However it is perfectly acceptable for candidates to write decimal values .
\n\n
METHOD 1
\nA1
\narea of triangle area of triangle R1
\narea of triangle area of triangle AG
\n
Note: Award at most A1R0 for responses that omit modulus sign.
\n
METHOD 2
\nstatement of fact that rotation leaves area unchanged R1
\nstatement of fact that reflection leaves area unchanged R1
\narea of triangle area of triangle AG
\n\n
[2 marks]
\nNote: For clarity, exact answers are used throughout this markscheme. However it is perfectly acceptable for candidates to write decimal values .
\n\n
attempt to find angles associated with values of elements in matrix (M1)
\n\nreflection (in ) (M1)
\nwhere A1
\nreflection in A1
\n\n
[4 marks]
\nThere were many good attempts at this problem. In most cases these good attempts were undermined by a key lack of understanding. Whilst candidates were able to find the correct matrices in part (a)(i), they then invariably went onto multiply the matrices in the wrong order in part (a)(ii). Whilst follow through marks were readily available after this, the incorrect matrix for then caused issues in part (c). If these candidates had multiplied correctly, it seems that many of them could have gained close to full marks on this question. At the same time there was a lack of precision in the description of the transformation in part (c). As a general point, it would also help candidates if they resolved the trig ratios in the matrices; writing or rather than, for example . Finally, there were many attempts in part (c) that suggested candidates had a good knowledge and understanding of the concepts of matrices and affine transformations.
\nThere were many good attempts at this problem. In most cases these good attempts were undermined by a key lack of understanding. Whilst candidates were able to find the correct matrices in part (a)(i), they then invariably went onto multiply the matrices in the wrong order in part (a)(ii). Whilst follow through marks were readily available after this, the incorrect matrix for then caused issues in part (c). If these candidates had multiplied correctly, it seems that many of them could have gained close to full marks on this question. At the same time there was a lack of precision in the description of the transformation in part (c). As a general point, it would also help candidates if they resolved the trig ratios in the matrices; writing or rather than, for example . Finally, there were many attempts in part (c) that suggested candidates had a good knowledge and understanding of the concepts of matrices and affine transformations.
\nThere were many good attempts at this problem. In most cases these good attempts were undermined by a key lack of understanding. Whilst candidates were able to find the correct matrices in part (a)(i), they then invariably went onto multiply the matrices in the wrong order in part (a)(ii). Whilst follow through marks were readily available after this, the incorrect matrix for then caused issues in part (c). If these candidates had multiplied correctly, it seems that many of them could have gained close to full marks on this question. At the same time there was a lack of precision in the description of the transformation in part (c). As a general point, it would also help candidates if they resolved the trig ratios in the matrices; writing or rather than, for example . Finally, there were many attempts in part (c) that suggested candidates had a good knowledge and understanding of the concepts of matrices and affine transformations.
\nThere were many good attempts at this problem. In most cases these good attempts were undermined by a key lack of understanding. Whilst candidates were able to find the correct matrices in part (a)(i), they then invariably went onto multiply the matrices in the wrong order in part (a)(ii). Whilst follow through marks were readily available after this, the incorrect matrix for then caused issues in part (c). If these candidates had multiplied correctly, it seems that many of them could have gained close to full marks on this question. At the same time there was a lack of precision in the description of the transformation in part (c). As a general point, it would also help candidates if they resolved the trig ratios in the matrices; writing or rather than, for example . Finally, there were many attempts in part (c) that suggested candidates had a good knowledge and understanding of the concepts of matrices and affine transformations.
\nThere were many good attempts at this problem. In most cases these good attempts were undermined by a key lack of understanding. Whilst candidates were able to find the correct matrices in part (a)(i), they then invariably went onto multiply the matrices in the wrong order in part (a)(ii). Whilst follow through marks were readily available after this, the incorrect matrix for then caused issues in part (c). If these candidates had multiplied correctly, it seems that many of them could have gained close to full marks on this question. At the same time there was a lack of precision in the description of the transformation in part (c). As a general point, it would also help candidates if they resolved the trig ratios in the matrices; writing or rather than, for example . Finally, there were many attempts in part (c) that suggested candidates had a good knowledge and understanding of the concepts of matrices and affine transformations.
\nThere were many good attempts at this problem. In most cases these good attempts were undermined by a key lack of understanding. Whilst candidates were able to find the correct matrices in part (a)(i), they then invariably went onto multiply the matrices in the wrong order in part (a)(ii). Whilst follow through marks were readily available after this, the incorrect matrix for then caused issues in part (c). If these candidates had multiplied correctly, it seems that many of them could have gained close to full marks on this question. At the same time there was a lack of precision in the description of the transformation in part (c). As a general point, it would also help candidates if they resolved the trig ratios in the matrices; writing or rather than, for example . Finally, there were many attempts in part (c) that suggested candidates had a good knowledge and understanding of the concepts of matrices and affine transformations.
