A1
[1 mark]
\nc.i.
\nLet .
\nThe graph of has horizontal tangents at the points where = and = , < .
\nFind .
\nFind the value of and the value of .
\nSketch the graph of .
\nHence explain why the graph of has a local maximum point at .
\nFind .
\nHence, use your answer to part (d)(i) to show that the graph of has a local minimum point at .
\nThe normal to the graph of at and the tangent to the graph of at intersect at the point (, ) .
\n\n
Find the value of and the value of .
\n(M1)A1
\n\n
[2 marks]
\ncorrect reasoning that (seen anywhere) (M1)
\n\n
valid approach to solve quadratic M1
\n, quadratic formula
\ncorrect values for
\n3, −5
\ncorrect values for and
\n= −5 and = 3 A1
\n[3 marks]
\n A1
[1 mark]
\nfirst derivative changes from positive to negative at A1
\nso local maximum at AG
\n[1 mark]
\nA1
\nsubstituting their into their second derivative (M1)
\n\n
(A1)
\n[3 marks]
\nis positive so graph is concave up R1
\nso local minimum at AG
\n[1 mark]
\nnormal to at is = −5 (seen anywhere) (A1)
\nattempt to find -coordinate at their value of (M1)
\n−10 (A1)
\ntangent at has equation = −10 (seen anywhere) A1
\nintersection at (−5, −10)
\n= −5 and = −10 A1
\n[5 marks]
\nAdam sets out for a hike from his camp at point A. He hikes at an average speed of 4.2 km/h for 45 minutes, on a bearing of 035° from the camp, until he stops for a break at point B.
\nAdam leaves point B on a bearing of 114° and continues to hike for a distance of 4.6 km until he reaches point C.
\nAdam’s friend Jacob wants to hike directly from the camp to meet Adam at point C .
\nFind the distance from point A to point B.
\nShow that is 101°.
\nFind the distance from the camp to point C.
\nFind .
\nFind the bearing that Jacob must take to point C.
\nJacob hikes at an average speed of 3.9 km/h.
\nFind, to the nearest minute, the time it takes for Jacob to reach point C.
\nA1
\nAB = 3.15 (km) A1
\n[2 marks]
\n66° or (180 − 114) A1
\n35 + 66 A1
\n= 101° AG
\n[2 marks]
\nattempt to use cosine rule (M1)
\nAC2 = 3.152 + 4.62 − 2 × 3.15 × 4.6 cos 101° (or equivalent) A1
\nAC = 6.05 (km) A1
\n\n
[3 marks]
\nvalid approach to find angle BCA (M1)
\neg sine rule
\ncorrect substitution into sine rule A1
\neg
\n= 30.7° A1
\n[3 marks]
\n= 48.267 (seen anywhere) A1
\nvalid approach to find correct bearing (M1)
\neg 48.267 + 35
\nbearing = 83.3° (accept 083°) A1
\n[3 marks]
\nattempt to use M1
\nor 0.065768 km/min (A1)
\nt = 93 (minutes) A1
\n[3 marks]
\nThe length, X mm, of a certain species of seashell is normally distributed with mean 25 and variance, .
\nThe probability that X is less than 24.15 is 0.1446.
\nA random sample of 10 seashells is collected on a beach. Let Y represent the number of seashells with lengths greater than 26 mm.
\nFind P(24.15 < X < 25).
\nFind , the standard deviation of X.
\nHence, find the probability that a seashell selected at random has a length greater than 26 mm.
\nFind E(Y).
\nFind the probability that exactly three of these seashells have a length greater than 26 mm.
\nA seashell selected at random has a length less than 26 mm.
\nFind the probability that its length is between 24.15 mm and 25 mm.
\nattempt to use the symmetry of the normal curve (M1)
\neg diagram, 0.5 − 0.1446
\nP(24.15 < X < 25) = 0.3554 A1
\n[2 marks]
\nuse of inverse normal to find z score (M1)
\nz = −1.0598
\ncorrect substitution (A1)
\n= 0.802 A1
\n[3 marks]
\nP(X > 26) = 0.106 (M1)A1
\n[2 marks]
\nrecognizing binomial probability (M1)
\nE(Y) = 10 × 0.10621 (A1)
\n= 1.06 A1
\n[3 marks]
\nP(Y = 3) (M1)
\n= 0.0655 A1
\n[2 marks]
\nrecognizing conditional probability (M1)
\ncorrect substitution A1
\n\n
= 0.398 A1
\n[3 marks]
\n\n
Let where , .
\nThe graph of has exactly one maximum point P.
\nThe second derivative of is given by . The graph of has exactly one point of inflexion Q.
\nShow that .
\nFind the x-coordinate of P.
\nShow that the x-coordinate of Q is .
\nThe region R is enclosed by the graph of , the x-axis, and the vertical lines through the maximum point P and the point of inflexion Q.
\nGiven that the area of R is 3, find the value of .
\nattempt to use quotient rule (M1)
\ncorrect substitution into quotient rule
\n(or equivalent) A1
\n, A1
\nAG
\n[3 marks]
\nM1
\n\n
(A1)
\nA1
\n[3 marks]
\nM1
\n\n
A1
\nA1
\nso the point of inflexion occurs at AG
\n[3 marks]
\nattempt to integrate (M1)
\n\n
(A1)
\nEITHER
\n= A1
\nso A1
\nOR
\nA1
\nso A1
\nTHEN
\n\n
A1
\nsetting their expression for area equal to 3 M1
\n\n
A1
\n[7 marks]
\nThe following diagram shows the graph of . The graph has a horizontal asymptote at . The graph crosses the -axis at and , and the -axis at .
\nOn the following set of axes, sketch the graph of , clearly showing any asymptotes with their equations and the coordinates of any local maxima or minima.
\n\n
no values below 1 A1
\nhorizontal asymptote at with curve approaching from below as A1
\n(±1,1) local minima A1
\n(0,5) local maximum A1
\nsmooth curve and smooth stationary points A1
\n[5 marks]
\nA continuous random variable X has the probability density function given by
\n.
\nFind P(0 ≤ X ≤ 3).
\nattempting integration by parts, eg
\n(M1)
\nP(0 ≤ X ≤ 3) (or equivalent) A1A1
\nNote: Award A1 for a correct and A1 for a correct .
\nattempting to substitute limits M1
\n(A1)
\nso P(0 ≤ X ≤ 3) (or equivalent) A1
\nA1
\n[7 marks]
\nThe plane П has the Cartesian equation
\nThe line L has the vector equation r . The acute angle between the line L and the plane П is 30°.
\nFind the possible values of .
\nrecognition that the angle between the normal and the line is 60° (seen anywhere) R1
\nattempt to use the formula for the scalar product M1
\ncos 60° = A1
\nA1
\n\n
attempt to square both sides M1
\n\n
(or equivalent) A1A1
\n[7 marks]
\nThe function is defined by . The graph of is shown in the following diagram.
\nFind the largest value of such that has an inverse function.
\nFor this value of , find an expression for , stating its domain.
\nattempt to differentiate and set equal to zero M1
\nA1
\nminimum at
\nA1
\n[3 marks]
\nNote: Interchanging and can be done at any stage.
\n(M1)
\nA1
\nas , R1
\nso A1
\ndomain of is , A1
\n[5 marks]
\nConsider a function , such that , 0 ≤ ≤ 10, .
\nThe function has a local maximum at the point (2, 21.8) , and a local minimum at (8, 10.2).
\nA second function is given by , 0 ≤ ≤ 10; , .
\nThe function passes through the points (3, 2.5) and (6, 15.1).
\nFind the period of .
\nFind the value of .
\nHence, find the value of (6).
\nFind the value of and the value of .
\nFind the value of for which the functions have the greatest difference.
\ncorrect approach A1
\neg (or equivalent)
\nperiod = 12 A1
\n[2 marks]
\nvalid approach (M1)
\neg
\n, or equivalent
\n= 16 A1
\n[2 marks]
\nattempt to substitute into their function (M1)
\n\n
(6) = 13.1 A1
\n[2 marks]
\nvalid attempt to set up a system of equations (M1)
\ntwo correct equations A1
\n,
\nvalid attempt to solve system (M1)
\n= 8.4; = 6.7 A1A1
\n[5 marks]
\nattempt to use to find maximum difference (M1)
\n= 1.64 A1
\n\n
[2 marks]
\nLet the roots of the equation be , and .
\nOn an Argand diagram, , and are represented by the points U, V and W respectively.
\nExpress in the form , where and .
\nFind , and expressing your answers in the form , where and .
\nFind the area of triangle UVW.
\nBy considering the sum of the roots , and , show that
\n.
\nattempt to find modulus (M1)
\nA1
\nattempt to find argument in the correct quadrant (M1)
\nA1
\nA1
\n\n
[5 marks]
\nattempt to find a root using de Moivre’s theorem M1
\nA1
\nattempt to find further two roots by adding and subtracting to the argument M1
\nA1
\nA1
\nNote: Ignore labels for , and at this stage.
\n\n
[5 marks]
\nMETHOD 1
attempting to find the total area of (congruent) triangles UOV, VOW and UOW M1
Area A1A1
\nNote: Award A1 for and A1 for
\n= (or equivalent) A1
\n\n
METHOD 2
\nUV2 (or equivalent) A1
\nUV (or equivalent) A1
\nattempting to find the area of UVW using Area = × UV × VW × sin for example M1
\nArea
\n= (or equivalent) A1
\n\n
[4 marks]
\n+ + = 0 R1
\nA1
\nconsideration of real parts M1
\n\n
explicitly stated A1
\nAG
\n[4 marks]
\nThe following table below shows the marks scored by seven students on two different mathematics tests.
\nLet L1 be the regression line of x on y. The equation of the line L1 can be written in the form x = ay + b.
\nLet L2 be the regression line of y on x. The lines L1 and L2 pass through the same point with coordinates (p , q).
\nFind the value of a and the value of b.
\nFind the value of p and the value of q.
\nJennifer was absent for the first test but scored 29 marks on the second test. Use an appropriate regression equation to estimate Jennifer’s mark on the first test.
\na = 1.29 and b = −10.4 A1A1
\n[2 marks]
\nrecognising both lines pass through the mean point (M1)
\np = 28.7, q = 30.3 A2
\n[3 marks]
\nsubstitution into their on equation (M1)
\n= 1.29082(29) − 10.3793
\n= 27.1 A1
\nNote: Accept 27.
\n[2 marks]
\nThe function is defined by .
\nFind the first two derivatives of and hence find the Maclaurin series for up to and including the term.
\nShow that the coefficient of in the Maclaurin series for is zero.
\nUsing the Maclaurin series for and , find the Maclaurin series for up to and including the term.
\nHence, or otherwise, find .
\nattempting to use the chain rule to find the first derivative M1
\nA1
\nattempting to use the product rule to find the second derivative M1
\n(or equivalent) A1
\nattempting to find , and M1
\n; ; A1
\nsubstitution into the Maclaurin formula M1
\nso the Maclaurin series for up to and including the term is A1
\n[8 marks]
\nMETHOD 1
\nattempting to differentiate M1
\n(or equivalent) A2
\nsubstituting into their M1
\n\n
so the coefficient of in the Maclaurin series for is zero AG
\n\n
METHOD 2
\nsubstituting into the Maclaurin series for (M1)
\n\n
substituting Maclaurin series for M1
\nA1
\ncoefficient of is A1
\nso the coefficient of in the Maclaurin series for is zero AG
\n\n
[4 marks]
\nsubstituting into the Maclaurin series for M1
\nA1
\nsubstituting into the Maclaurin series for M1
\n\n
A1
\nselecting correct terms from above M1
\n\n
A1
\n[6 marks]
\nMETHOD 1
\nsubstitution of their series M1
\nA1
\n\n
A1
\n\n
METHOD 2
\nuse of l’Hôpital’s rule M1
\n(or equivalent) A1
\nA1
\n\n
[3 marks]
\nIn a city, the number of passengers, X, who ride in a taxi has the following probability distribution.
\nAfter the opening of a new highway that charges a toll, a taxi company introduces a charge for passengers who use the highway. The charge is $ 2.40 per taxi plus $ 1.20 per passenger. Let T represent the amount, in dollars, that is charged by the taxi company per ride.
\nFind E(T).
\nGiven that Var(X) = 0.8419, find Var(T).
\nMETHOD 1
\nattempting to use the expected value formula (M1)
\n\n
($) (A1)
\nuse of (M1)
\n\n
($) A1
\n\n
METHOD 2
\nattempting to find the probability distribution for T (M1)
\n (A1)
attempting to use the expected value formula (M1)
\n\n
($) A1
\n\n
[4 marks]
\nMETHOD 1
\nusing Var(1.20 X + 2.40) = (1.20)2 Var(X) with Var(X) = 0.8419 (M1)
\nVar(T) = 1.21 A1
\nMETHOD 2
\nfinding the standard deviation for their probability distribution found in part (a) (M1)
\nVar(T) = (1.101…)2
\n= 1.21 A1
\nNote: Award M1A1 for Var(T) = (1.093…)2 = 1.20.
\n[2 marks]
\nTwo ships, A and B , are observed from an origin O. Relative to O, their position vectors at time t hours after midday are given by
\nrA =
\nrB =
\nwhere distances are measured in kilometres.
\nFind the minimum distance between the two ships.
\nattempting to find rB − rA for example (M1)
\nrB − rA =
\nattempting to find |rB − rA| M1
\ndistance A1
\nusing a graph to find the − coordinate of the local minimum M1
\nthe minimum distance between the ships is 2.81 (km) A1
\n[5 marks]
\nA metal sphere has a radius 12.7 cm.
\nFind the volume of the sphere expressing your answer in the form , and .
\nThe sphere is to be melted down and remoulded into the shape of a cone with a height of 14.8 cm.
\nFind the radius of the base of the cone, correct to 2 significant figures.
\n(or equivalent) A1
\n8580.24 (A1)
\nV = 8.58 × 103 A1
\n[3 marks]
\nrecognising volume of the cone is same as volume of their sphere (M1)
\n8580.24 (or equivalent) A1
\nr = 23.529
\nr = 24 (cm) correct to 2 significant figures A1
\n[3 marks]
\nThe complex numbers and satisfy the equations
\n\n
.
\nFind and in the form where , .
\nsubstituting into M1
\nA1
\nlet
\ncomparing real and imaginary parts of M1
\nto obtain and A1
\nattempting to solve for and ) M1
\nand so A1
\nhence A1
\n[7 marks]
\nConsider the graphs of and , .
\nFind the set of values for such that the two graphs have no intersection points.
\nMETHOD 1
\nsketching the graph of () M1
\nthe (oblique) asymptote has a gradient equal to 1
\nand so the maximum value of is 1 R1
\nconsideration of a straight line steeper than the horizontal line joining (−3, 0) and (0, 0) M1
\nso > 0 R1
\nhence 0 < ≤ 1 A1
\n\n
METHOD 2
\nattempting to eliminate to form a quadratic equation in M1
\n\n
A1
\n\n
EITHER
\nattempting to solve for M1
\n\n
OR
\nattempting to solve < 0 ie for M1
\n\n
THEN
\nA1
\na valid reason to explain why gives no solutions eg if ,
\nand so 0 < ≤ 1 R1
\n\n
\n
[5 marks]
\nThe following diagram shows part of a circle with centre O and radius 4 cm.
\nChord AB has a length of 5 cm and AÔB = θ.
\nFind the value of θ, giving your answer in radians.
\nFind the area of the shaded region.
\nMETHOD 1
\nattempt to use the cosine rule (M1)
\ncos θ = (or equivalent) A1
\nθ = 1.35 A1
\n\n
METHOD 2
\nattempt to split triangle AOB into two congruent right triangles (M1)
\nsin A1
\nθ = 1.35 A1
\n\n
[3 marks]
\nattempt to find the area of the shaded region (M1)
\nA1
\n= 39.5 (cm2) A1
\n[3 marks]
\nA large tank initially contains pure water. Water containing salt begins to flow into the tank The solution is kept uniform by stirring and leaves the tank through an outlet at its base. Let grams represent the amount of salt in the tank and let minutes represent the time since the salt water began flowing into the tank.
\nThe rate of change of the amount of salt in the tank, , is described by the differential equation .
\nShow that + 1 is an integrating factor for this differential equation.
\nHence, by solving this differential equation, show that .
\nSketch the graph of versus for 0 ≤ ≤ 60 and hence find the maximum amount of salt in the tank and the value of at which this occurs.
\nFind the value of at which the amount of salt in the tank is decreasing most rapidly.
\nThe rate of change of the amount of salt leaving the tank is equal to .
\nFind the amount of salt that left the tank during the first 60 minutes.
\nMETHOD 1
\nM1
\n\n
= A1
\nAG
\n\n
METHOD 2
\nattempting product rule differentiation on M1
\n\n
A1
\nso is an integrating factor for this differential equation AG
\n\n
[2 marks]
\n\n
attempting to multiply through by and rearrange to give (M1)
\nA1
\n\n
A1
\nattempting to integrate the RHS by parts M1
\n\n
A1
\nNote: Condone the absence of C.
\n\n
EITHER
\nsubstituting M1
\nA1
\nusing as the highest common factor of and M1
\n\n
OR
\nusing as the highest common factor of and giving
\n(or equivalent) M1A1
\nsubstituting M1
\n\n
THEN
\nAG
\n\n
[8 marks]
\n\n
graph starts at the origin and has a local maximum (coordinates not required) A1
\nsketched for 0 ≤ ≤ 60 A1
\ncorrect concavity for 0 ≤ ≤ 60 A1
\nmaximum amount of salt is 14.6 (grams) at = 6.60 (minutes) A1A1
\n[5 marks]
\nusing an appropriate graph or equation (first or second derivative) M1
\namount of salt is decreasing most rapidly at = 12.9 (minutes) A1
\n[2 marks]
\nEITHER
\nattempting to form an integral representing the amount of salt that left the tank M1
\n\n
A1
\n\n
OR
\nattempting to form an integral representing the amount of salt that entered the tank minus the amount of salt in the tank at = 60(minutes)
\namount of salt that left the tank is A1
\n\n
THEN
\n= 36.7 (grams) A2
\n[4 marks]
\nOn 1st January 2020, Laurie invests $P in an account that pays a nominal annual interest rate of 5.5 %, compounded quarterly.
\nThe amount of money in Laurie’s account at the end of each year follows a geometric sequence with common ratio, r.
\nFind the value of r, giving your answer to four significant figures.
\nLaurie makes no further deposits to or withdrawals from the account.
\nFind the year in which the amount of money in Laurie’s account will become double the amount she invested.
\n(M1)(A1)
\n1.056 A1
\n[3 marks]
\nEITHER
\nOR (M1)(A1)
\nNote: Award (M1) for substitution into loan payment formula. Award (A1) for correct substitution.
\nOR
\nPV = ±1
FV = 1
I% = 5.5
P/Y = 4
C/Y = 4
n = 50.756… (M1)(A1)
OR
\nPV = ±1
FV = 2
I% = 100(their (a) − 1)
P/Y = 1
C/Y = 1 (M1)(A1)
THEN
\n⇒ 12.7 years
\nLaurie will have double the amount she invested during 2032 A1
[3 marks]
\nA six-sided biased die is weighted in such a way that the probability of obtaining a “six” is .
\nThe die is tossed five times. Find the probability of obtaining at most three “sixes”.
\nThe die is tossed five times. Find the probability of obtaining the third “six” on the fifth toss.
\nrecognition of binomial (M1)
\nX ~ B(5, 0.7)
\nattempt to find P (X ≤ 3) M1
\n= 0.472 (= 0.47178) A1
\n[3 marks]
\nrecognition of 2 sixes in 4 tosses (M1)
\nP (3rd six on the 5th toss) A1
\n= 0.185 (= 0.18522) A1
\n[3 marks]
\nShow that .
\nVerify that and satisfy the equation .
\nHence, or otherwise, show that the exact value of .
\nUsing the results from parts (b) and (c) find the exact value of .
\nGive your answer in the form where , .
\nstating the relationship between and and stating the identity for M1
\nand
\n⇒ AG
\n[1 mark]
\nMETHOD 1
\nattempting to substitute for and using the result from (a) M1
\nLHS = A1
\n(= RHS) A1
\nso satisfies the equation AG
\nattempting to substitute for and using the result from (a) M1
\nLHS = A1
\nA1
\n(= RHS) A1
\nso satisfies the equation AG
\n\n
METHOD 2
\nlet and
\nattempting to find the sum of roots M1
\n\n
A1
\n(from part (a)) A1
\nattempting to find the product of roots M1
\nA1
\n= −1 A1
\nthe coefficient of and the constant term in the quadratic are and −1 respectively R1
\nhence the two roots are and AG
\n[7 marks]
\nMETHOD 1
\nand are roots of R1
\nNote: Award R1 if only is stated as a root of .
\nA1
\nattempting to solve their quadratic equation M1
\nA1
\n() R1
\nso AG
\n\n
METHOD 2
\nattempting to substitute into the identity for M1
\n\n
A1
\nattempting to solve their quadratic equation M1
\nA1
\nR1
\nso AG
\n[5 marks]
\nis the sum of the roots of R1
\nA1
\nA1
\nattempting to rationalise their denominator (M1)
\nA1A1
\n[6 marks]
\nThe following table below shows the marks scored by seven students on two different mathematics tests.
\nLet L1 be the regression line of x on y. The equation of the line L1 can be written in the form x = ay + b.
\nFind the value of a and the value of b.
\nLet L2 be the regression line of y on x. The lines L1 and L2 pass through the same point with coordinates (p , q).
\nFind the value of p and the value of q.
\na = 1.29 and b = −10.4 A1A1
\n[2 marks]
\nrecognising both lines pass through the mean point (M1)
\np = 28.7, q = 30.3 A2
\n[3 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]
\nThe displacement, in centimetres, of a particle from an origin, O, at time t seconds, is given by s(t) = t 2 cos t + 2t sin t, 0 ≤ t ≤ 5.
\nFind the maximum distance of the particle from O.
\nFind the acceleration of the particle at the instant it first changes direction.
\nuse of a graph to find the coordinates of the local minimum (M1)
\ns = −16.513... (A1)
\nmaximum distance is 16.5 cm (to the left of O) A1
\n[3 marks]
\nattempt to find time when particle changes direction eg considering the first maximum on the graph of s or the first t – intercept on the graph of s'. (M1)
\nt = 1.51986... (A1)
\nattempt to find the gradient of s' for their value of t, s\" (1.51986...) (M1)
\n=–8.92 (cm/s2) A1
\n[4 marks]
\nFind the acute angle between the planes with equations and .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nn and n (A1)(A1)
\nEITHER
\n(M1)
\n(A1)
\n\n
OR
\n(M1)
\n(A1)
\n\n
\n
THEN
\nA1
\n[5 marks]
\nThe function is defined by , 0 < < 3.
\nDraw a set of axes showing and values between −3 and 3. On these axes
\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 .
\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]
This question asks you to investigate regular -sided polygons inscribed and circumscribed in a circle, and the perimeter of these as tends to infinity, to make an approximation for .
\nLet represent the perimeter of any -sided regular polygon inscribed in a circle of radius 1 unit.
\nConsider an equilateral triangle ABC of side length, units, circumscribed about a circle of radius 1 unit and centre O as shown in the following diagram.
\nLet represent the perimeter of any -sided regular polygon circumscribed about a circle of radius 1 unit.
\nConsider an equilateral triangle ABC of side length, units, inscribed in a circle of radius 1 unit and centre O as shown in the following diagram.
\nThe equilateral triangle ABC can be divided into three smaller isosceles triangles, each subtending an angle of at O, as shown in the following diagram.
\nUsing right-angled trigonometry or otherwise, show that the perimeter of the equilateral triangle ABC is equal to units.
\nConsider a square of side length, units, inscribed in a circle of radius 1 unit. By dividing the inscribed square into four isosceles triangles, find the exact perimeter of the inscribed square.
\n\n
Find the perimeter of a regular hexagon, of side length, units, inscribed in a circle of radius 1 unit.
\n\n
Show that .
\nUse an appropriate Maclaurin series expansion to find and interpret this result geometrically.
\nShow that .
\nBy writing in the form , find .
\nUse the results from part (d) and part (f) to determine an inequality for the value of in terms of .
\nThe inequality found in part (h) can be used to determine lower and upper bound approximations for the value of .
\nDetermine the least value for such that the lower bound and upper bound approximations are both within 0.005 of .
\nMETHOD 1
\nconsider right-angled triangle OCX where CX
\nM1A1
\nA1
\nAG
\n\n
METHOD 2
\neg use of the cosine rule M1A1
\nA1
\nAG
\nNote: Accept use of sine rule.
\n\n
[3 marks]
\nwhere = side of square M1
\nA1
\nA1
\n[3 marks]
\n6 equilateral triangles ⇒ = 1 A1
\nA1
\n[2 marks]
\nin right-angled triangle M1
\nA1
\n\n
M1
\nAG
\n[3 marks]
\nconsider
\nuse of M1
\n(A1)
\nA1
\nA1
\nas polygon becomes a circle of radius 1 and R1
\n[5 marks]
\nconsider an -sided polygon of side length
\n2 right-angled triangles with angle at centre M1A1
\nopposite side M1A1
\nPerimeter AG
\n[4 marks]
\nconsider
\nR1
\nattempt to use L’Hopital’s rule M1
\nA1A1
\nA1
\n[5 marks]
\n\n
M1
\nA1
\n[2 marks]
\nattempt to find the lower bound and upper bound approximations within 0.005 of (M1)
\n= 46 A2
\n[3 marks]
\nThe function is defined by , .
\nWrite down the range of .
\nFind , stating its domain.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n≥ 3 A1
\n[1 mark]
\n(M1)
\nNote: Exchange of variables can take place at any point.
\n(A1)
\n, ≥ 3 A1A1
\nNote: Allow follow through from (a) for last A1 mark which is independent of earlier marks in (b).
\n[4 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]
\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.
\nDraw a scatter diagram for temperature difference from 37 °C () against heartbeat (). Use a scale of 2 cm for 0.1 °C on the horizontal axis, starting with −0.3 °C. Use a scale of 1 cm for 2 heartbeats per minute on the vertical axis, starting with 60 beats per minute.
\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.
\nUse your graphic display calculator to find the equation of the regression line on .
\nDraw the regression line on on the scatter diagram.
\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, axis labels, minimum , and minimum . Award (A0) if axes are reversed and follow through for their points.
\nAward (A3) for all eight points correctly plotted,
(A2) for six or seven points correctly plotted.
(A1) for four or five points correctly plotted.
Allow a tolerance of half a small square.
\nIf graph paper has not been used, award at most (A1)(A0)(A0)(A0).
\nIf accuracy cannot be determined award (A0)(A0)(A0)(A0).
\n[4 marks]
\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]
\n(G2)
\nNote: Award (G1) for , (G1) for 73.5.
\nAward a maximum of (G0)(G1) if the answer is not an equation.
\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]
\nSketch the graph of , stating the equations of any asymptotes and the coordinates of any points of intersection with the axes.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ncorrect shape: two branches in correct quadrants with asymptotic behaviour A1
\ncrosses at (4, 0) and A1A1
\nasymptotes at and A1A1
\n\n
[5 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\"/)
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n![\"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]
\nConsider 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]
\nColorado beetles are a pest, which can cause major damage to potato crops. For a certain Colorado beetle the amount of oxygen, in millilitres (ml), consumed each day increases with temperature as shown in the following table.
\nThis information has been used to plot a scatter diagram.
\nThe mean point has coordinates (20, 230).
\nFind the equation of the regression line of on .
\nDraw the regression line of on on the scatter diagram.
\nIn order to estimate the amount of oxygen consumed, this regression line is considered to be reliable for a temperature such that ≤ ≤ .
\nWrite down 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.
(A1)(A1) (C2)
\nNote: Award (A1) for ; (A1) for −80. Award at most (A1)(A0) if answer is not an equation. Award (A0)(A1)(ft) for .
\n[2 marks]
\n\n
(A1)(A1) (C2)
Note: Award (A1) for a straight line using a ruler passing through (20, 230); (A1) for correct -intercept. If a ruler has not been used, award at most (A0)(A1).
\n[2 marks]
\nAND (A1)(A1) (C2)
\nNote: Accept [10, 30] or 10 ≤ ≤ 30.
\n[2 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.
\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]
\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]
\nConsider the functions and
\nThe functions intersect at points P and Q. Part of the graph of and part of the graph of are shown on the diagram.
\nFind the range of f.
\nWrite down the x-coordinate of P and the x-coordinate of Q.
\nWrite down the values of x for which .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nOR OR (A1)(A1) (C2)
\nNote: Award (A1) for −2 and (A1) for completely correct mathematical notation, including weak inequalities. Accept .
\n[2 marks]
\n–1 and 1.52 (1.51839…) (A1)(A1) (C2)
\nNote: Award (A1) for −1 and (A1) for 1.52 (1.51839).
\n[2 marks]
\nOR . (A1)(ft)(A1)(ft) (C2)
\nNote: Award (A1)(ft) for both critical values in inequality or range statements such as .
\nAward the second (A1)(ft) for correct strict inequality statements used with their critical values. If an incorrect use of strict and weak inequalities has already been penalized in (a), condone weak inequalities for this second mark and award (A1)(ft).
\n[2 marks]
\nConsider the function defined by where is a positive constant.
\nThe function is defined by for .
\nShowing any and intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.
\n;
\nShowing any and intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.
\n;
\nShowing any and intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.
\n.
\nFind .
\nBy finding explain why is an increasing function.
\n![\"M17/5/MATHL/HP1/ENG/TZ2/09.a.i/M\"](\"../assets/uploads/tinymce_asset/asset/3512/Schermafbeelding_2017-08-09_om_08.15.01.png\"/)
A1 for correct shape
\nA1 for correct and intercepts and minimum point
\n[2 marks]
\n![\"M17/5/MATHL/HP1/ENG/TZ2/09.a.ii/M\"](\"../assets/uploads/tinymce_asset/asset/3513/Schermafbeelding_2017-08-09_om_08.17.28.png\"/)
A1 for correct shape
\nA1 for correct vertical asymptotes
\nA1 for correct implied horizontal asymptote
\nA1 for correct maximum point
\n[??? marks]
\n![\"M17/5/MATHL/HP1/ENG/TZ2/09.a.iii/M\"](\"../assets/uploads/tinymce_asset/asset/3514/Schermafbeelding_2017-08-09_om_08.20.22.png\"/)
A1 for reflecting negative branch from (ii) in the -axis
\nA1 for correctly labelled minimum point
\n[2 marks]
\nEITHER
\nattempt at integration by parts (M1)
\nA1A1
\nA1
\nA1
\nOR
\n\n
attempt at integration by parts (M1)
\nA1A1
\nA1
\n\n
\n
A1
\n[5 marks]
\n\n
M1A1A1
\n\n
Note: Method mark is for differentiating the product. Award A1 for each correct term.
\n\n
\n
both parts of the expression are positive hence is positive R1
\nand therefore is an increasing function (for ) AG
\n[4 marks]
\nLine 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 equation of in the form .
\nFind the -coordinate of point A.
\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]
\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.
\n 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]
\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.
Units 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]
\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.
\nattempt 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]
\nSketch the graphs of and on the following axes.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nstraight 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]
\nThe 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]
\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]
\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.
\n3 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]
\nA ladder on a fire truck has its base at point B which is 4 metres above the ground. The ladder is extended and its other end rests on a vertical wall at point C, 16 metres above the ground. The horizontal distance between B and C is 9 metres.
\nFind the angle of elevation from B to C.
\nA second truck arrives whose ladder, when fully extended, is 30 metres long. The base of this ladder is also 4 metres above the ground. For safety reasons, the maximum angle of elevation that the ladder can make is 70º.
\nFind the maximum height on the wall that can be reached by the ladder on the second truck.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(A1)(M1)
\nNote: Award (A1) for 12 seen, (M1) for correct substitution into tan (or equivalent). Accept equivalent methods, such as Pythagoras, to find BC and correct substitution into other trig ratios. If seen award (A0)(M1)(A0).
\n53.1° (53.1301…°) (A1) (C3)
\nNote: If radians are used the answer is 0.927295…; award at most (A1)(M1)(A0).
\n[3 marks]
\n(M1)(M1)
\nNote: Award (M1) for (or equivalent) and (M1) for adding 4.
\n32.2 (32.1907…) (m) (A1) (C3)
\nNote: If radians are used the answer is 27.2167…; award at most (M1)(M1)(A0).
\n[3 marks]
\nConsider the differential equation , with when .
\nUse Euler’s method, with step length , to find an approximate value of when .
\nSketch the isoclines for .
\nExpress in the form , where .
\nSolve the differential equation, for , giving your answer in the form .
\nSketch the graph of for .
\nWith reference to the curvature of your sketch in part (c)(iii), and without further calculation, explain whether you conjecture will be less than, equal to, or greater than your answer in part (a).
\n (M1)(A1)(A1)(A1)A1
\n
Note: Award A1 for each correct value.
For the intermediate values, accept answers that are accurate to 2 significant figures.
The final value must be accurate to 3 significant figures or better.
[5 marks]
\nattempt to solve (M1)
\n\n
\n
or
\n A1A1
[3 marks]
\nA1
\n[1 mark]
\nrecognition of homogeneous equation,
let M1
the equation can be written as
\n(A1)
\n\n
M1
\nNote: Award M1 for attempt to separate the variables.
\nfrom part (c)(i) M1
\nA1A1
\n\n
M1
\nNote: Award M1 for using initial conditions to find .
\nA1
\n\n
substituting M1
\nNote: This M1 may be awarded earlier.
\nA1
\n[10 marks]
\ncurve drawn over correct domain A1
\n\n
[1 mark]
\nthe sketch shows that is concave up A1
\nNote: Accept is increasing.
\nthis means the tangent drawn using Euler’s method will give an underestimate of the real value, so > estimate in part (a) R1
\nNote: The R1 is dependent on the A1.
\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]
\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]
\nThe derivative of a function is given by .
\nGiven that , find the value 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.
\ncorrect working (A1)
\neg −5 + (8 − 1)(3)
\nu8 = 16 A1 N2
\n\n
[2 marks]
\nMETHOD 1
\n(A1)
\nattempts to integrate (M1)
\nA1
\nuses to obtain and so M1
\nsubstitutes into their expression for (M1)
\nso A1
\n\n
METHOD 2
\n(A1)
\nattempts to integrate both sides (M1)
\nA1
\nM1
\nuses to find their value of (M1)
\n\nso A1
\n\n
[6 marks]
\nIn parts (b) and (c), denotes the number written in base , where . For example, .
\nState Fermat’s little theorem.
\nFind the remainder when is divided by .
\nConvert to base , where .
\nConsider the equation .
\nFind the value of .
\nEITHER
\nA1
\nwhere is prime A1
\n\n
OR
\nA1
\nwhere is prime and does not divide (or equivalent statement) A1
\n\n
[2 marks]
\n\n
(M1)(A1)
\n(M1)
\n(M1)A1
\nthe remainder is
\nNote: Award as above for using instead of .
\n[5 marks]
\nM1
\nA1
\nEITHER
\n\n
\n
\n
M1
\n\n
\n
OR
\nM1
\nTHEN
\nA1
\n\n
[4 marks]
\nthe equation can be written as
\nM1A1
\n(M1)
\nNote: The (M1) is for an attempt to solve the original equation.
\nA1
\n[4 marks]
\nSolve the equation . Give your answer in the form where .
\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
METHOD 1
\n\nuses (M1)
\n\nuses (M1)
\n\nA1
\nA1
\nA1
\n\n
METHOD 2
\nexpresses as and uses (M1)
\nA1
\nuses and (M1)
\nA1
\nA1
\n\n
METHOD 3
\nexpresses as and uses (M1)
\nA1
\nuses (M1)
\n\nA1
\nso A1
\n\n
[5 marks]
\nThe following table shows the probability distribution of a discrete random variable where .
\nFind the value of , 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.
\nuses (M1)
\n\nA1
\n\n
EITHER
\nattempts to factorize their quadratic M1
\n\n\n
OR
\nattempts use of the quadratic formula on their equation M1
\n\n\n
THEN
\nA1
\nrejects as this value leads to invalid probabilities, for example, R1
\nso A1
\n\n
Note: Award R0A1 if is stated without a valid reason given for rejecting .
\n\n
[6 marks]
\nThe first three terms of an arithmetic sequence are and .
\nShow that .
\nProve that the sum of the first terms of this arithmetic sequence is a square number.
\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.
\nEITHER
\nuses (M1)
\n\nA1
\n\n
OR
\nuses (M1)
\n\nA1
\n\n
THEN
\nso AG
\n\n
[2 marks]
\n(A1)
\nuses M1
\nA1
\n\nA1
\n\n
Note: The final A1 can be awarded for clearly explaining that is a square number.
\n\n
so sum of the first terms is a square number AG
\n\n
[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]
\nThe functions and are defined for by and , where .
\nGiven that and , find the value of and the value 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.
\n(M1)
\nA1
\n(M1)
\nA1
\na valid attempt to solve their two linear equations for and M1
\nso and A1
\n\n
[6 marks]
\nThe following diagram shows the graph of , and rectangle . The rectangle has a vertex at the origin , a vertex on the -axis at the point , a vertex on the -axis at the point and a vertex at point on the graph.
\nLet represent the perimeter of rectangle .
\nLet represent the area of rectangle .
\nShow that .
\nFind the dimensions of rectangle that has maximum perimeter and determine the value of the maximum perimeter.
\nFind an expression for in terms of .
\nFind the dimensions of rectangle that has maximum area.
\nDetermine the maximum area of rectangle .
\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(A1)
\nA1
\nso AG
\n\n
[2 marks]
\nMETHOD 1
\nEITHER
\nuses the axis of symmetry of a quadratic (M1)
\n\n\n
OR
\nforms (M1)
\n\n\n
THEN
\nA1
\nsubstitutes their value of into (M1)
\n\nA1
\nso the dimensions of rectangle of maximum perimeter are by
\n\n
EITHER
\nsubstitutes their value of into (M1)
\n\n\n
OR
\nsubstitutes their values of and into (M1)
\n\nA1
\nso the maximum perimeter is
\n\n
METHOD 2
\nattempts to complete the square M1
\nA1
\nA1
\nsubstitutes their value of into (M1)
\n\nA1
\nso the dimensions of rectangle of maximum perimeter are by
\nA1
\nso the maximum perimeter is
\n\n
[6 marks]
\nsubstitutes into (M1)
\nA1
\n\n
[2 marks]
\nA1
\nattempts to solve their for (M1)
\n\nA1
\nsubstitutes their (positive) value of into (M1)
\n\nA1
\n\n
[5 marks]
\nA1
\n\n
[1 mark]
\nA geometric sequence has and . Find the second term of the sequence.
\n, (M1)
\n(A1)
\n(A1)
\nvalid attempt to find (M1)
\nfor example:
\n\n
A1
\n[5 marks]
\nA random variable has probability density function
\n\n
\n
Consider the case where .
\nFind the value of
\nFind, in terms of , the probability that lies between 1 and 3.
\nSketch the graph of . State the coordinates of the end points and any local maximum or minimum points, giving your answers in terms of .
\n.
\n.
\nthe median of .
\n(M1)(A1)(A1)
\n\n
A1
\n[4 marks]
\n A4
award A1 for (0, 3), A1 for continuity at (2, 3), A1 for maximum at (3, 4), A1 for (5, 0)
\nNote: Award A3 if correct four points are not joined by a straight line and a quadratic curve.
\n[4 marks]
\n(M1)
\n(A1)
\n(M1)
\nA1
\n[4 marks]
\n(M1)(A1)
\n= 2.35 A1
\n[3 marks]
\nattempt to use (M1)
\n(A1)
\n\n
attempt to solve integral using GDC and/or analytically (M1)
\n\n
A1
\n[4 marks]
\nConsider the function , , where , , .
\nThe following graph shows the curve . It has asymptotes at and and meets the -axis at A.
\nOn the following axes, sketch the two possible graphs of giving the equations of any asymptotes in terms of and .
\nGiven that , and A has coordinates , determine the possible sets of values for , and .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\neither graph passing through (or touching) A A1
\ncorrect shape and vertical asymptote with correct equation for either graph A1
\ncorrect horizontal asymptote with correct equation for either graph A1
\ntwo completely correct sketches A1
\n\n
[4 marks]
\nA1
\nfrom horizontal asymptote, (M1)
\nA1
\nfrom vertical asymptote,
\n= 3, = −4 or = −3, = 4 A1
\n\n
[4 marks]
\nConsider the polynomial .
\nGiven that has a factor , find the value of .
\nHence or otherwise, factorize as a product of linear factors.
\n(M1)
\nA1
\nA1
\n[3 marks]
\n\n
equate coefficients of : (M1)
\n\n
\n
(A1)
\nA1
\n\n
Note: Allow part (b) marks if any of this work is seen in part (a).
\n\n
Note: Allow equivalent methods (eg, synthetic division) for the M marks in each part.
\n\n
[3 marks]
\nThe following diagram shows the graph of for .
\nA function is defined by for .
\nDescribe a sequence of transformations that transforms the graph of for to the graph of for .
\nState the range of .
\nFind an expression for , stating its domain.
\nFind the coordinates of the point(s) where the graphs of and intersect.
\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.
\nfor example,
\na reflection in the -axis (in the line ) A1
\na horizontal translation (shift) units to the left A1
\na vertical translation (shift) down by unit A1
\n\n
Note: Award A1 for each correct transformation applied in a correct position in the sequence. Do not accept use of the “move” for a translation.
\nNote: Award A1A1A1 for a correct alternative sequence of transformations. For example,
\na vertical translation (shift) down by unit, followed by a horizontal translation (shift) units to the left and then a reflection in the line .
\n\n
[3 marks]
\nrange is A1
\n\n
Note: Correct alternative notations include , or .
\n\n
[1 mark]
\nM1
\n\n
Note: Award M1 for interchanging and (can be done at a later stage).
\n\n
A1
\nA1
\nso A1
\ndomain is A1
\n\n
Note: Correct alternative notations include or .
\n\n
[5 marks]
\nthe point of intersection lies on the line
\n\n
EITHER
\nM1
\nattempts to solve their quadratic equation M1
\nfor example, or
\n\n
OR
\nM1
\n\nsubstitutes to obtain
\nattempts to solve their quadratic equation M1
\nfor example, or
\n\n
THEN
\nA1
\nas , the only solution is R1
\nso the coordinates of the point of intersection are A1
\n\n
Note: Award R0A1 if is stated without a valid reason given for rejecting .
\n\n
[5 marks]
\nThe function is defined by , for , , .
\nThe graph of has exactly one point of inflexion.
\nThe function is defined by , for .
\nFind the value of and the value of .
\nFind an expression for .
\nFind the -coordinate of the point of inflexion.
\nSketch the graph of for , showing the values of any axes intercepts, the coordinates of any local maxima and local minima, and giving the equations of any asymptotes.
\nFind the equations of all the asymptotes on the graph of .
\nBy considering the graph of , or otherwise, solve for .
\nattempt to solve e.g. by factorising (M1)
\nor vice versa A1
\n\n
[2 marks]
\nattempt to use quotient rule or product rule (M1)
\n\n
EITHER
\nA1A1
\n\n
Note: Award A1 for each term in the numerator with correct signs, provided correct denominator is seen.
\n\n
OR
\nA1A1
\n\n
Note: Award A1 for each term.
\n\n
[3 marks]
\nattempt to find the local min point on OR solve (M1)
\nA1
\n\n
[2 marks]
\n A1A1A1A1A1
\n
Note: Award A1 for both vertical asymptotes with their equations, award A1 for horizontal asymptote with equation, award A1 for each correct branch including asymptotic behaviour, coordinates of minimum and maximum points (may be seen next to the graph) and values of axes intercepts.
If vertical asymptotes are absent (or not vertical) and the branches overlap as a consequence, award maximum A0A1A0A1A1.
\n
[5 marks]
\nA1
\n(oblique asymptote has) gradient (A1)
\nappropriate method to find complete equation of oblique asymptote M1
\n\n
\n
\n
A1
\nNote: Do not award the final A1 if the answer is not given as an equation.
\n\n
[4 marks]
\nattempting to find at least one critical value (M1)
\nOR OR A1A1A1
\n\n
Note: Only penalize once for use of rather than .
\n\n
[4 marks]
\nIt is given that where and are positive integers.
\nGiven that is a factor of find the value of and the value of .
\nFactorize into a product of linear factors.
\nUsing your graph state the range of values of for which has exactly two distinct real roots.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
M1A1
\nA1
\nA1
\n[4 marks]
\n\n
attempt to equate coefficients (M1)
\n(A1)
\n\n
A1
\n\n
Note: Accept any equivalent valid method.
\n\n
[3 marks]
\nA1
\nA1A1
\n\n
Note: Award A1 for correct end points and A1 for correct inequalities.
\n\n
Note: If the candidate has misdrawn the graph and omitted the first minimum point, the maximum mark that may be awarded is A1FTA0A0 for seen.
\n\n
[3 marks]
\nThe following diagram shows a ball attached to the end of a spring, which is suspended from a ceiling.
\nThe height, metres, of the ball above the ground at time seconds after being released can be modelled by the function where .
\nFind the height of the ball above the ground when it is released.
\nFind the minimum height of the ball above the ground.
\nShow that the ball takes seconds to return to its initial height above the ground for the first time.
\nFor the first 2 seconds of its motion, determine the amount of time that the ball is less than metres above the ground.
\nFind the rate of change of the ball’s height above the ground when . Give your answer in the form where and .
\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.
\nattempts to find (M1)
\n\n(m) (above the ground) A1
\n\n
[2 marks]
\nEITHER
\nuses the minimum value of which is M1
\n(m)
\n\n
OR
\nthe amplitude of motion is (m) and the mean position is (m) M1
\n\n
OR
\nfinds , attempts to solve for and determines that the minimum height above the ground occurs at M1
\n(m)
\n\n
THEN
\n(m) (above the ground) A1
\n\n
[2 marks]
\nEITHER
\nthe ball is released from its maximum height and returns there a period later R1
\nthe period is A1
\n\n
OR
\nattempts to solve for M1
\n\nA1
\n\n
THEN
\nso it takes seconds for the ball to return to its initial position for the first time AG
\n\n
[2 marks]
\n(M1)
\n\n
A1
\n(A1)
\n\n
Note: Accept extra correct positive solutions for .
\nA1
\n\n
Note: Do not award A1 if solutions outside are also stated.
\nthe ball is less than metres above the ground for (s)
\n(s) A1
\n\n
[5 marks]
\nEITHER
attempts to find (M1)
\n\n
OR
\nrecognizes that is required (M1)
\n\n
THEN
\nA1
\nattempts to evaluate their (M1)
\n\nA1
\n\n
Note: Accept equivalent correct answer forms where . For example, .
\n\n
[4 marks]
\nLet f(x) = x4 + px3 + qx + 5 where p, q are constants.
\nThe remainder when f(x) is divided by (x + 1) is 7, and the remainder when f(x) is divided by (x − 2) is 1. Find the value of p and the value of q.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to substitute x = −1 or x = 2 or to divide polynomials (M1)
\n1 − p − q + 5 = 7, 16 + 8p + 2q + 5 = 1 or equivalent A1A1
\nattempt to solve their two equations M1
\np = −3, q = 2 A1
\n[5 marks]
\nThe function has a derivative given by where is a positive constant.
\nConsider , the population of a colony of ants, which has an initial value of .
\nThe rate of change of the population can be modelled by the differential equation , where is the time measured in days, , and is the upper bound for the population.
\nAt the population of the colony has doubled in size from its initial value.
\nThe expression for can be written in the form , where . Find and in terms of .
\nHence, find an expression for .
\nBy solving the differential equation, show that .
\nFind the value of , giving your answer correct to four significant figures.
\nFind the value of when the rate of change of the population is at its maximum.
\n(A1)
\nattempt to compare coefficients OR substitute and and solve (M1)
\nand A1
\n\n\n
[3 marks]
\nattempt to integrate their (M1)
\n\nA1A1
\n\n
Note: Award A1 for each correct term. Award A1A0 for a correct answer without modulus signs. Condone the absence of .
\n\n
[3 marks]
\nattempt to separate variables and integrate both sides M1
\n\nA1
\n\n
Note: There are variations on this which should be accepted, such as . Subsequent marks for these variations should be awarded as appropriate.
\n\n
EITHER
\nattempt to substitute into an equation involving M1
\nA1
\nA1
\n\nA1
\n\n
OR
\n\nA1
\nattempt to substitute M1
\nA1
\nA1
\n\n
THEN
\nattempt to rearrange and isolate M1
\nOR OR
\nOR A1
\n\n
AG
\n\n
[8 marks]
\nattempt to substitute (M1)
\n(A1)
\n\nA1
\n\n
Note: Award (M1)(A1)A0 for any other value of which rounds to
\n\n
[3 marks]
\nattempt to find the maximum of the first derivative graph OR zero of the second derivative graph OR that (M1)
\n\n(days) A2
\n\n
Note: Accept any value which rounds to .
\n\n
[3 marks]
\nThe following diagram shows part of the graph of for .
\nThe shaded region is the area bounded by the curve, the -axis and the lines and .
\nUsing implicit differentiation, find an expression for .
\nFind the equation of the tangent to the curve at the point .
\nFind the area of .
\nThe region is now rotated about the -axis, through radians, to form a solid.
\nBy writing as , show that the volume of the solid formed is .
\nvalid attempt to differentiate implicitly (M1)
\nA1A1
\nA1
\n[4 marks]
\nat (M1)
\nA1
\nhence equation of tangent is
\nOR (M1)A1
\nNote: Accept .
\n[4 marks]
\n(M1)
\n(A1)
\nA1
\n[3 marks]
\nuse of volume (M1)
\nA1
\n\n
Note: Condone absence of limits up to this point.
\nreasonable attempt to integrate (M1)
\nA1A1
\nNote: Award A1 for correct limits (not to be awarded if previous M1 has not been awarded) and A1 for correct integrand.
\nA1
\nAG
\nNote: Do not accept decimal answer equivalent to .
\n[6 marks]
\nConsider the polynomial .
\nSketch the graph of , stating clearly the coordinates of any maximum and minimum points and intersections with axes.
\nHence, or otherwise, state the condition on such that all roots of the equation are real.
\nshape A1
\n-axis intercepts at (−3, 0), (1, 0) and -axis intercept at (0, −51) A1A1
\nminimum points at (−1.62, −118) and (3.72, 19.7) A1A1
\nmaximum point at (2.40, 26.9) A1
\nNote: Coordinates may be seen on the graph or elsewhere.
\nNote: Accept −3, 1 and −51 marked on the axes.
\n[6 marks]
\nfrom graph, 19.7 ≤ ≤ 26.9 A1A1
\nNote: Award A1 for correct endpoints and A1 for correct inequalities.
\n[2 marks]
\nThe following diagram shows the graph of , .
\nFind the value of .
\nGiven that , determine the value of .
\nGiven that , find the domain and range of .
\n(A1)
\nA1
\n[2 marks]
\n(M1)
\nA1
\n[2 marks]
\ndomain is A1
\nrange is A1
\n[2 marks]
\nConsider the function .
\nShow that the graph of is concave up for .
\nSketch the graph of showing clearly any intercepts with the axes.
\nM1A1
\nfor concave up R1AG
\n\n
[3 marks]
\n![\"M17/5/MATHL/HP1/ENG/TZ1/B12.e.ii/M\"](\"../assets/uploads/tinymce_asset/asset/3506/Schermafbeelding_2017-08-08_om_16.48.38.png\"/)
-intercept at A1
\n-intercept at A1
\nstationary point of inflexion at with correct curvature either side A1
\n[3 marks]
\nUse l’Hôpital’s rule to determine the value 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.
\nattempts to apply l’Hôpital’s rule on M1
\nM1A1A1
\n\n
Note: Award M1 for attempting to use product and chain rule differentiation on the numerator, A1 for a correct numerator and A1 for a correct denominator. The awarding of A1 for the denominator is independent of the M1.
\nA1
\n\n
[5 marks]
\nConsider quadrilateral where is parallel to .
\nIn , , , and .
\nFind an expression for in terms of and .
\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.
\nMETHOD 1
\nfrom vertex , draws a line parallel to that meets at a point (M1)
\nuses the sine rule in M1
\nA1
\n(A1)
\nA1
\n\n
METHOD 2
\nlet the height of quadrilateral be
\nA1
\nattempts to find a second expression for M1
\n\n\nwrites as , multiplies through by and expands the RHS M1
\n\nA1
\nA1
\n\n
[5 marks]
\nThe polynomial is exactly divisible by each of , and .
\nFind the values of , and .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nsubstitute each of = 1,2 and 3 into the quartic and equate to zero (M1)
\n\n
or equivalent (A2)
\n\n
Note: Award A2 for all three equations correct, A1 for two correct.
\nattempting to solve the system of equations (M1)
\n= −7, = 17, = −17 A1
\nNote: Only award M1 when some numerical values are found when solving algebraically or using GDC.
\n\n
METHOD 2
\nattempt to find fourth factor (M1)
\nA1
\nattempt to expand M1
\n( = −7, = 17, = −17) A2
\nNote: Award A2 for all three values correct, A1 for two correct.
\nNote: Accept long / synthetic division.
\n[5 marks]
\nThis question asks you to explore the behaviour and key features of cubic polynomials of the form .
\n\n
Consider the function for and where is a parameter, .
\nThe graphs of for and are shown in the following diagrams.
\n
On separate axes, sketch the graph of showing the value of the -intercept and the coordinates of any points with zero gradient, for
\nHence, or otherwise, find the set of values of such that the graph of has
\nGiven that the graph of has one local maximum point and one local minimum point, show that
\nHence, for , find the set of values of such that the graph of has
\n.
\n.
\nWrite down an expression for .
\na point of inflexion with zero gradient.
\none local maximum point and one local minimum point.
\nno points where the gradient is equal to zero.
\nthe -coordinate of the local maximum point is .
\nthe -coordinate of the local minimum point is .
\nexactly one -axis intercept.
\nexactly two -axis intercepts.
\nexactly three -axis intercepts.
\nConsider the function for and where .
\nFind all conditions on and such that the graph of has exactly one -axis intercept, explaining your reasoning.
\n: positive cubic with correct -intercept labelled A1
\nlocal maximum point correctly labelled A1
\nlocal minimum point correctly labelled A1
\n\n
[3 marks]
\n: positive cubic with correct -intercept labelled A1
\nlocal maximum point correctly labelled A1
\nlocal minimum point correctly labelled A1
\n\n
Note: Accept the following exact answers:
Local maximum point coordinates .
Local minimum point coordinates .
\n
[3 marks]
\nA1
\n\n
Note: Accept (an expression).
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\nconsiders the number of solutions to their (M1)
\n\nA1
\n\n
[2 marks]
\nA1
\n\n
Note: The (M1) in part (c)(ii) can be awarded for work shown in either (ii) or (iii).
\n\n
[1 mark]
\nattempts to solve their for (M1)
\n(A1)
\n\n
Note: Award (A1) if either or is subsequently considered.
Award the above (M1)(A1) if this work is seen in part (c).
\n
correctly evaluates A1
\n\nthe -coordinate of the local maximum point is AG
\n\n
[3 marks]
\n\n
correctly evaluates A1
\n\nthe -coordinate of the local minimum point is AG
\n\n
[1 mark]
\nthe graph of will have one -axis intercept if
\nEITHER
\n(or equivalent reasoning) R1
\n\n
OR
\nthe minimum point is above the -axis R1
\n\n
Note: Award R1 for a rigorous approach that does not (only) refer to sketched graphs.
\n\n
THEN
\nA1
\n\n
Note: Condone . The A1 is independent of the R1.
\n\n
[2 marks]
\nthe graph of will have two -axis intercepts if
\nEITHER
\n(or equivalent reasoning) (M1)
\n\n
OR
\nevidence from the graph in part(a)(i) (M1)
\n\n
THEN
\nA1
\n\n
[2 marks]
\nthe graph of will have three -axis intercepts if
\nEITHER
\n(or equivalent reasoning) (M1)
\n\n
OR
\nreasoning from the results in both parts (e)(i) and (e)(ii) (M1)
\n\n
THEN
\nA1
\n\n
[2 marks]
\ncase 1:
\n(independent of the value of ) A1
\nEITHER
\ndoes not have two solutions (has no solutions or solution) R1
\n
OR
for R1
\n
OR
the graph of has no local maximum or local minimum points, hence any vertical translation of this graph () will also have no local maximum or local minimum points R1
\n
THEN
therefore there is only one -axis intercept AG
\n\n
Note: Award at most A0R1 if only is considered.
\n\n
case 2
is a local maximum point and is a local minimum point (A1)
\n\n
Note: Award (A1) for a correct -coordinate seen for either the maximum or the minimum.
\n\n
considers the positions of the local maximum point and/or the local minimum point (M1)
\n\n
EITHER
considers both points above the -axis or both points below the -axis
OR
considers either the local minimum point only above the -axis OR the local maximum point only below the -axis
THEN
(both points above the -axis) A1
\n(both points above the -axis) A1
\n\n
Note: Award at most (A1)(M1)A0A0 for case 2 if is not clearly stated.
\n\n
[6 marks]
\nThe lines and have the following vector equations where and .
\n\nThe plane has Cartesian equation where .
\n\n
Given that and have no points in common, find
\nShow that and are never perpendicular to each other.
\nthe value of .
\nthe condition on the value 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.
\nattempts to calculate (M1)
\nA1
\nsince for R1
\nso and are never perpendicular to each other AG
\n\n
[3 marks]
\n(since is parallel to , is perpendicular to the normal of and so)
\nR1
\n\nA1
\n\n
[2 marks]
\nsince there are no points in common, does not lie in
\n\n
EITHER
\nsubstitutes into (M1)
\n\n
OR
\n(M1)
\n\n
THEN
\nA1
\n\n
[2 marks]
\nLet , where , ,
\nGiven that is a factor of , find a relationship between , and .
\nGiven that is a factor of , write down the value of .
\nGiven that is a factor of , and that , find the values of and .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to substitute and set equal to zero, or use of long / synthetic division (M1)
\nA1
\n\n
\n
[2 marks]
\n0 A1
\n\n
[1 mark]
\nEITHER
\nattempt to solve (M1)
\n\n
\n
OR
\n(M1)
\ncomparing coefficients gives = 5, = 2
\n\n
THEN
\n= 105 A1
\n\n
= 50 A1
\n\n
[3 marks]
\nThe following shape consists of three arcs of a circle, each with centre at the opposite vertex of an equilateral triangle as shown in the diagram.
\nFor this shape, calculate
\nthe perimeter.
\nthe area.
\neach arc has length (M1)
\nperimeter is therefore (cm) A1
\n[2 marks]
\narea of sector, , is (A1)
\narea of triangle, , is (M1)(A1)
\nNote: area of segment, , is 3.261… implies area of triangle
\nfinding or or similar
\narea (cm2) (M1)A1
\n[5 marks]
\nIt is given that . (Do not prove this identity.)
\nUsing mathematical induction and the above identity, prove that 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.
\nlet be the proposition that for
\nconsidering :
\nand
\nso is true R1
\nassume is true, i.e. M1
\n\n
Note: Award M0 for statements such as “let ”.
\nNote: Subsequent marks after this M1 are independent of this mark and can be awarded.
\n\n
considering
\nM1
\nA1
\n\nM1
\n\n
Note: Award M1 for use of with and .
\n\n
A1
\nA1
\nis true whenever is true, is true, so is true for R1
\n\n
Note: Award the final R1 mark provided at least five of the previous marks have been awarded.
\n\n
[8 marks]
\nA function is defined by .
\nThe region is bounded by the curve , the -axis and the lines and . Let be the area of .
\nThe line divides into two regions of equal area.
\nLet be the gradient of a tangent to the curve .
\nSketch the curve , clearly indicating any asymptotes with their equations and stating the coordinates of any points of intersection with the axes.
\nShow that .
\nFind the value of .
\nShow that .
\nShow that the maximum value of 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.
\na curve symmetrical about the -axis with correct concavity that has a local maximum point on the positive -axis A1
\na curve clearly showing that as A1
\nA1
\nhorizontal asymptote (-axis) A1
\n\n
[4 marks]
\nattempts to find (M1)
\nA1
\n\n
Note: Award M1A0 for obtaining where .
\nNote: Condone the absence of or use of incorrect limits to this stage.
\n\n
(M1)
\nA1
\nAG
\n\n
[4 marks]
\nMETHOD 1
\nEITHER
\n\n(M1)
\n\n
OR
\n\n(M1)
\n\n\n
THEN
\nA1
\nA1
\nA1
\n\n
METHOD 2
\n\n(M1)
\nA1
\nA1
\nA1
\n\n
[4 marks]
\nattempts to find (M1)
\nA1
\nso AG
\n\n
[2 marks]
\nattempts product rule or quotient rule differentiation M1
\nEITHER
\nA1
\nOR
\nA1
\n\n
Note: Award A0 if the denominator is incorrect. Subsequent marks can be awarded.
\n\n
THEN
\nattempts to express their as a rational fraction with a factorized numerator M1
\n\nattempts to solve their for M1
\nA1
\nfrom the curve, the maximum value of occurs at R1
\n(the minimum value of occurs at )
\n\n
Note: Award R1 for any equivalent valid reasoning.
\n\n
maximum value of is A1
\nleading to a maximum value of AG
\n\n
[7 marks]
\nConsider the expansion of , where and .
\nThe coefficient of is four times the coefficient of . Find the value of .
\nattempt to find coefficients in binomial expansion (M1)
\ncoefficient of ; coefficient of A1A1
\nNote: Condone terms given rather than coefficients. Terms may be seen in an equation such as that below.
\n(A1)
\nattempt to solve equation using GDC or algebraically (M1)
\n\n
\n
\n
A1
\n[6 marks]
\nConsider the equation , where , , , and .
\nTwo of the roots of the equation are log26 and and the sum of all the roots is 3 + log23.
\nShow that 6 + + 12 = 0.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nis a root (A1)
\nis a root (A1)
\nsum of roots: M1
\nNote: Award M1 for use of is equal to the sum of the roots, do not award if minus is missing.
\nNote: If expanding the factored form of the equation, award M1 for equating to the coefficient of .
\n\n
product of roots: M1
\nA1
\nNote: Award M1A0 for
\n\n
\n
\n
EITHER
\nM1A1AG
\nNote: M1 is for a correct use of one of the log laws.
\nOR
\nM1A1AG
\nNote: M1 is for a correct use of one of the log laws.
\n\n
[7 marks]
\nThe random variable follows a normal distribution with mean and standard deviation .
\nThe avocados grown on a farm have weights, in grams, that are normally distributed with mean and standard deviation . Avocados are categorized as small, medium, large or premium, according to their weight. The following table shows the probability an avocado grown on the farm is classified as small, medium, large or premium.
\nThe maximum weight of a small avocado is grams.
\nThe minimum weight of a premium avocado is grams.
\nA supermarket purchases all the avocados from the farm that weigh more than grams.
\nFind the probability that an avocado chosen at random from this purchase is categorized as
\nFind .
\nFind the value of and of .
\nmedium.
\nlarge.
\npremium.
\nThe selling prices of the different categories of avocado at this supermarket are shown in the following table:
\n\n
\n
\n
\n
The supermarket pays the farm for the avocados and assumes it will then sell them in exactly the same proportion as purchased from the farm.
\nAccording to this model, find the minimum number of avocados that must be sold so that the net profit for the supermarket is at least .
\n(M1)
\nOR (A1)
\n\nA1
\n
Note: Do not award any marks for use of their answers from part (b).
\n
[3 marks]
\nand (seen anywhere) (A1)
\ncorrect equations (A1)(A1)
\n\nattempt to solve their equations involving z values (M1)
\n\nA1
\n\n
[5 marks]
\nnew sample space is (may be seen in (ii) or (iii)) (M1)
\nOR
\nA1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\nattempt to express revenue from avocados (M1)
\nOR
\ncorrect inequality or equation for net profit in terms of (A1)
\nOR
\nattempt to solve the inequality (M1)
\nsketch OR
\nA1
\n
Note: Only award follow through in part (d) for 3 probabilities which add up to 1. FT of probabilities from c) that do not add up to 1 should only be awarded M marks, where appropriate, in d).
\n
[4 marks]
\nThe quadratic equation has roots and such that . Without solving the equation, find the possible values of the real number .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
\nA1
\n(M1)
\n\n
A1
\nattempt to solve quadratic (M1)
\nA1
\n[6 marks]
\nLet , where and .
\nOne of the roots of is . Find the value of .
\nMETHOD 1
\none other root is A1
\nlet third root be (M1)
\nconsidering sum or product of roots (M1)
\nsum of roots A1
\nproduct of roots A1
\nhence A1
\n\n
METHOD 2
\none other root is A1
\nquadratic factor will be (M1)A1
\nM1
\ncomparing coefficients (M1)
\nhence A1
\n\n
METHOD 3
\nsubstitute into (M1)
\n(M1)A1
\nequating real or imaginary parts or dividing M1
\nor or A1
\nhence A1
\n\n
[6 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.
\nA1
\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]
\nRunners in an athletics club have season’s best times for the 100 m, which can be modelled by a normal distribution with mean 11.6 seconds and standard deviation 0.8 seconds. To qualify for a particular competition a runner must have a season’s best time of under 11 seconds. A runner from this club who has qualified for the competition is selected at random. Find the probability that he has a season’s best time of under 10.7 seconds.
\n\n
(M1)
\n(M1)
\n(M1)
\n(A1)
\n(A1)
\nA1
\nNote: Accept only 0.575.
\n[6 marks]
\nUse the binomial theorem to expand . Give your answer in the form where and are expressed in terms of and .
\nUse de Moivre’s theorem and the result from part (a) to show that .
\nUse the identity from part (b) to show that the quadratic equation has roots and .
\nHence find the exact value of .
\nDeduce a quadratic equation with integer coefficients, having roots and .
\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.
\nuses the binomial theorem on M1
\nA1
\nA1
\n\n
[3 marks]
\n(using de Moivre’s theorem with gives) (A1)
\nequates both the real and imaginary parts of and M1
\nand
\nrecognizes that (A1)
\nsubstitutes for and into M1
\n\ndivides the numerator and denominator by to obtain
\nA1
\nAG
\n\n
[5 marks]
\nsetting and putting in the numerator of gives M1
\nattempts to solve for M1
\n(A1)
\nA1
\n\n
Note: Do not award the final A1 if solutions other than are listed.
\n\n
finding the roots of corresponds to finding the roots of where R1
\nso the equation as roots and AG
\n\n
[5 marks]
\nattempts to solve for M1
\nA1
\nsince has the smaller value of the two roots R1
\n\n
Note: Award R1 for an alternative convincing valid reason.
\n\n
so A1
\n\n
[4 marks]
\nlet
\nuses where (M1)
\nM1
\nA1
\n\n
[3 marks]
\nEight boys and two girls sit on a bench. Determine the number of possible arrangements, given that
\nthe girls do not sit together.
\nthe girls do not sit on either end.
\nthe girls do not sit on either end and do not sit together.
\nMETHOD 1
\n(A1)(A1)A1
\nNote: Award A1 for , A1 for , A1 for final answer.
\n\n
METHOD 2
\n(A1)(A1)A1
\nNote: Award A1 for or equivalent, A1 for and A1 for answer.
\n\n
[3 marks]
\nMETHOD 1
\n(A1)A1
\nNote: Award (A1) for , A1 for final answer.
\n\n
METHOD 2
\n\n
Note: Award A1 for minus EITHER subtracted terms and A1 for final correct answer.
\n\n
[2 marks]
\nMETHOD 1
\n(A1)(A1)A1
\nNote: Award (A1) for , (A1) for , A1 for final answer. can be replaced by or which may be awarded the second A1.
\n\n
METHOD 2
\ntheir answer to (a) (A1)(A1)A1
\nNote: Award A1 for subtracting each of the terms and A1 for final answer.
\n\n
METHOD 3
\ntheir answer to (b) or equivalent (A1)A2
\nNote: Award A1 for the subtraction and A2 for final answer.
\n\n
[3 marks]
\nIn Lucy’s music academy, eight students took their piano diploma examination and achieved scores out of 150. For her records, Lucy decided to record the average number of hours per week each student reported practising in the weeks prior to their examination. These results are summarized in the table below.
\nFind Pearson’s product-moment correlation coefficient, , for these data.
\nThe relationship between the variables can be modelled by the regression equation . Write down the value of and the value of .
\nOne of these eight students was disappointed with her result and wished she had practised more. Based on the given data, determine how her score could have been expected to alter had she practised an extra five hours per week.
\nLucy asserts that the number of hours a student practises has a direct effect on their final diploma result. Comment on the validity of Lucy’s assertion.
\nLucy suspected that each student had not been practising as much as they reported. In order to compensate for this, Lucy deducted a fixed number of hours per week from each of the students’ recorded hours.
\nState how, if at all, the value of would be affected.
\nuse of GDC to give (M1)
\n\nA1
\n
Note: Award the (M1) for any correct value of , , or seen in part (a) or part (b).
[2 marks]
A1
\n
[1 mark]
attempt to find their difference (M1)
\nOR
\n\n\n
the student could have expected her score to increase by marks. A1
\n
Note: Accept an increase of or .
[2 marks]
Lucy is incorrect in suggesting there is a causal relationship.
\nThis might be true, but the data can only indicate a correlation. R1
\n
Note: Accept ‘Lucy is incorrect as correlation does not imply causation’ or equivalent.
[1 mark]
no effect A1
\n
[1 mark]
A body moves in a straight line such that its velocity, , after seconds is given by for .
\nThe following diagram shows the graph of against . Point is a local maximum and point is a local minimum.
\nThe body first comes to rest at time . Find
\nDetermine the coordinates of point and the coordinates of point .
\nHence, write down the maximum speed of the body.
\nthe value of .
\nthe distance travelled between and .
\nthe acceleration when .
\nFind the distance travelled in the first 30 seconds.
\nand A1A1A1A1
\n[4 marks]
\nmaximum speed is A1
\n[1 mark]
\n(M1)A1
\n[2 marks]
\n(M1)
\nA1
\n[2 marks]
\nat (M1)
\nA1
\nNote: Accept .
\n[2 marks]
\nattempt to integrate between 0 and 30 (M1)
\nNote: An unsupported answer of 38.6 can imply integrating from 0 to 30.
\n\n
EITHER
\n(A1)
\n\n
OR
\n(A1)
\n\n
THEN
\nA1
\n[3 marks]
\nThis question asks you to examine various polygons for which the numerical value of the area is the same as the numerical value of the perimeter. For example, a by rectangle has an area of and a perimeter of .
\n\n
For each polygon in this question, let the numerical value of its area be and let the numerical value of its perimeter be .
\nAn -sided regular polygon can be divided into congruent isosceles triangles. Let be the length of each of the two equal sides of one such isosceles triangle and let be the length of the third side. The included angle between the two equal sides has magnitude .
\nPart of such an -sided regular polygon is shown in the following diagram.
\nConsider a -sided regular polygon such that .
\nThe Maclaurin series for is
\nConsider a right-angled triangle with side lengths and , where , such that .
\nFind the side length, , where , of a square such that .
\nWrite down, in terms of and , an expression for the area, , of one of these isosceles triangles.
\nShow that .
\nUse the results from parts (b) and (c) to show that .
\nUse the Maclaurin series for to find .
\nInterpret your answer to part (e)(i) geometrically.
\nShow that .
\nBy using the result of part (f) or otherwise, determine the three side lengths of the only two right-angled triangles for which .
\nDetermine the area and perimeter of these two right-angled triangles.
\nand (A1)
\n(M1)
\n\nA1
\n\n
Note: Award A1M1A0 if both and are stated as final answers.
\n\n
[3 marks]
\nA1
\n\n
Note: Award A1 for a correct alternative form expressed in terms of and only.
\nFor example, using Pythagoras’ theorem, or or .
\n\n
[1 mark]
\nMETHOD 1
\nuses (M1)
\nA1
\nAG
\n\n
METHOD 2
\nuses Pythagoras’ theorem and (M1)
\n\nA1
\nAG
\n\n
METHOD 3
\nuses the cosine rule (M1)
\n\nA1
\nAG
\n\n
METHOD 4
\nuses the sine rule (M1)
\n\nA1
\nAG
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award M1 for equating correct expressions for and .
\n\n\n
A1
\nuses (seen anywhere in part (d) or in part (b)) (M1)
\n\nattempts to either factorise or divide their expression (M1)
\n\n(or equivalent) A1
\n\n
EITHER
\nsubstitutes (or equivalent) into (M1)
\nA1
\n
Note: Other approaches are possible. For example, award A1 for and M1 for substituting into .
OR
substitutes (or equivalent) into (M1)
\n\nA1
\n\n
THEN
\nAG
\n\n
[7 marks]
\nattempts to use the Maclaurin series for with (M1)
\n\n(or equivalent) A1
\n\nA1
\n\n
Note: Award a maximum of M1A1A0 if is not stated anywhere.
\n\n
[3 marks]
\n(as and )
\nthe polygon becomes a circle of radius R1
\n\n
Note: Award R1 for alternative responses such as:
the polygon becomes a circle of area OR
the polygon becomes a circle of perimeter OR
the polygon becomes a circle with .
Award R0 for polygon becomes a circle.
\n
[1 mark]
\nand (A1)(A1)
\nequates their expressions for and M1
\n\nM1
\n\n
Note: Award M1 for isolating or . This step may be seen later.
\n\n\n
M1
\n\n\n
Note: Award M1 for attempting to expand their RHS of either or .
\n\n
EITHER
\nA1
\n\n\n\n
OR
\n\nA1
\n\n\n
THEN
\nA1
\n\nAG
\n\n
Note: Award a maximum of A1A1M1M1M0A0A0 for attempting to verify.
For example, verifying that gains of the marks.
\n
[7 marks]
\nusing an appropriate method (M1)
\neg substituting values for or using divisibility properties
\nand A1A1
\n\n
Note: Award A1A0 for either one set of three correct side lengths or two sets of two correct side lengths.
\n\n
[3 marks]
\nand A1
\n\n
Note: Do not award A1FT.
\n\n
[1 mark]
\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\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]
\nA data set consisting of test scores has mean . One test score of requires a second marking and is removed from the data set.
\nFind the mean of the remaining test scores.
\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: Award M1 for use of .
\n\n
(A1)
\nnew (A1)
\nA1
\n\n
Note: Do not accept 15.
\n\n
[4 marks]
\n\n
The following diagram shows a circle with centre and radius .
\nPoints , and lie on the circumference of the circle.
\nChord has length and radians.
\nShow that arc has length .
\nShow that .
\nArc is twice the length of chord .
\nFind the value 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.
\nEITHER
\nuses the arc length formula (M1)
\narc length is A1
\n\n
OR
\nlength of arc is A1
\nthe sum of the lengths of arc and arc is A1
\n\n
THEN
\nso arc has length AG
\n\n
[2 marks]
\nuses the cosine rule (M1)
\nA1
\nso AG
\n\n
[2 marks]
\nA1
\nattempts to solve for (M1)
\nA1
\n\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 in radians.
\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]
\ntan () or equivalent (M1)
\n= 0.857 A1
\n[2 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]
\nProve the identity .
\nThe equation has two real roots, and .
\nConsider the equation , where and which has roots and .
Without solving , determine the values of and .
METHOD 1
\n\nattempts to expand M1
\n\n\nA1
\nAG
\n
Note: Condone the use of equals signs throughout.
\n
METHOD 2
\n\nattempts to factorise M1
\n\nA1
\nAG
\n
Note: Condone the use of equals signs throughout.
\n
METHOD 3
\n\nattempts to factorise M1
\n\nA1
\nAG
\n
Note: Condone the use of equals signs throughout.
[2 marks]
Note: Award a maximum of A1M0A0A1M0A0 for and found by using .
Condone, as appropriate, solutions that state but clearly do not use the values of and .
Special case: Award a maximum of A1M1A0A1M0A0 for and obtained by solving simultaneously for and from product of roots and sum of roots equations.
product of roots of
(seen anywhere) A1
\nconsiders by stating M1
\n
Note: Award M1 for attempting to substitute their value of into .
A1
\nsum of roots of
\n(seen anywhere) A1
\nconsiders and by stating M1
\n
Note: Award M1 for attempting to substitute their values of and into their expression. Award M0 for use of only.
A1
\n\n
[6 marks]
In an arithmetic sequence, and .
\nFind the common difference.
\nFind the first term.
\nFind the sum of the first terms.
\nvalid approach (M1)
\neg ,
\nA1 N2
\n[2 marks]
\nvalid approach (M1)
\neg , ,
\nA1 N2
\n[2 marks]
\ncorrect substitution into sum formula
\neg , (A1)
\nA1 N2
\n[2 marks]
\nThe following diagram shows a circle with centre and radius .
\nPoints and lie on the circumference of the circle, and .
\nThe perimeter of the shaded region is .
\nFind the value of .
\nHence, find the exact area of the non-shaded region.
\nminor arc has length (A1)
\nrecognition that perimeter of shaded sector is (A1)
\n\nA1
\n\n
[3 marks]
\nEITHER
(M1)
Area of non-shaded region (A1)
\n\n
OR
\narea of circle area of shaded sector (M1)
\n(A1)
\n\n
THEN
\narea A1
\n\n
[3 marks]
\nA particle moves in a straight line such that its velocity, , at time seconds is given by
\n.
\nThe particle’s acceleration is zero at .
\nFind the value of .
\nLet be the distance travelled by the particle from to and let be the distance travelled by the particle from to .
\nShow that .
\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.
\nattempts either graphical or symbolic means to find the value of when (M1)
\nA1
\n\n
[2 marks]
\nattempts to find the value of either or (M1)
\nand A1A1
\n\n
Note: Award as above for obtaining, for example, or
\nNote: Award a maximum of M1A1A0FT for use of an incorrect value of from part (a).
\n\n
so AG
\n\n
[3 marks]
\nA continuous random variable has a probability density function given by
\n\nThe median of this distribution is .
\nDetermine the value of .
\nGiven that , determine the value of .
\nrecognises that (M1)
\n\n\nA1
\n
[2 marks]
METHOD 1
\nattempts to find at least one endpoint (limit) both in terms of (or their ) and (M1)
\n\n
(A1)
\n
Note: Award (A1) for .
attempts to solve their equation for (M1)
\n
Note: The above (M1) is dependent on the first (M1).
A1
\n\n
METHOD 2
\n(M1)(A1)
\n\n
Note: Only award (M1) if at least one limit has been translated correctly.
\nNote: Award (M1)(A1) for .
\n
attempts to solve their equation for (M1)
A1
\n\n
METHOD 3
\nEITHER
\n(M1)(A1)
\n\n
Note: Only award (M1) if at least one limit has been translated correctly.
\nNote: Award (M1)(A1) for .
\n
OR
(M1)(A1)
\n
Note: Only award (M1) if at least one limit has been translated correctly.
Note: Award (M1)(A1) for .
\n
THEN
attempts to solve their equation for (M1)
\n
Note: The above (M1) is dependent on the first (M1).
A1
\n\n
[4 marks]
\nThis diagram shows a metallic pendant made out of four equal sectors of a larger circle of radius and four equal sectors of a smaller circle of radius .
The angle 20°.
![\"N17/5/MATHL/HP2/ENG/TZ0/03\"](\"../assets/uploads/tinymce_asset/asset/4120/Schermafbeelding_2018-02-08_om_11.16.43.png\"/)
Find the area of the pendant.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\narea = (four sector areas radius 9) + (four sector areas radius 3) (M1)
\n(A1)(A1)
\n\n
A1
\nMETHOD 2
\narea =
\n(area of circle radius 3) + (four sector areas radius 9) – (four sector areas radius 3) (M1)
\n(A1)(A1)
\n\n
Note: Award A1 for the second term and A1 for the third term.
\n\n
\n
A1
\n\n
Note: Accept working in degrees.
\n\n
[4 marks]
\nConsider two consecutive positive integers, and .
\nShow that the difference of their squares is equal to the sum of the two integers.
\nattempt to subtract squares of integers (M1)
\n\n\n
EITHER
\ncorrect order of subtraction and correct expansion of , seen anywhere A1A1
\n\n\n
OR
\ncorrect order of subtraction and correct factorization of difference of squares A1A1
\n\n\n
THEN
\nA1
\n\n
Note: Do not award final A1 unless all previous working is correct.
\n\n
which is the sum of and AG
\n\n
Note: If expansion and order of subtraction are correct, award full marks for candidates who find the sum of the integers as and then show that the difference of the squares (subtracted in the correct order) is .
\n\n
[4 marks]
\nConsider the curve given by where .
\nShow that .
\nHence find the equation of the tangent to at the point where .
\nMETHOD 1
\nattempts to differentiate implicitly including at least one application of the product rule (M1)
\n\nA1
\n
Note: Award (M1)A1 for implicitly differentiating and obtaining .
\n\n
A1
\nAG
\n\n
METHOD 2
\n\nattempts to differentiate implicitly including at least one application of the product rule (M1)
\nA1
\nor equivalent to the above, for example
\n\nA1
\nor equivalent to the above, for example
\n\nAG
\n\n
METHOD 3
\nattempt to differentiate implicitly including at least one application of the product rule M1
\n\nA1
\nA1
\nAG
\n\n
METHOD 4
\nlets and attempts to find where M1
\nA1
\nA1
\n\nAG
\n\n
[3 marks]
\nMETHOD 1
\nsubstitutes into (M1)
\nA1
\nsubstitutes and their non-zero value of into (M1)
\nA1
\nequation of the tangent is A1
\n\n
METHOD 2
\nsubstitutes into (M1)
\n\n
EITHER
correctly substitutes into A1
\nA1
\n
OR
correctly substitutes into A1
\nA1
\n
THEN
substitutes into (M1)
\n\nequation of the tangent is A1
\n\n
[5 marks]
\nThe following table shows the systolic blood pressures, mmHg, and the ages, years, of male patients at a medical clinic.
\nThe relationship between and can be modelled by the regression line of on with equation .
\nA ‐year‐old male patient enters the medical clinic for his appointment.
\nDetermine the value of Pearson’s product‐moment correlation coefficient, , for these data.
\nInterpret, in context, the value of found in part (a) (i).
\nFind the equation of the regression line of on .
\nUse the regression equation from part (b) to predict this patient’s systolic blood pressure.
\nA ‐year‐old male patient enters the medical clinic for his appointment.
\nExplain why the regression equation from part (b) should not be used to predict this patient’s systolic blood pressure.
\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.
\nA2
\n\n
[2 marks]
\nthe value of shows a (very) strong positive correlation between age and (systolic) blood pressure A1
\n\n
[1 mark]
\nA1A1
\n\n
Note: Only award marks for an equation. Award A1 for and A1 for . Award A1A0 for .
\n\n
[2 marks]
\n(mmHg) (M1)A1
\n\n
[2 marks]
\nthe regression equation should not be used because it involves extrapolation A1
\n\n
[1 mark]
\nLet , for . The point lies on the graph of .
\nLet . The point lies on the graph of and is the reflection of point in the line .
\nThe line is tangent to the graph of at .
\nWrite down the coordinates of .
\nGiven that , find the equation of in terms of , and .
\nThe line is tangent to the graph of at and has equation .
\nThe line passes through the point .
\nThe gradient of the normal to at is .
\n\n
Find the equation of in terms of .
\n(accept ) A2 N2
\n[2 marks]
\nNote: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may work with the equation of the line before finding .
\n\n
FINDING
\nvalid attempt to find an expression for in terms of (M1)
\n\n
(A1)
\n\n
FINDING THE EQUATION OF
\nEITHER
\nattempt to substitute tangent gradient and coordinates into equation of straight line (M1)
\neg
\ncorrect equation in terms of and (A1)
\neg
\nOR
\nattempt to substitute tangent gradient and coordinates to find
\neg
\n(A1)
\nTHEN (must be in terms of both and )
\nA1 N3
\nNote: Award A0 for final answers in the form
\n\n
[5 marks]
\nNote: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may find in terms of before finding a value for .
\n\n
FINDING
\nvalid approach to find the gradient of the tangent (M1)
\neg
\ncorrect application of log rule (seen anywhere) (A1)
\neg
\ncorrect equation (seen anywhere) A1
\neg
\n\n
FINDING
\ncorrect substitution of into equation (A1)
\neg
\n(seen anywhere) A1
\n\n
FINDING
\ncorrect substitution of their and into their (A1)
\neg
\nA1 N2
\n\n
Note: Award A0 for final answers in the form .
\n\n
[7 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.
M1A1A1
\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]
\nIn a class of students, are fluent in Spanish, are fluent in French, and are not fluent in either of these languages. The following Venn diagram shows the events “fluent in Spanish” and “fluent in French”.
\nThe values , , and represent numbers of students.
\nWrite down the value of .
\nFind the value of .
\nWrite down the value of and of .
\nA1 N1
\n[1 mark]
\nvalid approach (M1)
\neg , ,
\nA1 N2
\n[2 marks]
\nvalid approach for finding or (may be seen in part (b)) (M1)
\neg ,
\n, A1A1 N3
\n[3 marks]
\nShow that the equation may be written in the form .
\nHence, solve the equation .
\nMETHOD 1
\ncorrect substitution of A1
\n\nAG
\n\n
METHOD 2
\ncorrect substitution using double-angle identities A1
\n\n\nAG
\n\n
[1 mark]
\nEITHER
\nattempting to factorise M1
\nA1
\n\n
OR
\nattempting to use the quadratic formula M1
\nA1
\n\n
THEN
\n(A1)
\nA1A1
\n\n
[5 marks]
\nIn the expansion of , where , the coefficient of the term in is .
\nFind the possible values of .
\nEITHER
\nattempt to use the binomial expansion of (M1)
\n(or )
\nidentifying the correct term (or ) (A1)
\n\n
OR
\nattempt to use the general term (or ) (M1)
\n(or ) (A1)
\n\n
THEN
\n(or (seen anywhere) (A1)
\nA1
\nA1
\n\n
Note: If working shown, award M1A1A1A1A0 for .
\n\n
[5 marks]
\nLet . The point lies on the graph of .
\nFind the value of .
\nThe graph of is transformed to obtain the graph of .
\nDescribe this transformation.
\nvalid attempt to substitute coordinates (M1)
\neg
\ncorrect substitution (A1)
\neg ,
\nA1 N2
\n[3 marks]
\nvalid attempt to solve (M1)
\neg , ,
\ncorrect working A1
\neg , ,
\ntranslation or shift (do not accept move) of vector (accept left by and up by ) A1A1 N2
\n[4 marks]
\nConsider the function defined by for .
\nThe following diagram shows part of the graph of which crosses the -axis at point , with coordinates . The line is the tangent to the graph of at the point .
\nFind the exact value of .
\nGiven that the gradient of is , find the -coordinate of .
\n(M1)
\n\nOR (A1)
\nA1
\n\n
[3 marks]
\nattempt to differentiate (must include and/or ) (M1)
\nA1
\nsetting their derivative M1
\n\nOR (or equivalent) A1
\nvalid attempt to solve their quadratic (M1)
\nA1
\n\n
Note: Award A0 if the candidate’s final answer includes additional solutions (such as ).
\n\n
[6 marks]
\nThe following diagram shows triangle , with , and .
\nGiven that , find the area of the triangle.
\nGive your answer in the form where .
\nMETHOD 1
\nattempt to use the cosine rule to find the value of (M1)
\nA1
\n\nOR A1
\nattempt to find (seen anywhere) (M1)
\nOR or right triangle with side and hypotenuse
\n(A1)
\n
Note: The marks for finding may be awarded independently of the first three marks for finding .
correct substitution into the area formula using their value of (or ) and their value of (M1)
or
\nA1
\n\n
METHOD 2
\nattempt to find the height, , of the triangle in terms of (M1)
\nOR OR A1
\nequating their expressions for either or (M1)
\nOR (or equivalent) A1
\nOR A1
\ncorrect substitution into the area formula using their value of (or ) (M1)
\nOR
\nA1
\n\n
[7 marks]
\nThe quadratic equation , where , has real distinct roots.
\nFind the range of possible values 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.
\n\n
attempts to find an expression for the discriminant, , in terms of (M1)
\n(A1)
\n\n
Note: Award M1A1 for finding .
\n\n
attempts to solve for (M1)
\n\n
Note: Award M1 for attempting to solve for .
\n\n
A1A1
\n\n
Note: Award A1 for obtaining critical values and A1 for correct inequality signs.
\n\n
[5 marks]
\nThe height of water, in metres, in Dungeness harbour is modelled by the function , where is the number of hours after midnight, and and are constants, where and .
\nThe following graph shows the height of the water for hours, starting at midnight.
\nThe first high tide occurs at and the next high tide occurs hours later. Throughout the day, the height of the water fluctuates between and .
\nAll heights are given correct to one decimal place.
\nShow that .
\nFind the value of .
\nFind the value of .
\nFind the smallest possible value of .
\nFind the height of the water at .
\nDetermine the number of hours, over a 24-hour period, for which the tide is higher than metres.
\nA fisherman notes that the water height at nearby Folkestone harbour follows the same sinusoidal pattern as that of Dungeness harbour, with the exception that high tides (and low tides) occur minutes earlier than at Dungeness.
\nFind a suitable equation that may be used to model the tidal height of water at Folkestone harbour.
\nOR A1
\nAG
\n\n
[1 mark]
\nOR (M1)
\nA1
\n\n
[2 marks]
\nOR (M1)
\nA1
\n\n
[2 marks]
\nMETHOD 1
\nsubstituting and for example into their equation for (A1)
\n\nattempt to solve their equation (M1)
\nA1
\n\n
METHOD 2
\nusing horizontal translation of (M1)
\n(A1)
\nA1
\n\n
METHOD 3
\n(A1)
\nattempts to solve their for (M1)
\n\nA1
\n\n
[3 marks]
\nattempt to find when or , graphically or algebraically (M1)
\n\nA1
\n\n
[2 marks]
\nattempt to solve (M1)
\ntimes are and (A1)
\ntotal time is
\n\n(hours) A1
\n
Note: Accept .