\nThree towns, , and are represented as coordinates on a map, where the and axes represent the distances east and north of an origin, respectively, measured in kilometres.
\nTown is located at and town is located at . A road runs along the perpendicular bisector of . This information is shown in the following diagram.
\nFind the equation of the line that the road follows.
\nTown is due north of town and the road passes through town .
\nFind the -coordinate of town .
\nmidpoint A1
\n(M1)A1
\n
Note: Accept equivalent gradient statements including using midpoint.
\n
M1
\n
Note: Award M1 for finding the negative reciprocal of their gradient.
OR OR A1
\n
[5 marks]
\nsubstituting into their equation from part (a) (M1)
\n\nA1
\n
Note: Award M1A0 for as their final answer.
\n
[2 marks]
\nA large proportion of candidates seemed to be well drilled into finding the gradient of a line and the subsequent gradient of the normal. But without finding the coordinates of the midpoint of AB, no more marks were gained.
\n\n
Many candidates worked out the value of correctly (or “correct” following the value they found in part (a)) but then incorrectly gave their answer as a coordinate pair.
\nMackenzie conducted an experiment on the reaction times of teenagers. The results of the experiment are displayed in the following cumulative frequency graph.
\nUse the graph to estimate the
\nMackenzie created the cumulative frequency graph using the following grouped frequency table.
\nUpon completion of the experiment, Mackenzie realized that some values were grouped incorrectly in the frequency table. Some reaction times recorded in the interval should have been recorded in the interval .
\nmedian reaction time.
\ninterquartile range of the reaction times.
\nFind the estimated number of teenagers who have a reaction time greater than seconds.
\nDetermine the percentile of the reaction times from the cumulative frequency graph.
\nWrite down the value of .
\nWrite down the value of .
\nWrite down the modal class from the table.
\nUse your graphic display calculator to find an estimate of the mean reaction time.
\nSuggest how, if at all, the estimated mean and estimated median reaction times will change if the errors are corrected. Justify your response.
\nA1
\n\n
[1 mark]
\n(A1)(M1)
\n
Note: Award A1 for correct quartiles seen, M1 for subtraction of their quartiles.
A1
\n
[3 marks]
\n(people have reaction time ) (A1)
\n(people have reaction time ) A1
\n\n
[2 marks]
\nOR (A1)
\nA1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\nA2
\n\n
[2 marks]
\nthe mean will increase A1
\nbecause the incorrect reaction times are moving from a lower interval to a higher interval which will increase the numerator of the mean calculation R1
\n\n
the median will stay the same A1
\nbecause the median or middle of the data is greater than both intervals being changed R1
\n\n
Note: Do not award A1R0.
\n\n
[4 marks]
\nMost candidates were able to determine the median and interquartile range from the given graph. Some lost marks due to use of one significant figure values or because of incorrectly reading the quartiles as 0.75 and 0.25. Candidates were also able to find the estimated number of teenagers with reaction time greater than 0.4s in part (b), but determining the 90th percentile in part (c) proved to be more challenging. Most made a good attempt at completing the frequency table in part (d), but some used cumulative values from the graph incorrectly. Candidates who lost marks in part (d), were able to get “follow through” marks in parts (e) and (f). In part (e), most candidates were able to determine the modal class correctly. Not all candidates used the correct formula to find an estimate for the mean. Candidates who used their calculators usually obtained the correct answer. In part (g), few candidates were able to produce correct statements related to the changes of the mean and the median, and even fewer were able to support these statements with well-articulated reasons.
\nHyungmin designs a concrete bird bath. The bird bath is supported by a pedestal. This is shown in the diagram.
\nThe interior of the bird bath is in the shape of a cone with radius , height and a constant slant height of .
\nLet be the volume of the bird bath.
\nHyungmin wants the bird bath to have maximum volume.
\nWrite down an equation in and that shows this information.
\nShow that .
\nFind .
\nUsing your answer to part (c), find the value of for which is a maximum.
\nFind the maximum volume of the bird bath.
\nTo prevent leaks, a sealant is applied to the interior surface of the bird bath.
\nFind the surface area to be covered by the sealant, given that the bird bath has maximum volume.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(or equivalent) (A1)
\n
Note: Accept equivalent expressions such as or . Award (A0) for a final answer of or , or any further incorrect working.
[1 mark]
OR (M1)
\n
Note: Award (M1) for correct substitution in the volume of cone formula.
(AG)
Note: The final line must be seen, with no incorrect working, for the (M1) to be awarded.
[1 mark]
(A1)(A1)
\n
Note: Award (A1) for , (A1) for . Award at most (A1)(A0) if extra terms are seen. Award (A0) for the term .
[2 marks]
(M1)
\n
Note: Award (M1) for equating their derivative to zero. Follow through from part (c).