\n
[3 marks]
\nMETHOD 1
\nsubstitutes and into their equation for and attempts to solve for (M1)
\n\nA1
\n\n
METHOD 2
uses their horizontal translation (M1)
A1
\n\n
[2 marks]
\nConsider the curves and for .
\nFind the -coordinates of the points of intersection of the two curves.
\nFind the area, , of the region enclosed by the two curves.
\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.
\nattempts to solve (M1)
\nA1A1
\n\n
Note: Award A1A0 if additional solutions outside the domain are given.
\n\n
[3 marks]
\n(or equivalent) (M1)(A1)
\n\n
Note: Award M1 for attempting to form an integrand involving “top curve” − “bottom curve”.
\n\n
so A2
\n\n
[4 marks]
\nA large school has students from Year 6 to Year 12.
\nA group of students in Year 12 were randomly selected and surveyed to find out how many hours per week they each spend doing homework. Their results are represented by the following cumulative frequency graph.
\nThis same information is represented by the following table.
\nThere are students in Year 12 at this school.
\nFind the median number of hours per week these Year 12 students spend doing homework.
\nGiven that of these Year 12 students spend more than hours per week doing homework, find the value of .
\nFind the value of and the value of .
\nEstimate the number of Year 12 students that spend more than hours each week doing homework.
\nExplain why this sampling method might not provide an accurate representation of the amount of time all of the students in the school spend doing homework.
\nSuggest a more appropriate sampling method.
\nevidence of median position (M1)
\nstudents
\nmedian (hours) A1
\n\n
[2 marks]
\nrecognizing there are students in the top (M1)
\nstudents spent less than hours (A1)
\n(hours) A1
\n\n
[3 marks]
\nhours is students OR (M1)
\nA1
\nhours is students OR OR (A1)
\nA1
\n\n
[4 marks]
\nof the students OR spend more than hours doing homework (A1)
\nOR OR (A1)
\n(students) A1
\n\n
[3 marks]
\nonly year 12 students surveyed OR amount of homework might be different for different year levels R1
\n\n
[1 mark]
\nstratified sampling OR survey students in all years R1
\n\n
[1 mark]
\nConsider .
\nFind the value of .
\nHence or otherwise find the coefficient of the term in in the expansion of .
\nvalid approach (M1)
\neg ,
\nA1 N2
\n[2 marks]
\nvalid approach for expansion using (M1)
\neg ,
\nevidence of choosing correct term A1
\neg , ,
\ncorrect working for binomial coefficient (seen anywhere, do not accept factorials) A1
\neg , , , ,
\nA1 N2
\nNote: If there is clear evidence of adding instead of multiplying, award A1 for the correct working for binomial coefficient, but no other marks. For example, would earn M0A0A1A0.
\nDo not award final A1 for a final answer of , even if is seen previously. If no working shown, award N1 for .
\n[4 marks]
\nHelen and Jane both commence new jobs each starting on an annual salary of . At the start of each new year, Helen receives an annual salary increase of .
\nLet represent Helen’s annual salary at the start of her th year of employment.
\nAt the start of each new year, Jane receives an annual salary increase of of her previous year’s annual salary.
\nJane’s annual salary, , at the start of her th year of employment is given by .
\nAt the start of year , Jane’s annual salary exceeds Helen’s annual salary for the first time.
\nShow that .
\nGiven that follows a geometric sequence, state the value of the common ratio, .
\nFind the value of .
\nFor the value of found in part (c) (i), state Helen’s annual salary and Jane’s annual salary, correct to the nearest dollar.
\nFind Jane’s total earnings at the start of her th year of employment. Give your answer correct to the nearest dollar.
\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.
\nuses with and (M1)
\nA1
\nso AG
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nevidence of use of an appropriate table or graph or GDC numerical solve feature to find the value of such that (M1)
\n\n
EITHER
\nfor example, an excerpt from an appropriate table
\n (A1)
\n
OR
\nfor example, use of a GDC numerical solve feature to obtain (A1)
\n\n
Note: Award A1 for an appropriate graph. Condone use of a continuous graph.
\n\n
THEN
\nA1
\n\n
[3 marks]
\nA1
\nA1
\nHelen’s annual salary is and Jane’s annual salary is
\n\n
Note: Award A1 for a correct value and A1 for a correct value seen in part (c) (i).
\n\n
[2 marks]
\nat the start of the th year, Jane will have worked for years so the value of is required R1
\n\n
Note: Award R1 if is seen anywhere.
\n\n
uses with , and (M1)
\n\n
Note: Award M1 if is used.
\n\n
(A1)
\n\nJane’s total earnings are (correct to the nearest dollar)
\n\n
[4 marks]
\nConsider the function , with derivative where .
\nShow that the discriminant of is .
\nGiven that is an increasing function, find all possible values of .
\ncorrect substitution into (A1)
\neg ,
\ncorrect expansion of each term A1
\neg ,
\nAG N0
\n[2 marks]
\nvalid approach M1
\neg ,
\nrecognizing discriminant or M1
\neg , ,
\ntwo correct values for /endpoints (even if inequalities are incorrect) (A1)
\neg , and ,
\ncorrect interval A1 N2
\neg ,
\nNote: Candidates may work with an equation, then write the intervals with inequalities at the end. If inequalities are not seen until the candidate’s final correct answer, M0M0A1A1 may be awarded.
If candidate is working with incorrect inequalitie(s) at the beginning, then gets the correct final answer, award M0M0A1A0 or M1M0A1A0 or M0M1A1A0 in line with the markscheme.
[4 marks]
\nConsider the function defined by , for .
\nThe following diagram shows the graph of .
\nThe graph of touches the -axis at points and , as shown. The shaded region is enclosed by the graph of and the -axis, between the points and .
\nThe right cone in the following diagram has a total surface area of , equal to the shaded area in the previous diagram.
\nThe cone has a base radius of , height , and slant height .
\nFind the -coordinates of and .
\nShow that the area of the shaded region is .
\nFind the value of .
\nHence, find the volume of the cone.
\n(or setting their ) (M1)
\n(or )
\nA1A1
\n\n
[3 marks]
\nattempt to integrate (M1)
\nA1A1
\nsubstitute their limits into their integrated expression and subtract (M1)
\n\nA1
\narea AG
\n\n
[5 marks]
\nattempt to substitute into formula for surface area (including base) (M1)
\n(A1)
\n\n\nA1
\n\n
[3 marks]
\nvalid attempt to find the height of the cone (M1)
\ne.g.
\n(A1)
\nattempt to use with their values substituted M1
\n\nA1
\n\n
[4 marks]
\nConsider the function .
\nFind the coordinates where the graph of crosses the
\n-axis.
\n-axis.
\nWrite down the equation of the vertical asymptote of the graph of .
\nThe oblique asymptote of the graph of can be written as where .
\nFind the value of and the value of .
\nSketch the graph of for , clearly indicating the points of intersection with each axis and any asymptotes.
\nExpress in partial fractions.
\nHence find the exact value of , expressing your answer as a single logarithm.
\nNote: In part (a), penalise once only, if correct values are given instead of correct coordinates.
\n
attempts to solve (M1)
and A1
\n\n
[2 marks]
\nNote: In part (a), penalise once only, if correct values are given instead of correct coordinates.
\n\n
A1
\n\n
[1 mark]
\nA1
\n
Note: Award A0 for .
Award A1 in part (b), if is seen on their graph in part (d).
[1 mark]
\nMETHOD 1
\n\nattempts to expand (M1)
\n\nA1
\nequates coefficients of (M1)
\n\nA1
\n\n\n
METHOD 2
\nattempts division on M1
\nM1
\nA1
\nA1
\n\n\n
METHOD 3
\nA1
\nM1
\n\nequates coefficients of : (M1)
\n\nA1
\n\n\n
METHOD 4
\nattempts division on M1
\n\nA1
\nM1
\nA1
\n\n\n
[4 marks]
\n
two branches with approximately correct shape (for ) A1
\ntheir vertical and oblique asymptotes in approximately correct positions with both branches showing correct asymptotic behaviour to these asymptotes A1
\ntheir axes intercepts in approximately the correct positions A1
\n
Note: Points of intersection with the axes and the equations of asymptotes are not required to be labelled.
\n
[3 marks]
\nattempts to split into partial fractions: (M1)
\n\n\nA1
\nA1
\n\n\n
[3 marks]
\nattempts to integrate and obtains two terms involving ‘ln’ (M1)
\nA1
\nA1
\n\nA1
\n
Note: The final A1 is dependent on the previous two A marks.
\n
[4 marks]
\nLet , for . Find the values of for which .
\nMETHOD 1 – FINDING INTERVALS FOR x
\n\n
correct working (A1)
\neg ,
\nrecognizing (A1)
\none additional correct value for (ignoring domain and equation/inequalities) (A1)
\neg
\nthree correct values for A1A1
\neg
\nvalid approach to find intervals (M1)
\neg
correct intervals (must be in radians) A1A1 N2
\n,
\nNote: If working shown, award A1A0 if inclusion/exclusion of endpoints is incorrect. If no working shown award N1.
If working shown, award A1A0 if both correct intervals are given, and additional intervals are given. If no working shown award N1.
Award A0A0 if inclusion/exclusion of endpoints are incorrect and additional intervals are given.
\n
METHOD 2 – FINDING INTERVALS FOR
\n\n
correct working (A1)
\neg ,
\nrecognizing (A1)
\none additional correct value for (ignoring domain and equation/inequalities) (A1)
\neg
\nthree correct values for A1
\neg
\nvalid approach to find intervals (M1)
\neg
one correct interval for A1
\neg
\ncorrect intervals (must be in radians) A1A1 N2
\n,
\nNote: If working shown, award A1A0 if inclusion/exclusion of endpoints is incorrect. If no working shown award N1.
If working shown, award A1A0 if both correct intervals are given, and additional intervals are given. If no working shown award N1.
Award A0A0 if inclusion/exclusion of endpoints are incorrect and additional intervals are given.
\n
[8 marks]
\nLet and be normally distributed with and , .
\nFind so that .
\nIt is given that .
\nFind .
\nMETHOD 1
\nrecognizing that is midway between the means of and . (M1)
\neg ,
A1 N2
\n\n
METHOD 2
\nvalid attempt to compare distributions (M1)
\neg
\nA1 N2
\n\n
[2 marks]
\nvalid attempt to compare distributions (seen anywhere) (M1)
\neg is a horizontal translation of of units to the right,
\n\n
valid approach using symmetry (M1)
\neg
\ncorrect working (A1)
\neg
\nA1 N3
\n[4 marks]
\nParticle A travels in a straight line such that its displacement, metres, from a fixed origin after seconds is given by , for , as shown in the following diagram.
\nParticle A starts at the origin and passes through the origin again when .
\nParticle A changes direction when .
\nThe total distance travelled by particle A is given by .
\nFind the value of .
\nFind the value of .
\nFind the displacement of particle A from the origin when .
\nFind the distance of particle A from the origin when .
\nFind the value of .
\nA second particle, particle B, travels along the same straight line such that its velocity is given by , for .
\nWhen , the distance travelled by particle B is equal to .
\nFind the value of .
\nsetting (M1)
\n\n\n(accept ) A1
\n\n
Note: Award A0 if the candidate’s final answer includes additional solutions (such as ).
\n\n
[2 marks]
\nrecognition that when particle changes direction OR local maximum on graph of OR vertex of parabola (M1)
\n(accept ) A1
\n\n
[2 marks]
\nsubstituting their value of into OR integrating from to (M1)
\nA1
\n\n
[2 marks]
\nOR OR integrating from to (M1)
\nA1
\n\n
[2 marks]
\nforward backward OR OR (M1)
\nA1
\n\n
[2 marks]
\nMETHOD 1
\ngraphical method with triangles on graph M1
\n(A1)
\n(A1)
\nA1
\n\n
METHOD 2
\nrecognition that distance M1
\n\n(A1)
\n(A1)
\nA1
\n\n
[4 marks]
\nA small cuboid box has a rectangular base of length cm and width cm, where . The height is cm, where .
\nThe sum of the length, width and height is cm.
\nThe volume of the box is cm3.
\nWrite down an expression for in terms of .
\nFind an expression for in terms of .
\nFind .
\nFind the value of for which is a maximum.
\nJustify your answer.
\nFind the maximum volume.
\nA1 N1
\n[1 mark]
\ncorrect substitution into volume formula (A1)
\neg
\nA1 N2
\nNote: Award A0 for unfinished answers such as .
\n[2 marks]
\nA1A1 N2
\nNote: Award A1 for and A1 for .
\n[2 marks]
\nvalid approach to find maximum (M1)
\neg
\ncorrect working (A1)
\neg
\nA2 N2
\nNote: Award A1 for and .
\n[4 marks]
\nvalid approach to explain that is maximum when (M1)
\neg attempt to find , sign chart (must be labelled )
\ncorrect value/s A1
\neg , where and where
\ncorrect reasoning R1
\neg , is positive for and negative for
\nNote: Do not award R1 unless A1 has been awarded.
\nis maximum when AG N0
\n[3 marks]
\ncorrect substitution into their expression for volume A1
\neg ,
\n(cm3) A1 N1
\n[2 marks]
\nThe time, minutes, taken to complete a jigsaw puzzle can be modelled by a normal distribution with mean and standard deviation .
\nIt is found that of times taken to complete the jigsaw puzzle are longer than minutes.
\nUse in the remainder of the question.
\nSix randomly chosen people complete the jigsaw puzzle.
\nBy stating and solving an appropriate equation, show, correct to two decimal places, that .
\nFind the th percentile time to complete the jigsaw puzzle.
\nFind the probability that a randomly chosen person will take more than minutes to complete the jigsaw puzzle.
\nFind the probability that at least five of them will take more than minutes to complete the jigsaw puzzle.
\nHaving spent minutes attempting the jigsaw puzzle, a randomly chosen person had not yet completed the puzzle.
\nFind the probability that this person will take more than minutes to complete the jigsaw puzzle.
\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)
\nstates a correct equation, for example, A1
\nattempts to solve their equation (M1)
\nA1
\nthe solution to the equation is , correct to two decimal places AG
\n\n
[4 marks]
\nlet be the th percentile
\nattempts to use the inverse normal feature of a GDC to find (M1)
\n(mins) A1
\n\n
[2 marks]
\nevidence of identifying the correct area under the normal curve (M1)
\nNote: Award M1 for a clearly labelled sketch.
\nA1
\n\n
[2 marks]
\nlet represent the number of people out of the six who take more than minutes to complete the jigsaw puzzle
\n(M1)
\nfor example, or (A1)
\nA1
\n\n
[3 marks]
\nrecognizes that is required (M1)
\n\n
Note: Award M1 for recognizing conditional probability.
\n\n
(A1)
\nM1
\nA1
\n\n
[4 marks]
\nThree points and lie on the plane .
\nPlane has equation .
\nThe plane is given by . The line and the plane intersect at the point .
\nThe point lies on .
\nFind the vector and the vector .
\nHence find the equation of , expressing your answer in the form , where .
\nThe line is the intersection of and . Verify that the vector equation of can be written as .
\nShow that at the point .
\nHence find the coordinates of .
\nFind the reflection of the point in the plane .
\nHence find the vector equation of the line formed when is reflected in the plane .
\nattempts to find either or (M1)
\nand A1
\n\n
[2 marks]
\nMETHOD 1
\nattempts to find (M1)
\nA1
\n
EITHER
equation of plane is of the form (A1)
\nsubstitutes a valid point e.g to obtain a value of M1
\n\n
OR
attempts to use (M1)
\nA1
\n\n
THEN
A1
\n\n
METHOD 2
\nequation of plane is of the form A1
\nattempts to form equations for in terms of their parameters (M1)
\nA1
\neliminates at least one of their parameters (M1)
\nfor example,
\nA1
\n\n
[5 marks]
\nMETHOD 1
\nsubstitutes into their and (given) (M1)
\nand A1
\n
Note: Award (M1)A0 for correct verification using a specific value of .
so the vector equation of can be written as AG
\n
METHOD 2
EITHER
\nattempts to find M1
\n\n
OR
and M1
\n
THEN
substitutes into and
\nand A1
\nso the vector equation of can be written as AG
\n\n
METHOD 3
\nattempts to solve and (M1)
\nfor example, A1
\n
Note: Award A1 for substituting (or or ) into and and solving simultaneously. For example, solving and to obtain and .
so the vector equation of can be written as AG
\n\n
[2 marks]
\nsubstitutes the equation of into the equation of (M1)
\nA1
\nAG
\n\n
[2 marks]
\nhas coordinates A1
\n\n
[1 mark]
\nnormal to is (A1)
\n
Note: May be seen or used anywhere.
considers the line normal to passing through (M1)
A1
EITHER
finding the point on the normal line that intersects
attempts to solve simultaneously with plane (M1)
A1
\npoint is
\n
OR
(M1)
\n\nA1
\n
OR
attempts to find the equation of the plane parallel to containing and solve simultaneously with (M1)
\n\nA1
\n
THEN
so, another point on the reflected line is given by
\n(A1)
\nA1
\n\n
[7 marks]
\nEITHER
\nattempts to find the direction vector of the reflected line using their and (M1)
\n\n
OR
attempts to find their direction vector of the reflected line using a vector approach (M1)
\n\n
THEN
(or equivalent) A1
\n
Note: Award A0 for either '' or '' not stated. Award A0 for ''
\n
[2 marks]
\nSolve the equation .
\nattempt to use M1
\nA1
\n\n
EITHER
\nattempting to factorise M1
\nA1
\n\n
OR
\nattempting to use the quadratic formula M1
\nA1
\n\n
THEN
\n(A1)
\nA1A1
\n\n
[7 marks]
\nGiven any two non-zero vectors, and , show that .
\nMETHOD 1
\nuse of on the LHS (M1)
\nA1
\nM1
\nOR A1
\nAG
\n\n
METHOD 2
\nuse of on the RHS (M1)
\nA1
\nM1
\nOR A1
\nAG
\n\n
Note: If candidates attempt this question using cartesian vectors, e.g
\nand ,
\naward full marks if fully developed solutions are seen.
Otherwise award no marks.
\n
[4 marks]
\nThe temperature of water minutes after being poured into a cup can be modelled by where and are positive constants.
\nThe water is initially boiling at . When , the temperature of the water is .
\nShow that .
\nShow that .
\nFind the temperature of the water when .
\nSketch the graph of versus , clearly indicating any asymptotes with their equations and stating the coordinates of any points of intersection with the axes.
\nFind the time taken for the water to have a temperature of . Give your answer correct to the nearest second.
\nThe model for the temperature of the water can also be expressed in the form for and is a positive constant.
\nFind the exact value 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.
\nwhen A1
\nso AG
\n\n
[1 mark]
\ncorrect substitution of M1
\nor
\n\n
EITHER
\nA1
\nor A1
\n\n
OR
\nA1
\nA1
\n\n
THEN
\nAG
\n\n
[3 marks]
\nsubstitutes into (M1)
\nA1
\n\n
[2 marks]
\na decreasing exponential A1
\nstarting at labelled on the graph or stated A1
\nas A1
\nhorizontal asymptote labelled on the graph or stated A1
\n\n
Note: Award A0 for stating as the horizontal asymptote.
\n\n
[4 marks]
\nwhere A1
\n\n
EITHER
\nuses an appropriate graph to attempt to solve for (M1)
\n\n
OR
\nmanipulates logs to attempt to solve for e.g. (M1)
\nA1
\n\n
THEN
\ntemperature will be after minutes and seconds A1
\n\n
[4 marks]
\nMETHOD 1
\nsubstitutes , and into (M1)
\nA1
\nA1
\n\n
METHOD 2
\nwhere
\n\n
EITHER
\n(M1)
\n\n
OR
\n(M1)
\n\n
THEN
\nA1
\nA1
\n\n
[3 marks]
\nThe cubic equation where has roots and .
\nGiven that , find the value of .
\n(A1)
\n\n(A1)
\nM1
\nattempting to solve (or equivalent) for (M1)
\nA1
\n\n
Note: Award A0 for .
\n\n
[5 marks]
\nThe lines and have the following vector equations where .
\n\n\nShow that and do not intersect.
\nFind the minimum distance between and .
\nMETHOD 1
\nsetting at least two components of and equal M1
\n\n\n\nattempt to solve two of the equations eg. adding and M1
\ngives a contradiction (no solution), eg R1
\nso and do not intersect AG
\n\n
Note: For an error within the equations award M0M1R0.
Note: The contradiction must be correct to award the R1.
\n
METHOD 2
\nand are parallel, so and are either identical or distinct. R1
\nAttempt to subtract two position vectors from each line,
\ne.g. M1
\nA1
\n\n
[3 marks]
\nMETHOD 1
\nand are parallel (as is a multiple of )
\nlet be on and let be on
\nAttempt to find vector (M1)
\nDistance required is M1
\n(A1)
\nA1
\nminimum distance is A1
\n\n
METHOD 2
\nand are parallel (as is a multiple of )
\nlet be a fixed point on eg and let be a general point on
\nattempt to find vector (M1)
\nA1
\nM1
\n
EITHER
null A1
\n
OR
to obtain A1
\n
THEN
minimum distance is A1
\n\n
METHOD 3
\nlet be on and let be on (M1)
\n(or let be on and let be on )
\n(or ) A1
\n(or ) M1
\nor A1
\nminimum distance is A1
\n\n
[5 marks]
\nThe points and have position vectors and respectively.
\nPoint has position vector . Let be the origin.
\nFind, in terms of ,
\n.
\n.
\nGiven that , show that .
\nCalculate the area of triangle .
\ncorrect substitution into either or into (in (ii)) (A1)
\neg ,
\ncorrect expression A1 N1
\neg ,
\n[2 marks]
\ncorrect expression A1 N1
\neg ,
\n[1 mark]
\nfinding magnitudes (seen anywhere) A1A1
\neg ,
\ncorrect substitution of their values into formula for angle (A1)
\neg
\ncorrect substitution of their values into formula for angle (A1)
\neg
\nrecognizing that (seen anywhere) (M1)
\neg ,
\ncorrect working (without radicals) (A2)
\neg ,
\ncorrect working clearly leading to the required answer A1
\neg , , and ,
\nAG N0
\n[8 marks]
\nfinding magnitude of (seen anywhere) A1
\neg ,
\nvalid attempt to find (M1)
\neg , ,
\nfinding A1
\neg
\nvalid approach to find (seen anywhere) (M1)
\neg , , , ,
\ncorrect substitution of their values into (A1)
\neg ,
\narea is A1 N3
\n[6 marks]
\nIn this question you will explore some of the properties of special functions and and their relationship with the trigonometric functions, sine and cosine.
\n
Functions and are defined as and , where .
Consider and , such that .
\nUsing , find expressions, in terms of and , for
\nThe functions and are known as circular functions as the general point () defines points on the unit circle with equation .
\nThe functions and are known as hyperbolic functions, as the general point ( ) defines points on a curve known as a hyperbola with equation . This hyperbola has two asymptotes.
\nVerify that satisfies the differential equation .
\nShow that .
\n.
\n.
\nHence find, and simplify, an expression for .
\nShow that .
\nSketch the graph of , stating the coordinates of any axis intercepts and the equation of each asymptote.
\nThe hyperbola with equation can be rotated to coincide with the curve defined by .
\nFind the possible values of .
\nA1
\nA1
\nAG
\n
[2 marks]
METHOD 1
\n\nsubstituting and M1
\n\n(M1)
\nA1
\nAG
\n\n
METHOD 2
\n\nM1
\nM1A1
\nAG
\n
Note: Accept combinations of METHODS 1 & 2 that meet at equivalent expressions.
[3 marks]
substituting into the expression for (M1)
\nobtaining (A1)
\n\n
Note: The M1 can be awarded for the use of sine and cosine being odd and even respectively.
A1
\n
[3 marks]
substituting and attempt to simplify (M1)
\n\nA1
\n
[2 marks]
METHOD 1
\n\nsubstituting expressions found in part (c) (M1)
\nA1
\n\n
METHOD 2
\n\nM1
\nA1
\n
Note: Accept equivalent final answers that have been simplified removing all imaginary parts eg etc
[2 marks]
M1
\nA1
\nA1
\n
Note: Award A1 for a value of 1 obtained from either LHS or RHS of given expression.
M1
(hence ) AG
\n
Note: Award full marks for showing that .
[4 marks]
A1A1A1A1
Note: Award A1 for correct curves in the upper quadrants, A1 for correct curves in the lower quadrants, A1 for correct -intercepts of and (condone and ), A1 for and .
[4 marks]
attempt to rotate by in either direction (M1)
\n
Note: Evidence of an attempt to relate to a sketch of would be sufficient for this (M1).
attempting to rotate a particular point, eg (M1)
rotates to (or similar) (A1)
\nhence A1A1
\n
[5 marks]
The acceleration, , of a particle moving in a horizontal line at time seconds, , is given by where is the particle’s velocity and .
\nAt , the particle is at a fixed origin and has initial velocity .
\nInitially at , the particle moves in the positive direction until it reaches its maximum displacement from . The particle then returns to .
\nLet metres represent the particle’s displacement from and its maximum displacement from .
\nLet represent the particle’s velocity seconds before it reaches , where
\n.
\nSimilarly, let represent the particle’s velocity seconds after it reaches .
\nBy solving an appropriate differential equation, show that the particle’s velocity at time is given by .
\nShow that the time taken for the particle to reach satisfies the equation .
\nBy solving an appropriate differential equation and using the result from part (b) (i), find an expression for in terms of .
\nBy using the result to part (b) (i), show that .
\nDeduce a similar expression for in terms of .
\nHence, show that .
\n(A1)
\n(or equivalent / use of integrating factor) M1
\nA1
\n\n
EITHER
\nattempt to find with initial conditions M1
\n\n\nA1
\n\nA1
\nAG
\n\n
OR
\n\nAttempt to find with initial conditions M1
\n\n\nA1
\n\nA1
\nAG
\n\n
OR
\nA1
\n\nAttempt to find with initial conditions M1
\n\nA1
\nAG
\n\n
Note: condone use of modulus within the ln function(s)
\n\n
[6 marks]
\nrecognition that when M1
\nA1
\nAG
\n\n
Note: Award M1A0 for substituting into and showing that .
\n\n
[6 marks]
\n(M1)
\nA1
\n( so) A1
\n\nat
\nSubstituting into M1
\nA1
\n\n\n
[5 marks]
\nMETHOD 1
\n(M1)
\nA1
\nAG
\n\n
METHOD 2
\n\nM1
\nA1
\nAG
\n\n
[2 marks]
\nMETHOD 1
\n(A1)
\nA1
\n\n
METHOD 2
\n(A1)
\n\n\n
A1
\n\n
[2 marks]
\nMETHOD 1
\nA1
\nattempt to express as a square M1
\nA1
\nso AG
\n\n
METHOD 2
\nA1
\nAttempt to solve M1
\nminimum value of , (when ), hence R1
\nso AG
\n\n
[3 marks]
\nThe number of messages, , that six randomly selected teenagers sent during the month of October is shown in the following table. The table also shows the time, , that they spent talking on their phone during the same month.
\nThe relationship between the variables can be modelled by the regression equation .
\nWrite down the value of and of .
\nUse your regression equation to predict the number of messages sent by a teenager that spent minutes talking on their phone in October.
\nevidence of set up (M1)
\neg correct value for or (accept )
\n,
\n, (accept ) A1A1 N3
\n[3 marks]
\nvalid approach (M1)
\neg
\neg ( from 3 sf values) (A1)
\nnumber of messages (must be an integer) A1 N3
\n[3 marks]
\nA rocket is travelling in a straight line, with an initial velocity of m s−1. It accelerates to a new velocity of m s−1 in two stages.
\nDuring the first stage its acceleration, m s−2, after seconds is given by , where .
\nThe first stage continues for seconds until the velocity of the rocket reaches m s−1.
\nFind an expression for the velocity, m s−1, of the rocket during the first stage.
\nFind the distance that the rocket travels during the first stage.
\nDuring the second stage, the rocket accelerates at a constant rate. The distance which the rocket travels during the second stage is the same as the distance it travels during the first stage.
\nFind the total time taken for the two stages.
\nrecognizing that (M1)
\ncorrect integration A1
\neg
\nattempt to find using their (M1)
\neg
\nA1 N3
\n[4 marks]
\nevidence of valid approach to find time taken in first stage (M1)
\neg graph,
\nA1
\nattempt to substitute their and/or their limits into distance formula (M1)
\neg , ,
\n\n
distance is (m) A1 N3
\n[4 marks]
\nrecognizing velocity of second stage is linear (seen anywhere) R1
\neg graph, ,
\nvalid approach (M1)
\neg
\ncorrect equation (A1)
\neg
\ntime for stage two ( from 3 sf) A2
\n( from 3 sf)
\nseconds ( from 3 sf) A1 N3
\n[6 marks]
\nThe curve has equation .
\nShow that .
\nThe tangent to at the point Ρ is parallel to the -axis.
\nFind the -coordinate of Ρ.
\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.
\nattempts implicit differentiation on both sides of the equation M1
\nA1
\nA1
\nso AG
\n\n
[3 marks]
\nattempts to solve for (M1)
\nA1
\nattempts to solve for given their value of (M1)
\nA1
\n\n
[4 marks]
\nConsider the lines and with respective equations
\nand .
\nA third line, , has gradient .
\nFind the point of intersection of and .
\nWrite down a direction vector for .
\npasses through the intersection of and .
\nWrite down a vector equation for .
\nvalid approach (M1)
\neg , ,
\n(exact) A1 N2
\n[2 marks]
\n(or any multiple of ) A1 N1
\n[1 mark]
\nany correct equation in the form r = a + b (accept any parameter for ) where
a is a position vector for a point on , and b is a scalar multiple of A2 N2
eg r
\nNote: Award A1 for the form a + b, A1 for the form = a + b, A0 for the form r = b + a.
\n[2 marks]
\nIn this question you will be exploring the strategies required to solve a system of linear differential equations.
\n\n
Consider the system of linear differential equations of the form:
\nand ,
\nwhere and is a parameter.
\nFirst consider the case where .
\nNow consider the case where .
\nNow consider the case where .
\nFrom previous cases, we might conjecture that a solution to this differential equation is , and is a constant.
\nBy solving the differential equation , show that where is a constant.
\nShow that .
\nSolve the differential equation in part (a)(ii) to find as a function of .
\nBy differentiating with respect to , show that .
\nBy substituting , show that where is a constant.
\nHence find as a function of .
\nHence show that , where is a constant.
\nShow that .
\nFind the two values for that satisfy .
\nLet the two values found in part (c)(ii) be and .
\nVerify that is a solution to the differential equation in (c)(i),where is a constant.
\nMETHOD 1
\n\n(M1)
\nOR A1A1
\n
Note: Award A1 for and A1 for and .
AG
\n
METHOD 2
\nrearranging to AND multiplying by integrating factor M1
\nA1A1
\nAG
\n\n
[3 marks]
\nsubstituting into differential equation in M1
\n\nAG
\n\n
[1 mark]
\nintegrating factor (IF) is (M1)
\n(A1)
\n\n(A1)
\nA1
\n
Note: The first constant must be , and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.
\n
[4 marks]
\nA1
\n
EITHER
(M1)
\nA1
\n
OR
(M1)
\nA1
\n
THEN
AG
\n
[3 marks]
A1
\nM1
\nOR A1
\nAG
\n\n
[3 marks]
\nM1
\nA1
\n
Note: The first constant must be , and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.
\n
[2 marks]
\nMETHOD 1
\nsubstituting and their (iii) into M1(M1)
\nA1
\nAG
\nNote: Follow through from incorrect part (iii) cannot be awarded if it does not lead to the AG.
\n
METHOD 2
M1
\n\nA1
\n\nM1
\nAG
\n\n
[3 marks]
\nseen anywhere M1
\n\n
METHOD 1
\n\nattempt to eliminate M1
\n\nA1
\nAG
\n\n
METHOD 2
\nrewriting LHS in terms of and M1
\nA1
\nAG
\n\n
[3 marks]
\n(A1)
\n(M1)
\n(since ) A1
\nand are and (either order) A1
\n\n
[4 marks]
\nMETHOD 1
\n\n(A1)(A1)
\nM1
\nA1
\nAG
\n\n
METHOD 2
\n\n(A1)(A1)
\nM1
\nA1
\nAG
\n\n
[4 marks]
\nThe following diagram shows the graph of for , with asymptotes at and .
\nDescribe a sequence of transformations that transforms the graph of to the graph of for .
\nShow that where and .
\nVerify that for .
\nUsing mathematical induction and the result from part (b), prove that for .
\nEITHER
horizontal stretch/scaling with scale factor
Note: Do not allow ‘shrink’ or ‘compression’
followed by a horizontal translation/shift units to the left A2
Note: Do not allow ‘move’
OR
horizontal translation/shift unit to the left
\nfollowed by horizontal stretch/scaling with scale factor A2
\n
THEN
vertical translation/shift up by (or translation through A1
(may be seen anywhere)
\n
[3 marks]
\nlet and M1
\nand (A1)
\nA1
\nA1
\nso where and . AG
\n\n
[4 marks]
\nMETHOD 1
\n(or equivalent) A1
\nA1
\nA1
\nAG
\n\n
METHOD 2
\n(or equivalent) A1
\nConsider
\n\nA1
\nA1
\nAG
\n\n
METHOD 3
\n\n(or equivalent) A1
\nA1
\nA1
\n\n
[3 marks]
\nlet be the proposition that for
\nconsider
\nwhen and so is true R1
\nassume is true, ie. M1
\n\n
Note: Award M0 for statements such as “let ”.
Note: Subsequent marks after this M1 are independent of this mark and can be awarded.
\n
consider :
\n(M1)
\nA1
\nM1
\nA1
\n\n
Note: Award A1 for correct numerator, with factored. Denominator does not need to be simplified
\n\n
A1
\n\n
Note: Award A1 for denominator correctly expanded. Numerator does not need to be simplified. These two A marks may be awarded in any order
\n\n
A1
\n\n
Note: The word ‘arctan’ must be present to be able to award the last three A marks
\n\n
is true whenever is true and is true, so
\nis true for for R1
\n\n
Note: Award the final R1 mark provided at least four of the previous marks have been awarded.
Note: To award the final R1, the truth of must be mentioned. ‘ implies ’ is insufficient to award the mark.
\n
[9 marks]
\nConsider the identity , where .
\nFind the value of and the value of .
\nHence, expand in ascending powers of , up to and including the term in .
\nGive a reason why the series expansion found in part (b) is not valid 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.
\n\n\n
\n
EITHER
\nsubstitutes and attempts to solve for and substitutes and attempts to solve for (M1)
\n\n\n
OR
\nforms and and attempts to solve for and (M1)
\n\n
THEN
\nand A1A1
\n\n
[3 marks]
\nuses the binomial expansion on either or M1
\nA1
\nA1
\n\nso the expansion is (in ascending powers of ) A1
\n\n
[4 marks]
\n(is convergent) requires and is outside this so the expansion is not valid R1
\n\n
[1 mark]
\nLet , and .
\nFind .
\nLet be a point on the graph of . The tangent to the graph of at is parallel to the graph of .
\nFind the -coordinate of .
\nattempt to form composite (in any order) (M1)
\neg ,
\nA1 N2
\n[2 marks]
\nrecognizing that the gradient of the tangent is the derivative (M1)
\neg
\ncorrect derivative (seen anywhere) (A1)
\n\n
correct value for gradient of (seen anywhere) (A1)
\n,
\nsetting their derivative equal to (M1)
\n\n
\n
(exact), A1 N3
\n[5 marks]
\nProve by contradiction that is an irrational number.
\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.
\nassume there exist where such that M1A1
\n\n
Note: Award M1 for attempting to write the negation of the statement as an assumption. Award A1 for a correctly stated assumption.
\n\n
A1
\nA1
\n\n
EITHER
\nis a factor of but not a factor of R1
\n\n
OR
\nis a factor of but not a factor of R1
\n\n
OR
\nis odd and is even R1
\n\n
THEN
\nno (where ) satisfy the equation and this is a contradiction R1
\nso is an irrational number AG
\n\n
[6 marks]
\nAt a café, the waiting time between ordering and receiving a cup of coffee is dependent upon the number of customers who have already ordered their coffee and are waiting to receive it.
\nSarah, a regular customer, visited the café on five consecutive days. The following table shows the number of customers, , ahead of Sarah who have already ordered and are waiting to receive their coffee and Sarah’s waiting time, minutes.
\nThe relationship between and can be modelled by the regression line of on with equation .
\nFind the value of and the value of .
\nWrite down the value of Pearson’s product-moment correlation coefficient, .
\nInterpret, in context, the value of found in part (a)(i).
\nOn another day, Sarah visits the café to order a coffee. Seven customers have already ordered their coffee and are waiting to receive it.
\nUse the result from part (a)(i) to estimate Sarah’s waiting time to receive her coffee.
\nand
\nand A1A1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nrepresents the (average) increase in waiting time ( mins) per additional customer (waiting to receive their coffee) R1
\n\n
[1 mark]
\nattempt to substitute into their equation (M1)
\n\n(mins) A1
\n\n
[2 marks]
\nA biased coin is weighted such that the probability, , of obtaining a tail is . The coin is tossed repeatedly and independently until a tail is obtained.
\nLet be the event “obtaining the first tail on an even numbered toss”.
\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.
\nMETHOD 1
\nis the event “the first tail occurs on the nd, th, th, …, th toss”
\n(A1)
\n\n
Note: Award A1 for deducing that either head before a tail or heads before a tail or heads before a tail etc. is required. In other words, deduces heads before a tail.
\n\n
M1A1
\n\n
Note: Award M1 for attempting to form an infinite geometric series.
\nNote: Award A1 for .
\n\n
uses with and (M1)
\n\n
Note: Award M1 for using with and
\n\n
A1
\nA1
\n\n
METHOD 2
\nlet be the event “tail occurs on the first toss”
\nuses M1
\nconcludes that and so R1
\nA1
\n\n
Note: Award A1 for concluding: given that a tail is not obtained on the first toss, then is the probability that the first tail is obtained after a further odd number of tosses, .
\n\n\n
A1
\nattempts to solve for (M1)
\nA1
\n\n
[6 marks]
\nThe following diagram shows a right-angled triangle, , with , and .
\nThe points and lie on .
is perpendicular to .
is the arc of a circle, centred at .
The region is bounded by , and arc .
Find .
\nFind the area of .
\ncorrect working (A1)
\neg , ,
\n\n
(A1) N2
\n[2 marks]
\nNote: There may be slight differences in the final answer, depending on the approach the candidate uses in part (b). Accept a final answer that is consistent with their working.
\ncorrect area of sector (seen anywhere) (A1)
\neg , ,
\ncorrect expression (or value) for either or (seen anywhere) (A1)
\neg
\n\n
correct area of triangle (seen anywhere) (A1)
\neg , , (exact)
\nappropriate approach (seen anywhere) (M1)
\neg , their sector − their triangle
\n\n
area of shaded region (cm2) A1 N2
\n[5 marks]
\nThe points , , and are the vertices of a right-pyramid.
\nThe line passes through the point and is perpendicular to .
\nFind the vectors and .
\nUse a vector method to show that .
\nShow that the Cartesian equation of the plane that contains the triangle is .
\nFind a vector equation of the line .
\nHence determine the minimum distance, , from to .
\nFind the volume of right-pyramid .
\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
\nA1
\n\n
[2 marks]
\nattempts to use (M1)
\nA1
\nA1
\nso AG
\n\n
[3 marks]
\nattempts to find a vector normal to M1
\nfor example, leading to A1
\na vector normal to is
\n\n
EITHER
\nsubstitutes (or or ) into and attempts to find the value of
\nfor example, M1
\n\n
OR
\nattempts to use M1
\nfor example,
\n\n
THEN
\nleading to the Cartesian equation of as AG
\n\n
[3 marks]
\nA1
\n\n
[1 mark]
\nsubstitutes into (M1)
\n\nA1
\nshows a correct calculation for finding , for example, attempts to find
\nM1
\nA1
\n\n
[4 marks]
\nlet the area of triangle be
\n\n
EITHER
\nattempts to find , for example M1
\n\n\n
OR
\nattempts to find , for example M1
\n(where )
\n\n
THEN
\nA1
\nuses where is the area of triangle and M1
\n\n
A1
\n\n
[4 marks]
\nThe first two terms of a geometric sequence are and .
\nFind the value of .
\nFind the value of .
\nFind the least value of such that .
\nvalid approach (M1)
\neg , ,
\n(exact) A1 N2
\n[2 marks]
\ncorrect substitution (A1)
\neg
\nA1 N2
\n\n
[2 marks]
\ncorrect substitution into formula (A1)
\neg , , , sketch of and
\ncorrect inequality for or crossover values A1
\neg , and
\nA1 N2
\n[3 marks]
\nLet for .
\nSketch the graph of on the grid below.
\nFind the value of for which .
\n A1A1A1
\n
Note: Award A1 for a smooth concave down curve with generally correct shape. If first mark is awarded, award A1 for local maximum and -intercept in approximately correct position, award A1 for endpoints at and with approximately correct -coordinates.
\n\n
[3 marks]
\nrecognizing that at local maximum (M1)
\n\nA1
\n\n
[2 marks]
\nAn arithmetic sequence has first term and common difference .
\nGiven that the th term of the sequence is zero, find the value of .
\nLet denote the sum of the first terms of the sequence.
\nFind the maximum value of .
\nattempt to use (M1)
\n\nA1
\n\n
[2 marks]
\nMETHOD 1
\nattempting to express in terms of (M1)
\nuse of a graph or a table to attempt to find the maximum sum (M1)
\nA1
\n\n
METHOD 2
\n
EITHER
recognizing maximum occurs at (M1)
\n(A1)
\n
OR
attempting to calculate (M1)
\n(A1)
\n
THEN
A1
\n\n
[3 marks]
\nThe diagram below shows a triangular-based pyramid with base .
Edge is perpendicular to the edges and .
, , , ,
\nCalculate
\nevidence of choosing cosine rule (M1)
\neg
\ncorrect substitution to find (A1)
\neg
\nA2
\nappropriate approach to find (M1)
\neg ,
\n\n
A1 N3
\n[6 marks]
\nThe following table shows the probability distribution of a discrete random variable , where and .
\nShow that .
\nFind the difference between the greatest possible expected value and the least possible expected value.
\ncorrect approach A1
\neg ,
\nAG N0
\n[1 mark]
\ncorrect substitution into (A1)
\neg ,
\nvalid attempt to express in one variable M1
\neg , ,
\n,
\ncorrect value of greatest (A1)
\n(exact)
\nvalid attempt to find least value (M1)
\neg graph with minimum indicated, and
\nand if in terms of
\nand if in terms of
\ncorrect value of least (A1)
\neg (exact)
\ndifference (exact) A1 N2
\n[6 marks]
\nAt a school, of the students play a sport and of the students are involved in theatre. of the students do neither activity.
\nA student is selected at random.
\nAt the school of the students are girls, and of the girls are involved in theatre.
\nA student is selected at random. Let be the event “the student is a girl” and let be the event “the student is involved in theatre”.
\nFind the probability that the student plays a sport and is involved in theatre.
\nFind the probability that the student is involved in theatre, but does not play a sport.
\nFind .
\nDetermine if the events and are independent. Justify your answer.
\nEITHER
\nOR (M1)
\nOR
\n
OR
a clearly labelled Venn diagram (M1)
\n
THEN
(accept ) A1
\n\n
Note: To obtain the M1 for the Venn diagram all labels must be correct and in the correct sections. For example, do not accept in the area corresponding to .
\n\n
[2 marks]
\nEITHER
\nOR
\n(M1)
\n
OR
a clearly labelled Venn diagram including , and (M1)
\n
THEN
(accept ) A1
\n\n
[2 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nMETHOD 1
\nA1
\nand are not independent R1
\n\n
METHOD 2
\nA1
\nand are not independent R1
\n\n
Note: Do not award A0R1.
\n\n
[2 marks]
\nThis question asks you to explore some properties of polygonal numbers and to determine and prove interesting results involving these numbers.
\n
A polygonal number is an integer which can be represented as a series of dots arranged in the shape of a regular polygon. Triangular numbers, square numbers and pentagonal numbers are examples of polygonal numbers.
For example, a triangular number is a number that can be arranged in the shape of an equilateral triangle. The first five triangular numbers are and .
\nThe following table illustrates the first five triangular, square and pentagonal numbers respectively. In each case the first polygonal number is one represented by a single dot.
\nFor an -sided regular polygon, where , the th polygonal number is given by
\n, where .
\nHence, for square numbers, .
\nThe th pentagonal number can be represented by the arithmetic series
\n.
\nFor triangular numbers, verify that .
\nThe number is a triangular number. Determine which one it is.
\nShow that .
\nState, in words, what the identity given in part (b)(i) shows for two consecutive triangular numbers.
\nFor , sketch a diagram clearly showing your answer to part (b)(ii).
\nShow that is the square of an odd number for all .
\nHence show that for .
\nBy using a suitable table of values or otherwise, determine the smallest positive integer, greater than , that is both a triangular number and a pentagonal number.
\nA polygonal number, , can be represented by the series
\nwhere .
\nUse mathematical induction to prove that where .
\nOR A1
\nA1
\n\n
Note: Award A0A1 if only is seen.
\nDo not award any marks for numerical verification.
\n\n
so for triangular numbers, AG
\n\n
[2 marks]
\nMETHOD 1
\nuses a table of values to find a positive integer that satisfies (M1)
\nfor example, a list showing at least consecutive terms
\n\n
Note: Award (M1) for use of a GDC’s numerical solve or graph feature.
\n\n
(th triangular number) A1
\n\n
Note: Award A0 for . Award A0 if additional solutions besides are given.
\n\n
METHOD 2
\nattempts to solve for (M1)
\nOR
\n(th triangular number) A1
\n\n
Note: Award A0 for . Award A0 if additional solutions besides are given.
\n\n
[2 marks]
\nattempts to form an expression for in terms of M1
\n\n
EITHER
\n\nA1
\n\n
OR
\n\nA1
\n\n
THEN
\nAG
\n\n
[2 marks]
\nthe sum of the th and th triangular numbers
\nis the th square number A1
\n\n
[1 mark]
\n A1
\n
Note: Accept equivalent single diagrams, such as the one above, where the th and th triangular numbers and the th square number are clearly shown.
Award A1 for a diagram that show (a triangle with dots) and (a triangle with dots) and (a square with dots).
\n
[1 mark]
\nMETHOD 1
\nA1
\nattempts to expand their expression for (M1)
\n\nA1
\nand is odd AG
\n\n
METHOD 2
\nA1
\nattempts to expand their expression for (M1)
\n\nA1
\nand is odd AG
\n\n
Method 3
\n(where ) A1
\nattempts to expand their expression for (M1)
\n\n
now equates coefficients and obtains and
\nA1
\nand is odd AG
\n\n
[3 marks]
\nEITHER
\nand (A1)
\nsubstitutes their and their into M1
\nA1
\n\n
OR
\nand (A1)
\nsubstitutes their and their into M1
\nA1
\n\n
OR
\n\n(A1)
\nsubstitutes into their expression for M1
\n\nA1
\n\n
OR
\nattempts to find the arithmetic mean of terms (M1)
\nA1
\nmultiplies the above expression by the number of terms
\nA1
\n\n
THEN
\nso AG
\n\n
[3 marks]
\nMETHOD 1
\nforms a table of values that includes some values for (M1)
\nforms a table of values that includes some values for (M1)
\n\n
Note: Award (M1) if at least one value is correct. Award (M1) if at least one value is correct. Accept as above for values and values.
\n\n
for triangular numbers (A1)
\nfor pentagonal numbers (A1)
\n\n
Note: Award (A1) if is seen in or out of a table. Award (A1) if is seen in or out of a table. Condone the use of the same parameter for triangular numbers and pentagonal numbers, for example, for triangular numbers and for pentagonal numbers.
\n\n
(is a triangular number and a pentagonal number) A1
\n\n
Note: Award all five marks for seen anywhere with or without working shown.
\n\n
METHOD 2
\nEITHER
\nattempts to express as a quadratic in (M1)
\n(or equivalent)
\nattempts to solve their quadratic in (M1)
\n\n\n
OR
\nattempts to express as a quadratic in (M1)
\n(or equivalent)
\nattempts to solve their quadratic in (M1)
\n\n\n
THEN
\nfor triangular numbers (A1)
\nfor pentagonal numbers (A1)
\n(is a triangular number and a pentagonal number) A1
\n\n
METHOD 3
\n\nlet and so M1
\nA1
\nattempts to find the discriminant of their quadratic
\nand recognises that this must be a perfect square M1
\n\n\ndetermines that leading to and so A1
\n(is a triangular number and a pentagonal number) A1
\n\n
\n
METHOD 4
\n\nlet and so M1
\nA1
\nattempts to find the discriminant of their quadratic
\nand recognises that this must be a perfect square M1
\n\n\ndetermines that leading to and so A1
\n(is a triangular number and a pentagonal number) A1
\n\n
[5 marks]
\nNote: Award a maximum of R1M0M0A1M1A1A1R0 for a ‘correct’ proof using and .
\n\n
consider and
\nso true for R1
\n\n
Note: Accept and .
Do not accept one-sided considerations such as ' and so true for '.
Subsequent marks after this R1 are independent of this mark can be awarded.
\n
Assume true for , ie. M1
\n\n
Note: Award M0 for statements such as “let ”. The assumption of truth must be clear.
Subsequent marks after this M1 are independent of this mark and can be awarded.
\n
Consider
\n( can be represented by the sum
\nand so
\nM1
\nA1
\n\n
M1
\n(A1)
\nA1
\nhence true for and true true R1
\ntherefore true for all
\n\n
Note: Only award the final R1 if the first five marks have been awarded. Award marks as appropriate for solutions that expand both the LHS and (given) RHS of the equation.
\n\n
[8 marks]
\nPart (a) (i) was generally well done. Unfortunately, some candidates adopted numerical verification. Part (a) (ii) was generally well done with the majority of successful candidates using their GDC judiciously and disregarding n = −27 as a possible solution. A few candidates interpreted the question as needing to deal with P3(351).
\nAlthough part (b) (i) was generally well done, a significant number of candidates laboured unnecessarily to show the required result. Many candidates set their LHS to equal the RHS throughout the solution. Part (b) (ii) was generally not well done with many candidates unable to articulate clearly in words and symbols what the given identity shows for the sum of two consecutive triangular numbers. In part (b) (iii), most candidates were unable to produce a clear diagram illustrating the identity stated in part (b) (i).
\nPart (c) was reasonably well done. Most candidates were able to show algebraically that . A good number of candidates were then able to express as and conclude that is odd. Rather than making the connection that is a perfect square, many candidates attempted instead to analyse the parity of either or . As with part (b) (i), many candidates set their LHS to equal the RHS throughout the solution. A number of candidates unfortunately adopted numerical verification.
\nPart (d) was not answered as well as anticipated with many candidates not understanding what was
required. Instead of using the given arithmetic series to show that , a large number of
candidates used . Unfortunately, a number of candidates adopted numerical verification.
In part (e), the overwhelming majority of candidates who successfully determined that 210 is the smallest positive integer greater than 1 that is both triangular and pentagonal used a table of values. Unfortunately, a large proportion of these candidates seemingly spent quite a few minutes listing the first 20 triangular numbers and the first 12 pentagonal numbers. And it can be surmised that a number of these candidates constructed their table of values either without the use of a GDC or with the arithmetic functionality of a GDC rather than with a GDC's table of values facility. Candidates should be aware that a relevant excerpt from a table of values is sufficient evidence of correct working. A number of candidates started constructing a table of values but stopped before identifying 210. Disappointingly, a significant number of candidates attempted to solve for .
\nPart (f) proved beyond the reach of most with only a small number of candidates successfully proving the given result. A significant number of candidates were unable to show that the result is true for . A number of candidates established the validity of the base case for the RHS only while a number of other candidates attempted to prove the base case for r = 3. A large number of candidates did not state the inductive step correctly with the assumption of truth not clear. A number of candidates then either attempted to work backwards from the given result or misinterpreted the question and attempted to prove the result stated in the question stem rather than the result stated in the question. Some candidates who were awarded the first answer mark when considering the case were unable to complete the square or equivalent simplification correctly. Disappointingly, a significant number listed the steps involved in an induction proof without engaging in the actual proof.
\nConsider the differential equation
\n\nThe curve for has a gradient function given by
\n.
\nThe curve passes through the point .
\nUse the substitution to show that where is an arbitrary constant.
\nBy using the result from part (a) or otherwise, solve the differential equation and hence show that the curve has equation .
\nThe curve has a point of inflexion at where . Determine the coordinates of this point of inflexion.
\nUse the differential equation to show that the points of zero gradient on the curve lie on two straight lines of the form where the values of are to be determined.
\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
\nA1
\nintegrating the RHS, AG
\n\n
[3 marks]
\nEITHER
\nattempts to find M1
\n(A1)
\nsubstitutes their into M1
\n\nattempts to complete the square (M1)
\nA1
\nA1
\n\n
OR
\nattempts to find M1
\nA1
\nM1
\nattempts to complete the square (M1)
\nA1
\nA1
\n\n
THEN
\nwhen , (or ) and so M1
\nsubstitutes for into their expression M1
\n\nA1
\nso AG
\n\n
[9 marks]
\nMETHOD 1
\nEITHER
\na correct graph of (for approximately ) with a local minimum point below the -axis A2
\n\n
Note: Award M1A1 for .
\n\n
attempts to find the -coordinate of the local minimum point on the graph of (M1)
\nOR
\na correct graph of (for approximately ) showing the location of the -intercept A2
\n\n
Note: Award M1A1 for .
\n\n
attempts to find the -intercept (M1)
\nTHEN
\nA1
\nattempts to find (M1)
\nthe coordinates are A1
\n\n
METHOD 2
\nattempts implicit differentiation on to find M1
\n(or equivalent)
\n() A1
\nattempts to solve for where M1
\nA1
\nattempts to find (M1)
\nthe coordinates are A1
\n\n
[6 marks]
\nM1
\nattempts to solve for M1
\nor A1
\nand A1
\n\n
Note: Award M1 for stating , M1 for substituting into , A1 for and A1 for and .
\n\n
[4 marks]
\nLet , for . The following diagram shows the graph of .
\nThere are -intercepts at and at . There is a maximum at point where , and a point of inflexion at point where .
\nFind the value of .
\nWrite down the coordinates of .
\nFind the equation of the tangent to the graph of at .
\nFind the coordinates of .
\nFind the rate of change of at .
\nLet be the region enclosed by the graph of , the -axis and the lines and . The region is rotated 360º about the -axis. Find the volume of the solid formed.
\nevidence of valid approach (M1)
\neg ,
\n\n
A1 N2
\n[2 marks]
\n,
\nA2 N2
\n[2 marks]
\nvalid approach (M1)
\neg tangent at maximum point is horizontal,
\n(must be an equation) A1 N2
\n[2 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
\n(exact) (accept ) A1 N1
\n\n
METHOD 2 (analytical)
\nA1
\nvalid approach (M1)
\neg ,
\nA1 N1
\nsubstituting their value into (M1)
\neg
\n(exact) (accept ) A1 N1
\n\n
[5 marks]
\nrecognizing rate of change is (M1)
\neg ,
\nrate of change is (exact) A1 N2
\n[2 marks]
\nattempt to substitute either their limits or the function into volume formula (M1)
\neg , ,
\n\n
volume A2 N3
\n[3 marks]
\nA Gaussian integer is a complex number, , such that where . In this question, you are asked to investigate certain divisibility properties of Gaussian integers.
\nConsider two Gaussian integers, and , such that for some Gaussian integer .
\nNow consider two Gaussian integers, and .
\nThe norm of a complex number , denoted by , is defined by . For example, if then .
\nA Gaussian prime is a Gaussian integer, , that cannot be expressed in the form where are Gaussian integers with .
\nThe positive integer is a prime number, however it is not a Gaussian prime.
\nLet be Gaussian integers.
\nThe result from part (h) provides a way of determining whether a Gaussian integer is a Gaussian prime.
\nFind .
\nDetermine whether is a Gaussian integer.
\nOn an Argand diagram, plot and label all Gaussian integers that have a norm less than .
\nGiven that where , show that .
\nBy expressing the positive integer as a product of two Gaussian integers each of norm , show that is not a Gaussian prime.
\nVerify that is not a Gaussian prime.
\nWrite down another prime number of the form that is not a Gaussian prime and express it as a product of two Gaussian integers.
\nShow that .
\nHence show that is a Gaussian prime.
\nUse proof by contradiction to prove that a prime number, , that is not of the form is a Gaussian prime.
\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(Since and/or are not integers)
\nis not a Gaussian integer R1
\n\n
Note: Award R1 for correct conclusion from their answer.
\n\n
[3 marks]
\nplotted and labelled A1
\nplotted and labelled A1
\n\n
Note: Award A1A0 if extra points to the above are plotted and labelled.
\n\n
[2 marks]
\n(and as ) A1
\nthen AG
\n\n
[1 mark]
\nA1
\nand R1
\n(since are positive) R1
\nso is not a Gaussian prime, by definition AG
\n\n
[3 marks]
\n(A1)
\nA1
\nso is not a Gaussian prime AG
\n\n
[2 marks]
\nFor example, (M1)A1
\n\n
[2 marks]
\nMETHOD 1
\nLet and
\nLHS:
\nM1
\nA1
\nA1
\nA1
\nRHS:
\nM1
\nA1
\nLHS = RHS and so AG
\n\n
METHOD 2
\nLet and
\nLHS
\nM1
\nA1
\n\nM1A1
\nA1
\nA1
\n(= RHS) AG
\n\n
[6 marks]
\nwhich is a prime (in ) R1
\nif then R1
\nwe cannot have R1
\n\n
Note: Award R1 for stating that is not the product of Gaussian integers of smaller norm because no such norms divide
\n\n
so is a Gaussian prime AG
\n\n
[3 marks]
\nAssume is not a Gaussian prime
\nwhere are Gaussian integers and M1
\nM1
\nA1
\nIt cannot be from definition of Gaussian prime R1
\nhence R1
\nIf then which is a contradiction R1
\nhence a prime number, , that is not of the form is a Gaussian prime AG
\n\n
[6 marks]
\nThis question asks you to explore cubic polynomials of the form for and corresponding cubic equations with one real root and two complex roots of the form for .
\n\n
In parts (a), (b) and (c), let and .
\nConsider the equation for .
\nConsider the function for .
\nConsider the function for where and .
\nThe equation for has roots and where and .
\nOn the Cartesian plane, the points and represent the real and imaginary parts of the complex roots of the equation .
\n
The following diagram shows a particular curve of the form and the tangent to the curve at the point . The curve and the tangent both intersect the -axis at the point . The points and are also shown.
Consider the curve for . The points and are as defined in part (d)(ii). The curve has a point of inflexion at point .
\nConsider the special case where and .
\nGiven that and are roots of the equation, write down the third root.
\nVerify that the mean of the two complex roots is .
\nShow that the line is tangent to the curve at the point .
\nSketch the curve and the tangent to the curve at point , clearly showing where the tangent crosses the -axis.
\nShow that .
\nHence, or otherwise, prove that the tangent to the curve at the point intersects the -axis at the point .
\nDeduce from part (d)(i) that the complex roots of the equation can be expressed as .
\nUse this diagram to determine the roots of the corresponding equation of the form for .
\nState the coordinates of .
\nShow that the -coordinate of is .
\nYou are not required to demonstrate a change in concavity.
\nHence describe numerically the horizontal position of point relative to the horizontal positions of the points and .
\nSketch the curve for and .
\nFor and , state in terms of , the coordinates of points and .
\nA1
\n\n
[1 mark]
\nmean A1
\nAG
\n\n
[1 mark]
\nMETHOD 1
\nattempts product rule differentiation (M1)
\n\n
Note: Award (M1) for attempting to express as
\n\n
A1
\nA1
\n\n
Note: Where is correct, award A1 for solving and obtaining .
\n
EITHER
A1
\n
OR
A1
\n
OR
states the gradient of is also and verifies that lies on the line A1
\n
THEN
so is the tangent to the curve at AG
\n\n
Note: Award a maximum of (M0)A0A1A1 to a candidate who does not attempt to find .
\n\n
METHOD 2
\nsets to form (M1)
\n
EITHER
A1
\nattempts to solve a correct cubic equation (M1)
\n\n
OR
recognises that and forms A1
\nattempts to solve a correct quadratic equation (M1)
\n\n
THEN
is a double root R1
\nso is the tangent to the curve at AG
\n\n
Note: Candidates using this method are not required to verify that .
\n\n
[4 marks]
\na positive cubic with an -intercept , and a local maximum and local minimum in the first quadrant both positioned to the left of A1
\n\n
Note: As the local minimum and point A are very close to each other, condone graphs that seem to show these points coinciding.
For the point of tangency, accept labels such as or the point labelled from both axes. Coordinates are not required.
\n
a correct sketch of the tangent passing through and crossing the -axis at the same point as the curve A1
\n\n
Note: Award A1A0 if both graphs cross the -axis at distinctly different points.
\n\n
[2 marks]
\nEITHER
\n(M1)A1
\n
OR
attempts to find M1
\n\nA1
\n\n
THEN
AG
\n\n
[2 marks]
\nMETHOD 1
\n(A1)
\n(A1)
\nattempts to substitute their and into M1
\n\n
EITHER
A1
\nsets so M1
\nOR R1
\n
OR
sets so M1
\nOR R1
\n A1
THEN
so the tangent intersects the -axis at the point AG
\n\n
METHOD 2
\n(A1)
\n(A1)
\nattempts to substitute their and into and attempts to find M1
\n\n
EITHER
A1
\nsets so M1
\nOR R1
\n
OR
sets so M1
\nOR R1
\nA1
\n\n
METHOD 3
\n(A1)
\nthe line through parallel to the tangent at has equation
A1
sets to form M1
\nA1
\nA1
\nsince there is a double root , this parallel line through is the required tangent at R1
\n\n
[6 marks]
\nEITHER
\n (since ) R1
Note: Accept .
OR
and R1
\n
THEN
hence the complex roots can be expressed as AG
\n\n
[1 mark]
\n(seen anywhere) A1
\n
EITHER
attempts to find the gradient of the tangent in terms of and equates to (M1)
OR
substitutes and to form (M1)
\n
OR
substitutes and into (M1)
\n
THEN
roots are (seen anywhere) and A1A1
\n\n
Note: Award A1 for and A1 for . Do not accept coordinates.
\n\n
[4 marks]
\nA1
\n\n
Note: Accept “ and ”.
Do not award A1FT for .
\n
[1 mark]
\nattempts to find M1
\n\nsets and correctly solves for A1
\nfor example, obtaining leading to
\nso AG
\n
Note: Do not award A1 if the answer does not lead to the AG.
\n
[2 marks]
\npoint is of the horizontal distance (way) from point to point A1
\n
Note: Accept equivalent numerical statements or a clearly labelled diagram displaying the numerical relationship.
Award A0 for non-numerical statements such as “ is between and , closer to ”.
\n
[1 mark]
\n(A1)
\na positive cubic with no stationary points and a non-stationary point of inflexion at A1
\n
Note: Graphs may appear approximately linear. Award this A1 if a change of concavity either side of is apparent.
Coordinates are not required and the -intercept need not be indicated.
\n
[2 marks]
\nA1
\n\n
[1 mark]
\nPart (a) (i) was generally well done with a significant majority of candidates using the conjugate root theorem to state as the third root. A number of candidates, however, wasted considerable time attempting an algebraic method to determine the third root. Part (a) (ii) was reasonably well done. A few candidates however attempted to calculate the product of and .
\nPart (b) was reasonably well done by a significant number of candidates. Most were able to find a correct expression for and a good number of those candidates were able to determine that . Candidates that did not determine the equation of the tangent had to state that the gradient of is also 1 and verify that the point (4,3) lies on the line. A few candidates only met one of those requirements. Weaker candidates tended to only verify that the point (4,3) lies on the curve and the tangent line without attempting to find .
\nPart (c) was not answered as well as anticipated. A number of sketches were inaccurate and carelessly drawn with many showing both graphs crossing the x-axis at distinctly different points.
\nPart (d) (i) was reasonably well done by a good number of candidates. Most successful responses involved use of the product rule. A few candidates obtained full marks by firstly expanding , then differentiating to find and finally simplifying to obtain the desired result. A number of candidates made elementary mistakes when differentiating. In general, the better candidates offered reasonable attempts at showing the general result in part (d) (ii). A good number gained partial credit by determining that and/or . Only the very best candidates obtained full marks by concluding that as or , then when .
\nIn general, only the best candidates were able to use the result to deduce that the complex roots of the equation can be expressed as . Although given the complex roots , a significant number of candidates attempted, with mixed success, to use the quadratic formula to solve the equation .
\nIn part (f) (i), only a small number of candidates were able to determine all the roots of the equation. Disappointingly, a large number did not state as a root. Some candidates determined that but were unable to use the diagram to determine that . Of the candidates who determined all the roots in part (f) (i), very few gave the correct coordinates for C2 . The most frequent error was to give the y-coordinate as .
\nOf the candidates who attempted part (g) (i), most were able to find an expression for and a reasonable number of these were then able to convincingly show that . It was very rare to see a correct response to part (g) (ii). A few candidates stated that P is between R and A with some stating that P was closer to A. A small number restated in words.
\nOf the candidates who attempted part (h) (i), most were able to determine that . However, most graphs were poorly drawn with many showing a change in concavity at rather than at . In part (h) (ii), only a very small number of candidates determined that A and P coincide at (r,0).
\nThis question asks you to investigate some properties of the sequence of functions of the form , −1 ≤ ≤ 1 and .
\nImportant: When sketching graphs in this question, you are not required to find the coordinates of any axes intercepts or the coordinates of any stationary points unless requested.
\nFor odd values of > 2, use your graphic display calculator to systematically vary the value of . Hence suggest an expression for odd values of describing, in terms of , the number of
\nFor even values of > 2, use your graphic display calculator to systematically vary the value of . Hence suggest an expression for even values of describing, in terms of , the number of
\nThe sequence of functions, , defined above can be expressed as a sequence of polynomials of degree .
\nConsider .
\nOn the same set of axes, sketch the graphs of and for −1 ≤ ≤ 1.
\nlocal maximum points;
\nlocal minimum points;
\nOn a new set of axes, sketch the graphs of and for −1 ≤ ≤ 1.
\nlocal maximum points;
\nlocal minimum points.
\nSolve the equation and hence show that the stationary points on the graph of occur at where and 0 < < .
\nUse an appropriate trigonometric identity to show that .
\nUse an appropriate trigonometric identity to show that .
\nHence show that , .
\nHence express as a cubic polynomial.
\ncorrect graph of A1
\ncorrect graph of A1
\n[2 marks]
\ngraphical or tabular evidence that has been systematically varied M1
\neg = 3, 1 local maximum point and 1 local minimum point
\n= 5, 2 local maximum points and 2 local minimum points
\n= 7, 3 local maximum points and 3 local minimum points (A1)
\nlocal maximum points A1
\n[3 marks]
\nlocal minimum points A1
\nNote: Allow follow through from an incorrect local maximum formula expression.
\n[1 mark]
\ncorrect graph of A1
\ncorrect graph of A1
\n[2 marks]
\ngraphical or tabular evidence that has been systematically varied M1
\neg = 2, 0 local maximum point and 1 local minimum point
\n= 4, 1 local maximum points and 2 local minimum points
\n= 6, 2 local maximum points and 3 local minimum points (A1)
\nlocal maximum points A1
\n[3 marks]
\nlocal minimum points A1
\n[1 mark]
\n\n
M1A1
\nNote: Award M1 for attempting to use the chain rule.
\nM1
\nA1
\nleading to
\n( and 0 < < ) AG
\n[4 marks]
\n\n
M1
\nstating that A1
\nso AG
\n[2 marks]
\n\n
\n
A1
\nuse of cos(A + B) = cos A cos B − sin A sin B leading to M1
\nAG
\n[2 marks]
\nA1
\nM1
\nA1
\nAG
\n[3 marks]
\n(M1)
\n\n
A1
\n[2 marks]
\nThis question investigates the sum of sine and cosine functions
\nThe expression can be written in the form , where and and .
\nThe expression can be written in the form , where and and .
\nIn general, the expression can be written in the form , where and and .
\nConjecture an expression, in terms of and , for
\nThe expression can also be written in the form .
\nLet
\nSketch the graph , for
\nWrite down the amplitude of this graph
\nWrite down the period of this graph
\nUse your answers from part (a) to write down the value of , and .
\nFind the value of .
\nFind , giving the answer to 3 significant figures.
\nComment on your answer to part (c)(i).
\nBy considering the graph of , find the value of , , and .
\n.
\n.
\n.
\n.
\nShow that .
\nShow that .
\nHence prove your conjectures in part (e).
\n A1
[1 mark]
\n5 A1
\n[1 mark]
\nA1
\n[1 mark]
\n, , A1
\n[1 mark]
\nmaximum at M1
\nSo A1
\n[2 marks]
\n0.644 A1
\n[1 mark]
\nit appears that A1
\n[1 mark]
\n M1
A1
\nand A1
\nmaximum at M1
\nSo C = −0.395 A1
\n[5 marks]
\nA1
\n[1 mark]
\nA1
\n[1 mark]
\nA1
\n[1 mark]
\nA1
\n[1 mark]
\nEITHER
\nuse of a right triangle and Pythgoras’ to show the missing side length is M1A1
\nOR
\nUse of , leading to the required result M1A1
\n[2 marks]
\nEITHER
\nuse of a right triangle, leading to the required result. M1
\nOR
\nUse of , leading to the required result. M1
\n[1 mark]
\nM1
\nM1A1
\nSo , and A1
\nAnd M1
\nSo A1
\n[6 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]
\nLet for . Use partial fractions to find .
\nM1M1A1
\nM1A1A1
\nM1A1
\n[8 marks]
\nConsider the integral for .
\nVery briefly, explain why the value of this integral must be negative.
\nExpress the function in partial fractions.
\nUse parts (a) and (b) to show that .
\nThe numerator is negative but the denominator is positive. Thus the integrand is negative and so the value of the integral will be negative. R1AG
\n[1 mark]
\nM1M1A1
\nM1A1
\nA1
\n[6 marks]
\nM1A1A1
\nHence R1AG
\n[4 marks]
\nConsider the rectangle OABC such that AB = OC = 10 and BC = OA = 1 , with the points P , Q and R placed on the line OC such that OP = , OQ = and OR = , such that 0 < < < < 10.
\nLet be the angle APO, be the angle AQO and be the angle ARO.
\nConsider the case when and QR = 1.
\nFind an expression for in terms of .
\nShow that .
\nBy sketching the graph of as a function of , determine the range of values of for which there are possible values of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1
\nuse of tan (M1)
\n(A1)
\nA1
\n\n
METHOD 2
\nAP (A1)
\nuse of sin, cos, sine rule or cosine rule using the correct length of AP (M1)
\nor A1
\n\n
[3 marks]
\nQR = 1 ⇒ (A1)
\nNote: This may be seen anywhere.
\n\n
\n
attempt to use compound angle formula for tan M1
\n(A1)
\n(M1)
\nor A1
\nM1
\nNote: Award M1 for multiplying top and bottom by .
\n\n
AG
\n\n
[6 marks]
\nincreasing function with positive -intercept A1
\nNote: Accept curves which extend beyond the domain shown above.
\n\n
(0.618 <) < 9 (A1)
\n⇒ range is (0 <) < 4.68 (A1)
\n0 < < 4.68 A1
\n\n
[4 marks]
\n\n
Let .
\nExpress in partial fractions.
\nUse part (a) to show that is always decreasing.
\nUse part (a) to find the exact value of , giving the answer in the form , .
\nM1A1
\nM1A1
\nA1A1
\n\n
[6 marks]
\nM1A1
\nThis is always negative so function is always decreasing. R1AG
\n[3 marks]
\nM1A1
\nA1A1
\n[4 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.
\n(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]
\nExplain why any integer can be written in the form or or or , where .
\nHence prove that the square of any integer can be written in the form or , where .
\nUpon division by 4 M1
\nany integer leaves a remainder of 0, 1, 2 or 3. R1
\nHence, any integer can be written in the form or or or , where AG
\n[2 marks]
\nM1A1
\nM1A1
\nA1
\nA1
\nHence, the square of any integer can be written in the form or , where . AG
\n[6 marks]
\nThis question will investigate power series, as an extension to the Binomial Theorem for negative and fractional indices.
\nA power series in is defined as a function of the form where the .
\nIt can be considered as an infinite polynomial.
\nThis is an example of a power series, but is only a finite power series, since only a finite number of the are non-zero.
\nWe will now attempt to generalise further.
\nSuppose can be written as the power series .
\nExpand using the Binomial Theorem.
\nConsider the power series
\nBy considering the ratio of consecutive terms, explain why this series is equal to and state the values of for which this equality is true.
\nDifferentiate the equation obtained part (b) and hence, find the first four terms in a power series for .
\nRepeat this process to find the first four terms in a power series for .
\nHence, by recognising the pattern, deduce the first four terms in a power series for , .
\nBy substituting , find the value of .
\nBy differentiating both sides of the expression and then substituting , find the value of .
\nRepeat this procedure to find and .
\nHence, write down the first four terms in what is called the Extended Binomial Theorem for .
\nWrite down the power series for .
\nHence, using integration, find the power series for , giving the first four non-zero terms.
\nM1A1
\n[2 marks]
\nIt is an infinite GP with R1A1
\nM1A1AG
\n[4 marks]
\n\n
A1
\nA1
\n\n
[2 marks]
\nA1
\nA1
\n[2 marks]
\nA1A1A1
\n[3 marks]
\nA1
\n[1 mark]
\nA1
\nA1
\n[2 marks]
\nA1
\nA1
\nA1
\nA1
\n[4 marks]
\nA1
\n[1 mark]
\nM1A1
\n[2 marks]
\nM1A1
\nPutting R1
\nSo A1
\n[4 marks]
\nThe function is defined by .
\n\n
The function satisfies the equation .
\nShow that .
\nBy differentiating the above equation twice, show that
\n\n
where and denote the 3rd and 4th derivative of respectively.
\nHence show that the Maclaurin series for up to and including the term in is .
\nUse this series approximation for with to find an approximate value for .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1
\nNote: Award M1 for an attempt at chain rule differentiation.
Award M0A0 for .
AG
\n[2 marks]
\ndifferentiating gives M1A1
\ndifferentiating again gives M1A1
\nNote: Award M1 for an attempt at product rule differentiation of at least one product in each of the above two lines.
Do not penalise candidates who use poor notation.
AG
\n[4 marks]
\nattempting to find one of , or by substituting into relevant differential equation(s) (M1)
\nNote: Condone found by calculating at .
\n\n
and A1
\nand so A1
\nNote: Only award the above A1, for correct first differentiation in part (b) leading to stated or seen from use of the general Maclaurin series.
Special Case: Award (M1)A0A1 if is stated without justification or found by working backwards from the general Maclaurin series.
so the Maclaurin series for up to and including the term in is AG
\n[3 marks]
\nsubstituting into M1
\nthe series approximation gives a value of
\nso
\nA1
\nNote: Accept 9.76.
\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.
A1A1 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 , for . The following diagram shows part of the graph of .
\n![\"N17/5/MATME/SP1/ENG/TZ0/08\"](\"../assets/uploads/tinymce_asset/asset/4145/Schermafbeelding_2018-02-11_om_09.25.10.png\"/)
The graph of crosses the -axis at the origin and at the point .
\nThe line intersects the graph of at another point Q, as shown in the following diagram.
\n![\"N17/5/MATME/SP1/ENG/TZ0/08.c.d\"](\"../assets/uploads/tinymce_asset/asset/4146/Schermafbeelding_2018-02-11_om_09.27.48.png\"/)
Find the area of the region enclosed by the graph of and the line .
\nvalid approach (M1)
\neg, splitting area into triangles and integrals
\ncorrect integration (A1)(A1)
\neg
\nsubstituting their limits into their integrated function and subtracting (in any order) (M1)
\neg
\n\n
Note: Award M0 for substituting into original or differentiated function.
\n\n
area A2 N3
\n[6 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.
\nevidence 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]
\nThis question will explore connections between complex numbers and regular polygons.
\nThe diagram below shows a sector of a circle of radius 1, with the angle subtended at the centre being . A perpendicular is drawn from point to intersect the -axis at . The tangent to the circle at intersects the -axis at .
\nBy considering the area of two triangles and the area of the sector show that .
\nHence show that .
\nLet . Working in modulus/argument form find the solutions to this equation.
\nRepresent these solutions on an Argand diagram. Let their positions be denoted by placed in order in an anticlockwise direction round the circle, starting on the positive -axis. Show the positions of and .
\nShow that the length of the line segment is .
\nHence, write down the total length of the perimeter of the regular sided polygon .
\nUsing part (b) find the limit of this perimeter as .
\nFind the total area of this sided polygon.
\nUsing part (b) find the limit of this area as .
\nArea triangle A1
\nArea sector A1
\nArea triangle A1
\nSo looking at the diagram M1
\nAG
\n[5 marks]
\nHence and as we have M1R1
\nAG
\n[2 marks]
\nM1A1M1A1
\nA1A1
\nA1
\nA1
\n[8 marks]
\n A1
[1 mark]
\nBisecting the triangle to form two right angle triangles M1
\n
Length of where M1A1A1
\nSo length is AG
\n[4 marks]
\nLength of perimeter is A1
\n[1 mark]
\nas M1A1
\n[2 marks]
\nArea of so total area is . M1A1A1
\n[3 marks]
\nas M1A1
\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]
\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]
\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]
\nLet and , for .
\nThe graph of can be obtained from the graph of by two transformations:
\n\n
Let , for . The following diagram shows the graph of and the line .
\n![\"M17/5/MATME/SP2/ENG/TZ1/10.b.c\"](\"../assets/uploads/tinymce_asset/asset/3546/Schermafbeelding_2017-08-14_om_10.34.27.png\"/)
The graph of intersects the graph of at two points. These points have coordinates 0.111 and 3.31 correct to three significant figures.
\nWrite down the value of ;
\nWrite down the value of ;
\nWrite down the value of .
\nFind .
\nHence, find the area of the region enclosed by the graphs of and .
\nLet be the vertical distance from a point on the graph of to the line . There is a point on the graph of where is a maximum.
\nFind the coordinates of P, where .
\nA1 N1
\n\n
Note: Accept , , and , 2.31 as candidate may have rewritten as equal to .
\n\n
[1 mark]
\nA1 N1
\n\n
Note: Accept , , and , 2.31 as candidate may have rewritten as equal to .
\n\n
[1 mark]
\nA1 N1
\n\n
Note: Accept , , and , 2.31 as candidate may have rewritten as equal to .
\n\n
[1 mark]
\n2.72409
\n2.72 A2 N2
\n[2 marks]
\nrecognizing area between and equals 2.72 (M1)
\neg![\"M17/5/MATME/SP2/ENG/TZ1/10.b.ii/M\"](\"../assets/uploads/tinymce_asset/asset/3552/Schermafbeelding_2017-08-14_om_17.00.04.png\"/)
recognizing graphs of and are reflections of each other in (M1)
\negarea between and equals between and
\n\n
5.44819
\n5.45 A1 N3
\n[??? marks]
\nvalid attempt to find (M1)
\negdifference in -coordinates,
\ncorrect expression for (A1)
\neg
\nvalid approach to find when is a maximum (M1)
\negmax on sketch of , attempt to solve
\n0.973679
\nA2 N4
\nsubstituting their value into (M1)
\n2.26938
\nA1 N2
\n[7 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\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.
\nM1A1
\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]
\nWrite down and simplify the first three terms, in ascending powers of , in the Extended Binomial expansion of .
\nBy substituting find a rational approximation to .
\nM1A1A1
\n[3 marks]
\nM1A1A1
\n[3 marks]
\nLet
\nShow that has no vertical asymptotes.
\nFind the equation of the horizontal asymptote.
\nFind the exact value of , giving the answer in the form .
\nM1A1
\nSo the denominator is never zero and thus there are no vertical asymptotes. (or use of discriminant is negative) R1
\n[3 marks]
\nso the equation of the horizontal asymptote is M1A1
\n[2 marks]
\nM1A1A1
\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.
valid 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]
\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 = 7 (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]
\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.
recognizing 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]
\nThis question will investigate methods for finding definite integrals of powers of trigonometrical functions.
\nLet .
\n\n
Let
\nLet .
\nFind the exact values of , and .
\nUse integration by parts to show that .
\nExplain where the condition was used in your proof.
\nHence, find the exact values of and .
\nUse the substitution to show that .
\nHence, find the exact values of and
\nFind the exact values of and .
\nUse the fact that to show that .
\nExplain where the condition was used in your proof.
\nHence, find the exact values of and .
\nM1A1
\nM1A1
\nM1A1
\n[6 marks]
\n\n
\n
M1A1A1
\nM1A1
\nAG
\n[6 marks]
\nneed so that in R1
\n[1 mark]
\nA1A1
\n[2 marks]
\nA1
\nM1A1A1AG
\n[4 marks]
\nA1A1
\n[2 marks]
\nA1
\nM1A1
\n[3 marks]
\nM1
\nA1A1AG
\n[3 marks]
\nneed so that the powers of tan in are not negative R1
\n\n
[1 mark]
\nA1
\nA1
\n[2 marks]
\nConsider and for ≥ 0. The first time the graphs of and intersect is at .
\nThe set of all non-zero values that satisfy can be described as an arithmetic sequence, where ≥ 1.
\nFind the two smallest non-zero values of for which .
\nAt point P, the graphs of and intersect for the 21st time. Find the coordinates of P.
\nThe following diagram shows part of the graph of reflected in the -axis. It also shows part of the graph of and the point P.
\nFind an expression for the area of the shaded region. Do not calculate the value of the expression.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ncorrect working (A1)
\neg ,
\n(seen anywhere) (A1)
\ncorrect working (ignore additional values) (A1)
\neg ,
\n= 2, 10 A1A1 N1N1
\n[5 marks]
\nvalid approach (M1)
\neg first intersection at ,
\ncorrect working A1
\neg , ,
\nP(154, ) (accept and ) A1A1 N3
\n[4 marks]
\nvalid attempt to find upper boundary (M1)
\neg half way between and , , 154 + 4, , at least two values of new sequence {6, 14, ...}
\nupper boundary at (seen anywhere) (A1)
\ncorrect integral expression (accept missing ) A1A1 N4
\neg , ,
\nNote: Award A1 for two correct limits and A1 for correct integrand. The A1 for correct integrand may be awarded independently of all the other marks.
\n[4 marks]
\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]
\nOlivia’s house consists of four vertical walls and a sloping roof made from two rectangles. The height, , from the ground to the base of the roof is 4.5 m.
\nThe base angles of the roof are and .
\nThe house is 10 m long and 5 m wide.
\nThe length is approximately 2.84 m.
\nOlivia decides to put solar panels on the roof. The solar panels are fitted to both sides of the roof.
\nEach panel is 1.6 m long and 0.95 m wide. All the panels must be arranged in uniform rows, with the shorter edge of each panel parallel to or . Each panel must be at least 0.3 m from the edge of the roof and the top of the roof, .
\nOlivia estimates that the solar panels will cover an area of 29 m2.
\nFind the length , giving your answer to four significant figures.
\nFind the total area of the two rectangles that make up the roof.
\nFind the maximum number of complete panels that can be fitted to the whole roof.
\nFind the percentage error in her estimate.
\nOlivia investigates arranging the panels, such that the longer edge of each panel is parallel to or .
State whether this new arrangement will allow Olivia to fit more solar panels to the roof. Justify your answer.
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
180° − 27° − 26° (M1)
\nNote: Award (M1) for correct working to find angle or 127 seen.
\n(M1)(A1)
\nNote: Award (M1) for substitution into sine rule formula and (A1) for correct substitution.
\n2.74450 (m) (A1)
\n( =) 2.745 (m) (A1)(ft)(G4)
\nNote: The final (A1)(ft) is for correctly rounding their unrounded to 4 sf. If 2.745 is given as the final answer, the unrounded answer need not be seen, award (M1)(M1)(A1)(A2). For all other answers, the unrounded answer must be seen to an accuracy greater than 4 sf.
\nAward (G3) for a final answer of 2.74450…(m) with no working. If radians are used then award at most (M1)(M1)(A1)(A0)(A1)(ft) for an answer of 3.920 (m).
\n[5 marks]
\nUnits are required in this question part.
\n\n
10 × 2.84 + 10 × 2.74450… (M1)(M1)
\nNote: Award (M1) for finding their area of each rectangle and (M1) for adding their areas.
\nOR
\n10 × (2.84 + 2.74450…) (M1)(M1)
\nNote: Award (M1) for adding and their . Award (M1) for multiplying their total area by 10.
\n55.8 (55.8450…) m2 (A1)(ft)(G3)
\nNote: Follow through from their in part (a).
\n[3 marks]
\n(M1)
\nNote: Award (M1) for correct calculation of the number of panels on the long side.
\nOR (M1)
\nNote: Award (M1) for correct calculation of the number of panels on either short side with no further incorrect working.
\n20 (A1)(ft)(G2)
\nNote: Follow through from part (a). Do not award (M0)(M1)(A1)(ft).
\n[3 marks]
\n20 × 1.6 × 0.95 (= 30.4) (M1)
\nNote: Award (M1) for their 20 × 1.6 × 0.95 or 30.4 seen. Follow through from their 20 in part (c). Award (M0) if their 20 is not an integer.
\n(M1)
\nNote: Award (M1) for correct substitution of their 30.4 into the percentage error formula. Their 30.4 must be exact.
\nfound. Accept a method in two steps where “×100” is implicit from their answer.
\nThe second (M1) is contingent on the first (M1) being awarded, eg do not award (M0)(M1)(A0).
\n4.61 (%) (4.60526 (%)) (A1)(ft)(G3)
\nNote: Follow through from their answer to part (c). Percentage sign is not required.
\nAward (G2) for an unsupported final answer of 4.61.
\n[3 marks]
\n1 × 9 (array) OR 18 (total panels) (R1)(ft)
\nNote: Award (R1) for one correct array seen (1 × 9) or total number of panels (18). Working is not required, but award (R0) for incorrect working seen. Correct working is as follows.
Reasoning may compare both sides of the roof or just one side; accept correct comparisons with part (c) values. Follow through from their treatment of tolerances in part (c) and maximum number of panels.
Award (R0) for any approach with no clearance or for any method which includes further incorrect working.
No (new arrangement will mean fewer solar panels) (A1)(ft)
\nNote: Follow through from their maximum number of panels in part (c). Do not award (R0)(A1)(ft).
\n[2 marks]
\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.
METHOD 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 following table shows the probability distribution of a discrete random variable , in terms of an angle .
\n![\"M17/5/MATME/SP1/ENG/TZ1/10\"](\"../assets/uploads/tinymce_asset/asset/3528/Schermafbeelding_2017-08-11_om_09.10.36.png\"/)
Show that .
\nGiven that , find .
\nLet , for . The graph of between 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.
\nevidence of summing to 1 (M1)
\neg
\ncorrect equation A1
\neg
\ncorrect equation in A1
\neg
\nevidence of valid approach to solve quadratic (M1)
\negfactorizing equation set equal to
\ncorrect working, clearly leading to required answer A1
\neg
\ncorrect reason for rejecting R1
\neg is a probability (value must lie between 0 and 1),
\n\n
Note: Award R0 for without a reason.
\n\n
AG N0
\nvalid approach (M1)
\negsketch of right triangle with sides 3 and 4,
\ncorrect working
\n(A1)
\negmissing side
\nA1 N2
\n[3 marks]
\nattempt to substitute either limits or the function into formula involving (M1)
\neg
\ncorrect substitution of both limits and function (A1)
\neg
\ncorrect integration (A1)
\neg
\nsubstituting their limits into their integrated function and subtracting (M1)
\neg
\n\n
Note: Award M0 if they substitute into original or differentiated function.
\n\n
(A1)
\neg
\nA1 N3
\n\n
[6 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]
\nLet and .
\nThe graphs of and intersect at and , where .
\nFind the value of and of .
\nHence, find the area of the region enclosed by the graphs of and .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid attempt to find the intersection (M1)
\neg, sketch, one correct answer
\n\n
A1A1 N3
\n[3 marks]
\nattempt to set up an integral involving subtraction (in any order) (M1)
\neg
\n0.537667
\nA2 N3
\n[3 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 width of a rectangular garden is 4.5 metres shorter than its length, which is metres.
\nThe perimeter of the garden is 111 m.
\nWrite down an expression for the width of the garden in terms of .
\nWrite down an equation for the perimeter of the garden in terms 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(A1) (C1)
\nNote: Accept OR width.
\n[1 mark]
\n(A1)(ft) (C1)
\nNote: Follow through from their expression for the width from part (a).
\n[1 mark]
\n(or equivalent) (M1)
\nNote: Award (M1) for correctly removing the parentheses and collecting terms. This may be seen in part (b).
\n() 30 (A1)(ft) (C2)
\nNote: Follow through from their equation from part (b) provided .
\n[2 marks]
\nLet , 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.
\nvalid 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]
\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 .
\nHence find .
\nWrite down an expression for the area of .
\nHence find the exact area of .
\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]
\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.
\nlet 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]
\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]
\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]
\nThe volume of a hemisphere, V, is given by the formula
\nV = ,
\nwhere S is the total surface area.
\nThe total surface area of a given hemisphere is 350 cm2.
\nCalculate the volume of this hemisphere in cm3.
\nGive your answer correct to one decimal place.
\nWrite down your answer to part (a) correct to the nearest integer.
\nWrite down your answer to part (b) in the form a × 10k , where 1 ≤ a < 10 and k ∈ .
\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 substitution of 350 into volume formula.
\n\n
= 473.973… (A1)
\n= 474 (cm3) (A1)(ft) (C3)
\n\n
Note: The final (A1)(ft) is awarded for rounding their answer to 1 decimal place provided the unrounded answer is seen.