OR
sketch of (M1)
\n
Note: Award (M1) for a labelled sketch of with the curve/axes correctly labelled or the -intercept explicitly indicated.
(A1)(ft)
Note: An unsupported is awarded no marks. Graphing the function is not an acceptable method and (M0)(A0) should be awarded. Follow through from part (c). Given the restraints of the question, is not possible.
[2 marks]
(M1)
\nOR
\n(M1)
\n
Note: Award (M1) for substituting their in the volume formula.
(A1)(ft)(G2)
Note: Follow through from part (d).
[2 marks]
(A1)(ft)(M1)
\n
Note: Award (A1) for their correct radius seen .
Award (M1) for correctly substituted curved surface area formula for a cone.
(A1)(ft)(G2)
Note: Follow through from parts (a) and (d).
[3 marks]
A cafe makes litres of coffee each morning. The cafe’s profit each morning, , measured in dollars, is modelled by the following equation
\n\nwhere is a positive constant.
\nThe cafe’s manager knows that the cafe makes a profit of when litres of coffee are made in a morning.
\nThe manager of the cafe wishes to serve as many customers as possible.
\nFind an expression for in terms of and .
\nHence find the maximum value of in terms of . Give your answer in the form , where is a constant.
\nFind the value of .
\nUse the model to find how much coffee the cafe should make each morning to maximize its profit.
\nSketch the graph of against , labelling the maximum point and the -intercepts with their coordinates.
\nDetermine the maximum amount of coffee the cafe can make that will not result in a loss of money for the morning.
\nattempt to expand given expression (M1)
\n\nM1A1
\n
Note: Award M1 for power rule correctly applied to at least one term and A1 for correct answer.
\n
[3 marks]
\nequating their to zero (M1)
\n\n\n(A1)
\nsubstituting their back into given expression (M1)
\n\nA1
\n\n
[4 marks]
\nsubstituting into given expression and equating to M1
\n\nA1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\n A1A1A1
Note: Award A1 for graph drawn for positive indicating an increasing and then decreasing function, A1 for maximum labelled and A1 for graph passing through the origin and , marked on the -axis or whose coordinates are given.
\n
[3 marks]
\nsetting their expression for to zero OR choosing correct -intercept on their graph of (M1)
\nlitres A1
\n\n
[2 marks]
\nThe ticket prices for a concert are shown in the following table.
\nLet the number of adult tickets sold be , the number of child tickets sold be , and the number of student tickets sold be .
\nWrite down three equations that express the information given above.
\nFind the number of each type of ticket sold.
\nA1
\nA1
\nA1
\n
Note: Condone other labelling if clear, e.g. (adult), (child) and (student). Accept equivalent, distinct equations e.g. .
\n
[3 marks]
\nA1A1
\n
Note: Award A1 for all three correct values seen, A1 for correctly labelled as or .
Accept answers written in words: e.g. adult tickets.
\n
[2 marks]
\nMany candidates had at least two of the three equations written down correctly. The interpretation of the phrase “twice as many adult tickets sold as child tickets” was enigmatic. Consequently, was a popular but erroneous answer.
\n\n
Too many candidates spent considerable time attempting to solve three equations with three unknowns by hand with pages of working rather than using their GDC.
\nLet .
\nLet and , where .
\nThe current, , in an AC circuit can be modelled by the equation where is the frequency and is the phase shift.
\nTwo AC voltage sources of the same frequency are independently connected to the same circuit. If connected to the circuit alone they generate currents and . The maximum value and the phase shift of each current is shown in the following table.
\nWhen the two voltage sources are connected to the circuit at the same time, the total current can be expressed as .
\nPlot the position of on an Argand Diagram.
\nExpress in the form , where , giving the exact value of and the exact value of .
\nFind in the form .
\nHence find in the form , where and .
\nFind the maximum value of .
\nFind the phase shift of .
\n A1
\n
[1 mark]
\nA1A1
\n
Note: Accept an argument of . Do NOT accept answers that are not exact.
\n
[2 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nM1
\n(A1)
\nattempt extract real part using cis form (M1)
\nOR A1
\n\n
[4 marks]
\n(M1)
\n(M1)
\n\nmax A1
\n\n
[3 marks]
\nphase shift A1
\n\n
[1 mark]
\nThis question produced the weakest set of responses on the paper. There seemed a general lack of confidence when tackling a problem involving complex numbers. Whilst most candidates could represent a complex number on the complex plane, far fewer had the ability to move between the different forms of complex numbers. This is clearly an area of the course that needs more attention when being taught. Part (c) is challenging but it should be noted that a candidate who has answered parts (a) and (b) with confidence should find this both straightforward, and also an example of a type of problem that is mentioned in the syllabus guidance.