\n\n
[3 marks]
\n474 (cm3) (A1)(ft) (C1)
\nNote: Follow through from part (a).
\n\n
[1 mark]
\n4.74 × 102 (cm3) (A1)(ft)(A1)(ft) (C2)
\n\n
Note: Follow through from part (b) only.
\nAward (A0)(A0) for answers of the type 0.474 × 103.
\n\n
[2 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]
\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]
\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.
valid 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]
\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.
\nattempt 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]
\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]
\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 project mark.
\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.
\nShow that the point M (, ) lies on the regression line y on x.
\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.
\nIn his final IB examination Jerome scored 65.
\nCalculate the percentage error in Jerome’s estimated examination score.
\n14 (G1)
\n\n
[1 mark]
\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(14) + 41.75 (M1)
\nNote: Award (M1) for their correct substitution into their regression line. Follow through from parts (a)(i) and (b)(i).
\n\n
= 54
\nand so the mean point lies on the regression line (A1)
\n(accept 54 is , the mean value of the y data)
\nNote: Do not award (A1) unless the conclusion is explicitly stated and the 54 seen. The (A1) can be awarded only if their conclusion is consistent with their equation and it lies on the line.
\nThe use of 41.8 as their c value precludes awarding (A1).
\n\n
OR
\n54 = 0.875(14) + 41.75 (M1)
\n54 = 54
\nNote: Award (M1) for their correct substitution into their regression line. Follow through from parts (a)(i) and (b)(i).
\n\n
and so the mean point lies on the regression line (A1)
\nNote: Do not award (A1) unless the conclusion is explicitly stated. Follow through from part (a).
\nThe use of 41.8 as their c value precludes the awarding of (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]
\n(M1)
\nNote: Award (M1) for correct substitution into percentage error formula. Follow through from part (c)(i).
\n\n
= 12.9 (%)(12.9230…) (A1)(ft)(G2)
\nNote: Follow through from part (c)(i). Condone use of percentage symbol.
Award (G0) for an answer of −12.9 with no working.
\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
A 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]
\nAll 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.
\nattempt 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]
\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]
\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.
\ndifferentiating 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]
\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.
\nattempt 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]
\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.
\nCalculate the curved surface area of the cone which has been removed.
\nCalculate the curved surface area of the remaining solid.
\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]
\n(M1)
\nNote: Award (M1) for their correct substitution into curved surface area of a cone formula.
\n(A1)(ft) (C2)
\nNote: Follow through from part (a).
\n[2 marks]
\n(M1)
\nNote: Award (M1) for their correct substitution into curved surface area of a cone formula and for subtracting their part (b).
\n(A1)(ft) (C2)
\nNote: Follow through from part (b).
\n[2 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 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]
\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]
\nLet .
\nFind all the intercepts of the graph of with both the and axes.
\nWrite down the equation of the vertical asymptote.
\nAs the graph of approaches an oblique straight line asymptote.
\nDivide by to find the equation of this asymptote.
\nintercept on the axes is (0, −6) A1
\nM1
\nintercepts on the axes are A1A1
\n[4 marks]
\nA1
\n[1 mark]
\nM1A1
\nSo equation of asymptote is M1A1
\n[4 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]
\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.
\nlet 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.
METHOD 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]
\nConsider the function defined by .
\nThe curvature at any point on a graph is defined as .
\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.
\nThe curvature at any point on a graph is defined as .
\nFind 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.
\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]
\nLet .
\nFind the co-ordinates of all stationary points.
\nWrite down the equation of the vertical asymptote.
\nWith justification, state if each stationary point is a minimum, maximum or horizontal point of inflection.
\nM1
\nM1
\nStationary points are A1A1
\n[4 marks]
\nA1
\n[1 mark]
\nLooking at the nature table
\n M1A1
is a max and is a min A1A1
\n[4 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]
\nA park in the form of a triangle, ABC, is shown in the following diagram. AB is 79 km and BC is 62 km. Angle AC is 52°.
\nCalculate the length of side AC in km.
\nCalculate the area of the park.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(AC2 =) 622 + 792 − 2 × 62 × 79 × cos(52°) (M1)(A1)
\nNote: Award (M1) for substituting in the cosine rule formula, (A1) for correct substitution.
\n63.7 (63.6708…) (km) (A1) (C3)
\n[3 marks]
\n× 62 × 79 × sin(52°) (M1)(A1)
\nNote: Award (M1) for substituting in the area of triangle formula, (A1) for correct substitution.
\n1930 km2 (1929.83…km2) (A1) (C3)
\n[3 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]
\n\n
For example 1, 2, 3, 4, 5, 6, 100 where A1A1
\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]
\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]
\nJulio is making a wooden pencil case in the shape of a large pencil. The pencil case consists of a cylinder attached to a cone, as shown.
\nThe cylinder has a radius of r cm and a height of 12 cm.
\nThe cone has a base radius of r cm and a height of 10 cm.
\nFind an expression for the slant height of the cone in terms of r.
\nThe total external surface area of the pencil case rounded to 3 significant figures is 570 cm2.
\nUsing your graphic display calculator, calculate the value of r.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(slant height2 =) 102 + r 2 (M1)
\nNote: For correct substitution of 10 and r into Pythagoras’ Theorem.
\n(A1) (C2)
\n[2 marks]
\n(M1)(M1)(M1)
\nNote: Award (M1) for correct substitution in curved surface area of cylinder and area of the base, (M1) for their correct substitution in curved surface area of cone, (M1) for adding their 3 surface areas and equating to 570. Follow through their part (a).
\n= 4.58 (4.58358...) (A1)(ft) (C4)
\nNote: Last line must be seen to award final (A1). Follow through from part (a).
\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 principal of a high school is concerned about the effect social media use might be having on the self-esteem of her students. She decides to survey a random sample of 9 students to gather some data. She wants the number of students in each grade in the sample to be, as far as possible, in the same proportion as the number of students in each grade in the school.
\nThe number of students in each grade in the school is shown in table.
\nIn order to select the 3 students from grade 12, the principal lists their names in alphabetical order and selects the 28th, 56th and 84th student on the list.
\nOnce the principal has obtained the names of the 9 students in the random sample, she surveys each student to find out how long they used social media the previous day and measures their self-esteem using the Rosenberg scale. The Rosenberg scale is a number between 10 and 40, where a high number represents high self-esteem.
\nState the name for this type of sampling technique.
\nShow that 3 students will be selected from grade 12.
\nCalculate the number of students in each grade in the sample.
\nState the name for this type of sampling technique.
\nCalculate Pearson’s product moment correlation coefficient, .
\nInterpret the meaning of the value of in the context of the principal’s concerns.
\nExplain why the value of makes it appropriate to find the equation of a regression line.
\nAnother student at the school, Jasmine, has a self-esteem value of 29.
\nBy finding the equation of an appropriate regression line, estimate the time Jasmine spent on social media the previous day.
\nStratified sampling A1
\n[1 mark]
\nThere are 260 students in total A1
\nM1A1
\nSo 3 students will be selected. AG
\n[3 marks]
\ngrade 9 , grade 10 , grade 11 A2
\n[2 marks]
\nSystematic sampling A1
\n[1 mark]
\nA2
\n[2 marks]
\nThe negative value of indicates that more time spent on social media leads to lower self-esteem, supporting the principal’s concerns. R1
\n[1 mark]
\nbeing close to –1 indicates there is strong correlation, so a regression line is appropriate. R1
\n[1 mark]
\nFind the regression line of on . M1
\nA1
\nhours M1A1
\n[4 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 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]
\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]
\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]
\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]
\nConsider the curve defined by the equation .
\nFind the volume of the solid formed when the region bounded by the curve, the -axis for and the -axis for is rotated through about the -axis.
\nUse of
\n(M1)(A1)
\n\n
Note: Condone absence of limits or incorrect limits for M mark.
\nDo not condone absence of or multiples of .
\n\n
A1
\n[3 marks]
\nThis question investigates some applications of differential equations to modeling population growth.
\nOne model for population growth is to assume that the rate of change of the population is proportional to the population, i.e. , where , is the time (in years) and is the population
\nThe initial population is 1000.
\nGiven that , use your answer from part (a) to find
\nConsider now the situation when is not a constant, but a function of time.
\nGiven that , find
\nAnother model for population growth assumes
\nShow that the general solution of this differential equation is , where .
\nthe population after 10 years
\nthe number of years it will take for the population to triple.
\n\n
the solution of the differential equation, giving your answer in the form .
\nthe number of years it will take for the population to triple.
\nShow that , where .
\nSolve the differential equation , giving your answer in the form .
\nGiven that the initial population is 1000, and , find the number of years it will take for the population to triple.
\nM1A1
\nA1A1
\nA1
\n, where AG
\n[5 marks]
\nwhen
\nA1
\nA1
\n[2 marks]
\nM1
\nyears A1
\n[2 marks]
\nA1
\n[1 mark]
\nM1
\nA1A1
\nA1
\nwhen
\nM1
\n\n
[5 marks]
\nM1
\nA1
\nUse of quadratic formula or GDC graph or GDC polysmlt M1
\nyears A1
\n[4 marks]
\n, where is the constant of proportionality A1
\nSo A1
\nAG
\n[2 marks]
\nM1
\nM1
\nA1
\nA1
\n\n
A1A1
\n, where M1
\n, where A1
\nM1
\nA1
\n[10 marks]
\nM1
\nA1
\nM1
\nyears A1
\n[4 marks]
\nThe speed of light is kilometres per second. The average distance from the Sun to the Earth is 149.6 million km.
\nA light-year is the distance light travels in one year and is equal to million km. Polaris is a bright star, visible from the Northern Hemisphere. The distance from the Earth to Polaris is 323 light-years.
\nCalculate the time, in minutes, it takes for light from the Sun to reach the Earth.
\nFind the distance from the Earth to Polaris in millions of km. Give your answer in the form with and .
\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]
\n(M1)
\n\n
Note: Award (M1) for multiplying 323 by , seen with any power of 10; therefore only penalizing incorrect power of 10 once.
\n\n
(A1)(A1) (C3)
\n\n
Note: Award (A1) for 3.06.
\nAward (A1) for
\nAward (A0)(A0) for answers of the type:
\n\n
[3 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]
\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
State what 13 represents in the given diagram.
\nWrite down the interquartile range for this data.
\nWrite down the approximate number of snacks whose amount of sugar ranges from 18 to 20 grams.
\nThe 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.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
median (A1) (C1)
\n[1 mark]
\n18 − 12 (A1)
\nNote: Award (M1) for correct quartiles seen.
\n6 (g) (A1) (C2)
\n[2 marks]
\n125 (A1) (C1)
\n[1 mark]
\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 type of candy is packaged in a right circular cone that has volume and vertical height 8 cm.
\n![\"M17/5/MATSD/SP1/ENG/TZ1/09\"](\"../assets/uploads/tinymce_asset/asset/3569/Schermafbeelding_2017-08-15_om_11.14.55.png\"/)
Find the radius, , of the circular base of the cone.
\nFind the slant height, , of the cone.
\nFind the curved surface area of the cone.
\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 cone formula.
\n\n
(A1) (C2)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution into Pythagoras’ theorem.
\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 their correct substitutions into curved surface area of a cone formula.
\n\n
(A1)(ft) (C2)
\n\n
Note: Follow through from parts (a) and (b). Accept from use of 3 sf values.
\n\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.
\nGiving your answer to one decimal place, write down the value of
\n(i) the median level of Vitamin C content of the oranges in the sample;
\n(ii) the lower quartile;
\n(iii) the upper quartile.
\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\"/)
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(i) 32.5 (A1)
\n(ii) 31.9 (A1)
\n(iii) 33.1 (A1) (C3)
\n\n
Note: Answers must be given correct to 1 decimal place.
\n\n
[3 marks]
\n![\"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]
\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 following diagram shows triangle ABC, with AB = 6 and AC = 8.
\nGiven that find the value of .
\nFind the area of triangle ABC.
\nvalid approach using Pythagorean identity (M1)
\n(or equivalent) (A1)
\nA1
\n[3 marks]
\n(or equivalent) (A1)
\narea A1
\n[2 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]
\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]
\nThe 3rd term of an arithmetic sequence is 1407 and the 10th term is 1183.
\nCalculate the number of positive terms in the sequence.
\n1471 + (n − 1)(−32) > 0 (M1)
\n⇒ n <
\nn < 46.96… (A1)
\nso 46 positive terms A1
\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]
\nLet A and B be events such that , and .
\nFind .
\nattempt to substitute into (M1)
\nNote: Accept use of Venn diagram or other valid method.
\n(A1)
\n(seen anywhere) A1
\nattempt to substitute into (M1)
\n\n
A1
\n[5 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.
(A1)
\nsolving, (M1)A1
\n\n
[3 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nShow that , where .
\nHence, or otherwise, prove that the sum of the squares of any two consecutive odd integers is even.
\nattempting to expand the LHS (M1)
\nLHS A1
\n(= RHS) AG
\n[2 marks]
\nMETHOD 1
\nrecognition that and represent two consecutive odd integers (for ) R1
\nA1
\nvalid reason eg divisible by 2 (2 is a factor) R1
\nso the sum of the squares of any two consecutive odd integers is even AG
\n\n
METHOD 2
\nrecognition, eg that and represent two consecutive odd integers (for ) R1
\nA1
\nvalid reason eg divisible by 2 (2 is a factor) R1
\nso the sum of the squares of any two consecutive odd integers is even AG
\n[3 marks]
\nLet . Given that , find .
\nattempt to integrate (M1)
\n\n
(A1)
\nEITHER
\nA1
\nOR
\nA1
\nTHEN
\ncorrect substitution into their integrated function (must have C) (M1)
\n\n
A1
\n[5 marks]
\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]
\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 value of , the Pearson’s product–moment correlation coefficient, for these data.
\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(G2)
\n\n
Note: Award (G1) for 0.973, without minus sign.
\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]
\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 volume of one 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.
(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]
\nThe functions and are defined such that and .
\nShow that .
\nGiven that , find the value of .
\nattempt to form composition M1
\ncorrect substitution A1
\nAG
\n[2 marks]
\nattempt to substitute 4 (seen anywhere) (M1)
\ncorrect equation (A1)
\n= 19 A1
\n[3 marks]
\nGiven that , find in terms of .
\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\n
\n
\n
M1
\nhence A1
\n[5 marks]
\nShow that .
\nHence or otherwise solve for .
\nattempting to use the change of base rule M1
\nA1
\nA1
\nAG
\n[3 marks]
\n\n
M1
\n(or equivalent) A1
\nuse of (M1)
\n\n
A1
\nA1
\nNote: Award A0 if solutions other than are included.
\n[5 marks]
\nShow that where .
\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
Solve the simultaneous equations
\n\n
.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
use of at least one “log rule” applied correctly for the first equation M1
\n\n
\n
\n
A1
\nuse of at least one “log rule” applied correctly for the second equation M1
\n\n
\n
\n
A1
\nattempt to eliminate (or ) from their two equations M1
\n\n
\n
\n
A1
\nor A1
\nNote: , values do not have to be “paired” to gain either of the final two A marks.
\n[7 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]
\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]
\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]
\nA large company surveyed 160 of its employees to find out how much time they spend traveling to work on a given day. The results of the survey are shown in the following cumulative frequency diagram.
\nOnly 10% of the employees spent more than minutes traveling to work.
\nThe results of the survey can also be displayed on the following box-and-whisker diagram.
\nFind the median number of minutes spent traveling to work.
\nFind the number of employees whose travelling time is within 15 minutes of the median.
\nFind the value of .
\nWrite down the value of .
\nFind the value of .
\nHence, find the interquartile range.
\nTravelling times of less than minutes are considered outliers.
\nFind the value of .
\nevidence of median position (M1)
\n80th employee
\n40 minutes A1
\n[2 marks]
\nvalid attempt to find interval (25–55) (M1)
\n18 (employees), 142 (employees) A1
\n124 A1
\n[3 marks]
\nrecognising that there are 16 employees in the top 10% (M1)
\n144 employees travelled more than minutes (A1)
\n= 56 A1
\n[3 marks]
\n= 70 A1
\n[1 mark]
\nrecognizing is first quartile value (M1)
\n40 employees
\n= 33 A1
\n[2 marks]
\n47 − 33 (M1)
\nIQR = 14 A1
\n[2 marks]
\nattempt to find 1.5 × their IQR (M1)
\n33 − 21
\n12 (A1)
\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.
\n
(M1)
\nM1A1
\n\n
(A1)
\nA1
\n[5 marks]
\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]
\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]
\nA calculator fits into a cuboid case with height 29 mm, width 98 mm and length 186 mm.
\nFind the volume, in cm3, of this calculator case.
\nevidence of 10 mm = 1 cm (A1)
\nNote: Award (A1) for dividing their volume from part (a) or part (b) by 1000.
\n529 (cm3) (528.612 (cm3)) (A1)(ft) (C2)
\nNote: Follow through from parts (a) or (b). Accept answers written in scientific notation.
\n[2 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.
METHOD 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]
\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 distance from A to C.
\nFind the size of angle .
\nFind the size of angle .
\nFind the size of angle .
\nThe length of path AD is 287 m.
\nFind the area of the region ABCD.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
AC2 = 1852 + 2502 − 2 × 185 × 250 × cos(125°) (M1)(A1)
\nNote: Award (M1) for substitution in the cosine formula; (A1) for correct substitution.
\n387 (387.015…) (m) (A1)(G2)
\nNote: If radians are used the answer is 154 (154.471…), award at most (M1)(A1)(A0).
\n[3 marks]
\n(M1)(A1)(ft)
\nOR
\n(M1)(A1)(ft)
\nNote: Award (M1) for substitution in the sine or cosine formulas; (A1)(ft) for correct substitution.
\n(31.9478…°) (A1)(ft)(G2)
\nNote: Follow through from part (a).
\n[3 marks]
\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]
\n(M1)(M1)(M1)
\nNote: Award (M1) for substitution in the area formula for either triangle; (M1) for correct substitution for both areas; (M1) for adding their two areas;
\n18942.8… + 44383.9…
\n63300 (m2) (63326.8… (m2)) (A1)(ft)(G3)
\nNote: Follow through from parts (a) and (b)(ii).
\nOR
\n\n
M1M1M1
\nNote: Award (M1) for substitution in the area formula for either triangle; (M1) for correct substitution for both areas; (M1) for adding their two areas;
\n26446.4… + 36869.3…
\n63300 (63315.8…) (m2) (A1)(ft)(G3)
\n[4 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.
\n (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]
\nA sphere with diameter 3 474 000 metres can model the shape of the Moon.
\nUse this model to calculate the circumference of the Moon in kilometres. Give your full calculator display.
\nGive your answer to part (a) correct to three significant figures.
\nWrite your answer to part (b) in the form , where 1 ≤ < 10 , .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)(M1)
\nNote: Award (M1) for correct numerator and (M1) for dividing by 1000 OR equivalent, such as ie diameter.
Do not accept use of area formula ie .
10 913.89287… (km) (A1) (C3)
\n[3 marks]
\n\n
10 900 (km) (A1)(ft) (C1)
\nNote: Follow through from part (a).
\n[1 mark]
\n\n
1.09 × 104 (A1)(ft)(A1)(ft) (C2)
\nNote: Follow through from part (b) only. Award (A1)(ft) for 1.09, and (A1)(ft) × 104. Award (A0)(A0) for answers of the type: 10.9 × 103.
\n[2 marks]
\n\n
Express the binomial coefficient as a polynomial in .
\nHence find the least value of for which .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\nA1
\nor equivalent A1
\n[3 marks]
\nattempt to solve (M1)
\n(A1)
\nNote: Allow equality.
\nA1
\n[3 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.
\n= {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]
\nThree girls and four boys are seated randomly on a straight bench. Find the probability that the girls sit together and the boys sit together.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\ntotal number of arrangements 7! (A1)
\nnumber of ways for girls and boys to sit together (M1)(A1)
\n\n
Note: Award M1A0 if the 2 is missing.
\n\n
probability M1
\n\n
Note: Award M1 for attempting to write as a probability.
\n\n
\n
A1
\n\n
Note: Award A0 if not fully simplified.
\n\n
METHOD 2
\n(M1)A1A1
\n\n
Note: Accept or .
\n\n
(M1)A1
\n\n
Note: Award A0 if not fully simplified.
\n\n
[5 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]
\nThe coefficient of in the expansion of is equal to the coefficient of in the expansion of . Find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1
\n(M1)
\n(A1)
\nM1A1
\n\n
(A1)
\nA1
\nMETHOD 2
\nattempt to expand both binomials M1
\nA1
\nA1
\nM1
\n(A1)
\nA1
\n[6 marks]
\nIn a trial examination session a candidate at a school has to take 18 examination papers including the physics paper, the chemistry paper and the biology paper. No two of these three papers may be taken consecutively. There is no restriction on the order in which the other examination papers may be taken.
\nFind the number of different orders in which these 18 examination papers may be taken.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nconsideration of all papers
\nall papers may be sat in 18! ways A1
\nnumber of ways of positioning “pairs” of science subjects A1
\nbut this includes two copies of each “triple” (R1)
\nnumber of ways of positioning “triplets” of science subjects A1
\nhence number of arrangements is M1A1
\n\n
METHOD 2
\nconsideration of all the non-science papers (M1)
\nhence all non-science papers can be sat in 15! ways A1
\nthere are ways of positioning the three science papers (M1)A1
\nhence the number of arrangements is (M1)A1
\nMETHOD 3
\nconsideration of all papers
\nall papers may be sat in 18! ways A1
\nnumber of ways of positioning exactly two science subjects M1A1
\nnumber of ways of positioning “triplets” of science subjects A1
\nhence number of arrangements is M1A1
\n\n
[6 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.
\nUse your graphic display calculator to find the -value 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.
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 (0.208093…) (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]
\nA team of four is to be chosen from a group of four boys and four girls.
\nFind the number of different possible teams that could be chosen.
\nFind the number of different possible teams that could be chosen, given that the team must include at least one girl and at least one boy.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\n(A1)
\n(M1)
\n= 70 A1
\n\n
METHOD 2
\nrecognition that they need to count the teams with 0 boys, 1 boy… 4 boys M1
\n\n
(A1)
\n= 70 A1
\n\n
[3 marks]
\nEITHER
\nrecognition that the answer is the total number of teams minus the number of teams with all girls or all boys (M1)
\n70 − 2
\nOR
\nrecognition that the answer is the total of the number of teams with 1 boy,
\n2 boys, 3 boys (M1)
\n\n
\n
THEN
\n= 68 A1
\n\n
[2 marks]
\nConsider the curves and defined as follows
\n,
\n,
\nUsing implicit differentiation, or otherwise, find for each curve in terms of and .
\nLet P(, ) be the unique point where the curves and intersect.
\nShow that the tangent to at P is perpendicular to the tangent to at P.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(M1)
\nNote: M1 is for use of both product rule and implicit differentiation.
\n\n
A1
\nNote: Accept
\n\n
(M1)
\nA1
\nNote: Accept
\n\n
[4 marks]
\nsubstituting and for and M1
\nproduct of gradients at P is or equivalent reasoning R1
\nNote: The R1 is dependent on the previous M1.
\n\n
so tangents are perpendicular AG
\n\n
[2 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]
\nDifferentiate from first principles the function .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\n\n
M1
\n(A1)
\nA1
\ncancelling M1
\n\n
then
\nA1
\nNote: Final A1 dependent on all previous marks.
\n\n
METHOD 2
\n\n
M1
\n(A1)
\nA1
\ncancelling M1
\n\n
then
\nA1
\nNote: Final A1 dependent on all previous marks.
\n\n
[5 marks]
\n\n
Sila 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.
\nthe statistic.
\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(8.68507…) (G2)
\n[2 marks]
\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]
\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.
\nattempt 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]
\nTwo distinct lines, and , intersect at a point . In addition to , four distinct points are marked out on and three distinct points on . A mathematician decides to join some of these eight points to form polygons.
\nThe line has vector equation r1 , and the line has vector equation r2 , .
\nThe point has coordinates (4, 6, 4).
\nThe point has coordinates (3, 4, 3) and lies on .
\nThe point has coordinates (−1, 0, 2) and lies on .
\nFind how many sets of four points can be selected which can form the vertices of a quadrilateral.
\nFind how many sets of three points can be selected which can form the vertices of a triangle.
\nVerify that is the point of intersection of the two lines.
\nWrite down the value of corresponding to the point .
\nWrite down and .
\nLet be the point on with coordinates (1, 0, 1) and be the point on with parameter .
\nFind the area of the quadrilateral .
\nappreciation that two points distinct from need to be chosen from each line M1
\n\n
=18 A1
\n[2 marks]
\nEITHER
\nconsider cases for triangles including or triangles not including M1
\n(A1)(A1)
\nNote: Award A1 for 1st term, A1 for 2nd & 3rd term.
\nOR
\nconsider total number of ways to select 3 points and subtract those with 3 points on the same line M1
\n(A1)(A1)
\nNote: Award A1 for 1st term, A1 for 2nd & 3rd term.
\n56−10−4
\nTHEN
\n= 42 A1
\n[4 marks]
\nMETHOD 1
\nsubstitution of (4, 6, 4) into both equations (M1)
\nand A1A1
\n(4, 6, 4) AG
\nMETHOD 2
\nattempting to solve two of the three parametric equations M1
\nand A1
\ncheck both of the above give (4, 6, 4) M1AG
\nNote: If they have shown the curve intersects for all three coordinates they only need to check (4,6,4) with one of \"\" or \"\".
\n[3 marks]
\nA1
\n[1 mark]
\n, A1A1
\nNote: Award A1A0 if both are given as coordinates.
\n[2 marks]
\nMETHOD 1
\narea triangle M1
\nA1
\nA1
\nEITHER
\n, (M1)
\narea triangle area triangle (M1)A1
\nA1
\nOR
\nhas coordinates (−11, −12, −2) A1
\narea triangle M1A1
\nNote: A1 is for the correct vectors in the correct formula.
\nA1
\nTHEN
\narea of
\nA1
\n\n
METHOD 2
\nhas coordinates (−11, −12, −2) A1
\narea M1
\nNote: Award M1 for use of correct formula on appropriate non-overlapping triangles.
\nNote: Different triangles or vectors could be used.
\n, A1
\nA1
\n, A1
\nA1
\nNote: Other vectors which might be used are , , .
\nNote: Previous A1A1A1A1 are all dependent on the first M1.
\nvalid attempt to find a value of M1
\nNote: M1 independent of triangle chosen.
\narea
\nA1
\nNote: accept or equivalent.
\n\n
[8 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]
\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]
\nThe curve is given by the equation .
\nAt the point (1, 1) , show that .
\nHence find the equation of the normal to at the point (1, 1).
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nattempt to differentiate implicitly M1
\nA1A1
\nNote: Award A1 for each term.
\nattempt to substitute , into their equation for M1
\n\n
A1
\nAG
\n[5 marks]
\nattempt to use gradient of normal (M1)
\n\n
so equation of normal is or A1
\n[2 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 functions defined for , given by and .
\nHence, or otherwise, find .
\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 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]
\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.
\nJohn purchased the bicycle in 2008.
\nJustify why John should not insure his bicycle in 2019.
\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]
\n2019 is the 12th year/term (M1)
\nNote: Award (M1) for 12 seen.
\n75.59 (value of bicycle) AND 81.5 (cost of insurance policy) (A1)(ft)
\nNote: Award (A1) for both sequences’ 12th term seen. The value of the bicycle will follow through from their common ratio in part (b). Do not award (M0)(A1).
\nthe cost of the insurance policy is greater than the value of the bicycle (R1)(ft)
\nNote: Award (R1)(ft) for a reason consistent with their cost of insurance policy and their value of the bicycle. Follow through within this part. Award (R0) if the correct values are not explicitly seen. Accept the following contextualized reasons: “the insurance is not worth it\", \"the values are too close\", \"insurance is as much as the value of the bike\", but only if their cost of insurance is greater than the value of the bicycle.
\nOR
\n75.59 < 81.5 (R1)(ft)
\nNote: Award (R1)(ft) for a correct numerical comparison showing their cost of insurance policy is greater than their value of the bicycle. Follow through within this part.
\n[3 marks]
\nUse the method of mathematical induction to prove that is divisible by 9 for .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
let be the proposition that is divisible by 9
\nshowing true for A1
\niefor
\nwhich is divisible by 9, therefore is true
\nassume is true so M1
\n\n
Note: Only award M1 if “truth assumed” or equivalent.
\n\n
consider
\n\n
M1
\nA1
\nwhich is divisible by 9 R1
\n\n
Note: Award R1 for either the expression or the statement above.
\n\n
since is true and true implies is true, therefore (by the principle of mathematical induction) is true for R1
\n\n
Note: Only award the final R1 if the 2 M1s have been awarded.
\n\n
[6 marks]
\nLet
\nFind .
\nFind .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1A1
\nNote: M1 is for use of the chain rule.
\n[2 marks]
\nattempt at integration by parts M1
\n\n
(A1)
\nA1
\nusing integration by substitution or inspection (M1)
\nA1
\nNote: Award A1 for or equivalent.
\nNote: Condone lack of limits to this point.
\nattempt to substitute limits into their integral M1
\nA1
\n[7 marks]
\nConsider the function .
\nDetermine whether is an odd or even function, justifying your answer.
\nBy using mathematical induction, prove that
\nwhere .
\nHence or otherwise, find an expression for the derivative of with respect to .
\nShow that, for , the equation of the tangent to the curve at is .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\neven function A1
\nsince and is a product of even functions R1
\nOR
\neven function A1
\nsince R1
\n\n
Note: Do not award A0R1.
\n\n
[2 marks]
\nconsider the case
\nM1
\nhence true for R1
\nassume true for , ie, M1
\n\n
Note: Do not award M1 for “let ” or “assume ” or equivalent.
\n\n
consider :
\n(M1)
\nA1
\nA1
\nA1
\nso true and true true. Hence true for all R1
\n\n
Note: To obtain the final R1, all the previous M marks must have been awarded.
\n\n
[8 marks]
\nattempt to use (or correct product rule) M1
\nA1A1
\n\n
Note: Award A1 for correct numerator and A1 for correct denominator.
\n\n
[3 marks]
\n(M1)(A1)
\n(A1)
\nA1
\nA1
\nA1
\n\n
Note: This A mark is independent from the previous marks.
\n\n
M1A1
\nAG
\n[8 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.
METHOD 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]
\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]
\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]
\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]
\nProve by mathematical induction that , where .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
show true for (M1)
\nA1
\nhence true for
\nassume true for M1
\nconsider for (M1)
\nA1
\nor any correct expression with a visible common factor (A1)
\nor any correct expression with a common denominator (A1)
\n\n
\n
Note: At least one of the above three lines or equivalent must be seen.
\n\n
or equivalent A1
\n\n
Result is true for . If result is true for it is true for . Hence result is true for all . Hence proved by induction. R1
\n\n
Note: In order to award the R1 at least [5 marks] must have been awarded.
\n\n
[9 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.
\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.
\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]
\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]
\nSolve the inequality .
\nUse mathematical induction to prove that for , .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1A1
\n\n
Note: Award A1 for −0.414, 2.41 and A1 for correct inequalities.
\n[2 marks]
\ncheck for ,
\n16 > 9 so true when A1
\nassume true for
\nM1
\nNote: Award M0 for statements such as “let ”.
\nNote: Subsequent marks after this M1 are independent of this mark and can be awarded.
\nprove true for
\n\n
M1
\n(M1)
\n(from part (a)) A1
\nwhich is true for ≥ 3 R1
\nNote: Only award the A1 or the R1 if it is clear why. Alternate methods are possible.
\n\n
hence if true for true for , true for so true for all ≥ 3 R1
\nNote: Only award the final R1 provided at least three of the previous marks are awarded.
\n[7 marks]
\nGiven that can be expressed in the form , find the values of the constants , and .
\nHence find .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
A1
\nA1
\n[2 marks]
\nM1M1
\nNote: Award M1 for dividing by to get , M1 for separating the and 1.
\n(M1)A1A1
\nNote: Award (M1)A1 for integrating , A1 for the other two terms.
\n[5 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.
![\"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]
\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]
\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]
\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]
\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]
\nUse the principle of mathematical induction to prove that
\n, where .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
if
\nM1
\nhence true for
\nassume true for M1
\nNote: Assumption of truth must be present. Following marks are not dependent on the first two M1 marks.
\nso
\nif
\n\n
M1A1
\nfinding a common denominator for the two fractions M1
\n\n
A1
\nhence if true for then also true for , as true for , so true (for all ) R1
\nNote: Award the final R1 only if the first four marks have been awarded.
\n[7 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.
Find the value of x.
\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]
\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]
\nUse mathematical induction to prove that for where .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
Let be the statement: for some where consider the case M1
\nbecause . Therefore is true R1
\nassume is true for some
\nM1
\nNote: Assumption of truth must be present. Following marks are not dependent on this M1.
\nEITHER
\nconsider M1
\nA1
\nis true (as ) R1
\nOR
\nmultiply both sides by (which is positive) M1
\n\n
A1
\nis true (as ) R1
\nTHEN
\nis true is true is true so true for all (or equivalent) R1
\nNote: Only award the last R1 if at least four of the previous marks are gained including the A1.
\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.
(M1)(A1)
\nA1
\nA1
\n[4 marks]
\n(M1)
\n= 12 A1
\n[2 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]
\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]
Use mathematical induction to prove that , for .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
consider . and therefore true for R1
\nNote: There must be evidence that has been substituted into both expressions, or an expression such LHS=RHS=1 is used. “therefore true for ” or an equivalent statement must be seen.
\n\n
assume true for , (so that ) M1
\nNote: Assumption of truth must be present.
\n\n
consider
\n(M1)
\nA1
\nM1
\nNote: M1 is for factorising
\n\n
\n
\n
so if true for , then also true for , and as true for then true for all R1
\nNote: Only award final R1 if all three method marks have been awarded.
Award R0 if the proof is developed from both LHS and RHS.
\n
[6 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 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]
\nLet S be the sum of the roots found in part (a).
\nFind the roots of which satisfy the condition , expressing your answers in the form , where , .
\nShow that Re S = Im S.
\nBy writing as , find the value of cos in the form , where , and are integers to be determined.
\nHence, or otherwise, show that S = .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n\n
use of De Moivre’s theorem (M1)
\n(A1)
\n, (A1)
\n1, 2, 3, 4, 5
\nor or or or A2
\nNote: Award A1 if additional roots are given or if three correct roots are given with no incorrect (or additional) roots.
\n\n
[5 marks]
\nRe S =
\nIm S = A1
\nNote: Award A1 for both parts correct.
\nbut , , , and M1A1
\n⇒ Re S = Im S AG
\nNote: Accept a geometrical method.
\n\n
[4 marks]
\nM1A1
\n\n
A1
\n\n
[3 marks]
\n\n
(M1)
\nNote: Allow alternative methods eg .
\n(A1)
\nRe S =
\nRe S = A1
\nA1
\n\n
S = Re(S)(1 + i) since Re S = Im S, R1
\nS = AG
\n\n
[4 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]
\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 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.
\nThe average body weight of mice is 0.023 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).
\nState whether it is valid to use the regression line to estimate the average weight of the brain of mice. 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.
(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]
\nNot valid (A1)
\nthe mouse is smaller/lighter/weighs less than the cat (lightest mammal) (R1)
\nOR
\nas it would mean the mouse’s brain is heavier than the whole mouse (R1)
\nOR
\n0.023 kg is outside the given data range. (R1)
\nOR
\nExtrapolation (R1)
\n\n
Note: Do not award (A1)(R0). Do not accept percentage error as a reason for validity.
\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 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]
\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]
\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]
\nLet be one of the non-real solutions of the equation .
\nConsider the complex numbers and , where .
\nDetermine the value of
\n(i) ;
\n(ii) .
\nShow that .
\nFind the values of that satisfy the equation .
\nSolve the inequality .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(i) METHOD 1
\nA1
\nas R1
\nMETHOD 2
\nsolutions of are A1
\nverification that the sum of these roots is 0 R1
\n(ii) A2
\n[4 marks]
\nM1A1
\nEITHER
\nM1
\nA1
\nOR
\nM1
\nA1
\nOR
\nsubstitution by in any form M1
\nnumerical values of each term seen A1
\nTHEN
\nAG
\n[4 marks]
\n(M1)(A1)
\nA1
\n(M1)
\nA1
\n[5 marks]
\nM1A1
\nM1
\nA1
\nM1
\nA1
\n[6 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]
\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]
\nBy using the substitution , show that .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
EITHER
\n\n
M1A1
\n\n
M1A1
\nOR
\n\n
M1A1
\n\n
M1A1
\nTHEN
\n(M1)
\n\n
A1
\nM1
\n\n
Note: This M1 may be seen anywhere, including a sketch of an appropriate triangle.
\n\n
so AG
\n[7 marks]
\nUsing the substitution show that .
\nHence find the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
let
\n(A1)
\nM1
\n\n
Note: The method mark is for an attempt to substitute for both and .
\n\n
(or equivalent) A1
\nwhen and when M1
\nAG
\n[4 marks]
\nM1
\nA1
\nA1
\n[3 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]
\nUsing the substitution , find .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)
\nvalid attempt to write integral in terms of and M1
\nA1
\n\n
(A1)
\nor equivalent A1
\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\"/)
The researcher finds that 10% of the leaves have a length greater than cm.
\nWrite down the median length of these leaves.
\nWrite down the number of leaves with a length less than or equal to 8 cm.
\nUse the graph to find the value of .
\nBefore measuring, the researcher estimated to be approximately 9.5 cm. Find the percentage error in her estimate.
\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]
\nor equivalent (M1)
\n\n
Note: Award (M1) for a horizontal line drawn through the cumulative frequency value of 180 and meeting the curve (or the corresponding vertical line from 10.5 cm).
\n\n
(A1) (C2)
\n\n
Note: Accept an error of ±0.1.
\n\n
[2 marks]
\n(M1)
\n\n
Notes: Award (M1) for their correct substitution into the percentage error formula.
\n\n
(A1)(ft) (C2)
\n\n
Notes: Follow through from their answer to part (c)(i).
\nAward (A1)(A0) for an answer of with or without working.
\n\n
[2 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]
\nUse the substitution to find .
\nHence find the value of , expressing your answer in the form arctan , where .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(accept or equivalent) A1
\nsubstitution, leading to an integrand in terms of M1
\nor equivalent A1
\n= 2 arctan A1
\n[4 marks]
\n\n
\n
= arctan 3 − arctan 1 A1
\ntan(arctan 3 − arctan 1) = (M1)
\ntan(arctan 3 − arctan 1) =
\narctan 3 − arctan 1 = arctan A1
\n[3 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.
\n(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]
\nThe function is defined by , .
\nUse integration by parts to find .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nwrite as (M1)
\nM1A1
\n(M1)(A1)
\nA1
\n\n
METHOD 2
\nlet M1
\n\n
A1
\nM1
\nA1
\n\n
M1A1
\n\n
METHOD 3
\nSetting up and M1
\nM1A1
\nM1A1
\nA1
\n\n
\n
[6 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]
\nFind
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt at integration by parts with and M1
\nA1A1
\n\n
Note: Award A1 for and A1 for .
\n\n
solving by substitution with or inspection (M1)
\nA1
\n[5 marks]
\nLet ,
\nwhere and .
\nCalculate the value of . Write down your full calculator display.
\nWrite your answer to part (a)
\n(i) correct to two decimal places;
\n(ii) correct to three significant figures.
\nWrite your answer to part (b)(ii) 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)
\n\n
Note: Award (M1) for correct substitution into formula.
\n\n
(A1) (C2)
\n\n
Note: Accept .
\n\n
[2 marks]
\n(i) 0.04 (A1)(ft)
\n(ii) 0.0391 (A1)(ft) (C2)
\n\n
Note: Follow through from part (a).
\n\n
[2 marks]
\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 random variables X , Y follow a bivariate normal distribution with product moment correlation coefficient ρ.
\nA random sample of 11 observations on X, Y was obtained and the value of the sample product moment correlation coefficient, r, was calculated to be −0.708.
\nThe covariance of the random variables U, V is defined by
\nCov(U, V) = E((U − E(U))(V − E(V))).
\nState suitable hypotheses to investigate whether or not a negative linear association exists between X and Y.
\nDetermine the p-value.
\nState your conclusion at the 1 % significance level.
\nShow that Cov(U, V) = E(UV) − E(U)E(V).
\nHence show that if U, V are independent random variables then the population product moment correlation coefficient, ρ, is zero.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
H0 : ρ = 0; H1 : ρ < 0 A1
\n[1 mark]
\n(M1)
\ndegrees of freedom = 9 (A1)
\nP(T < −3.0075...) = 0.00739 A1
\nNote: Accept any answer that rounds to 0.0074.
\n[3 marks]
\nreject H0 or equivalent statement R1
\nNote: Apply follow through on the candidate’s p-value.
\n[1 mark]
\nCov(U, V) + E((U − E(U))(V − E(V)))
\n= E(UV − E(U)V − E(V)U + E(U)E(V)) M1
\n= E(UV) − E(E(U)V) − E(E(V)U) + E(E(U)E(V)) (A1)
\n= E(UV) − E(U)E(V) − E(V)E(U) + E(U)E(V) A1
\nCov(U, V) = E(UV) − E(U)E(V) AG
\n[3 marks]
\nE(UV) = E(U)E(V) (independent random variables) R1
\n⇒Cov(U, V) = E(U)E(V) − E(U)E(V) = 0 A1
\nhence, ρ = A1AG
\nNote: Accept the statement that Cov(U,V) is the numerator of the formula for ρ.
\nNote: Only award the first A1 if the R1 is awarded.
\n[3 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]
\nPassengers of Flyaway Airlines can purchase tickets for either Business Class or Economy Class.
\nOn one particular flight there were 154 passengers.
\nLet be the number of Business Class passengers and be the number of Economy Class passengers on this flight.
\nOn this flight, the cost of a ticket for each Business Class passenger was 320 euros and the cost of a ticket for each Economy Class passenger was 85 euros. The total amount that Flyaway Airlines received for these tickets was .
\nThe airline’s finance officer wrote down the total amount received by the airline for these tickets as .
\nUse the above information to write down an equation in and .
\nUse the information about the cost of tickets to write down a second equation in and .
\nFind the value of and the value of .
\nFind the percentage error.
\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) (C1)
\n[1 mark]
\n(A1)(ft)(A1)(ft) (C2)
\n\n
Note: Follow through from parts (a) and (b) irrespective of working seen, but only if both values are positive integers.
\nAward (M1)(A0) for a reasonable attempt to solve simultaneous equations algebraically, leading to at least one incorrect or missing value.
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution into percentage error formula.
\n\n
(A1) (C2)
\n[2 marks]
\nConsider the complex number .
\nExpress in the form , where .
\nFind the exact value of the modulus of .
\nFind the argument of , giving your answer to 4 decimal places.
\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]
\nattempt to use (M1)
\nor equivalent A1
\nNote: A1 is only awarded for the correct exact value.
\n[2 marks]
\nEITHER
\narg = arg(2 + 7i) − arg(6 + 2i) (M1)
\nOR
\narg = arctan (M1)
\nTHEN
\narg = 0.9707 (radians) (= 55.6197 degrees) A1
\nNote: Only award the last A1 if 4 decimal places are given.
\n[2 marks]
\nFind the roots of the equation , . Give your answers in Cartesian form.
\nOne of the roots satisfies the condition .
\nGiven that , express in the form , where , .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\n\n
writing (M1)
\nNote: Award M1 for an attempt to find cube roots of using modulus-argument form.
\ncube roots (M1)
\ni.e. A2
\nNote: Award A2 for all 3 correct, A1 for 2 correct.
\nNote: Accept and .
\n\n
METHOD 2
\n\n
M1
\nM1
\nA2
\nNote: Award A2 for all 3 correct, A1 for 2 correct.
\nNote: Accept and .
\n\n
[4 marks]
\n\n
M1
\n\n
\n
A1
\nA1
\nNote: Accept .
\n[3 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]
\nLast year a South American candy factory sold 4.8 × 108 spherical sweets. Each sweet has a diameter of 2.5 cm.
\nThe factory is producing an advertisement showing all of these sweets placed in a straight line.
\nThe advertisement claims that the length of this line is x times the length of the Amazon River. The length of the Amazon River is 6400 km.
\nFind the length, in cm, of this line. Give your answer in the form a × 10k , where 1 ≤ a < 10 and k ∈ .
\nWrite down the length of the Amazon River in cm.
\nFind the value of x.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
4.8 × 108 × 2.5 (M1)
\nNote: Award (M1) for multiplying by 2.5.
\n1.2 × 109 (cm) (A1)(ft)(A1)(ft) (C3)
\nNote: Award (A0)(A0) for answers of the type 12 × 108.
\n[3 marks]
\n640 000 000 (cm) (6.4 × 108 (cm)) (A1)
\n[1 mark]
\n(M1)
\nNote: Award (M1) for division by 640 000 000.
\n= 1.88 (1.875) (A1)(ft) (C3)
\nNote: Follow through from part (a) and part (b)(i).