\nThis question produced the weakest set of responses on the paper. There seemed a general lack of confidence when tackling a problem involving complex numbers. Whilst most candidates could represent a complex number on the complex plane, far fewer had the ability to move between the different forms of complex numbers. This is clearly an area of the course that needs more attention when being taught. Part (c) is challenging but it should be noted that a candidate who has answered parts (a) and (b) with confidence should find this both straightforward, and also an example of a type of problem that is mentioned in the syllabus guidance.
\nThis question produced the weakest set of responses on the paper. There seemed a general lack of confidence when tackling a problem involving complex numbers. Whilst most candidates could represent a complex number on the complex plane, far fewer had the ability to move between the different forms of complex numbers. This is clearly an area of the course that needs more attention when being taught. Part (c) is challenging but it should be noted that a candidate who has answered parts (a) and (b) with confidence should find this both straightforward, and also an example of a type of problem that is mentioned in the syllabus guidance.
\nThis question produced the weakest set of responses on the paper. There seemed a general lack of confidence when tackling a problem involving complex numbers. Whilst most candidates could represent a complex number on the complex plane, far fewer had the ability to move between the different forms of complex numbers. This is clearly an area of the course that needs more attention when being taught. Part (c) is challenging but it should be noted that a candidate who has answered parts (a) and (b) with confidence should find this both straightforward, and also an example of a type of problem that is mentioned in the syllabus guidance.
\nThis question produced the weakest set of responses on the paper. There seemed a general lack of confidence when tackling a problem involving complex numbers. Whilst most candidates could represent a complex number on the complex plane, far fewer had the ability to move between the different forms of complex numbers. This is clearly an area of the course that needs more attention when being taught. Part (c) is challenging but it should be noted that a candidate who has answered parts (a) and (b) with confidence should find this both straightforward, and also an example of a type of problem that is mentioned in the syllabus guidance.
\nThis question produced the weakest set of responses on the paper. There seemed a general lack of confidence when tackling a problem involving complex numbers. Whilst most candidates could represent a complex number on the complex plane, far fewer had the ability to move between the different forms of complex numbers. This is clearly an area of the course that needs more attention when being taught. Part (c) is challenging but it should be noted that a candidate who has answered parts (a) and (b) with confidence should find this both straightforward, and also an example of a type of problem that is mentioned in the syllabus guidance.
\nA student investigating the relationship between chemical reactions and temperature finds the Arrhenius equation on the internet.
\n\nThis equation links a variable with the temperature , where and are positive constants and .
\nThe Arrhenius equation predicts that the graph of against is a straight line.
\nWrite down
\nThe following data are found for a particular reaction, where is measured in Kelvin and is measured in :
\nFind an estimate of
\nShow that is always positive.
\nGiven that and , sketch the graph of against .
\n(i) the gradient of this line in terms of ;
\n(ii) the -intercept of this line in terms of .
\nFind the equation of the regression line for on .
\n.
\nIt is not required to state units for this value.
\n.
\nIt is not required to state units for this value.
\nattempt to use chain rule, including the differentiation of (M1)
\nA1
\nthis is the product of positive quantities so must be positive R1
\n
Note: The R1 may be awarded for correct argument from their derivative. R1 is not possible if their derivative is not always positive.
\n
[3 marks]
\n A1A1A1
\n
Note: Award A1 for an increasing graph, entirely in first quadrant, becoming concave down for larger values of , A1 for tending towards the origin and A1 for asymptote labelled at .
\n\n
[3 marks]
\ntaking of both sides OR substituting and (M1)
\nOR (A1)
\n
(i) so gradient is A1
(ii) -intercept is A1
\n
Note: The implied (M1) and (A1) can only be awarded if both correct answers are seen. Award zero if only one value is correct and no working is seen.
\n\n
[4 marks]
\nan attempt to convert data to and (M1)
\ne.g. at least one correct row in the following table
\nline is A1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nattempt to rearrange or solve graphically (M1)
\nA1
\nNote: Accept an value of … from use of value.
\n[2 marks]
\nThis question caused significant difficulties for many candidates and many did not even attempt the question. Very few candidates were able to differentiate the expression in part (a) resulting in difficulties for part (b). Responses to parts (c) to (e) illustrated a lack of understanding of linearizing a set of data. Those candidates that were able to do part (d) frequently lost a mark as their answer was given in x and y.
\nScott purchases food for his dog in large bags and feeds the dog the same amount of dog food each day. The amount of dog food left in the bag at the end of each day can be modelled by an arithmetic sequence.
\nOn a particular day, Scott opened a new bag of dog food and fed his dog. By the end of the third day there were cups of dog food remaining in the bag and at the end of the eighth day there were cups of dog food remaining in the bag.
\nFind the number of cups of dog food
\nIn , Scott spent on dog food. Scott expects that the amount he spends on dog food will increase at an annual rate of .
\nfed to the dog per day.
\nremaining in the bag at the end of the first day.
\nCalculate the number of days that Scott can feed his dog with one bag of food.
\nDetermine the amount that Scott expects to spend on dog food in . Round your answer to the nearest dollar.
\nCalculate the value of .
\nDescribe what the value in part (d)(i) represents in this context.
\nComment on the appropriateness of modelling this scenario with a geometric sequence.
\nEITHER
\n\n(M1)(A1)
\n
Note: Award M1 for attempting to use the arithmetic sequence term formula, A1 for both equations correct. Working for M1 and A1 can be found in parts (i) or (ii).
(cups/day) A1
\n
Note: Answer must be written as a positive value to award A1.
OR
(M1)(A1)
\n
Note: Award M1 for attempting a calculation using the difference between term and term ; A1 for a correct substitution.
(cups/day) A1
\n
[3 marks]
\n(cups) A1
\n\n
[1 mark]
\nattempting to substitute their values into the term formula for arithmetic sequence equated to zero (M1)
\n\ndays A1
\n
Note: Follow through from part (a) only if their answer is positive.
[2 marks]
(M1)(A1)
\n
Note: Award M1 for attempting to use the geometric sequence term formula; A1 for a correct substitution
\n
A1
\n
Note: The answer must be rounded to a whole number to award the final A1.
[3 marks]
A1
\n
[1 mark]
EITHER
\nthe total cost (of dog food) R1
\nfor years beginning in OR years before R1
\n
OR
the total cost (of dog food) R1
\nfrom to (inclusive) OR from to (the start of) R1
\n
[2 marks]
EITHER
According to the model, the cost of dog food per year will eventually be too high to keep a dog.
OR
The model does not necessarily consider changes in inflation rate.
OR
The model is appropriate as long as inflation increases at a similar rate.
OR
The model does not account for changes in the amount of food the dog eats as it ages/becomes ill/stops growing.
OR
The model is appropriate since dog food bags can only be bought in discrete quantities. R1
Note: Accept reasonable answers commenting on the appropriateness of the model for the specific scenario. There should be a reference to the given context. A reference to the geometric model must be clear: either “model” is mentioned specifically, or other mathematical terms such as “increasing” or “discrete quantities” are seen. Do not accept a contextual argument in isolation, e.g. “The dog will eventually die”.
[1 mark]
Parts (a) and (b) were mostly well answered, but some candidates ignored the context and did not give the number of dog food cups per day as a positive number. Most candidates considered geometric sequence in part (c) correctly, and used the correct formula for the nth term, although they used an incorrect value for n at times. Some candidates used the finance application incorrectly. The sum in part (d) was calculated correctly by some candidates, although many seemed unfamiliar with sigma notation and with calculating summations using GDC. In part (d), most candidates interpreted correctly that the sum represented the cost of dog food for 10 years but did not identify the specific 10-year period. Part (e) was not answered well – often candidates made very general and abstract statements devoid of any contextual references.
\nA shock absorber on a car contains a spring surrounded by a fluid. When the car travels over uneven ground the spring is compressed and then returns to an equilibrium position.
\nThe displacement, , of the spring is measured, in centimetres, from the equilibrium position of . The value of can be modelled by the following second order differential equation, where is the time, measured in seconds, after the initial displacement.
\n\nThe differential equation can be expressed in the form , where is a matrix.
\nGiven that , show that .
\nWrite down the matrix .
\nFind the eigenvalues of matrix .
\nFind the eigenvectors of matrix .
\nGiven that when the shock absorber is displaced and its velocity is zero, find an expression for in terms of .
\nA1
\nR1
\n
Note: If no explicit reference is made to , or equivalent, award A0R1 if second line is seen. If used instead of , award A0R0.
AG
\n
[2 marks]
\nA1
\n\n
[1 mark]
\n(M1)
\n(A1)
\nA1
\n\n
[3 marks]
\n(M1)
\n\nA1
\n\n\nA1
\n
Note: Award M1 for a valid attempt to find either eigenvector. Accept equivalent forms of the eigenvectors.
Do not award FT for eigenvectors that do not satisfy both rows of the matrix.
\n
[3 marks]
\nM1A1
\n(M1)
\n\n(M1)
\nA1
\nA1
\n
Note: Do not award the final A1 if the answer is given in the form .
\n
[6 marks]
\nThere were many good attempts at this problem, although simple errors often complicated things. In part (a) an explicit statement of the relationship between the second derivative of and the first derivative of was often issing. Then in part (b) there seemed to be confusion about the matrix, with the correct values often placed in the wrong row or column of the matrix. Despite these errors, candidates made good attempts at finding eigenvalues and eigenvectors. It is to be noted that an error in solving the quadratic equation to find the eigenvectors means that follow-through marks are unlikely to be awarded since the eigenvectors are not reasonable answers and will not be consistent with the eigenvalues. Candidates need to take real care at this point of a question in part (c)(i). A significant number of candidates did not write down the final answer correctly, leaving their final answer in vector form, rather than “ ….” as asked for in the question.