\n[2 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
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]
\nLet , , and let .
\nShow the points represented by and on the following Argand diagram.
\nFind an expression in terms of θ for .
\nFind an expression in terms of θ for .
\nHence or otherwise find the value of θ for which .
\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]
\n(or any equivalent) A1
\n[1 mark]
\n(M1)
\n(or any equivalent) A1
\n[2 marks]
\nMETHOD 1
\nif then , ( odd) (M1)
\n\n
(A1)
\nA1
\n\n
METHOD 2
\nM1
\n\n
A1
\nA1
\nNote: Accept any equivalent, eg .
\n\n
\n
[3 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]
\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]
\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.
\nFind the value of when .
\nFind the function, , that would define this footpath on the map.
\nState the domain of .
\nFind the coordinates of the bridges relative to the centre of Orangeton.
\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.
\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 given function.
\n(A1)(G2)
\n[2 marks]
\n(M1)
\nNote: Award (M1) for correct substitution into gradient formula. Accept equivalent forms such as .
\n(1.75) (A1)(ft)
\n(A1)(ft)(G3)
\nNote: Follow through from part (a).
\n[3 marks]
\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(0.360, 1.34) ((0.359947…, 1.33669)) (A1)(A1)
\n(3.63, 1.01) ((3.63066…, 1.00926…)) (A1)(A1)
\nNote: Award (A1)(A1) for each correct coordinate pair. Accept correct answers in the form of , etc. Award at most (A0)(A1)(A1)(A1)ft if one or both parentheses are omitted.
\n[4 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]
\nIn the Canadian city of Ottawa:
\n\n
The total population of Ottawa is .
\nCalculate the percentage of the population of Ottawa that speak English but not French.
\nCalculate the number of people in Ottawa that speak both English and French.
\nWrite down your answer to part (b) in the form where and k .
\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 subtracting 36 from 97.
\n\n
OR
\n![\"M17/5/MATSD/SP1/ENG/TZ1/02.a/M\"](\"../assets/uploads/tinymce_asset/asset/3575/Schermafbeelding_2017-08-15_om_12.54.01.png\"/)
(M1)
\n\n
Note: Award (M1) for 61 and 36 seen in the correct places in the Venn diagram.
\n\n
(A1) (C2)
\n\n
Note: Accept 61.0 (%).
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for multiplying 0.36 (or equivalent) by .
\n\n
(A1) (C2)
\n[2 marks]
\n(A1)(ft)(A1)(ft) (C2)
\n\n
Note: Award (A1)(ft) for 3.55 (3.546) must match part (b), and (A1)(ft) .
\nAward (A0)(A0) for answers of the type: . Follow through from part (b).
\n\n
[2 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 $60.
\nThe tea-shop’s profit increases by 10 % every week.
\nCalculate the café’s total profit for the first 12 weeks.
\nCalculate the tea-shop’s total profit for the first 12 weeks.
\n(M1)(A1)(ft)
\nNote: Award (M1) for substituting the arithmetic series formula, (A1)(ft) for correct substitution. Follow through from their first term and common difference in part (a).
\n= ($) 1380 (A1)(ft)(G2)
\n[3 marks]
\n\n
(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]
\nConsider the function .
\nThe function has one local maximum at and one local minimum at .
\nWrite down the -intercept of the graph of .
\nSketch the graph of for −3 ≤ ≤ 3 and −4 ≤ ≤ 12.
\nDetermine the range of for ≤ ≤ .
\n−1 (A1)
\nNote: Accept (0, −1).
\n[1 mark]
\n (A1)(A1)(A1)(A1)
Note: Award (A1) for correct window and axes labels, −3 to 3 should be indicated on the -axis and −4 to 12 on the -axis.
(A1)) for smooth curve with correct cubic shape;
(A1) for -intercepts: one close to −3, the second between −1 and 0, and third between 1 and 2; and -intercept at approximately −1;
(A1) for local minimum in the 4th quadrant and maximum in the 2nd quadrant, in approximately correct positions.
Graph paper does not need to be used. If window not given award at most (A0)(A1)(A0)(A1).
[4 marks]
\n(A1)(ft)(A1)(ft)(A1)
\nNote: Award (A1) for −1.27 seen, (A1) for 1.33 seen, and (A1) for correct weak inequalities with their endpoints in the correct order. For example, award (A0)(A0)(A0) for answers like . Accept in place of . Accept alternative correct notation such as [−1.27, 1.33].
\nFollow through from their and values from part (g) only if their and values are between −4 and 12. Award (A0)(A0)(A0) if their values from (g) are given as the endpoints.
\n[3 marks]
\nSolve the equation .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 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]
\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]
\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]
\nConsider the numbers and .
\nCalculate . Give your full calculator display.
\nWrite down your answer to part (a) correct to two decimal places;
\nWrite down your answer to part (a) correct to three significant figures.
\nWrite your answer to part (b)(ii) in the form , where .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nOR (M1)
\n\n
Note: Award (M1) for correct substitution into given expression.
\n\n
44664.59503 (A1) (C2)
\n\n
Note: Award (A1) for a correct answer with at least 8 digits.
\nAccept 44664.5950301.
\n\n
[2 marks]
\n44664.60 (A1)(ft) (C1)
\n\n
Note: For a follow through mark, the answer to part (a) must be to at least 3 decimal places.
\n\n
[1 mark]
\n44700 (A1)(ft) (C1)
\n\n
Notes: Answer to part (a) must be to at least 4 significant figures.
\nAccept any equivalent notation which is correct to 3 significant figures.
\nFor example or .
\nFollow through from part (a).
\n\n
[1 mark]
\n(A1)(ft)(A1)(ft) (C2)
\n\n
Notes: Award (A1)(ft) for 4.47 and (A1)(ft) for .
\nAward (A0)(A0) for answers such as .
\nFollow through from part (b)(ii) only.
\n\n
[2 marks]
\nShow that .
\nShow that .
\nHence or otherwise find in the form where , .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nM1A1
\nNote: Do not award the M1 for just .
\nNote: Do not award A1 if correct expression is followed by incorrect working.
\nAG
\n[2 marks]
\nM1
\nNote: M1 is for an attempt to change both terms into sine and cosine forms (with the same argument) or both terms into functions of .
\n\n
A1A1
\nNote: Award A1 for numerator, A1 for denominator.
\nM1
\nAG
\nNote: Apply MS in reverse if candidates have worked from RHS to LHS.
\nNote: Alternative method using and in terms of .
\n[4 marks]
\nMETHOD 1
\nA1
\nNote: Award A1 for correct expression with or without limits.
\nEITHER
\nor (M1)A1A1
\nNote: Award M1 for integration by inspection or substitution, A1 for , A1 for completely correct expression including limits.
\nM1
\nNote: Award M1 for substitution of limits into their integral and subtraction.
\n(A1)
\nOR
\nlet M1
\n\n
A1A1
\nNote: Award A1 for correct limits even if seen later, A1 for integral.
\nor A1
\nM1
\nTHEN
\n\n
Note: Award M1 for both putting the expression over a common denominator and for correct use of law of logarithms.
\n(M1)A1
\n\n
METHOD 2
\nA1A1
\nA1A1(A1)
\nM1
\nM1A1
\nA1
\n\n
\n
[9 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]
\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\"/)
Write down the equation of the axis of symmetry for this graph.
\nFind the value of .
\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.
(A1)(A1) (C2)
\n\n
Note: Award (A1) for constant, (A1) for the constant being 3.
\nThe answer must be an equation.
\n\n
[2 marks]
\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]
\nOR (A1)(A1)
\n\n
Note: Award (A1) for two correct interval endpoints, (A1) for left endpoint excluded and right endpoint included.
\n\n
[2 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]
\nThe function is defined by , where 0 ≤ ≤ 5. The curve is shown on the following graph which has local maximum points at A and C and touches the -axis at B and D.
\nUse integration by parts to show that , .
\nHence, show that , .
\nFind the -coordinates of A and of C , giving your answers in the form , where , .
\nFind the area enclosed by the curve and the -axis between B and D, as shaded on the diagram.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1
\nattempt at integration by parts with , M1
\nA1
\n= M1A1
\n=
\nM1
\nAG
\n\n
\n
METHOD 2
\nattempt at integration by parts with , M1
\nA1
\nM1A1
\n\n
M1
\nAG
\n\n
METHOD 3
\nattempt at use of table M1
\neg
\n A1A1
Note: A1 for first 2 lines correct, A1 for third line correct.
\nM1
\nM1
\nAG
\n\n
[5 marks]
\nM1A1
\nA1
\nAG
\nNote: Do not accept solutions where the RHS is differentiated.
\n\n
[3 marks]
\nM1A1
\nNote: Award M1 for an attempt at both the product rule and the chain rule.
\n(M1)
\nNote: Award M1 for an attempt to factorise or divide by .
\ndiscount (as this would also be a zero of the function)
\n\n
(M1)
\n(at A) and (at C) A1A1
\nNote: Award A1 for each correct answer. If extra values are seen award A1A0.
\n\n
[6 marks]
\n\n
or A1
\nNote: The A1may be awarded for work seen in part (c).
\nM1
\nM1(A1)A1
\nNote: Award M1 for substitution of the end points and subtracting, (A1) for and and A1 for a completely correct answer.
\n\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]
\nLet where .
\nExpress in terms of sin and cos .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)A1
\n(M1)A1
\nNote: The two M1s can be awarded for observation or for expanding.
\nA1
\n[5 marks]
\nA comet orbits the Sun and is seen from Earth every 37 years. The comet was first seen from Earth in the year 1064.
\nFind the year in which the comet was seen from Earth for the fifth time.
\nDetermine how many times the comet has been seen from Earth up to the year 2014.
\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 substitution.
\n\n
(A1) (C3)
\n[3 marks]
\n(M1)
\n\n
Note: Award (M1) for a correct substitution into arithmetic sequence formula.
\nAccept an equation.
\n\n
(A1)
\n26 (times) (A1) (C3)
\n\n
Note: Award the final (A1) for the correct rounding down of their unrounded answer.
\n\n
OR
\n(M1)
\n\n
Note: Award (M1) for a correct substitution into a linear model (where ).
\nAccept an equation or weak inequality.
\nAccept for (M1).
\n\n
(A1)
\n\n
26 (times) (A1) (C3)
\n\n
Note: Award the final (A1) for adding 1 to the correct rounding down of their unrounded answer.
\n\n
[3 marks]
\nVioleta plans to grow flowers in a rectangular plot. She places a fence to mark out the perimeter of the plot and uses 200 metres of fence. The length of the plot is metres.
\n![\"M17/5/MATSD/SP2/ENG/TZ2/05\"](\"../assets/uploads/tinymce_asset/asset/3610/Schermafbeelding_2017-08-16_om_17.40.47.png\"/)
Violeta places the fence so that the area of the plot is maximized.
\nBy selling her flowers, Violeta earns 2 Bulgarian Levs (BGN) per square metre of the plot.
\nShow that the width of the plot, in metres, is given by .
\nWrite down the area of the plot in terms of .
\nFind the value of that maximizes the area of the plot.
\nShow that Violeta earns 5000 BGN from selling the flowers grown on the plot.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(or equivalent) (M1)
\nOR
\n(or equivalent) (M1)
\n\n
Note: Award (M1) for a correct expression leading to (the does not need to be seen). The 200 must be seen for the (M1) to be awarded. Do not accept substituted in the perimeter of the rectangle formula.
\n\n
(AG)
\n[1 mark]
\nOR (or equivalent) (A1)
\n[1 mark]
\nORORgraphical method (M1)
\n\n
Note: Award (M1) for use of axis of symmetry formula or first derivative equated to zero or a sketch graph.
\n\n
(A1)(ft)(G2)
\n\n
Note: Follow through from part (b), provided x is positive and less than 100.
\n\n
[2 marks]
\n(M1)(M1)
\n\n
Note: Award (M1) for substituting their into their formula for area (accept “” for the substituted formula), and (M1) for multiplying by 2. Award at most (M0)(M1) if their calculation does not lead to 5000 (BGN), although the 5000 (BGN) does not need to be seen explicitly.
\nSubstitution of 50 into area formula may be seen in part (c).
\n\n
5000 (BGN) (AG)
\n[2 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.
\nC represents the complex number A2
\nD represents the complex number A2
\n[4 marks]
\nConsider the functions and defined on the domain by and .
\nThe following diagram shows the graphs of and
\nFind the -coordinates of the points of intersection of the two graphs.
\nFind the exact area of the shaded region, giving your answer in the form , where , .
\nAt the points A and B on the diagram, the gradients of the two graphs are equal.
\nDetermine the -coordinate of A on the graph of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n\n
attempt to form a quadratic in M1
\nA1
\n\n
valid attempt to solve their quadratic M1
\n\n
A1
\nA1A1
\nNote: Ignore any “extra” solutions.
\n[6 marks]
\nconsider (±) M1
\nA1
\nNote: Ignore lack of or incorrect limits at this stage.
\nattempt to substitute their limits into their integral M1
\n\n
\n
A1A1
\n[5 marks]
\nattempt to differentiate both functions and equate M1
\nA1
\nattempt to solve for M1
\n\n
\n
A1
\nM1
\nA1
\n[6 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]
\nDetermine the roots of the equation , , giving the answers in the form where .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nA1
\n(M1)
\nA1
\n\n
\n
A2
\n\n
Note: Award A1A0 for one correct root.
\n\n
so roots are and M1A1
\n\n
Note: Award M1 for subtracting 2i from their three roots.
\n\n
METHOD 2
\n\n
M1A1
\n\n
and M1A1
\nand
\nor
\nif
\nA1
\n\n
if
\n\n
\n
\n
\n
\n
A1A1
\nso roots are and
\n\n
METHOD 3
\n\n
attempt to factorise: M1
\nA1
\nA1
\nA1
\nM1
\n\n
\n
A1A1
\n\n
Special Case:
\nNote: If a candidate recognises that (anywhere seen), and makes no valid progress in finding three roots, award A1 only.
\n\n
[7 marks]
\nA function is given by .
\nThe graph of the function intersects the graph of .
\nExpand the expression for .
\nFind .
\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(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![\"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 the geometric sequence .
\nWrite down the common ratio of the sequence.
\nFind the value of .
\nFind the smallest value of for which is less than .
\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(M1)
\n\n
Note: Award (M1) for their correct substitution into the geometric sequence formula. Accept a list of their five correct terms.
\n\n
(A1)(ft) (C2)
\n\n
Note: Follow through from their common ratio from part (a).
\n\n
[2 marks]
\n(M1)(M1)
\n\n
Notes: Award (M1) for their correct substitution into the geometric sequence formula with a variable in the exponent, (M1) for comparing their expression with .
\nAccept an equation.
\n\n
(A1)(ft) (C3)
\n\n
Note: Follow through from their common ratio from part (a). “” must be a positive integer for the (A1) to be awarded.
\n\n
[3 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 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.
\nSketch the graph of the function g (x) = 10x + 40 on the same axes.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(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]
\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]
\ncorrect line, y = 10x + 40, seen on sketch (A1)(A1)
\nNote: Award (A1) for straight line with positive gradient, award (A1) for x-intercept and y-intercept in approximately the correct positions. Award at most (A0)(A1) if ruler not used. If the straight line is drawn on different axes to part (a), award at most (A0)(A1).
\n\n
[2 marks]
\nSolve , giving your answers in the form
\nwhere , , .
\nwhere , .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
\nNote: Accept all answers in the form .
\nOR (M1)A1
\nNote: Accept all answers in the form .
\nNote: Award M1A0 for correct answers in the incorrect form, eg .
\n[3 marks]
\n, A1A1
\n\n
[2 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.
\nUse your graphic display calculator to find the Pearson’s product–moment correlation coefficient, .
\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]
\n(G2)
\n\n
Note: Award (G2) for 0.99 seen. Award (G1) for 0.98 or 0.989. Do not accept 1.00.
\n\n
[2 marks]
\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 random variable has probability density function given by
\n\n
Given that , show that
\nState the mode of .
\nFind .
\nHence show that .
\n.
\n.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nmode is 0 A1
\n[1 mark]
\nattempt at integration by parts (M1)
\n\n
A1
\nA1
\n[3 marks]
\nA1
\n\n
A1
\n(M1)
\nNote: This line can be seen (or implied) anywhere.
\nNote: Do not allow FT A marks from bi to bii.
\n\n
AG
\n[3 marks]
\nattempt to use product rule to differentiate M1
\nA2
\nNote: Award A2 for all terms correct, A1 for 4 correct terms.
\nA1
\nNote: Award A1 for equivalent combination of correct terms over a common denominator.
\nAG
\n[4 marks]
\nM1
\n\n
A1A1
\nNote: Award A1 for first term, A1 for next 3 terms.
\nA1
\nA1
\nAG
\n[5 marks]
\nThe probability distribution of a discrete random variable, , is given by the following table, where and are constants.
\nFind the value of .
\nGiven that , find the value of .
\n(M1)
\nA1
\n[2 marks]
\nattempt to find (M1)
\nA1
\nA1
\nNote: Do not allow FT in part (b) if their is outside the range .
\n[3 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 .
\nFor the graph of ,
\nFind .
\nShow that, if , then .
\nfind the coordinates of the -intercept.
\nshow that there are no -intercepts.
\nsketch the graph, showing clearly any asymptotic behaviour.
\nShow that .
\nThe area enclosed by the graph of and the line can be expressed as . Find the value of .
\nattempt to use quotient rule (or equivalent) (M1)
\nA1
\n\n
[2 marks]
\n\n
simplifying numerator (may be seen in part (i)) (M1)
\nor equivalent quadratic equation A1
\n\n
EITHER
\nuse of quadratic formula
\nA1
\n\n
OR
\nuse of completing the square
\nA1
\n\n
THEN
\n(since is outside the domain) AG
\n\n
Note: Do not condone verification that .
\nDo not award the final A1 as follow through from part (i).
\n\n
[3 marks]
\n(0, 4) A1
\n[1 mark]
\nA1
\noutside the domain R1
\n[2 marks]
\n A1A1
award A1 for concave up curve over correct domain with one minimum point in the first quadrant
award A1 for approaching asymptotically
[2 marks]
\nvalid attempt to combine fractions (using common denominator) M1
\nA1
\n\n
AG
\n[2 marks]
\nM1
\n( or) A1
\n\n
area under the curve is M1
\n\n
Note: Ignore absence of, or incorrect limits up to this point.
\n\n
A1
\n\n
A1
\narea is or M1
\n\n
A1
\n\n
\n
[7 marks]
\nThe first three terms of a geometric sequence are .
\nFind the value of , the common ratio of the sequence.
\nFind the value of for which .
\nFind the sum of the first 30 terms of the sequence.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
OR (M1)
\n\n
Note: Award (M1) for dividing any by .
\n\n
(A1) (C2)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for their correct substitution into geometric sequence formula.
\n\n
(A1)(ft) (C2)
\n\n
Note: Follow through from part (a).
\nAward (A1)(A0) for or with or without working.
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution into geometric series formula.
\n\n
(A1)(ft) (C2)
\n[2 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.
\n3 (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]
\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]
\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.
\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]
\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]
\nA hydraulic hammer drives a metal post vertically into the ground by striking the top of the post. The distance that the post is driven into the ground, by the strike of the hammer, is .
\nThe distances form a geometric sequence.
\nThe distance that the post is driven into the ground by the first strike of the hammer, , is 64 cm.
\nThe distance that the post is driven into the ground by the second strike of the hammer, , is 48 cm.
\nFind the value of the common ratio for this sequence.
\nFind the distance that the post is driven into the ground by the eighth strike of the hammer.
\nFind the total depth that the post has been driven into the ground after 10 strikes of the hammer.
\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 geometric sequence formula.
\n\n
(A1) (C2)
\n[2 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution into geometric sequence formula or list of eight values using their . Follow through from part (a), only if answer is positive.
\n\n
(A1)(ft) (C2)
\n\n
[2 marks]
\n(M1)
\n\n
Note: Award (M1) for correct substitution into geometric series formula. Follow through from part (a), only if answer is positive.
\n\n
(A1)(ft) (C2)
\n[2 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]
\nJuan buys a bicycle in a sale. He gets a discount of 30% off the original price and pays 560 US dollars (USD).
\nTo buy the bicycle, Juan takes a loan of 560 USD for 6 months at a nominal annual interest rate of 75%, compounded monthly. Juan believes that the total amount he will pay will be less than the original price of the bicycle.
\nCalculate the original price of the bicycle.
\nCalculate the difference between the original price of the bicycle and the total amount Juan will pay.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(or equivalent) (M1)
\n\n
Note: Award (M1) for dividing 560 by 0.7 or equivalent.
\n\n
(A1) (C2)
\n[2 marks]
\n(M1)(A1)
\n\n
Note: Award (M1) for substitution into interest formula, (A1) for their correct substitution.
\n\n
OR
\n\n
\n
\n
\n
(A1)(M1)
\n\n
Note: Award (A1) for seen, (M1) for all other entries correct.
\nOR
\n\n
\n
\n
\n
(A1)(M1)
\n\n
Note: Award (A1) for seen, (M1) for all other entries correct.
\n\n
(A1)
\n\n
Note: Award (A3) for 805.678… (806) seen without working.
\n\n
(Juan spends) 5.68 (USD) (5.67828… USD) (more than the original price) (A1)(ft) (C4)
\n[4 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]
\nPoints (0 , 0 , 10) , (0 , 10 , 0) , (10 , 0 , 0) , ( , , ) form the vertices of a tetrahedron.
\nConsider the case where the faces and are perpendicular.
\nThe following diagram shows the graph of against . The maximum point is shown by .
\nShow that and find a similar expression for .
\nHence, show that, if the angle between the faces and is , then .
\nFind the two possible coordinates of .
\nComment on the positions of in relation to the plane .
\nAt , find the value of and the value of .
\nFind the equation of the horizontal asymptote of the graph.
\nA1
\nA1
\nAG
\nA1
\n\n
[3 marks]
\nattempt to find a scalar product M1
\n\n
OR A1
\nattempt to find magnitude of either or M1
\nA1
\n\n
A1
\nNote: Award A1 for any intermediate step leading to the correct answer.
\nAG
\nNote: Do not allow FT marks from part (a)(i).
\n[8 marks]
\nor M1A1
\ncoordinates are (0, 0, 0) and A1
\nNote: Do not allow column vectors for the final A mark.
\n[3 marks]
\ntwo points are mirror images in the plane
or opposite sides of the plane
or equidistant from the plane
or the line connecting the two Vs is perpendicular to the plane R1
[1 mark]
\ngeometrical consideration or attempt to solve (M1)
\nor A1A1
\n[3 marks]
\nM1
\nhence the asymptote has equation A1
\n[2 marks]
\nConsider the equation , where .
\nSolve the equation, giving the solutions in the form , where .
\nThe solutions form the vertices of a polygon in the complex plane. Find the area of the polygon.
\nMETHOD 1
\n(A1)
\n(A1)
\nfirst solution is A1
\nvalid attempt to find all roots (De Moivre or +/− their components) (M1)
\nother solutions are , , A1
\n\n
METHOD 2
\n\n
\n
attempt to expand and equate both reals and imaginaries. (M1)
\n\n
and (A1)
\nfirst solution is A1
\nvalid attempt to find all roots (De Moivre or +/− their components) (M1)
\nother solutions are , , A1
\n\n
[5 marks]
\ncomplete method to find area of ‘rectangle' (M1)
\nA1
\n[2 marks]
\nTommaso plans to compete in a regional bicycle race after he graduates, however he needs to buy a racing bicycle. He finds a bicycle that costs 1100 euro (EUR). Tommaso has 950 EUR and invests this money in an account that pays 5 % interest per year, compounded monthly.
\nThe cost of the bicycle, , can be modelled by , where is the number of years since Tommaso invested his money.
\nDetermine the amount that he will have in his account after 3 years. Give your answer correct to two decimal places.
\nFind the difference between the cost of the bicycle and the amount of money in Tommaso’s account after 3 years. Give your answer correct to two decimal places.
\nAfter complete months Tommaso will, for the first time, have enough money in his account to buy the bicycle.
\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 substitution in the compound interest formula: (A1) for correct substitution.
\nOR
\nN = 3
I% = 5
PV = 950
P/Y = 1
C/Y = 12 (A1)(M1)
Note: Award (A1) for C/Y = 12 seen, (M1) for other correct entries.
\nOR
\nN = 36
I% = 5
PV = 950
P/Y = 12
C/Y = 12 (A1)(M1)
Note: Award (A1) for C/Y = 12 seen, (M1) for other correct entries.
\n1103.40 (EUR) (A1)(G3)
\nNote: Answer must be given to 2 decimal places.
\n[3 marks]
\n(20 × 3 + 1100) − 1103.40 (M1)(M1)
\nNote: Award (M1) for correct substitution into cost of bike function, (M1) for subtracting their answer to part (a). This subtraction may be implied by their final answer (follow through from their part (a) for this implied subtraction).
\n55.60 (EUR) (A1)(ft)(G3)
\nNote: Follow through from part (a). The answer must be two decimal places.
\n[3 marks]
\nMETHOD 1
\n(M1)(M1)
\nNote: Award (M1) for their correct substitution in the compound interest formula with a variable in the exponent; (M1) for comparing their expressions provided variables are the same (not an expression with for years and another with representing months). Award at most (M0)(M1)(A0)(M1)(A0) for substitution of an integer in both expressions and comparison of the results. Accept inequality.
\n( =) 4.52157… (years) (A1)(ft)
\n4.52157… × 12 (= 54.2588…) (M1)
\nNote: Award (M1) for multiplying their value for by 12. This may be implied.
\n= 55 (months) (A1)(ft)(G4)
\n\n
METHOD 2
\n(M1)(M1)(M1)
\nNote: Award (M1) for their correct substitution in the compound interest formula with a variable in the exponent to solve; (M1) for comparing their expressions provided variables are the same; (M1) for converting years to months in these expressions. Award at most (M0)(M1)(A0)(M1)(A0) for substitution of an integer in both expressions and comparison of the results. Accept inequality.
\n= 54.2588… (months) (A1)(ft)
\n= 55 (months) (A1)(ft)(G4)
\n\n
METHOD 3
\n (M1)(M1)
Note: Award (M1) for each graph drawn.
\n( =) 4.52157… (years) (A1)(ft)
\n4.52157… × 12 (= 54.2588…) (M1)
\nNote: Award (M1) for multiplying their value for by 12. This may be implied.
\nIf the graphs drawn are in terms of months, leading to a value of 54.2588…, award (M1)(M1)(M1)(A1), consistent with METHOD 2.
\n= 55 (months) (A1)(ft)(G4)
\nNote: Follow through for a compound interest formula consistent with their part (a). The final (A1)(ft) can only be awarded for correct answer, or their correct answer following through from previous parts and only if value is rounded up. For example, do not award (M0)(M0)(A0)(M1)(A1)(ft) for an unsupported “5 years × 12 = 60” or similar.
\n\n
[5 marks]
\nConsider the function , where . The derivative of is denoted by .
\n\n
Prove, by mathematical induction, that , .
\nA1
\nNote: This must be obtained from the candidate differentiating .
\nA1
\n(hence true for )
\n\n
assume true for : M1
\n\n
Note: Award M1 if truth is assumed. Do not allow “let ”.
\nconsider :
\n\n
attempt to differentiate M1
\nA1
\n\n
A1
\n\n
True for and true implies true for .
\nTherefore the statement is true for all R1
\nNote: Do not award final R1 if the two previous M1s are not awarded. Allow full marks for candidates who use the base case .
\n\n
[7 marks]
\nWrite in the form , where .
\nHence, find the value of .
\nattempt to complete the square or multiplication and equating coefficients (M1)
\nA1
\n, ,
\n[2 marks]
\nuse of their identity from part (a) (M1)
\nor A1
\nNote: Condone lack of, or incorrect limits up to this point.
\n(M1)
\n(A1)
\nA1
\n[5 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\"/)
State whether length of trout is a continuous or discrete variable.
\nWrite down the modal class.
\nAny trout with length 40 cm or less is returned to the lake.
\nCalculate the percentage of the fisherman’s catch that is returned to the lake.
\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/TZ1/01.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]
\ncontinuous (A1) (C1)
\n[1 mark]
\n(A1) (C1)
\n\n
Note: Accept equivalent notation such as or .
\nAward (A0) for “60-70” (incorrect notation).
\n\n
[1 mark]
\n(M1)
\n\n
Note: Award (M1) for their 4 divided by their 22.
\n\n
(A1)(ft) (C2)
\n\n
Note: Follow through from their part (a). Do not accept 0.181818….
\n\n
[2 marks]
\nGiven that , find the value of .
\nseen (A1)
\nattempt at using limits in an integrated expression (M1)
\n(A1)
\nSetting their equation M1
\nNote: their equation must be an integrated expression with limits substituted.
\nA1
\nA1
\nNote: Do not award final A1 for .
\n[6 marks]
\nA straight line, , has vector equation r .
\nThe plane , has equation .
\nShow that the angle between and is independent of both and .
\na vector normal to is (A1)
\nNote: Allow any scalar multiple of , including
\nattempt to find scalar product (or vector product) of direction vector of line with any scalar multiple of M1
\n(or ) A1
\n(if is the angle between the line and the normal to the plane)
\n(or ) A1
\nor A1
\nthis is independent of and , hence the angle between the line and the plane, , is also independent of and R1
\nNote: The final R mark is independent, but is conditional on the candidate obtaining a value independent of and .
\n[6 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
In the following diagram, the points , , and are on the circumference of a circle with centre and radius . is a diameter of the circle. , and .
\nGiven that , show that .
\nShow that .
\nBy considering triangle , show that .
\nBy considering triangle , find another expression for in terms of and .
\nUse your answers to part (c) to show that .
\nR1
\nAG
\nNote: Accept arguments using the unit circle or graphical/diagrammatical considerations.
\n[1 mark]
\nA1
\nvalid method to find (M1)
\nfor example:
\nA1
\nhence AG
\n[3 marks]
\n, A1A1
\napplying cosine rule (M1)
\nA1
\n\n
AG
\n[4 marks]
\n(A1)
\nattempt to use cosine rule on (M1)
\n\n
A1
\n[3 marks]
\n(M1)(A1)
\nA1
\nNote: Award A1 for any correct intermediate step seen using only two terms.
\nAG
\nNote: Do not award the final A1 if follow through is being applied.
\n[3 marks]
\nThree planes have equations:
\n\n
, where .
\n\n
Find the set of values of and such that the three planes have no points of intersection.
\nattempt to eliminate a variable (or attempt to find det ) M1
\n(or det )
\n(or two correct equations in two variables) A1
\n(or solving det )
\n(or attempting to reduce to one variable, e.g. ) M1
\nA1A1
\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]
\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]
\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.
\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]
\nLet . 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]
\nand are acute angles such that and .
\nShow that .
\nattempt to use (may be seen later) M1
\nattempt to use any double angle formulae (seen anywhere) M1
\nattempt to find either or (seen anywhere) M1
\n(A1)
\nA1
\nA1
\nA1
\nSo
\nAG
\n[7 marks]
\nThe function is defined by , for .
\nThe function is defined by
\nExpress in the form where A, B are constants.
\nA1A1
\n[2 marks]
\nLet and .
\nFind the values of so that has no real roots.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1 – (discriminant)
\ncorrect expression for (A1)
\neg
\nevidence of discriminant (M1)
\neg
\ncorrect substitution into discriminant of (A1)
\neg
\nrecognizing discriminant is negative (M1)
\neg
\ncorrect working (must be correct inequality) (A1)
\neg
\nA1 N3
\n\n
METHOD 2 – (transformation of vertex of )
\nvalid approach for finding vertex (M1)
\neg
\ncorrect vertex of (A1)
\neg
\ncorrect vertex of (A1)
\neg
\ncorrect vertex of (A1)
\neg
\nrecognizing when vertex is above -axis (M1)
\neg , sketch
\nA1 N3
\n\n
METHOD 3 – (transformation of )
\nrecognizing vertical reflection of (M1)
\neg , sketch
\ncorrect expression for (A1)
\neg
\nvalid approach for finding vertex of (M1)
\neg
\ncorrect coordinate of vertex of (A1)
\neg
\nrecognizing when vertex is above -axis (M1)
\neg , sketch
\nA1 N3
\n\n
[6 marks]
\nLet , for , where .
\nPoint lies on the graph of .
\nFind the value of .
\nThe -intercept of the graph of is .
\nOn the following grid, sketch the graph of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nattempt to substitute coordinates (in any order) into (M1)
\neg
\nfinding (seen anywhere) (A1)
\neg
\nA1 N2
\n[3 marks]
\n A1A1A1 N3
Note: Award A1 for correct shape of logarithmic function (must be increasing and concave down).
Only if the shape is correct, award the following:
A1 for being asymptotic to
A1 for curve including both points in circles.
[3 marks]
\nEach 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]
\nThe fastest recorded speeds of eight animals are shown in the following table.
\nState whether speed is a continuous or discrete variable.
\nWrite down the median speed for these animals.
\nWrite down the range of the animal speeds.
\nFor these eight animals find the mean speed.
\nFor these eight animals write down the standard deviation.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
continuous (A1) (C1)
\n[1 mark]
\n75.5 (km h−1) (A1) (C1)
\nNote: Answer must be exact.
\n[1 mark]
\n294 (km h−1) (A1) (C1)
\n[1 mark]
\nOR (M1)
\nNote: Award (M1) for correct sum divided by 8.
\n89.9 (89.875)(km h−1) (A1) (C2)
\n[2 marks]
\n84.6 (84.5597…)(km h−1) (A1) (C1)
\nNote: If the response to part (d)(i) is awarded zero marks, a correct response to part (d)(ii) is awarded (C2).
\n[1 mark]
\nThe function is defined by , where .
\nBy finding a suitable number of derivatives of , find the first two non-zero terms in the Maclaurin series for .
\nHence or otherwise, find .
\n\n
M1A1
\nNote: Award M1A0 for
\nA1
\nEITHER
\nA1
\nOR
\nA1
\nTHEN
\nsubstitute into or any of its derivatives (M1)
\n, and A1
\n\n
the Maclaurin series is
\n(M1)A1
\n[8 marks]
\nMETHOD 1
\nM1
\n(M1)
\nA1
\nNote: Condone the omission of +… in their working.
\n\n
METHOD 2
\nindeterminate form, using L’Hôpital’s rule
\nM1
\nindeterminate form, using L’Hôpital’s rule again
\nM1
\nNote: Award M1 only if their previous expression is in indeterminate form.
\nA1
\nNote: Award FT for use of their derivatives from part (a).
\n\n
[3 marks]
\nPoints 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]
\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]
\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]
\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]
\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]
\nLucy 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]
\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]
\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 .
\nHence, find the number of vases that will maximize the profit.
\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]
\n(M1)
\nNote: Award (M1) for equating their derivative to zero.
\nOR
\nsketch of their derivative (approximately correct shape) with -intercept seen (M1)
\n(A1)(ft)
\nNote: Award (C2) for seen without working.
\n23 (A1)(ft) (C3)
\nNote: Follow through from part (b).
\n[3 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]
\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]
\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
\nFind the mean.
\nFind the standard deviation.
\nthe mean.
\nthe standard deviation.
\nA ninth student also takes the test.
\nExplain why the median is unchanged.
\n6.75 A1
\n[1 mark]
\n2.22 A1
\n[1 mark]
\n8.75 A1
\n[1 mark]
\n2.22 A1
\n[1 mark]
\nthe order is 3, 4, 6, 7, 7, 8, 9, 10
\nmedian is currently 7 A1
\nNote: This can be indicated by a diagram/list, rather than actually stated.
\nwith 9 numbers the middle value (median) will be the 5th value R1
\nwhich will correspond to 7 regardless of whether the position of the median moves up or down R1
\nNote: Accept answers using data 5, 6, 8, 9, 9, 10, 11, 12 (ie from part (b)).
\n[3 marks]
\nConsider the curve y = 5x3 − 3x.
\nThe curve has a tangent at the point P(−1, −2).
\nFind the equation of this tangent. Give your answer in the form y = mx + c.
\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]
\nConsider the expansion of , where .
\nThe coefficient of the term in 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.
valid approach for expansion (must have correct substitution for parameters, but accept an incorrect value for ). (M1)
\neg
\nvalid attempt to identify correct term (M1)
\neg
\nidentifying correct term (may be indicated in expansion) (A1)
\neg
\ncorrect term or coefficient in binominal expansion (A1)
\neg
\ncorrect equation in (A1)
\neg
\n(exact) A1 N3
\n\n
Note: Do not award A1 if additional answers given.
\n[6 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]
\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.
\n 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]
\nAn 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]
\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 equation of the axis of symmetry of the graph of .
\nWrite down the value 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.
(A1)(A1) (C2)
\n\n
Note: Award (A1) for (a constant) and (A1) for .
\n\n
[2 marks]
\n(A1) (C1)
\n[1 mark]
\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]
\nFiona 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]
\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]
\nLet .
\nFind, in terms of b, the solutions of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
EITHER
\nor or … (M1)(A1)
\nNote: Award M1 for any one of the above, A1 for having final two.
\nOR
\n (M1)(A1)
Note: Award M1 for one of the angles shown with b clearly labelled, A1 for both angles shown. Do not award A1 if an angle is shown in the second quadrant and subsequent A1 marks not awarded.
\nTHEN
\nor (A1)(A1)
\nA1
\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.
The 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]
\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]
\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.
\n(M1)A1
\n[2 marks]
\n(M1)
\nA1
\n[2 marks]
\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]
\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]
\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. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.
(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]
Events 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]
\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]
\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]
\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]
\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 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. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.
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]
Consider 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. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.
(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]
Barry is at the top of a cliff, standing 80 m above sea level, and observes two yachts in the sea.
“Seaview” is at an angle of depression of 25°.
“Nauti Buoy” is at an angle of depression of 35°.
The following three dimensional diagram shows Barry and the two yachts at S and N.
X lies at the foot of the cliff and angle 70°.
![\"N17/5/MATHL/HP2/ENG/TZ0/05\"](\"../assets/uploads/tinymce_asset/asset/4121/Schermafbeelding_2018-02-08_om_11.45.43.png\"/)
Find, to 3 significant figures, the distance between the two yachts.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nattempt to use tan, or sine rule, in triangle BXN or BXS (M1)
\n(A1)
\n(A1)
\nAttempt to use cosine rule M1
\n° (A1)
\nA1
\n\n
Note: Award final A1 only if the correct answer has been given to 3 significant figures.
\n\n
[6 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]
\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. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.
(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]
Jean-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. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.
(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]
Find 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 differential equation where when .
\nShow that is an integrating factor for this differential equation.
\nSolve the differential equation giving your answer in the form .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nintegrating factor (M1)
\n(M1)
\n\n
Note: Award M1 for use of for example or .
\n\n
integrating factor A1
\nA1
\n\n
Note: Award A1 for where .
\n\n
AG
\n\n
METHOD 2
\nM1A1
\nM1A1
\n\n
Note: Award M1 for attempting to express in the form .
\n\n
so is an integrating factor for this differential equation AG
\n[4 marks]
\n(or equivalent) (M1)
\n\n
A1
\n(M1)A1
\n\n
Note: Award M1 for using an appropriate substitution.
\n\n
Note: Condone the absence of .
\n\n
substituting M1
\n\n
Note: Award M1 for attempting to find their value of .
\n\n
A1
\n[6 marks]
\nLet be the set . Let be the set .
\nA function is defined by .
\nLet be the set .
\nA function is defined by .
\n(i) Sketch the graph of and hence justify whether or not is a bijection.
\n(ii) Show that is a group under the binary operation of multiplication.
\n(iii) Give a reason why is not a group under the binary operation of multiplication.
\n(iv) Find an example to show that is not satisfied for all .
\n(i) Sketch the graph of and hence justify whether or not is a bijection.
\n(ii) Show that for all .
\n(iii) Given that and are both groups, explain whether or not they are isomorphic.
\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/TZ0/SG/M/02.a.i\"](\"../assets/uploads/tinymce_asset/asset/3304/Schermafbeelding_2017-03-02_om_12.36.51.png\"/)
A1
\n
Notes: Award A1 for general shape, labelled asymptotes, and showing that .
\n\n
graph shows that it is injective since it is increasing or by the horizontal line test R1
\ngraph shows that it is surjective by the horizontal line test R1
\n\n
Note: Allow any convincing reasoning.
\n\n
so is a bijection A1
\n(ii) closed since non-zero real times non-zero real equals non-zero real A1R1
\nwe know multiplication is associative R1
\nidentity is 1 A1
\ninverse of is A1
\nhence it is a group AG
\n(iii) does not have an identity A2
\nhence it is not a group AG
\n(iv) whereas is one counterexample A2
\nhence statement is not satisfied AG
\n[13 marks]
\n![\"N16/5/MATHL/HP3/ENG/TZ0/SG/M/02.b\"](\"../assets/uploads/tinymce_asset/asset/3305/Schermafbeelding_2017-03-02_om_12.52.46.png\"/)
award A1 for general shape going through (0, 1) and with domain A1
\ngraph shows that it is injective since it is increasing or by the horizontal line test and graph shows that it is surjective by the horizontal line test R1
\n\n
Note: Allow any convincing reasoning.
\n\n
so is a bijection A1
\n(ii) and M1A1
\nhence AG
\n(iii) since is a bijection and the homomorphism rule is obeyed R1R1
\nthe two groups are isomorphic A1
\n[8 marks]
\nConsider the differential equation where and is a positive integer, .
\nSolve the differential equation given that when . Give your answer in the form .
\nShow that the -coordinate(s) of the points on the curve where satisfy the equation .
\nDeduce the set of values for such that there are two points on the curve where . 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.
METHOD 1
\n(M1)
\nintegrating factor M1
\n(A1)
\n= A1
\n(M1)
\n\n
A1
\nNote: Condone the absence of C.
\n\n
substituting , M1
\nNote: Award M1 for attempting to find their value of C.
\nA1
\n[8 marks]
\n\n
METHOD 2
\nput so that M1(A1)
\nsubstituting, M1
\n(A1)
\nM1
\n\n
A1
\nNote: Condone the absence of C.
\n\n
substituting , M1
\nNote: Award M1 for attempting to find their value of C.
\nA1
\n[8 marks]
\nMETHOD 1
\nfind and solve for
\nM1
\nA1
\n\n
Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.
\nAG
\n\n
METHOD 2
\nsubstitute and their into the differential equation and solve for
\nM1
\nA1
\n\n
Note: Award a maximum of M1A0 if a candidate’s answer to part (a) is incorrect.
\nAG
\n[2 marks]
\n\n
there are two solutions for when is odd (and A1
\nif is even there are two solutions (to )
\nand if is odd there is only one solution (to ) R1
\nNote: Only award the R1 if both cases are considered.
\n[4 marks]
\nConsider the differential equation , where .
\nConsider the family of curves which satisfy the differential equation , where .
\nGiven that , use Euler’s method with step length = 0.25 to find an approximation for . Give your answer to two significant figures.
\nSolve the equation for .
\nFind the percentage error when is approximated by the final rounded value found in part (a). Give your answer to two significant figures.
\nFind the equation of the isocline corresponding to , where , .
\nShow that such an isocline can never be a normal to any of the family of curves that satisfy the differential equation.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to apply Euler’s method (M1)
\n\n
(A1)(A1)
Note: Award A1 for correct values, A1 for first three correct values.
\n\n
= 3.3 A1
\n\n
[4 marks]
\nMETHOD 1
\n(M1)
\n\n
(A1)
\n(M1)
\n\n
A1
\nM1
\nA1
\n\n
METHOD 2
\nM1
\n(A1)
\nM1
\n\n
\n
A1
\nM1
\nA1
\n\n
[6 marks]
\n\n
percentage error (M1)(A1)
\n= 2.5% A1
\n\n
[3 marks]
\nA1
\n\n
\n
[1 mark]
\ngradient of isocline equals gradient of normal (M1)
\nor A1
\nA1
\nR1
\nno solution AG
\nNote: Accept alternative reasons for no solutions.
\n\n
[4 marks]
\nConsider the differential equation
\n\n
Use the substitution to show that the general solution of this differential equation is
\n\n
Hence, or otherwise, solve the differential equation
\n\n
given that when . Give your answer in the form .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
M1
\nthe differential equation becomes
\nA1
\nA1
\nintegrating, Constant AG
\n[3 marks]
\nEITHER
\n(A1)
\nM1A1
\nA1
\n\n
Note: A1 is for correct factorization.
\n\n
A1
\nOR
\nA1
\nM1
\n(A1)
\n\n
Note: A1 is for correct factorization.
\n\n
A1A1
\nTHEN
\nsubstitute or when (M1)
\ntherefore A1
\n\n
Note: This A1 can be awarded anywhere in their solution.
\n\n
substituting for ,
\nM1
\n\n
Note: Award for correct substitution of into their expression.
\n\n
(A1)
\n\n
Note: Award for any rearrangement of a correct expression that has in the numerator.
\n\n
A1
\n\n
[10 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.
\nShow that submarine B travels in the same direction as originally planned.
\nFind the value of t when submarine B passes through P.
\nFind an expression for the distance between the two submarines in terms of t.