\nThere were many good attempts at this problem, although simple errors often complicated things. In part (a) an explicit statement of the relationship between the second derivative of and the first derivative of was often issing. Then in part (b) there seemed to be confusion about the matrix, with the correct values often placed in the wrong row or column of the matrix. Despite these errors, candidates made good attempts at finding eigenvalues and eigenvectors. It is to be noted that an error in solving the quadratic equation to find the eigenvectors means that follow-through marks are unlikely to be awarded since the eigenvectors are not reasonable answers and will not be consistent with the eigenvalues. Candidates need to take real care at this point of a question in part (c)(i). A significant number of candidates did not write down the final answer correctly, leaving their final answer in vector form, rather than “ ….” as asked for in the question.
\nThere were many good attempts at this problem, although simple errors often complicated things. In part (a) an explicit statement of the relationship between the second derivative of and the first derivative of was often issing. Then in part (b) there seemed to be confusion about the matrix, with the correct values often placed in the wrong row or column of the matrix. Despite these errors, candidates made good attempts at finding eigenvalues and eigenvectors. It is to be noted that an error in solving the quadratic equation to find the eigenvectors means that follow-through marks are unlikely to be awarded since the eigenvectors are not reasonable answers and will not be consistent with the eigenvalues. Candidates need to take real care at this point of a question in part (c)(i). A significant number of candidates did not write down the final answer correctly, leaving their final answer in vector form, rather than “ ….” as asked for in the question.
\nThere were many good attempts at this problem, although simple errors often complicated things. In part (a) an explicit statement of the relationship between the second derivative of and the first derivative of was often issing. Then in part (b) there seemed to be confusion about the matrix, with the correct values often placed in the wrong row or column of the matrix. Despite these errors, candidates made good attempts at finding eigenvalues and eigenvectors. It is to be noted that an error in solving the quadratic equation to find the eigenvectors means that follow-through marks are unlikely to be awarded since the eigenvectors are not reasonable answers and will not be consistent with the eigenvalues. Candidates need to take real care at this point of a question in part (c)(i). A significant number of candidates did not write down the final answer correctly, leaving their final answer in vector form, rather than “ ….” as asked for in the question.
\nThere were many good attempts at this problem, although simple errors often complicated things. In part (a) an explicit statement of the relationship between the second derivative of and the first derivative of was often issing. Then in part (b) there seemed to be confusion about the matrix, with the correct values often placed in the wrong row or column of the matrix. Despite these errors, candidates made good attempts at finding eigenvalues and eigenvectors. It is to be noted that an error in solving the quadratic equation to find the eigenvectors means that follow-through marks are unlikely to be awarded since the eigenvectors are not reasonable answers and will not be consistent with the eigenvalues. Candidates need to take real care at this point of a question in part (c)(i). A significant number of candidates did not write down the final answer correctly, leaving their final answer in vector form, rather than “ ….” as asked for in the question.
\nA geneticist uses a Markov chain model to investigate changes in a specific gene in a cell as it divides. Every time the cell divides, the gene may mutate between its normal state and other states.
\nThe model is of the form
\n\nwhere is the probability of the gene being in its normal state after dividing for the time, and is the probability of it being in another state after dividing for the time, where .
\nMatrix is found to be .
\nThe gene is in its normal state when . Calculate the probability of it being in its normal state
\nWrite down the value of .
\nWhat does represent in this context?
\nFind the eigenvalues of .
\nFind the eigenvectors of .
\nwhen .
\nin the long term.
\nA1
\n\n
[1 mark]
\nthe probability of mutating from ‘not normal state’ to ‘normal state’ A1
\n
Note: The A1 can only be awarded if it is clear that transformation is from the mutated state.
\n
[1 mark]
\n(M1)
\n
Note: Award M1 for an attempt to find eigenvalues. Any indication that has been used is sufficient for the (M1).
OR (A1)
A1
\n\n
[3 marks]
\nOR (M1)
\n
Note: Award M1 can be awarded for attempting to find either eigenvector.
OR
and A1A1
\n
Note: Accept any multiple of the given eigenvectors.
\n
[3 marks]
\nOR (M1)
\n
Note: Condone omission of the initial state vector for the M1.
A1
\n
[2 marks]
\n(A1)
\n
Note: Award A1 for OR seen.
A1
\n
[2 marks]
\nThere was some difficulty in interpreting the meaning of the values in the transition matrix, but most candidates did well with the rest of the question. In part (d) there was frequently evidence of a correct method, but a failure to identify the correct probabilities. It was surprising to see a significant number of candidates diagonalizing the matrix in part (d) and this often led to errors. Clearly this was not necessary.