\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.
rA = 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]
\nA1
\nNote: Accept use of cross product equalling zero.
\nhence in the same direction AG
\n[1 mark]
\nM1
\nNote: The M1 can be awarded for any one of the resultant equations.
\nA1
\n[2 marks]
\nrA − rB = (M1)(A1)
\n(A1)
\nNote: Accept rA − rB.
\ndistance M1A1
\n\n
[5 marks]
\nminimum when (M1)
\nt = 3.83 A1
\n[2 marks]
\n0.511 (km) A1
\n[1 mark]
\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.
\nMETHOD 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]
\nA 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]
\nIn an arithmetic sequence, the first term is 3 and the second term is 7.
\nFind the common difference.
\nFind the tenth term.
\nFind the sum of the first ten terms of the sequence.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to subtract terms (M1)
\neg
\nA1 N2
\n[2 marks]
\ncorrect approach (A1)
\neg
\nA1 N2
\n[2 marks]
\ncorrect substitution into sum (A1)
\neg
\nA1 N2
\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.
\nvalid 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]
\nConsider the differential equation , given that when .
\nShow that is an integrating factor for this differential equation.
\nHence solve this differential equation. Give the answer in the form .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nattempting to find an integrating factor (M1)
\n(M1)A1
\nIF is (M1)A1
\nAG
\nMETHOD 2
\nmultiply by the integrating factor
\nM1A1
\nleft hand side is equal to the derivative of
\nA3
\n[5 marks]
\n(M1)
\n\n
A1A1
\n\n
M1A1
\nA1
\n[6 marks]
\nThe points A, B, C and D have position vectors a, b, c and d, relative to the origin O.
\nIt is given that .
\nThe position vectors , , and are given by
\na = i + 2j − 3k
\nb = 3i − j + pk
\nc = qi + j + 2k
\nd = −i + rj − 2k
\nwhere p , q and r are constants.
\nThe point where the diagonals of ABCD intersect is denoted by M.
\nThe plane cuts the x, y and z axes at X , Y and Z respectively.
\nExplain why ABCD is a parallelogram.
\nUsing vector algebra, show that .
\nShow that p = 1, q = 1 and r = 4.
\nFind the area of the parallelogram ABCD.
\nFind the vector equation of the straight line passing through M and normal to the plane containing ABCD.
\nFind the Cartesian equation of .
\nFind the coordinates of X, Y and Z.
\nFind YZ.
\na pair of opposite sides have equal length and are parallel R1
\nhence ABCD is a parallelogram AG
\n[1 mark]
\nattempt to rewrite the given information in vector form M1
\nb − a = c − d A1
\nrearranging d − a = c − b M1
\nhence AG
\nNote: Candidates may correctly answer part i) by answering part ii) correctly and then deducing there
are two pairs of parallel sides.
[3 marks]
\nEITHER
\nuse of (M1)
\nA1A1
\nOR
\nuse of (M1)
\nA1A1
\nTHEN
\nattempt to compare coefficients of i, j, and k in their equation or statement to that effect M1
\nclear demonstration that the given values satisfy their equation A1
p = 1, q = 1, r = 4 AG
[5 marks]
\nattempt at computing (or equivalent) M1
\nA1
\narea (M1)
\n= 15 A1
\n[4 marks]
\nvalid attempt to find (M1)
\nA1
\nthe equation is
\nr = or equivalent M1A1
\nNote: Award maximum M1A0 if 'r = …' (or equivalent) is not seen.
\n[4 marks]
\nattempt to obtain the equation of the plane in the form ax + by + cz = d M1
\n11x + 10y + 2z = 25 A1A1
\nNote: A1 for right hand side, A1 for left hand side.
\n[3 marks]
\nputting two coordinates equal to zero (M1)
\nA1
\n[2 marks]
\nM1
\nA1
\n[4 marks]
\nA simple model to predict the population of the world is set up as follows. At time years the population of the world is , which can be assumed to be a continuous variable. The rate of increase of due to births is 0.056 and the rate of decrease of due to deaths is 0.035.
\nShow that .
\nFind a prediction for the number of years it will take for the population of the world to double.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
A1
\nAG
\n[1 mark]
\nMETHOD 1
\n\n
attempt to separate variables M1
\nA1
\nA1
\nEITHER
\n\n
A1
\nNote: This A1 is independent of the following marks.
\nOR
\n\n
A1
\nNote: This A1 is independent of the following marks.
\nTHEN
\n(M1)
\nyears A1
\nNote: If a candidate writes , so then award the final A1.
\n\n
METHOD 2
\n\n
attempt to separate variables M1
\nA1A1
\nNote: Award A1 for correct integrals and A1 for correct limits seen anywhere. Do not penalize use of in place of .
\nA1
\n(M1)
\nA1
\n\n
[6 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 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.
\nattempt 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]
\nConsider the differential equation , where .
\nSolve the differential equation and show that a general solution is where is a positive constant.
\nProve that there are two horizontal tangents to the general solution curve and state their equations, in terms of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
let M1
\n(A1)
\n(M1)
\n(A1)
\nNote: Or equivalent attempt at simplification.
\nA1
\n(M1)
\n(A1)
\nA1A1
\nNote: Award A1 for LHS and A1 for RHS and a constant.
\nM1
\nNote: Award M1 for substituting . May be seen at a later stage.
\nA1
\nNote: Award A1 for any correct equivalent equation without logarithms.
\nAG
\n[11 marks]
\nMETHOD 1
\n\n
(for horizontal tangents) M1
\n\n
EITHER
\nusing M1
\nA1
\nNote: Award M1A1 for .
\nOR
\nusing implicit differentiation of
\nM1
\nNote: Accept differentiation of .
\nA1
\nTHEN
\ntangents at A1A1
\nhence there are two tangents AG
\n\n
METHOD 2
\n\n
M1A1
\nthis is a circle radius centre A1
\nhence there are two tangents AG
\ntangents at A1A1
\n\n
[5 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]
\nUse l’Hôpital’s rule to determine the value of
\n\n
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to use l’Hôpital’s rule, M1
\nor A1A1
\n\n
Note: Award A1 for numerator A1 for denominator.
\n\n
this gives 0/0 so use the rule again (M1)
\nor A1A1
\n\n
Note: Award A1 for numerator A1 for denominator.
\n\n
A1
\n\n
Note: This A1 is dependent on all previous marks being awarded, except when the first application of L’Hopital’s does not lead to 0/0, when it should be awarded for the correct limit of their derived function.
\n\n
[7 marks]
\nUse L’Hôpital’s rule to determine the value of
\n\n
Hence find .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
M1A1A1
\nA1
\n= −3 A1
\n\n
[5 marks]
\nis of the form
\napplying l’Hôpital´s rule (M1)
\n(A1)
\n= −3 A1
\n\n
[3 marks]
\nUsing L’Hôpital’s rule, find .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n
M1A1A1
\nNote: Award M1 for attempt at differentiation using l'Hopital's rule, A1 for numerator, A1 for denominator.
\n\n
METHOD 1
\nusing l’Hopital’s rule again
\nA1A1
\nEITHER
\nA1A1
\nNote: Not all terms in numerator need to be written in final fraction. Award A1 for . However, if the terms are written, they
must be correct to award A1.
attempt to substitute M1
\n\n
OR
\n(M1)A1
\nA1
\nTHEN
\nA1
\n\n
METHOD 2
\nM1
\nA1
\nM1A1
\nattempt to substitute M1
\n\n
A1
\n\n
[9 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]
\nIn an arithmetic sequence, the first term is 8 and the second term is 5.
\nFind the common difference.
\nFind the tenth term.
\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]
\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 .
\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, 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 first two terms of an infinite geometric sequence are u1 = 18 and u2 = 12sin2 θ , where 0 < θ < 2 , and θ ≠ .
\nFind an expression for r in terms of θ.
\nFind the values of θ which give the greatest value of the sum.
\nvalid approach (M1)
\neg
\nA1 N2
\n[2 marks]
\n\n
METHOD 1 (using differentiation)
\nrecognizing (seen anywhere) (M1)
\nfinding any correct expression for (A1)
\neg
\ncorrect working (A1)
\neg sin 2θ = 0
\nany correct value for sin−1(0) (seen anywhere) (A1)
\neg 0, , … , sketch of sine curve with x-intercept(s) marked both correct values for 2θ (ignore additional values) (A1)
\n2θ = , 3 (accept values in degrees)
\nboth correct answers A1 N4
\nNote: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.
\n
METHOD 2 (using denominator)
\nrecognizing when S∞ is greatest (M1)
\neg 2 + cos 2θ is a minimum, 1−r is smallest
correct working (A1)
eg minimum value of 2 + cos 2θ is 1, minimum r =
\ncorrect working (A1)
\neg
\nEITHER (using cos 2θ)
\nany correct value for cos−1(−1) (seen anywhere) (A1)
\neg , 3, … (accept values in degrees), sketch of cosine curve with x-intercept(s) marked
\nboth correct values for 2θ (ignore additional values) (A1)
\n2θ = , 3 (accept values in degrees)
\nOR (using sinθ)
\nsinθ = ±1 (A1)
\nsin−1(1) = (accept values in degrees) (seen anywhere) A1
\nTHEN
\nboth correct answers A1 N4
\nNote: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.
[6 marks]
\nABCD is a parallelogram, where = –i + 2j + 3k and = 4i – j – 2k.
\nFind the area of the parallelogram ABCD.
\nBy using a suitable scalar product of two vectors, determine whether is acute or obtuse.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
i j – 7k M1A1
\n\n
A1
\n[3 marks]
\nMETHOD 1
\nM1A1
\n\n
considering the sign of the answer
\n, therefore angle is obtuse M1
\n(as it is a parallelogram), is acute A1
\n[4 marks]
\nMETHOD 2
\nM1A1
\nconsidering the sign of the answer M1
\nis acute A1
\n[4 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]
\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
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* 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]
\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 exact value of .
\nShow that .
\nAn infinite geometric series is given as , .
\nFind the largest value of such that .
\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]
\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 .
\nattempt 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 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]
\nThe acute angle between the vectors 3i − 4j − 5k and 5i − 4j + 3k is denoted by θ.
\nFind cos θ.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
cos θ = (M1)
\nA1A1
\nNote: A1 for correct numerator and A1 for correct denominator.
\nA1
\n[4 marks]
\n\n
The following diagram shows the graph of a function , with domain .
\n![\"N17/5/MATME/SP1/ENG/TZ0/03\"](\"../assets/uploads/tinymce_asset/asset/4143/Schermafbeelding_2018-02-11_om_09.13.25.png\"/)
The points and lie on the graph of .
\nOn the grid, sketch the graph of .
\n![\"N17/5/MATME/SP1/ENG/TZ0/03.c/M\"](\"../assets/uploads/tinymce_asset/asset/4150/Schermafbeelding_2018-02-11_om_10.32.42.png\"/)
A1A1A1 N3
\n
Notes: Award A1 for both end points within circles,
\nA1 for images of and within circles,
\nA1 for approximately correct reflection in , concave up then concave down shape (do not accept line segments).
\n\n
[3 marks]
\nSolve , for .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ncorrect application of (A1)
\neg
\ncorrect equation without logs A1
\neg
\nrecognizing double-angle identity (seen anywhere) A1
\neg
\nevaluating (A1)
\ncorrect working A1
\negand , one correct final answer
\n(do not accept additional values) A2 N0
\n[7 marks]
\nThe points A, B and C have the following position vectors with respect to an origin O.
\ni + j – 2k
\ni – j + 2k
\ni + 3j + 3k
\nThe plane Π contains the points O, A and B and the plane Π contains the points O, A and C.
\nFind the vector equation of the line (BC).
\nDetermine whether or not the lines (OA) and (BC) intersect.
\nFind the Cartesian equation of the plane Π, which passes through C and is perpendicular to .
\nShow that the line (BC) lies in the plane Π.
\nVerify that 2j + k is perpendicular to the plane Π.
\nFind a vector perpendicular to the plane Π.
\nFind the acute angle between the planes Π and Π.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
= (i + 3j + 3k) (2i j + 2k) = i + 4j + k (A1)
\nr = (2i j + 2k) + (i + 4j + k)
\n(or r = (i + 3j + 3k) + (i + 4j + k) (M1)A1
\n\n
Note: Do not award A1 unless r = or equivalent correct notation seen.
\n\n
[3 marks]
\nattempt to write in parametric form using two different parameters AND equate M1
\n\n
\n
A1
\nattempt to solve first pair of simultaneous equations for two parameters M1
\nsolving first two equations gives (A1)
\nsubstitution of these two values in third equation (M1)
\nsince the values do not fit, the lines do not intersect R1
\n\n
Note: Candidates may note that adding the first and third equations immediately leads to a contradiction and hence they can immediately deduce that the lines do not intersect.
\n\n
[6 marks]
\nMETHOD 1
\nplane is of the form r (2i + j 2k) = d (A1)
\nd = (i + 3j + 3k) (2i + j 2k) = 1 (M1)
\nhence Cartesian form of plane is A1
\nMETHOD 2
\nplane is of the form (A1)
\nsubstituting (to find gives ) (M1)
\nhence Cartesian form of plane is A1
\n[3 marks]
\nMETHOD 1
\nattempt scalar product of direction vector BC with normal to plane M1
\n(i + 4j + k) (2i + j 2k)
\nA1
\nhence BC lies in Π AG
\nMETHOD 2
\nsubstitute eqn of line into plane M1
\n\n
\n
A1
\nhence BC lies in Π AG
\n\n
Note: Candidates may also just substitute into the plane since they are told C lies on .
\n\n
Note: Do not award A1FT.
\n\n
[2 marks]
\nMETHOD 1
\napplying scalar product to and M1
\n(2j + k) (2i + j 2k) = 0 A1
\n(2j + k) (2i j + 2k) =0 A1
\nMETHOD 2
\nattempt to find cross product of and M1
\nplane Π has normal = 8j 4k A1
\nsince 8j 4k = 4(2j + k), 2j + k is perpendicular to the plane Π R1
\n[3 marks]
\nplane Π has normal = 9i 8j + 5k A1
\n[1 mark]
\nattempt to use dot product of normal vectors (M1)
\n(M1)
\n(A1)
\n\n
Note: Accept . acute angle between planes A1
\n\n
[4 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.
\nattempt 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 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]
\nThe following diagram shows the graph of a function , for .
\nThe points and lie on the graph of . There is a minimum point at .
\nLet .
\nWrite down the range of .
\nWrite down the domain 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]
\ncorrect interval A2 N2
\neg, from to 3
\n[2 marks]
\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 . 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 null 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]
\nLet 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.
\nattempt 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]
\nLet .
\nThe function can also be expressed in the form .
\n(i) Write down the value of .
\n(ii) Find the value of .
\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 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]
\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. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.
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]
Let for .
\nWrite down the equation of the horizontal asymptote to the graph of f.
\nvalid approach (M1)
eg
y = −4 (must be an equation) A1 N2
\n[2 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]
Hafizah 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. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.
(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]
In 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.
20 (A1) (C1)
\n[1 mark]
\n(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]
\nLet , for .
\nFor the graph of , find the -intercept.
\nHence or otherwise, write down .
\nvalid method (M1)
\neg (0), sketch of graph
\n-intercept is (exact), −0.333, A1 N2
\n\n
[2 marks]
\nvalid approach (M1)
\neg recognizing that is related to the horizontal asymptote,
\ntable with large values of , their value from (a)(iii), L’Hopital’s rule .
\nA1 N2
\n\n
[2 marks]
\nThe graph of for is shown in the following diagram.
\nWrite down the value of .
\nWrite down the value of .
\nLet for . On the axes above, sketch the graph of .
\n\n
A1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\n M1A1A1
\n
Note: Award M1 for an attempt to apply any vertical stretch or vertical translation, A1 for a correct horizontal line segment between and (located roughly at ),
A1 for a correct concave down parabola including max point at and for correct end points at and (within circles). Points do not need to be labelled.
\n
[3 marks]
\nConsider 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]
Let 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]
\nThe points A and B are given by and .
\nThe plane Π is defined by the equation .
\nFind a vector equation of the line L passing through the points A and B.
\nFind the coordinates of the point of intersection of the line L with the plane Π.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)
\n\n
r = or r = M1A1
\n\n
Note: Award M1A0 if r = is not seen (or equivalent).
\n\n
[3 marks]
\nsubstitute line L in M1
\n\n
(A1)
\n\n
r =
\nso coordinate is A1
\n\n
Note: Accept coordinate expressed as position vector .
\n\n
[3 marks]
\nLet f(x) = ln x − 5x , for x > 0 .
\nSolve f '(x) = f \"(x).
\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]
\nThe diameter of a spherical planet is .
\nWrite down the radius of the planet.
\nThe volume of the planet can be expressed in the form where and .
\nFind the value of and the value of .
\nOR (accept ) A1
\n\n
[1 mark]
\nOR (A1)
\nOR (A1)
\nOR A1
\n\n
[3 marks]
\nConsider the vectors a = and b = .
\nFind the possible values of p for which a and b are parallel.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1 (eliminating k)
\nrecognizing parallel vectors are multiples of each other (M1)
\neg a = kb, = k, , 3k = p + 1 and 2kp = 8
\ncorrect working (must be quadratic) (A1)
\neg 2p2 + 2p = 24, p2 + p – 12,
\nvalid attempt to solve their quadratic equation (M1)
\neg factorizing, formula, completing the square
\nevidence of correct working (A1)
\neg (p + 4)(p – 3),
\np = –4, p = 3 A1A1 N4
\n\n
METHOD 2 (solving for k)
\nrecognizing parallel vectors are multiples of each other (M1)
\neg a = kb, = k, 3k = p + 1 and 2kp = 8
\ncorrect working (must be quadratic) (A1)
\neg 3k2 – k = 4, 3k2 – k – 4, 4k2 = 3 – k
\none correct value for k (A1)
\neg k = –1, k = , k =
\nsubstituting their value(s) of k (M1)
\neg , and ,
\np = –4, p = 3 A1A1 N4
\n\n
METHOD 3 (working with angles and cosine formula)
\nrecognizing angle between parallel vectors is 0 and/or 180° M1
\neg cos θ = ±1,
\ncorrect substitution of scalar product and magnitudes into equation (A1)
\neg ,
\ncorrect working (must include both ± ) (A1)
\neg ,
\ncorrect quartic equation (A1)
\neg , , ,
\np = –4, p = 3 A2 N4
\n\n
[6 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]
\nConsider a triangle OAB such that O has coordinates (0, 0, 0), A has coordinates (0, 1, 2) and B has coordinates (2, 0, − 1) where < 0.
\nLet M be the midpoint of the line segment [OB].
\nFind, in terms of , a Cartesian equation of the plane Π containing this triangle.
\nFind, in terms of , the equation of the line L which passes through M and is perpendicular to the plane П.
\nShow that L does not intersect the -axis for any negative value of .
\n\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(M1)A1
\n(0, 0, 0) on Π so (M1)A1
\n\n
METHOD 2
\nusing equation of the form (M1)
\n(0, 1, 2) on Π ⇒
\n(2, 0, − 1) on Π ⇒ (M1)A1
\nNote: Award (M1)A1 for both equations seen.
\nsolve for , and (M1)
\nA1
\n\n
[5 marks]
\nM has coordinates (A1)
\nr = M1A1
\nNote: Award M1A0 if r = (or equivalent) is not seen.
\nNote: Allow equivalent forms such as .
\n\n
[3 marks]
\nMETHOD 1
\n(M1)
\nNote: Award M1 for either or or both.
\nand A1
\nattempt to eliminate M1
\n(A1)
\nA1
\nEITHER
\nconsideration of the signs of LHS and RHS (M1)
\nthe LHS is negative and the RHS must be positive (or equivalent statement) R1
\nOR
\n\n
\n
M1
\nno real solutions R1
\nTHEN
\nso no point of intersection AG
\n\n
METHOD 2
\n(M1)
\nNote: Award M1 for either or or both.
\nand A1
\nattempt to eliminate M1
\n(A1)
\nA1
\nconsideration of the signs of LHS and RHS (M1)
\nthere are no real solutions (or equivalent statement) R1
\nso no point of intersection AG
\n\n
[7 marks]
\nConsider an arithmetic sequence where . Find the value of the first term, , and the value of the common difference, .
\nMETHOD 1 (finding first, from )
\n(A1)
\nA1
\nOR (may be seen with their value of ) (A1)
\nattempt to substitute their (M1)
\nA1
\n\n
METHOD 2 (solving simultaneously)
\n(A1)
\nOR OR (A1)
\nattempt to solve linear or simultaneous equations (M1)
\nA1A1
\n\n
[5 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.
\nShow that .
\nFind the area of triangle OBC in terms of and θ.
\nGiven that the area of triangle OBC is of the area of sector OAB, find θ.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nA1
\nAG N0
\n[1 mark]
\nvalid approach (M1)
\neg , , ,
\narea (must be in terms of and θ) A1 N2
\n[2 marks]
\nvalid attempt to express the relationship between the areas (seen anywhere) (M1)
\neg OCB = OBA , ,
\ncorrect equation in terms of θ only A1
\neg ,
\nvalid attempt to solve their equation (M1)
\neg sketch, −0.830017, 0
\n0.830017
\nθ = 0.830 A1 N2
\nNote: Do not award final A1 if additional answers given.
\n[4 marks]
\nA research student weighed lizard eggs in grams and recorded the results. The following box and whisker diagram shows a summary of the results where and are the lower and upper quartiles respectively.
\nThe interquartile range is grams and there are no outliers in the results.
\nFind the minimum possible value of .
\nHence, find the minimum possible value of .
\nattempt to use definition of outlier
\n(M1)
\n(, accept ) OR A1
\nminimum value of A1
\n\n
[3 marks]
\nattempt to use interquartile range (M1)
\n(may be seen in part (a)) OR (accept )
\nminimum value of A1
\n\n
[2 marks]
\nConsider the functions and where .
\nThe graphs of and have a common tangent at .
\nFind .
\nShow that .
\nHence, show that .
\nA1
\n\n
[1 mark]
\nOR (may be seen anywhere) A1
\n\n
Note: The derivative of must be explicitly seen, either in terms of or .
\n\n
recognizing (M1)
\n\nOR A1
\n\n
Note: The final A1 is dependent on one of the previous marks being awarded.
\n\n
AG
\n\n
[3 marks]
\n(M1)
\n\ncorrect equation in
\n\n
EITHER
\nA1
\nA1
\n\n
OR
\nA1
\nA1
\n\n
THEN
\nAG
\n\n
[3 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]
\nThe following diagram shows triangle PQR.
\n![\"M17/5/MATME/SP1/ENG/TZ1/03\"](\"../assets/uploads/tinymce_asset/asset/3530/Schermafbeelding_2017-08-11_om_09.36.55.png\"/)
Find PR.
\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 sine rule (M1)
\neg
\ncorrect substitution A1
\neg
\n(A1)(A1)
\ncorrect working A1
\neg
\ncorrect answer A1 N3
\neg
\nMETHOD 2 (using height of ΔPQR)
\nvalid approach to find height of ΔPQR (M1)
\neg
\n(A1)
\nA1
\ncorrect working A1
\neg
\ncorrect working (A1)
\neg
\ncorrect answer A1 N3
\neg
\n[6 marks]
\nConsider the graph of the function , 0 ≤ < . The graph of intersects the line exactly twice, at point A and point B. This is shown in the following diagram.
\nConsider the graph of , 0 ≤ < , where > 0.
\nFind the greatest value of such that the graph of does not intersect the line .
\nrecognizing period of is larger than the period of (M1)
\neg sketch of with larger period (may be seen on diagram), A at ,
\nimage of A when , , ,
\ncorrect working (A1)
\neg , ,
\nA1 N2
\n[3 marks]
\nThe histogram shows the lengths of 25 metal rods, each measured correct to the nearest cm.
\nThe upper quartile is 4 cm.
\nWrite down the modal length of the rods.
\nFind the median length of the rods.
\nCalculate the lower quartile.
\nCalculate the interquartile range.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
3 (A1) (C1)
[1 mark]
\nmedian is 13th position (M1)
\nCF: 2, 6, 14, 20, 23, 25 (M1)
\nmedian = 3 (A1) (C3)
\n\n
[3 marks]
\n2.5 (A1) (C1)
\n\n
Note: Award (A1)(ft) if the sum of their parts (c)(i) and (c)(ii) is 4.
\n\n
[1 mark]
\n1.5 (A1)(ft) (C1)
\n\n
Note: Award (A1)(ft) if the sum of their parts (c)(i) and (c)(ii) is 4.
\n\n
[1 mark]
\nGiven that a b b c 0 prove that a c sb where s is a scalar.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\na b = b c
\n(a b) (b c) = 0
\n(a b) + (c b) = 0 M1A1
\n(a + c) b = 0 A1
\n(a + c) is parallel to b a + c = sb R1AG
\n\n
Note: Condone absence of arrows, underlining, or other otherwise “correct” vector notation throughout this question.
\n\n
Note: Allow “is in the same direction to”, for the final R mark.
\n\n
METHOD 2
\na b = b c M1A1
\n\n
\n
\n
A1
\n\n
\n
\n
A1
\na + c = sb AG
\n[4 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 .
\nJill plays the game nine times. Find the probability that she wins exactly two prizes.
\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]
\nvalid approach (M1)
\neg
\n0.279081
\n0.279 A1 N2
\n[2 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]
\nConsider the vectors a i j k, b j k.
\nFind a b.
\nHence find the Cartesian equation of the plane containing the vectors a and b, and passing through the point .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
a b i j k (M1)A1
\n[2 marks]
\nMETHOD 1
\nM1
\n(M1)
\nA1
\n\n
METHOD 2
\nM1A1
\nA1
\n[3 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.
evidence 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]
\nFind the Cartesian equation of plane Π containing the points and and perpendicular to the plane .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\n(A1)
\nM1A1
\nA1
\nM1
\nequation of plane Π is or equivalent A1
\nMETHOD 2
\nlet plane Π be
\nattempt to form one or more simultaneous equations: M1
\n(1) A1
\n(2)
\n(3) A1
\n\n
Note: Award second A1 for equations (2) and (3).
\n\n
attempt to solve M1
\nEITHER
\nusing GDC gives (A1)
\nequation of plane Π is or equivalent A1
\nOR
\nrow reduction M1
\nequation of plane Π is or equivalent A1
\n[6 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.
correct 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]
\nThe following figure shows a square based pyramid with vertices at O(0, 0, 0), A(1, 0, 0), B(1, 1, 0), C(0, 1, 0) and D(0, 0, 1).
\nThe Cartesian equation of the plane , passing through the points B , C and D , is .
\nThe plane passes through O and is normal to the line BD.
\ncuts AD and BD at the points P and Q respectively.
\nFind the Cartesian equation of the plane , passing through the points A , B and D.
\nFind the angle between the faces ABD and BCD.
\nFind the Cartesian equation of .
\nShow that P is the midpoint of AD.
\nFind the area of the triangle OPQ.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognising normal to plane or attempting to find cross product of two vectors lying in the plane (M1)
\nfor example, (A1)
\nA1
\n[3 marks]
\nEITHER
\nM1A1
\nOR
\nM1A1
\nNote: M1 is for an attempt to find the scalar or vector product of the two normal vectors.
\nA1
\nangle between faces is A1
\n[4 marks]
\nor (A1)
\n(M1)
\nA1
\n[3 marks]
\nMETHOD 1
\nline AD : (r =) M1A1
\nintersects when M1
\nso A1
\nhence P is the midpoint of AD AG
\n\n
METHOD 2
\nmidpoint of AD is (0.5, 0, 0.5) (M1)A1
\nsubstitute into M1
\n0.5 + 0.5 − 0.5 = 0 A1
\nhence P is the midpoint of AD AG
\n[4 marks]
\nMETHOD 1
\nA1A1A1
\nA1
\narea A1
\n\n
METHOD 2
\nline BD : (r =)
\n(A1)
\nA1
\narea = M1
\nA1
\nNote: This A1 is dependent on M1.
\narea = A1
\n[5 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]
\nThe plane П1 contains the points P(1, 6, −7) , Q(0, 1, 1) and R(2, 0, −4).
\nThe Cartesian equation of the plane П2 is given by .
\nThe Cartesian equation of the plane П3 is given by .
\nConsider the case that П3 contains .
\nFind the Cartesian equation of the plane containing P, Q and R.
\nGiven that П1 and П2 meet in a line , verify that the vector equation of can be given by r .
\nGiven that П3 is parallel to the line , show that .
\nShow that .
\nGiven that П3 is equally inclined to both П1 and П2, determine two distinct possible Cartesian equations for П3.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
\nfor example
\n, A1A1
\n= 33i + 11j + 11k (M1)A1
\nr.n = a.n
\n(M1)
\nor equivalent A1
\n\n
METHOD 2
\nassume plane can be written as M1
\nsubstituting each set of coordinates gives the system of equations:
\n\n
\n
A1
\nsolving by GDC (M1)
\n, , A1A1A1
\nor equivalent
\n\n
[6 marks]
\nMETHOD 1
\nsubstitution of equation of line into both equations of planes M1
\nA1
\nA1
\n\n
METHOD 2
\nadding Π1 and Π2 gives M1
\ngiven A1
\nA1
\n⇒r AG
\n\n
METHOD 3
\nn1 × n2 = A1
\nR1
\ncommon point and A1
\n\n
[3 marks]
\nnormal to П3 is perpendicular to direction of
\nA1
\n⇒ AG
\n[1 mark]
\nsubstituting into П3: M1
\nA1
\nAG
\n[2 marks]
\nattempt to find scalar products for П1 and П3, П2 and П3.
\nand equating M1
\nM1
\nNote: Accept .
\nA1
\nattempt to solve , , M1
\nA1
\nhence equation is
\nfor second equation:
\n(M1)
\n\n
attempt to solve , , M1
\n⇒, , A1
\nhence equation is
\n[7 marks]
\nThe diagram shows a circle, centre O, with radius 4 cm. Points A and B lie on the circumference of the circle and AÔB = θ , where 0 ≤ θ ≤ .
\nFind the area of the shaded region, in terms of θ.
\nThe area of the shaded region is 12 cm2. Find the value of θ.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach to find area of segment (M1)
\neg area of sector – area of triangle,
\ncorrect substitution (A1)
\neg
\narea = 80 – 8 sinθ, 8(θ – sinθ) A1 N2
\n[3 marks]
\nsetting their area expression equal to 12 (M1)
\neg 12 = 8(θ – sinθ)
\n2.26717
\nθ = 2.27 (do not accept an answer in degrees) A2 N3
\n[3 marks]
\nThe following diagram shows a right triangle ABC. Point D lies on AB such that CD bisects AĈB.
\nAĈD = θ and AC = 14 cm
\nGiven that , find the value of .
\nFind the value of .
\nHence or otherwise, find .
\nvalid approach (M1)
\neg labelled sides on separate triangle,
\ncorrect working (A1)
\neg missing side is 4,
\nA1 N3
\n[3 marks]
\ncorrect substitution into (A1)
\neg , ,
\nA1 N2
\n[2 marks]
\ncorrect working (A1)
\neg ,
\n(cm) A1 N2
\n[2 marks]
\nThe following diagram shows a circle with centre O and radius r cm.
\nThe points A and B lie on the circumference of the circle, and = θ. The area of the shaded sector AOB is 12 cm2 and the length of arc AB is 6 cm.
\nFind the value of r.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of correctly substituting into circle formula (may be seen later) A1A1
eg
attempt to eliminate one variable (M1)
eg
correct elimination (A1)
eg
correct equation (A1)
eg
correct working (A1)
eg
r = 4 (cm) A1 N2
\n[7 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]
\nFind 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.
\nMETHOD 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 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.
\nequating 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]
\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
\nShow that .
\nHence or otherwise, solve for .
\n\n
Note: Do not award the final A1 for proofs which work from both sides to find a common expression other than .
\n\n
METHOD 1 (LHS to RHS)
\nattempt to use double angle formula for or M1
\nLHS OR
\nOR
\n\nA1
\nRHS AG
\n\n
METHOD 2 (RHS to LHS)
\nRHS
\nattempt to use double angle formula for or M1
\nA1
\nLHS AG
\n\n
[2 marks]
\nattempt to factorise M1
\nA1
\nrecognition of OR (M1)
\none correct reference angle seen anywhere, accept degrees (A1)
\nOR (accept )
\n\n
Note: This (M1)(A1) is independent of the previous M1A1.
\n\n
A2
\n\n
Note: Award A1 for any two correct (radian) answers.
Award A1A0 if additional values given with the four correct (radian) answers.
Award A1A0 for four correct answers given in degrees.
\n
[6 marks]
\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 .
\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]
\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
Let , where and . The line meets the graph of at exactly one point.
\nThe function can be expressed in the form , where .
\nThe function can also be expressed in the form , where .
\nShow that .
\nFind the value of and the value of .
\nFind the value of and the value of .
\nHence find the values of where the graph of is both negative and increasing.
\nMETHOD 1 (discriminant)
\n(M1)
\n\nrecognizing (seen anywhere) M1
\n(do not accept only in quadratic formula for ) A1
\nvalid approach to solve quadratic for (M1)
\nOR
\nboth solutions A1
\nwith a valid reason R1
\nthe two graphs would not intersect OR
\nAG
\n\n
METHOD 2 (equating slopes)
\n(seen anywhere) (M1)
\nA1
\nequating slopes, (seen anywhere) M1
\n\nA1
\nsubstituting their value (M1)
\n\nA1
\n\nAG
\n\n
METHOD 3 (using )
\n(M1)
\n\nattempt to find -coord of vertex using (M1)
\nA1
\nA1
\nsubstituting their value (M1)
\n\nA1
\n\nAG
\n\n
[6 marks]
\n(A1)
\nand OR and A1
\n\n
[2 marks]
\nattempt to use valid approach (M1)
\nOR
\nA1A1
\n\n
[3 marks]
\nEITHER
\nrecognition to (may be seen on sketch) (M1)
\n\n
OR
\nrecognition that and (M1)
\n\n
THEN
\nA1A1
\n\n
Note: Award A1 for two correct values, A1 for correct inequality signs.
\n\n
[3 marks]
\nLet for .
\nConsider the function defined by for and its graph .
\nShow that .
\nThe graph of has a horizontal tangent at point . Find the coordinates of .
\nGiven that , show that is a local maximum point.
\nSolve for .
\nSketch the graph of , showing clearly the value of the -intercept and the approximate position of point .
\nattempt to use quotient or product rule (M1)
\nOR A1
\ncorrect working A1
\nOR cancelling OR
\nAG
\n\n
[3 marks]
\n(M1)
\n\n(A1)
\nA1
\nsubstitution of their to find (M1)
\n\nA1
\n\n\n
[5 marks]
\n(M1)
\nA1
\nwhich is negative R1
\nhence is a local maximum AG
\n\n
Note: The R1 is dependent on the previous A1 being awarded.
\n\n
[3 marks]
\n(A1)
\nA1
\n\n
[2 marks]
\n A1A1A1
\n
\n
Note: Award A1 for one -intercept only, located at
\n A1 for local maximum, , in approximately correct position
A1 for curve approaching -axis as (including change in concavity).
\n
[3 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]
\nLet , where is acute.
\nFind .
\nFind .
\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)
\negright triangle,
\ncorrect working (A1)
\negmissing side is 2,
\nA1 N2
\n[3 marks]
\ncorrect substitution into formula for (A1)
\neg
\nA1 N2
\n[2 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]
\nGiven that , where , 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
\ncorrect substitution into formula for or (A1)
\neg , , ,
\nor (may be seen in substitution) A2
\nrecognizing 4 is double angle of 2 (seen anywhere) (M1)
\neg , , ,
\ncorrect substitution of their value of and/or into formula for (A1)
\neg , , , , ,
\nA1 N2
\n\n
\n
METHOD 2
\nrecognizing 4 is double angle of 2 (seen anywhere) (M1)
\neg
\ndouble angle identity for 2 (M1)
\neg , ,
\ncorrect expression for in terms of and/or (A1)
\neg , ,
\ncorrect substitution for and/or A1
\neg , ,
\ncorrect working (A1)
\neg , ,
\nA1 N2
\n\n
[6 marks]
\nA biased four-sided die, , is rolled. Let be the score obtained when die is rolled. The probability distribution for is given in the following table.
\nA second biased four-sided die, , is rolled. Let be the score obtained when die is rolled.
The probability distribution for is given in the following table.
Find the value of .
\nHence, find the value of .
\nState the range of possible values of .
\nHence, find the range of possible values of .
\nHence, find the range of possible values for .
\nAgnes and Barbara play a game using these dice. Agnes rolls die once and Barbara rolls die once. The probability that Agnes’ score is less than Barbara’s score is .
\nFind the value of .
\nrecognising probabilities sum to (M1)
\n\nA1
\n\n
[2 marks]
\nvalid attempt to find (M1)
\n\nA1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nattempt to find a value of (M1)
\nOR OR
\nA1
\n\n
[2 marks]
\n( OR ) (A1)
\none correct boundary value A1
\nOR
\nOR
\nOR
\nOR
\nOR
\nA1
\n\n
[3 marks]
\nMETHOD 1
\nevidence of choosing at least four correct outcomes from
\n(M1)
\nOR OR (A1)
\nsolving for either or M1
\nOR OR
\nOR
\n\n
EITHER two correct values
\nand A1A1
\n\n
OR one correct value
\nOR A1
\nsubstituting their value for or A1
\nOR
\n\n
THEN
\nA1
\n\n
METHOD 2 (solving for )
\nevidence of choosing at least four correct outcomes from
\n(M1)
\nOR OR (A1)
\nrearranging to make the subject M1
\n\n
M1
\nA1
\n\n
A1
\n\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]
\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]
\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]
\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 .
\nHence or otherwise, find the total distance along the road where the signal from the tower can reach cellular phones.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
recognizing TR =32 (seen anywhere, including diagram) A1
\ncorrect working A1
\neg ,
\nAG N0
\n\n
[2 marks]
\nNote: There are many approaches to this question, depending on which triangle the candidate has used, and whether they used the cosine rule and/or the sine rule. Please check working carefully and award marks in line with the markscheme.
\n\n
METHOD 1
\ncorrect values for (seen anywhere) A1A1
\n= 9.02007, 46.5628
\nrecognizing the need to find difference in values of (M1)
\neg 46.5 − 9.02,
\n37.5427
\n37.5 (km) A1 N2
\n\n
METHOD 2
\ncorrect use of sine rule in ΔSRT
\neg , = 54.0835° (A1)
\nrecognizing isosceles triangle (seen anywhere) (M1)
\neg , two sides of 32
\ncorrect working to find distance A1
\neg ,
\n,
\n37.5427
\n37.5 (km) A1 N2
\n\n
[4 marks]
\n\n
Triangle ABC has a = 8.1 cm, b = 12.3 cm and area 15 cm2. Find the largest possible perimeter of triangle ABC.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\ncorrect substitution into the formula for area of a triangle (A1)
\neg 15 = × 8.1 × 12.3 × sin C
\ncorrect working for angle C (A1)
\neg sin C = 0.301114, 17.5245…, 0.305860
\nrecognizing that obtuse angle needed (M1)
\neg 162.475, 2.83573, cos C < 0
\nevidence of choosing the cosine rule (M1)
\neg a2 = b2 + c2 − 2bc cos(A)
\ncorrect substitution into cosine rule to find c (A1)
\neg c2 = (8.1)2 + (12.3)2 − 2(8.1)(12.3) cos C
\nc = 20.1720 (A1)
\n8.1 + 12.3 + 20.1720 = 40.5720
\nperimeter = 40.6 A1 N4
\n[7 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 .
\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
valid approach (M1)
\neg B − A, AO + OB,
\nA1 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
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]
\nIt is given that , where . Find the exact value of .
\nMETHOD 1
\nattempt to use a right angled triangle M1
\ncorrect placement of all three values and seen in the triangle (A1)
\n(since puts in the second quadrant) R1
\nA1
\nNote: Award M1A1R0A0 for seen as the final answer
The R1 should be awarded independently for a negative value only given as a final answer.
\n
METHOD 2
\nAttempt to use M1
\n\n(A1)
\n\n(since puts in the second quadrant) R1
\nA1
\nNote: Award M1A1R0A0 for seen as the final answer
The R1 should be awarded independently for a negative value only given as a final answer.
\n
METHOD 3
\n\nattempt to use M1
\n\n(A1)
\n\n(since puts in the second quadrant) R1
\n\nA1
\n\n
Note: Award M1A1R0A0 for seen as the final answer
The R1 should be awarded independently for a negative value only given as a final answer.
\n
[4 marks]
\nConsider the quartic equation .
\nTwo of the roots of this equation are and , where .
\nFind the possible values of .
\nMETHOD 1
\nother two roots are and A1
\nsum of roots and product of roots A1
\nattempt to set sum of four roots equal to or OR
attempt to set product of four roots equal to M1
A1
\n\nA1
\n\nattempt to solve simultaneous equations (M1)
\nor A1A1
\n\n
METHOD 2
\nother two roots are and A1
\nA1
\n\nA1
\nAttempt to equate coefficient of and constant with the given quartic equation M1
\nand A1
\nattempt to solve simultaneous equations (M1)
\nor A1A1
\n\n
[8 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.
−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]
\nUse l’Hôpital’s rule to find .
\nattempt to differentiate numerator and denominator M1
\n\nA1A1
\n\n
Note: A1 for numerator and A1 for denominator. Do not condone absence of limits.
\n\n
attempt to substitute (M1)
\nA1
\n\n
Note: Award a maximum of M1A1A0M1A1 for absence of limits.
\n\n
[5 marks]
\nA farmer has six sheep pens, arranged in a grid with three rows and two columns as shown in the following diagram.
\nFive sheep called Amber, Brownie, Curly, Daisy and Eden are to be placed in the pens. Each pen is large enough to hold all of the sheep. Amber and Brownie are known to fight.
\nFind the number of ways of placing the sheep in the pens in each of the following cases:
\nEach pen is large enough to contain five sheep. Amber and Brownie must not be placed in the same pen.
\nEach pen may only contain one sheep. Amber and Brownie must not be placed in pens which share a boundary.
\nMETHOD 1
\nB has one less pen to select (M1)
\n
EITHER
A and B can be placed in ways (A1)
\nC, D, E have choices each (A1)
\n
OR
A (or B), C, D, E have choices each (A1)
\nB (or A) has only choices (A1)
\n
THEN
A1
\n\n
METHOD 2
\ntotal number of ways (A1)
\nnumber of ways with Amber and Brownie together (A1)
\nattempt to subtract (may be seen in words) (M1)
\n\nA1
\n\n
[4 marks]
\nMETHOD 1
\ntotal number of ways (A1)
\nnumber of ways with Amber and Brownie sharing a boundary
\n(A1)
\nattempt to subtract (may be seen in words) (M1)
\nA1
\n\n
METHOD 2
\ncase 1: number of ways of placing A in corner pen
\n\nFour corners total no of ways is (A1)
\ncase 2: number of ways of placing A in the middle pen
\n\ntwo middle pens so (A1)
\nattempt to add (may be seen in words) (M1)
\ntotal no of ways
\nA1
\n\n
[4 marks]
\nConsider the line defined by the Cartesian equation .
\nConsider a second line defined by the vector equation , where and .
\nShow that the point lies on .
\nFind a vector equation of .
\nFind the possible values of when the acute angle between and is .
\nIt is given that the lines and have a unique point of intersection, , when .
\nFind the value of , and find the coordinates of the point in terms of .
\nA1
\nthe point lies on . AG
\n\n
[1 mark]
\nattempt to set equal to a parameter or rearrange cartesian form (M1)
\nOR
\ncorrect direction vector or equivalent seen in vector form (A1)
\n(or equivalent) A1
\n\n
Note: Award A0 if is omitted.
\n\n
[3 marks]
\nattempt to use the scalar product formula (M1)
\n(A1)(A1)
\n\n
Note: Award A1 for LHS and A1 for RHS
\n\n
A1A1
\n\n
Note: Award A1 for LHS and A1 for RHS
\n\n
A1
\nM1
\nattempt to solve their quadratic
\nA1
\n\n
[8 marks]
\nMETHOD 1
\nattempt to equate the parametric forms of and (M1)
\nA1
\nattempt to solve equations by eliminating or (M1)
\nor
\nSolutions exist unless
\nA1
\n\n
Note: This A1 is independent of the following marks.
\n\n
or A1
\nhas coordinates A2
\n\n
Note: Award A1 for any two correct coordinates seen or final answer in vector form.
\n\n
\n
METHOD 2
\nno unique point of intersection implies direction vectors of and parallel
\nA1
\n\n
Note: This A1 is independent of the following marks.
\n\n
attempt to equate the parametric forms of and (M1)
\nA1
\nattempt to solve equations by eliminating or (M1)
\nor
\nor A1
\nhas coordinates A2
\n\n
Note: Award A1 for any two correct coordinates seen or final answer in vector form.
\n\n
[7 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]
\nFind .
\nGiven and , find .
\ncorrect integration A1A1A1
\n\n
Note: Award A1 for , A1 for and A1 for
\n\n
[3 marks]
\nrecognition that (M1)
\n(A1)
\n\nA1
\n\n
[3 marks]
\nThe following table shows the data collected from an experiment.