\nLoreto is a manager at the Da Vinci health centre. If the mean rate of patients arriving at the health centre exceeds per minute then Loreto will employ extra staff. It is assumed that the number of patients arriving in any given time period follows a Poisson distribution.
\nLoreto performs a hypothesis test to determine whether she should employ extra staff. She finds that patients arrived during a randomly selected -hour clinic.
\nLoreto is also concerned about the average waiting time for patients to see a nurse. The health centre aims for at least of patients to see a nurse in under minutes.
\nLoreto assumes that the waiting times for patients are independent of each other and decides to perform a hypothesis test at a significance level to determine whether the health centre is meeting its target.
\nLoreto surveys patients and finds that of them waited more than minutes.
\nWrite down null and alternative hypotheses for Loreto’s test.
\nUsing the data from Loreto’s sample, perform the hypothesis test at a significance level to determine if Loreto should employ extra staff.
\nWrite down null and alternative hypotheses for this test.
\nPerform the test, clearly stating the conclusion in context.
\nlet be the random variable “number of patients arriving in a minute”, such that
\nA1
\nA1
\nNote: Allow a value of for . Award at most A0A1 if it is not clear that it is the population mean being referred to e.g
The number of patients is equal to 1.5 every minute
The number of patients exceeds 1.5 every minute.
Referring to the “expected” number of patients or the use of or is sufficient for A1A1.
\n
[2 marks]
\nunder let be the number of patients in hours
\n(A1)
\n(M1)A1
\nsince R1
\n(reject )
\nLoreto should employ more staff A1
\n\n
[5 marks]
\n: The probability of a patient waiting less than minutes is A1
\n: The probability of a patient waiting less than minutes is less than A1
\n\n
[2 marks]
\nUnder let be the number of patients waiting more than minutes
\n(A1)
\n(M1)A1
\nsince R1
\n(fail to reject )
\ninsufficient evidence to suggest they are not meeting their target A1
\n
Note: Do not accept “they are meeting target” for the A1. Accept use of and and any consistent use of a random variable, appropriate -value and significance level.
[5 marks]
\nIn part (a) there was a general lack of consistency in how candidates wrote down their null and alternative hypotheses. It was surprising how many candidates solved a Poisson rather than to find their -value. This suggests a lack of understanding of the nature of distributions or more specifically the concepts of hypothesis testing. In part (b), which was challenging, there were issues for many candidates in interpreting the situation. This is understandable since it was difficult, but as previously mentioned interpretation is a general issue in the paper. When writing down the conclusion of the tests, there was often very loose use of the terms accept/reject and candidates seemed unclear of the significance and importance of the correct use of these terms.
\nIn part (a) there was a general lack of consistency in how candidates wrote down their null and alternative hypotheses. It was surprising how many candidates solved a Poisson rather than to find their -value. This suggests a lack of understanding of the nature of distributions or more specifically the concepts of hypothesis testing. In part (b), which was challenging, there were issues for many candidates in interpreting the situation. This is understandable since it was difficult, but as previously mentioned interpretation is a general issue in the paper. When writing down the conclusion of the tests, there was often very loose use of the terms accept/reject and candidates seemed unclear of the significance and importance of the correct use of these terms.
\nIn part (a) there was a general lack of consistency in how candidates wrote down their null and alternative hypotheses. It was surprising how many candidates solved a Poisson rather than to find their -value. This suggests a lack of understanding of the nature of distributions or more specifically the concepts of hypothesis testing. In part (b), which was challenging, there were issues for many candidates in interpreting the situation. This is understandable since it was difficult, but as previously mentioned interpretation is a general issue in the paper. When writing down the conclusion of the tests, there was often very loose use of the terms accept/reject and candidates seemed unclear of the significance and importance of the correct use of these terms.
\nIn part (a) there was a general lack of consistency in how candidates wrote down their null and alternative hypotheses. It was surprising how many candidates solved a Poisson rather than to find their -value. This suggests a lack of understanding of the nature of distributions or more specifically the concepts of hypothesis testing. In part (b), which was challenging, there were issues for many candidates in interpreting the situation. This is understandable since it was difficult, but as previously mentioned interpretation is a general issue in the paper. When writing down the conclusion of the tests, there was often very loose use of the terms accept/reject and candidates seemed unclear of the significance and importance of the correct use of these terms.
\nAt an archery tournament, a particular competition sees a ball launched into the air while an archer attempts to hit it with an arrow.
\nThe path of the ball is modelled by the equation
\n\nwhere is the horizontal displacement from the archer and is the vertical displacement from the ground, both measured in metres, and is the time, in seconds, since the ball was launched.
\nIn this question both the ball and the arrow are modelled as single points. The ball is launched with an initial velocity such that and .
\nAn archer releases an arrow from the point . The arrow is modelled as travelling in a straight line, in the same plane as the ball, with speed and an angle of elevation of .
\nFind the initial speed of the ball.
\nFind the angle of elevation of the ball as it is launched.
\nFind the maximum height reached by the ball.
\nAssuming that the ground is horizontal and the ball is not hit by the arrow, find the coordinate of the point where the ball lands.
\nFor the path of the ball, find an expression for in terms of .
\nDetermine the two positions where the path of the arrow intersects the path of the ball.
\nDetermine the time when the arrow should be released to hit the ball before the ball reaches its maximum height.
\n(M1)
\nA1
\n\n
[2 marks]
\n(M1)
\nOR ( OR ) A1
\n
Note: Accept or from use of .
\n
[2 marks]
\n(M1)
\n
Note: The M1 might be implied by a correct graph or use of the correct equation.
\n
METHOD 1 – graphical Method
\nsketch graph (M1)
\n
Note: The M1 might be implied by correct graph or correct maximum (eg ).
max occurs when A1
METHOD 2 – calculus
differentiating and equating to zero (M1)
A1
\n\n
METHOD 3 – symmetry
\nline of symmetry is (M1)
\nA1
\n\n
[3 marks]
\nattempt to solve (M1)
\n(or ) (A1)
\n A1
Note: Do not award the final A1 if is also seen.
\n\n
[3 marks]
\nMETHOD 1
\nM1A1
\nA1
\n
METHOD 2
A1
\nwhen so M1A1
\n\n\n
METHOD 3
\nif
\n
M1A1
solving simultaneously, A1
\n()
\n\n
METHOD 4
use quadratic regression on M1A1
A1
\n
Note: Question asks for expression; condone omission of \"\".
\n
[3 marks]
\ntrajectory of arrow is (A1)
\nintersecting and their answer to (d) (M1)
\nA1
\nA1
\n\n
[4 marks]
\nwhen (A1)
\nattempt to find the distance from point of release to intersection (M1)
\n\ntime for arrow to get there is (A1)
\nso the arrow should be released when
\nA1
\n\n
[4 marks]
\nThis question was found to be the most difficult on the paper. There were a good number of good solutions to parts (a) and part (b), frequently with answers just written down with no working. Part (c) caused some difficulties with confusing variables. The most significant difficulties started with part (d) and became greater to the end of the question. Few candidates were able to work through the final two parts.
\nAn environmental scientist is asked by a river authority to model the effect of a leak from a power plant on the mercury levels in a local river. The variable measures the concentration of mercury in micrograms per litre.
\nThe situation is modelled using the second order differential equation
\n\nwhere is the time measured in days since the leak started. It is known that when and .
\nIf the mercury levels are greater than micrograms per litre, fishing in the river is considered unsafe and is stopped.
\nThe river authority decides to stop people from fishing in the river for longer than the time found from the model.
\nShow that the system of coupled first order equations:
\n\n\ncan be written as the given second order differential equation.
\nFind the eigenvalues of the system of coupled first order equations given in part (a).
\nHence find the exact solution of the second order differential equation.
\nSketch the graph of against , labelling the maximum point of the graph with its coordinates.
\nUse the model to calculate the total amount of time when fishing should be stopped.
\nWrite down one reason, with reference to the context, to support this decision.
\ndifferentiating first equation. M1
\n\nsubstituting in for M1
\n\ntherefore AG
\n
Note: The AG line must be seen to award the final M1 mark.
\n
[2 marks]
\nthe relevant matrix is (M1)
\n
Note: is also possible.
(this has characteristic equation) (A1)
A1
\n\n
[3 marks]
\nEITHER
\nthe general solution is M1
\n
Note: Must have constants, but condone sign error for the M1.
so M1A1
\n
OR
\nattempt to find eigenvectors (M1)
\nrespective eigenvectors are and (or any multiple)
\n(M1)A1
\n\n
THEN
\nthe initial conditions become:
\n\nM1
\nthis is solved by
\nso the solution is A1
\n\n
[5 marks]
\n A1A1
\n
Note: Award A1 for correct shape (needs to go through origin, have asymptote at and a single maximum; condone ). Award A1 for correct coordinates of maximum.
\n\n
[2 marks]
\nintersecting graph with (M1)
\nso the time fishing is stopped between and (A1)
\ndays A1
\n\n
[3 marks]
\nAny reasonable answer. For example:
\nThere are greater downsides to allowing fishing when the levels may be dangerous than preventing fishing when the levels are safe.
\nThe concentration of mercury may not be uniform across the river due to natural variation / randomness.
\nThe situation at the power plant might get worse.
\nMercury levels are low in water but still may be high in fish. R1
\n\n
Note: Award R1 for a reasonable answer that refers to this specific context (and not a generic response that could apply to any model).
\n\n
[1 mark]
\nMany candidates did not get this far, but the attempts at the question that were seen were generally good. The greater difficulties were seen in parts (e) and (f), but this could be a problem with time running out.
\n