\nThe data is also represented on the following scatter diagram.
\nThe relationship between and can be modelled by the regression line of on with equation , where .
\nWrite down the value of and the value of .
\nUse this model to predict the value of when .
\nWrite down the value of and the value of .
\nDraw the line of best fit on the scatter diagram.
\nA1A1
\n\n
[2 marks]
\nattempt to substitute into their equation (M1)
\n\n\nA1
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\n A1A1
Note: Award marks as follows:
\nA1 for a straight line going through
\nA1 for intercepting the -axis between their (when their line is extended), which includes all the data for .
\nIf the candidate does not use a ruler, award A0A1 where appropriate.
\n\n
[2 marks]
\nLet for .
\nShow that .
\nUse mathematical induction to prove that for .
\nLet .
\nConsider the function defined by for .
\nIt is given that the term in the Maclaurin series for has a coefficient of .
\nFind the possible values of .
\nattempt to use the chain rule M1
\nA1
\nA1
\nAG
\n\n
Note: Award M1A0A0 for or equivalent seen
\n\n
[3 marks]
\nlet
\nR1
\n\n
Note: Award R0 for not starting at . Award subsequent marks as appropriate.
\n\n
assume true for , (so ) M1
\n\n
Note: Do not award M1 for statements such as “let ” or “ is true”. Subsequent marks can still be awarded.
\n\n
consider
\nM1
\n(or equivalent) A1
\n\n
EITHER
\n(or equivalent) A1
\nA1
\n\n
Note: Award A1 for
\n\n
A1
\n\n\n
Note: Award A1 for leading coefficient of .
\n\n
A1
\n\n
OR
\nNote: The following A marks can be awarded in any order.
\n\n\n
A1
\n\n
Note: Award A1 for isolating correctly.
\n\n
A1
\n\n
Note: Award A1 for multiplying top and bottom by or .
\n\n
A1
\n\n
Note: Award A1 for leading coefficient of .
\n\n
A1
\n\n\n
\n
THEN
\nsince true for , and true for if true for , the statement is true for all, by mathematical induction R1
\n\n
Note: To obtain the final R1, at least four of the previous marks must have been awarded.
\n\n
[9 marks]
\nMETHOD 1
\n\nusing product rule to find (M1)
\nA1
\nA1
\nsubstituting into M1
\nA1
\n\nequating coefficient to M1
\n\nA1
\n\nor A1
\n\n
METHOD 2
\nEITHER
\nattempt to find (M1)
\n\n\n\nA1
\n\n
OR
\nattempt to apply binomial theorem for rational exponents (M1)
\n\nA1
\n\n
THEN
\n(A1)
\n(M1)
\ncoefficient of is A1
\nattempt to set equal to and solve M1
\n\nA1
\n\nor A1
\n\n
METHOD 3
\nand (A1)
\n\nequating coefficient to M1
\n\nusing product rule to find and (M1)
\n\nA1
\nsubstituting into M1
\n\nA1
\nA1
\n\nor A1
\n\n
[8 marks]
\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]
\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.
valid 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]
\nA company produces bags of sugar whose masses, in grams, can be modelled by a normal distribution with mean and standard deviation . A bag of sugar is rejected for sale if its mass is less than grams.
\nFind the probability that a bag selected at random is rejected.
\nEstimate the number of bags which will be rejected from a random sample of bags.
\nGiven that a bag is not rejected, find the probability that it has a mass greater than grams.
\nNote: In this question, do not penalise incorrect use of strict inequality signs.
\nLet mass of a bag of sugar
\n\n
evidence of identifying the correct area (M1)
\n\nA1
\n\n
[2 marks]
\nNote: In this question, do not penalise incorrect use of strict inequality signs.
\nLet mass of a bag of sugar
\n\n\n
A1
\n\n
Note: Accept .
\n\n
[1 mark]
\nNote: In this question, do not penalise incorrect use of strict inequality signs.
\nLet mass of a bag of sugar
\n\n
recognition that is required (M1)
\n\n(A1)
\n\nA1
\n\n
[3 marks]
\nA particle moves in a straight line. The velocity, , of the particle at time seconds is given by , for .
\nThe following diagram shows the graph of .
\nFind the smallest value of for which the particle is at rest.
\nFind the total distance travelled by the particle.
\nFind the acceleration of the particle when .
\nrecognising (M1)
\n\n(sec) A1
\n\n
Note: Do not award A1 if additional values are given.
\n\n
[2 marks]
\nOR (A1)
\n\nA1
\n\n
[2 marks]
\nrecognizing acceleration at is given by (M1)
\nacceleration
\nA1
\n\n
[2 marks]
\nA Ferris wheel with diameter metres rotates at a constant speed. The lowest point on the wheel is metres above the ground, as shown on the following diagram. is a point on the wheel. The wheel starts moving with at the lowest point and completes one revolution in minutes.
\nThe height, metres, of above the ground after minutes is given by , where .
\nFind the values of , and .
\namplitude is (A1)
\nA1
\nA1
\nOR (M1)
\nA1
\n\n
[5 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.
\n(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]
\nConsider the expansion of , where .
\nGiven that the coefficient of is , find the value of .
\nMETHOD 1
\nproduct of a binomial coefficient, a power of (and a power of ) seen (M1)
\nevidence of correct term chosen (A1)
\nOR
\nequating their coefficient to or their term to (M1)
\n\n
EITHER
\n(A1)
\n\n
OR
\n(A1)
\n\n
THEN
\nA1
\n\n
METHOD 2
product of a binomial coefficient, and a power of OR seen (M1)
\nevidence of correct term chosen (A1)
\n\n
equating their coefficient to or their term to (M1)
\n(A1)
\nA1
\n\n
[5 marks]
\nTwo friends Amelia and Bill, each set themselves a target of saving . They each have to invest.
\nAmelia invests her in an account that offers an interest rate of per annum compounded annually.
\nA third friend Chris also wants to reach the target. He puts his money in a safe where he does not earn any interest. His system is to add more money to this safe each year. Each year he will add half the amount added in the previous year.
\nFind the value of Amelia’s investment after years to the nearest hundred dollars.
\nDetermine the number of years required for Amelia’s investment to reach the target.
\nBill invests his in an account that offers an interest rate of per annum compounded monthly, where is set to two decimal places.
\nFind the minimum value of needed for Bill to reach the target after years.
\nShow that Chris will never reach the target if his initial deposit is .
\nFind the amount Chris needs to deposit initially in order to reach the target after years. Give your answer to the nearest dollar.
\nEITHER
\n(A1)
\n(A1)
\n\n
OR
\n
(A1)
(A1)
\n
THEN
\nA1
\n\n
[3 marks]
\nEITHER
\n(A1)
\n\n
OR
\n
(A1)
\n
THEN
\n(years) A1
\n\n
[2 marks]
\nMETHOD 1
\nattempt to substitute into compound interest formula (condone absence of compounding periods) (M1)
\n\n(A1)
\nA1
\n\n
METHOD 2
\n
(M1)(A1)
\n
Note: Award M1 for an attempt to use a financial app in their technology, award A1 for
\n\n
A1
\n\n
[3 marks]
\nrecognising geometric series (seen anywhere) (M1)
\n(A1)
\n\n
EITHER
\nconsidering (M1)
\nA1
\ncorrect reasoning that R1
\n\n
Note: Accept only if has been calculated.
\n\n
OR
\nconsidering for a large value of (M1)
\n\n
Note: Award M1 only if the candidate gives a valid reason for choosing a value of , where .
\n\n
correct value of for their A1
\nvalid reason why Chris will not reach the target, which involves their choice of , their value of and Chris’ age OR using two large values of to recognize asymptotic behaviour of as . R1
\n\n
Note: Do not award the R mark without the preceding A mark.
\n\n
THEN
\nTherefore, Chris will never reach the target. AG
\n\n
[5 marks]
\nrecognising geometric sum M1
\n(A1)
\n\nA1
\n\n
[3 marks]
\nNote: In this question, distance is in metres and time is in seconds.
\nTwo particles and start moving from a point A at the same time, along different straight lines.
\nAfter seconds, the position of is given by r = .
\nTwo seconds after leaving A, is at point B.
\nTwo seconds after leaving A, is at point C, where .
\nFind the coordinates of A.
\nFind ;
\nFind .
\nFind .
\nHence or otherwise, find the distance between and two seconds after they leave A.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nrecognizing at A (M1)
\nA is A1 N2
\n[2 marks]
\nMETHOD 1
\nvalid approach (M1)
\neg
\ncorrect approach to find (A1)
\neg
\nA1 N2
\nMETHOD 2
\nrecognizing is two times the direction vector (M1)
\ncorrect working (A1)
\neg
\nA1 N2
\n[3 marks]
\ncorrect substitution (A1)
\neg
\nA1 N2
\n[2 marks]
\nMETHOD 1 (vector approach)
\nvalid approach involving and (M1)
\neg
\nfinding scalar product and (A1)(A1)
\nscalar product
\n\n
substitution of their scalar product and magnitudes into cosine formula (M1)
\neg
\nA1 N2
\n\n
METHOD 2 (triangle approach)
\nvalid approach involving cosine rule (M1)
\neg
\nfinding lengths AC and BC (A1)(A1)
\n\n
substitution of their lengths into cosine formula (M1)
\neg
\nA1 N2
\n[5 marks]
\nNote: Award relevant marks for working seen to find BC in part (c) (if cosine rule used in part (c)).
\n\n
METHOD 1 (using cosine rule)
\nrecognizing need to find BC (M1)
\nchoosing cosine rule (M1)
\neg
\ncorrect substitution into RHS A1
\neg
\ndistance is 9 A1 N2
\n\n
METHOD 2 (finding magnitude of )
\nrecognizing need to find BC (M1)
\nvalid approach (M1)
\negattempt to find or , or
\ncorrect working A1
\neg
\ndistance is 9 A1 N2
\n\n
METHOD 3 (finding coordinates and using distance formula)
\nrecognizing need to find BC (M1)
\nvalid approach (M1)
\negattempt to find coordinates of B or C, or
\ncorrect substitution into distance formula A1
\neg
\ndistance is 9 A1 N2
\n[4 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 median.
\nComplete the table.
\nWrite down the mid-interval value for the 100 < x ≤ 150 group.
\nUsing the table, calculate an estimate for the mean number of people being followed on the social media website by these 160 students.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
180 (A1) (C1)
[1 mark]
36, 24 (A1)(A1) (C2)
\nNote: Award (A0)(A1) for two incorrect values that add up to 60.
\n[2 marks]
\n125 (accept 125.5) (A1)
\n(M1)
\nNote: Award (M1) for correct substitution of their mid-interval values, multiplied by their frequencies, into mean formula.
\n=156 (155.625) (A1)(ft) (C3)
\nNote: Follow through from parts (b) and (c)(i).
\n[3 marks]
\nThe following diagram shows the quadrilateral ABCD.
\nAB = 6.73 cm, BC = 4.83 cm, BĈD = 78.2° and CD = 3.80 cm.
\nFind BD.
\nThe area of triangle ABD is 18.5 cm2. Find the possible values of θ.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
choosing cosine rule (M1)
\neg
\ncorrect substitution into RHS (A1)
\neg , ,
\n\n
5.50111
\n5.50 (cm) A1 N2
\n[3 marks]
\ncorrect substitution for area of triangle ABD (A1)
\neg
\ncorrect equation A1
\neg ,
\n88.0023, 91.9976, 1.53593, 1.60566
\nθ = 88.0 (degrees) or 1.54 (radians)
\nθ = 92.0 (degrees) or 1.61 (radians) A1A1 N2
\n[4 marks]
\nThe following diagram shows a triangle ABC.
\n![\"N17/5/MATME/SP2/ENG/TZ0/01\"](\"../assets/uploads/tinymce_asset/asset/4158/Schermafbeelding_2018-02-12_om_10.20.56.png\"/)
50° and 112°
\nFind BC.
\nFind the area of triangle ABC.
\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)
\neg
\n4.13102
\nA1 N2
\n[3 marks]
\ncorrect working (A1)
\neg, 18°,
\ncorrect substitution into area formula (A1)
\neg
\n3.19139
\nA1 N2
\n[3 marks]
\nTwo straight fences meet at point and a field lies between them.
\nA horse is tied to a post, , by a rope of length metres. Point is on one fence and point is on the other, such that and radians. This is shown in the following diagram.
\nThe length of the arc shown in the diagram is .
\nA new fence is to be constructed between points and which will enclose the field, as shown in the following diagram.
\nPoint is due west of and . The bearing of from is .
\nWrite down an expression for in terms of .
\nShow that the area of the field that the horse can reach is .
\nThe area of field that the horse can reach is . Find the value of .
\nHence, find the size of .
\nFind the size of .
\nFind the length of new fence required.
\nA1
\n\n
[1 mark]
\nrecognising sum of area of sector and area of triangle required (M1)
\nA1
\n(substitution seen anywhere) A1
\nOR A1
\narea AG
\n\n
[4 marks]
\n(M1)
\n\nA1
\n\n
[2 marks]
\nOR (M1)
\n\nA1
\n\n
[2 marks]
\n(A1)
\nA1
\n\n
[2 marks]
\nchoosing sine rule (M1)
\nOR A1
\nA1
\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.
correct 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.
correct 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]
\nThe vectors a = and b = are perpendicular to each other.
\n\n
Find the value of .
\nGiven that c = a + 2b, find c.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nevidence of scalar product M1
\nega b,
\nrecognizing scalar product must be zero (M1)
\nega b
\ncorrect working (must involve combining terms) (A1)
\neg
\nA1 N2
\n[4 marks]
\nattempt to substitute their value of (seen anywhere) (M1)
\negb = , 2b =
\ncorrect working (A1)
\neg
\nc = A1 N2
\n[3 marks]
\nConsider the function defined by for .
\nThe graph of and the line intersect at point .
\nThe line has a gradient of and is a tangent to the graph of at the point .
\nThe shaded region is enclosed by the graph of and the lines and .
\nFind the -coordinate of .
\nFind the exact coordinates of .
\nShow that the equation of is .
\nFind the -coordinate of the point where intersects the line .
\nHence, find the area of .
\nThe line is tangent to the graphs of both and the inverse function .
\nFind the shaded area enclosed by the graphs of and and the line .
\nAttempt to find the point of intersection of the graph of and the line (M1)
\n\nA1
\n\n
[2 marks]
\nA1
\nattempt to set the gradient of equal to (M1)
\n\nhas coordinates (accept () A1A1
\n\n
Note: Award A1 for each value, even if the answer is not given as a coordinate pair.
\nDo not accept or as a final value for . Do not accept or as a final value for .
\n\n
[4 marks]
\nattempt to substitute coordinates of (in any order) into an appropriate equation (M1)
\nOR A1
\nequation of is AG
\n\n
[2 marks]
\nA1
\n\n
[1 mark]
\nappropriate method to find the sum of two areas using integrals of the difference of two functions (M1)
\n\n
Note: Allow absence of incorrect limits.
\n\n
(A1)(A1)
\n\n
Note: Award A1 for one correct integral expression including correct limits and integrand.
Award A1 for a second correct integral expression including correct limits and integrand.
\n\n
A1
\n\n
[4 marks]
\nby symmetry (M1)
\nA1
\n\n
Note: Accept any answer that rounds to (but do not accept ).
\n\n
[2 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]
\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.
\ncorrect 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]
\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.
\nCalculate the value of k.
\nHence, complete the histogram.
\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]
\n(M1)(A1)(ft)
\nNote: Award (M1) for substitution into the mean formula and equating to 12, (A1)(ft) for their correct substitutions.
\n(k =) 7 (A1)(ft) (C4)
\nNote: Follow through from part (b)(i) and their mid-interval values, consistent with part (a). Do not award final (A1) if answer is not an integer.
\n[3 marks]
\n(A1)(ft) (C1)
Note: Follow through from their part (b)(ii) but only if the value is between 1 and 10, inclusive.
\n[1 mark]
\n\n
A line passes through the points and .
\nGiven that and are perpendicular, show that .
\nFind .
\nHence, write down a vector equation for .
\nA second line , has equation r = .
\nGiven that and are perpendicular, show that .
\nThe lines and intersect at . Find .
\nFind a unit vector in the direction of .
\nHence or otherwise, find one point on which is units from C.
\nvalid approach (M1)
\neg
\nA1 N2
\n[2 marks]
\nany correct equation in the form r = a + tb (any parameter for ) A2 N2
\nwhere a is or , and b is a scalar multiple of
\n\n
egr = , r = , r = j + 8k + t(3i + 4j – 6k)
\n\n
Note: Award A1 for the form a + tb, A1 for the form L = a + tb, A0 for the form r = b + ta.
\n\n
[2 marks]
\nvalid approach (M1)
\neg
\nchoosing correct direction vectors (may be seen in scalar product) A1
\neg and
\ncorrect working/equation A1
\neg
\nAG N0
\n[3 marks]
\nvalid approach (M1)
\neg
\none correct equation (must be different parameters if both lines used) (A1)
\neg
\none correct value A1
\neg
\nvalid approach to substitute their or value (M1)
\neg
\nA1 N3
\n[5 marks]
\n(A1)
\nA1 N2
\n[2 marks]
\nMETHOD 1 (using unit vector)
\nvalid approach (M1)
\neg
\ncorrect working (A1)
\neg
\none correct point A1 N2
\neg
\nMETHOD 2 (distance between points)
\nattempt to use distance between and (M1)
\neg
\nsolving leading to or (A1)
\none correct point A1 N2
\neg
\n[3 marks]
\nConsider the planes and with the following equations.
\n\n\nFind a Cartesian equation of the plane which is perpendicular to and and passes through the origin .
\nFind the coordinates of the point where , and intersect.
\nattempt to find a vector perpendicular to and using a cross product (M1)
\n\n(A1)
\nequation is A1
\n\n
[3 marks]
\nattempt to solve simultaneous equations in variables (M1)
\nA1
\n\n
[2 marks]
\nA continuous random variable has the probability density function given by
\n\nwhere .
\nShow that .
\nFind the value of .
\nrecognition of the need to integrate (M1)
\n\n\n
EITHER
\n(or equivalent) (A1)
\n\nA1
\n\n
OR
\n(A1)
\nA1
\n\n
THEN
\nattempt to use correct limits for their integrand and set equal to M1
\nOR
\nA1
\nAG
\n\n
[5 marks]
\nattempt to solve (M1)
\n\nA1
\n\n
[2 marks]
\nConsider the complex numbers and , where .
\nSuppose that .
\nFind the modulus of .
\nFind the argument of in terms of .
\nFind the minimum value of .
\nFor the value of found in part (i), find the value of .
\nA1
\n\n
[1 mark]
\nattempt to find (M1)
\n\nA1
\n\n
[2 marks]
\nis a multiple of (M1)
\nis a multiple of (M1)
\nA1
\n\n
[3 marks]
\nA1
\n\n
[1 mark]
\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]
\nTwo boats and travel due north.
\nInitially, boat is positioned metres due east of boat .
\nThe distances travelled by boat and boat , after seconds, are metres and metres respectively. The angle is the radian measure of the bearing of boat from boat . This information is shown on the following diagram.
\nShow that .
\nAt time , the following conditions are true.
\nBoat has travelled metres further than boat .
Boat is travelling at double the speed of boat .
The rate of change of the angle is radians per second.
Find the speed of boat at time .
\nOR A1
\nAG
\n\n
Note: may be identified as a length on a diagram, and not written explicitly.
\n\n
[1 mark]
\nattempt to differentiate with respect to (M1)
\nA1
\nattempt to set speed of equal to double the speed of (M1)
\n\nA1
\nOR (A1)
\n\n
Note: This A1 can be awarded independently of previous marks.
\n\n\n
So the speed of boat is A1
\n\n
Note: Accept from the use of inexact values.
\n\n
[6 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.
valid 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]
\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]
\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]
\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 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]
\nConsider the following frequency table.
\n![\"M17/5/MATME/SP2/ENG/TZ1/01\"](\"../assets/uploads/tinymce_asset/asset/3542/Schermafbeelding_2017-08-14_om_09.52.57.png\"/)
Write down the mode.
\nFind the value of the range.
\nFind the mean.
\nFind the variance.
\nA1 N1
\n[1 mark]
\nvalid approach (M1)
\neg, interval 2 to 11
\nA1 N2
\n[2 marks]
\n7.14666
\nA2 N2
\n[2 marks]
\nrecognizing that variance is (M1)
\neg
\n\n
A1 N2
\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]
\nTen students were asked for the distance, in km, from their home to school. Their responses are recorded below.
\n0.3 0.4 3 3 3.5 5 7 8 8 10
\nThe following box-and-whisker plot represents this data.
\nFor these data, find the mean distance from a student’s home to school.
\nFind the value of .
\nFind the interquartile range.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nevidence of finding (M1)
\neg ,
\n(exact) A1 N2
\n[2 marks]
\n= 4.25 (exact) A1 N1
\n[1 mark]
\nvalid approach (M1)
\neg Q3 − Q1 3 − 8 , 3 to 8
\nIQR = 5 A1 N2
\n[2 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.
\n![\"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 group of 10 girls recorded the number of hours they spent watching television during a particular week. Their results are summarized in the box-and-whisker plot below.
\nThe group of girls watched a total of 180 hours of television.
\nA group of 20 boys also recorded the number of hours they spent watching television that same week. Their results are summarized in the table below.
\nThe following week, the group of boys had exams. During this exam week, the boys spent half as much time watching television compared to the previous week.
\nFor this exam week, find
\nThe range of the data is 16. Find the value of .
\nFind the value of the interquartile range.
\nFind the mean number of hours that the girls in this group spent watching television that week.
\nFind the total number of hours the group of boys spent watching television that week.
\nFind the mean number of hours that all 30 girls and boys spent watching television that week.
\nthe mean number of hours that the group of boys spent watching television.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nvalid approach (M1)
\neg 16 + 8, − 8
\n24 (hours) A1 N2
\n[2 marks]
\nvalid approach (M1)
\neg 20 − 15, , 15 − 20
\nIQR = 5 A1 N2
\n[2 marks]
\ncorrect working (A1)
\neg , ,
\nmean = 18 (hours) A1 N2
\n[2 marks]
\nattempt to find total hours for group B (M1)
\neg
\ngroup B total hours = 420 (seen anywhere) A1 N2
\n[2 marks]
\nattempt to find sum for combined group (may be seen in working) (M1)
\neg 180 + 420, 600
\ncorrect working (A1)
\neg ,
\nmean = 20 (hours) A1 N2
\n[3 marks]
\nvalid approach to find the new mean (M1)
\neg ,
\nmean (= 10.5) hours A1 N2
\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.
\n15 (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]
\nA 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.
![\"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]
\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]
\nThe functions and are defined for by and , where .
\nFind the range of .
\nGiven that for all , determine the set of possible values for .
\nattempting to find the vertex (M1)
\nOR OR
\nrange is A1
\n\n
[2 marks]
\nMETHOD 1
\n(A1)
\n
EITHER
relating to the range of OR attempting to find (M1)
\n(A1)
\n
OR
attempting to find the discriminant of (M1)
\n(A1)
\n
THEN
A1
\n\n
METHOD 2
\nvertical reflection followed by vertical shift (M1)
\nnew vertex is (A1)
\n(A1)
\nA1
\n\n
[4 marks]
\nA data set has n items. The sum of the items is 800 and the mean is 20.
\nThe standard deviation of this data set is 3. Each value in the set is multiplied by 10.
\nFind n.
\nWrite down the value of the new mean.
\nFind the value of the new variance.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct approach (A1)
\neg
\n40 A1 N2
\n[2 marks]
\n200 A1 N1
\n[1 mark]
\nMETHOD 1
\nrecognizing variance = σ 2 (M1)
\neg 32 = 9
\ncorrect working to find new variance (A1)
\neg σ 2 × 102, 9 × 100
\n900 A1 N3
\n\n
METHOD 2
\nnew standard deviation is 30 (A1)
\nrecognizing variance = σ 2 (M1)
\neg 32 = 9, 302
\n900 A1 N3
\n[3 marks]
\nA lampshade, in the shape of a cone, has a wireframe consisting of a circular ring and four straight pieces of equal length, attached to the ring at points A, B, C and D.
\nThe ring has its centre at point O and its radius is 20 centimetres. The straight pieces meet at point V, which is vertically above O, and the angle they make with the base of the lampshade is 60°.
\nThis information is shown in the following diagram.
\n![\"M17/5/MATSD/SP1/ENG/TZ2/03\"](\"../assets/uploads/tinymce_asset/asset/3580/Schermafbeelding_2017-08-15_om_17.16.13.png\"/)
Find the length of one of the straight pieces in the wireframe.
\nFind the total length of wire needed to construct this wireframe. Give your answer in centimetres correct to the nearest millimetre.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nOR (M1)
\n\n
Note: Award (M1) for correct substitution into a correct trig. ratio.
\n(A1) (C2)
\n[2 marks]
\n(M1)(M1)
\n\n
Note: Award (M1) for correct substitution in the circumference of the circle formula, (M1) for adding 4 times their answer to part (a) to their circumference of the circle.
\n\n
285.6637… (A1)(ft)
\n\n
Note: Follow through from part (a). This (A1) may be implied by a correct rounded answer.
\n\n
285.7 (cm) (A1)(ft) (C4)
\n\n
Notes: Award (A1)(ft) for rounding their answer (consistent with their method) to the nearest millimetre, irrespective of unrounded answer seen.
\nThe final (A1)(ft) is not dependent on any of the previous M marks. It is for rounding their unrounded answer correctly.
\n\n
[4 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.
\nvalid 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]
\nAll living plants contain an isotope of carbon called carbon-14. When a plant dies, the isotope decays so that the amount of carbon-14 present in the remains of the plant decreases. The time since the death of a plant can be determined by measuring the amount of carbon-14 still present in the remains.
\nThe amount, , of carbon-14 present in a plant years after its death can be modelled by where and are positive constants.
\nAt the time of death, a plant is defined to have units of carbon-14.
\nThe time taken for half the original amount of carbon-14 to decay is known to be years.
\nShow that .
\nShow that .
\nFind, correct to the nearest years, the time taken after the plant’s death for of the carbon-14 to decay.
\nA1
\nAG
\n\n
[1 mark]
\ncorrect substitution of values into exponential equation (M1)
\nOR
\n
EITHER
A1
\nOR A1
\n
OR
A1
\nA1
\n
THEN
AG
\n
Note: There are many different ways of showing that which involve showing different steps. Award full marks for at least two correct algebraic steps seen.
\n
[3 marks]
\nif of the carbon-14 has decayed, remains ie, units remain (A1)
\n\n
EITHER
using an appropriate graph to attempt to solve for (M1)
\n
OR
manipulating logs to attempt to solve for (M1)
\n\n\n
THEN
(years) (correct to the nearest 10 years) A1
\n\n
[3 marks]
\nThe final examination results obtained by a group of 3200 Biology students are summarized on the cumulative frequency graph.
\n350 of the group obtained the highest possible grade in the examination.
\nThe grouped frequency table summarizes the examination results of this group of students.
\nFind the median of the examination results.
\nFind the interquartile range.
\nFind the final examination result required to obtain the highest possible grade.
\nWrite down the modal class.
\nWrite down the mid-interval value of the modal class.
\nCalculate an estimate of the mean examination result.
\nCalculate an estimate of the standard deviation, giving your answer correct to three decimal places.
\nThe teacher sets a grade boundary that is one standard deviation below the mean.
\nUse the cumulative frequency graph to estimate the number of students whose final examination result was below this grade boundary.
\n60 (A2)
\n\n
[2 marks]
\n68 − 48 (A1)(M1)
\nNote: Award (A1) for two correct quartiles seen, (M1) for finding the difference between their two quartiles.
\n\n
= 20 (A1)(ft)(G3)
\n\n
[3 marks]
\n3200 − 350 = 2850 (M1)
\nNote: Award (M1) for 2850 seen. Follow through from their 3200.
\n\n
(Top grade boundary =) 76 (A1)(ft)(G2)
\n\n
[2 marks]
\n60 < x ≤ 80 (A1)(A1)
\nNote: Award (A1) for 60, 80 seen, (A1) for correct strict and weak inequalities.
\n\n
[2 marks]
\n70 (A1)(ft)
\nNote: Follow through from part (c)(i).
\n\n
[1 mark]
\n57.2 (57.1875) (A2)(ft)
\nNote: Follow through from part (c)(ii).
\n\n
[2 marks]
\n18.496 (A1)
\nNote: Award (A0) for 18.499.
\n\n
[1 mark]
\n57.2 − 18.5 (M1)
\n= 38.7 (38.6918…) (A1)(ft)
\nNote: Award (M1) for subtracting their standard deviation from their mean. Follow through from part (d) even if no working is shown.
\n\n
450 (students) (A1)(ft)(G2)
\nNote: Accept any answer within the range of 450 to 475, inclusive. Follow through from part (d), adjusting the acceptable range as necessary.
\n\n
[3 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.
\nFind the median number of hours worked by the employees.
\nWrite down the number of employees who worked 50 hours or less.
\nFind the amount of money an employee earned for working 40 hours;
\nFind the amount of money an employee earned for working 43 hours.
\nFind the number of employees who earned £200 or less.
\nOnly 10 employees earned more than £. Find the value of .
\nevidence of median position (M1)
\neg80th employee
\n40 hours A1 N2
\n[2 marks]
\n130 employees A1 N1
\n[1 mark]
\n£320 A1 N1
\n[1 mark]
\nsplitting into 40 and 3 (M1)
\neg3 hours more,
\ncorrect working (A1)
\neg
\n£350 A1 N3
\n[3 marks]
\nvalid approach (M1)
\neg200 is less than 320 so 8 pounds/hour, ,
\n18 employees A2 N3
\n[3 marks]
\nvalid approach (M1)
\neg
\n60 hours worked (A1)
\ncorrect working (A1)
\neg
\nA1 N3
\n[4 marks]
\nA flat horizontal area, ABC, is such that AB = 100 m , BC = 50 m and angle AĈB = 43.7° as shown in the diagram.
\nShow that the size of angle BÂC is 20.2°, correct to 3 significant figures.
\nCalculate the area of triangle ABC.
\nFind the length of AC.
\nA vertical pole, TB, is constructed at point B and has height 25 m.
\nCalculate the angle of elevation of T from, M, the midpoint of the side AC.
\n\n
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\n(M1)(A1)
\nNote: Award (M1) for substitution into sine rule formula, (A1) for correct substitution.
\n\n
BAC = 20.2087… = 20.2° (A1)(AG)
\nNote: Award (A1) only if both the correct unrounded and rounded answers are seen.
\n\n
[3 marks]
\n\n
units are required in part (b)
\n\n
(A1)(M1)(A1)
\nNote: Award (A1) for 116.1 or unrounded value or 116 seen, (M1) for substitution into area of triangle formula, (A1) for correct substitution.
\n\n
\n
= 2250 m2 (2245.06… m2) (A1)(G3)
\nNote: The answer is 2250 m2; the units are required. Use of 20.2087… gives 2245.23….
\n\n
[4 marks]
\n(M1)(A1)(ft)
\nNote: Award (M1) for substitution into sine rule formula, (A1)(ft) for their correct substitution. Follow through from their 116.1.
\n\n
AC = 130 (m) (129.982… (m)) (A1)(ft)(G2)
\nNote: Use of 20.2087… gives 129.992….
\n\n
OR
\nAC2 = 1002 + 502 −2(100)(50) cos (116.1) (M1)(A1)(ft)
\nNote: Award (M1) for substitution into cosine rule formula, (A1)(ft) for their correct substitution. Follow through from their 116.1.
\n\n
AC = 130 (m) (129.997… (m)) (A1)(ft)(G2)
\nNote: Award (M1) for substitution into cosine rule formula, (A1)(ft) for their correct substitution.
\n\n
[3 marks]
\n\n
BM2 = 1002 + 652 − 2(100)(65) cos (20.2°) (M1)(A1)(ft)
\nOR
\nBM2 = 502 + 652 − 2(50)(65) cos (43.7°) (M1)(A1)(ft)
\nNote: Award (M1) for substitution into cosine rule formula, (A1)(ft) for correct substitution, including half their AC.
\n\n
BM = 45.0 (44.9954… OR 45.0079…) (A1)(ft)
\nNote: Use of 20.2052… gives 45. Award (G2) for 45.0 seen without working.
\n\n
(M1)
\nNote: Award (M1) for correct substitution into tangent formula.
\n\n
TB = 29.1° (29.0546…°) (A1)(ft)(G4)
\nNote: Follow through within part (d) provided their BM is seen. Use of 44.9954 gives 29.0570… and use of 45.0079… gives 29.0503…. Follow through from their AC in part (c).
\n\n
[5 marks]
\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.
(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]
\nA continuous random variable has the probability density function given by
\n\nwhere .
\nShow that .
\nShow that .
\nM1
\nA1
\nleading to AG
\n\n
[2 marks]
\nMETHOD 1
\nuse of M1
\n\n\nA1
\nM1
\n
EITHER
A1
\n
OR
A1
\n
THEN
so AG
\n\n
METHOD 2
\nuse of M1
\n\n\nA1
\nM1
\n
EITHER
A1
\n
OR
A1
\n
THEN
so AG
\n\n
[6 marks]
\nThe six blades of a windmill rotate around a centre point . Points and and the base of the windmill are on level ground, as shown in the following diagram.
\nFrom point the angle of elevation of point is radians.
\nAn observer walks metres from point to point .
\nThe observer keeps walking until he is standing directly under point . The observer has a height of metres, and as the blades of the windmill rotate, the end of each blade passes metres over his head.
\nOne of the blades is painted a different colour than the others. The end of this blade is labelled point . The height , in metres, of point above the ground can be modelled by the function , where is in seconds and . When , point is at its maximum height.
\nGiven that point is metres from the base of the windmill, find the height of point above the ground.
\nFind the angle of elevation of point from point .
\nFind the length of each blade of the windmill.
\nFind the value of and the value of .
\nIf the observer stands directly under point for one minute, point will pass over his head times.
\nFind the value of .
\n(M1)
\n\nA1
\n\n
[2 marks]
\nOR (A1)
\n\n(radians) (accept ) A1
\n\n
[2 marks]
\n(or equivalent) (A1)
\n\n() A1
\n\n
[2 marks]
\nMETHOD 1
\nrecognition that blade length = amplitude, (M1)
\nA1
\ncentre of windmill = vertical shift, (M1)
\nA1
\n\n
METHOD 2
\nattempting to form two equations in terms of and (M1)(M1)
\n\nA1
\nA1
\n\n
[4 marks]
\nappropriate working towards finding the period (M1)
\n\nrotations per minute (M1)
\n(must be an integer) (accept , , ) A1
\n\n
[3 marks]
\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 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(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]
\nEight runners compete in a race where there are no tied finishes. Andrea and Jack are two of the eight competitors in this race.
\nFind the total number of possible ways in which the eight runners can finish if Jack finishes
\nin the position immediately after Andrea.
\nin any position after Andrea.
\nJack and Andrea finish in that order (as a unit) so we are considering the arrangement of objects (M1)
\nways A1
\n\n
[2 marks]
\nMETHOD 1
\nthe number of ways that Andrea finishes in front of Jack is equal to the number of ways that Jack finishes in front of Andrea (M1)
\ntotal number of ways is 8! (A1)
\nways A1
\n\n
METHOD 2
\nthe other six runners can finish in ways (A1)
\nwhen Andrea finishes first, Jack can finish in different positions
\nwhen Andrea finishes second, Jack can finish in different positions etc
\nways (A1)
\nhence there are ways
\nways A1
\n\n
[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]
\nConsider where .
\nShow that .
\nattempt to use the complex conjugate of their denominator M1
\nA1
\nM1A1
\n
Note: Award M1 for expanding the numerator and A1 for a correct numerator. Condone either an incorrect denominator or the absence of a denominator.
using to simplify the numerator (M1)
AG
\n\n
[5 marks]
\nWrite down the first three terms of the binomial expansion of in ascending powers of .
\nBy using the Maclaurin series for and the result from part (a), show that the Maclaurin series for up to and including the term in is .
\nBy using the Maclaurin series for and the result from part (b), find .
\nA1
\n
Note: Accept and .
\n
[1 mark]
\n(M1)
\nor (M1)
\nA1
\nA1
\nso the Maclaurin series for up to and including the term in is AG
\n
Note: Condone the absence of ‘…’
\n
[4 marks]
\nM1
\nA1
\n\nA1
\n\n
Note: Condone missing ‘lim’ and errors in higher derivatives.
Do not award M1 unless is replaced by in .
\n
[3 marks]
\nAC is a vertical communications tower with its base at C.
\nThe tower has an observation deck, D, three quarters of the way to the top of the tower, A.
\n![\"N16/5/MATSD/SP1/ENG/TZ0/11\"](\"../assets/uploads/tinymce_asset/asset/3322/Schermafbeelding_2017-03-06_om_10.32.25.png\"/)
From a point B, on horizontal ground 250 m from C, the angle of elevation of D is 48°.
\nCalculate CD, the height of the observation deck above the ground.
\nCalculate the angle of depression from A to B.
\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 the tangent ratio.
\n\n
(A1) (C2)
\n[2 marks]
\n(M1)(M1)(M1)
\n\n
Note: Award (M1) for multiplying their part (a), (M1) for substitution into the tangent ratio, (M1) for correct substitution.
\n\n
OR
\n(M1)(M1)(M1)
\n\n
Note: Award (M1) for multiplying their part (a), (M1) for substitution into the tangent ratio, (M1) for subtracting from 90 and for correct substitution.
\n\n
(A1)(ft) (C4)
\n\n
Note: Follow through from part (a).
\n\n
[4 marks]
\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]
\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°.
\nAngle APB is acute.
\nOn the diagram, draw and label with an x the angle of depression of B from P.
\nFind the size of angle APB.
\nFind the size of 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]
\n(M1)(A1)
\nNote: Award (M1) for substitution into sine rule, (A1) for correct substitution.
\n(angle APB =) 82.2° (82.2473…°) (A1) (C3)
\n[3 marks]
\n180 − 48 − 82.2473… (M1)
\n49.8° (49.7526…°) (A1)(ft) (C2)
\nNote: Follow through from parts (a) and (b).
\n[2 marks]
\nSpeedWay airline flies from city to city . The flight time is normally distributed with a mean of minutes and a standard deviation of minutes.
\nA flight is considered late if it takes longer than minutes.
\nThe flight is considered to be on time if it takes between and minutes. The probability that a flight is on time is .
\nDuring a week, SpeedWay has flights from city to city . The time taken for any flight is independent of the time taken by any other flight.
\nCalculate the probability a flight is not late.
\nFind the value of .
\nCalculate the probability that at least of these flights are on time.
\nGiven that at least of these flights are on time, find the probability that exactly flights are on time.
\nSpeedWay increases the number of flights from city to city to flights each week, and improves their efficiency so that more flights are on time. The probability that at least flights are on time is .
\nA flight is chosen at random. Calculate the probability that it is on time.
\nvalid approach (M1)
\neg ,
\n\n
A1 N2
\n[2 marks]
\nvalid approach (M1)
\neg
\ncorrect working (A1)
\neg
\n\n
(minutes) A1 N3
\n[3 marks]
\nevidence of recognizing binomial distribution (seen anywhere) (M1)
\neg ,
\nevidence of summing probabilities from to (M1)
\neg ,
\n\n
A1 N2
\n[3 marks]
\nfinding (seen anywhere) A1
\neg
\nrecognizing conditional probability (M1)
\neg , ,
\ncorrect working (A1)
\neg
\n\n
A1 N1
\nNote: Exception to the FT rule: if the candidate uses an incorrect value for the probability that a flight is on time in (i) and working shown, award full FT in (ii) as appropriate.
\n[4 marks]
\ncorrect equation (A1)
\neg
\nvalid attempt to solve (M1)
\neg graph
\n\n
A1 N1
\n[3 marks]
\nThe flight times, minutes, between two cities can be modelled by a normal distribution with a mean of minutes and a standard deviation of minutes.
\nOn a particular day, there are flights scheduled between these two cities.
\nGiven that of the flight times are longer than minutes, find the value of .
\nFind the probability that a randomly selected flight will have a flight time of more than minutes.
\nGiven that a flight between the two cities takes longer than minutes, find the probability that it takes less than minutes.
\nFind the expected number of flights that will have a flight time of more than minutes.
\nFind the probability that more than of the flights on this particular day will have a flight time of more than minutes.
\nuse of inverse normal to find -score (M1)
\n\n(A1)
\n\nA1
\n\n
[3 marks]
\nevidence of identifying the correct area under the normal curve (M1)
\n\nA1
\n\n
[2 marks]
\nrecognition that is required (M1)
\n(M1)(A1)
\n\nA1
\n\n
[4 marks]
\nrecognition of binomial probability (M1)
\nor (A1)
\n\n(flights) A1
\n\n
[3 marks]
\n(M1)
\n(A1)
\n\nA1
\n\n
[3 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]
\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.
(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 function is defined by , where and .
\nThe region enclosed by the graph of , the -axis, the -axis and the line is rotated about the -axis to form a solid of revolution.
\nPedro wants to make a small bowl with a volume of based on the result from part (a). Pedro’s design is shown in the following diagrams.
\nThe vertical height of the bowl, , is measured along the -axis. The radius of the bowl’s top is and the radius of the bowl’s base is . All lengths are measured in .
\nFor design purposes, Pedro investigates how the cross-sectional radius of the bowl changes.
\nShow that the volume of the solid formed is cubic units.
\nFind the value of that satisfies the requirements of Pedro’s design.
\nFind .
\nFind .
\nBy sketching the graph of a suitable derivative of , find where the cross-sectional radius of the bowl is decreasing most rapidly.
\nState the cross-sectional radius of the bowl at this point.
\nattempt to use (M1)
\n\n
EITHER
applying integration by recognition (M1)
\nA3
\n
OR
(A1)
\nattempt to express the integral in terms of (M1)
\nwhen and when
\n(A1)
\nA1
\n\n
OR
\n(A1)
\nattempt to express the integral in terms of (M1)
\nwhen and when
\n(A1)
\nA1
\n
Note: Accept equivalent working with indefinite integrals and original limits for .
\n
THEN
\nA1
\nso the volume of the solid formed is cubic units AG
\n
Note: Award (M1)(A0)(M0)(A0)(A0)(A1) when is obtained from GDC
\n
[6 marks]
\na valid algebraic or graphical attempt to find (M1)
\n\n(as ) A1
\n
Note: Candidates may use their GDC numerical solve feature.
\n
[2 marks]
\nattempting to find
\nwith (M1)
\nA1
\n\n
[2 marks]
\nattempting to find
\nwith (M1)
\nA1
\n\n
[2 marks]
\nEITHER
\nrecognising to graph (M1)
\nNote: Award M1 for attempting to use quotient rule or product rule differentiation.
\n
for graph decreasing to the local minimum A1
before increasing towards the -axis A1
\n\n
OR
\nrecognising to graph (M1)
\nNote: Award M1 for attempting to use quotient rule or product rule differentiation.
\nfor , graph increasing towards and beyond the -intercept A1
recognising for maximum rate (A1)
\n\n
THEN
\nA1
\n\n
Note: Only award A marks if either graph is seen.
\n[4 marks]
\nattempting to find (M1)
\nthe cross-sectional radius at this point is A1
\n\n
[2 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.
\nA1A1A1 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 quadrilateral ABCD represents a park, where , and . Angle DAB is 70° and angle DCB is 110°. This information is shown in the following diagram.
\n![\"M17/5/MATSD/SP2/ENG/TZ2/04\"](\"../assets/uploads/tinymce_asset/asset/3609/Schermafbeelding_2017-08-16_om_17.35.40.png\"/)
A straight path through the park joins the points B and D.
\nA new path, CE, is to be built such that E is the point on BD closest to C.
\nThe section of the park represented by triangle DCE will be used for a charity race. A track will be marked along the sides of this section.
\nFind the length of the path BD.
\nShow that angle DBC is 48.7°, correct to three significant figures.
\nFind the area of the park.
\nFind the length of the path CE.
\nCalculate the total length of the track.
\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 cosine rule, (A1) for correct substitution.
\n\n
(A1)(G2)
\n\n
[3 marks]
\n(M1)(A1)(ft)
\n\n
Note: Award (M1) for substituted sine rule, (A1)(ft) for correct substitution.
\nFollow through from their answer to part (a).
\n\n
(A1)(ft)
\n(AG)
\n\n
Notes: Award the final (A1)(ft) only if both their unrounded answer and 48.7° is seen. Follow through from their answer to part (a), only if their unrounded answer rounds to 48.7°.
\n\n
[3 marks]
\n(A1)(M1)(M1)
\n\n
Note: Award (A1) for 21.3° (21.2615…) seen, (M1) for substitution into (at least) one area of triangle formula in the form , (M1) for their correct substitutions and adding the two areas.
\n\n
(A1)(ft)(G3)
\n\n
Notes: Follow through from their answers to part (a). Accept from use of 48.7384…
\n\n
[4 marks]
\n(M1)
\n(A1)(ft)(G2)
\n\n
Note: Follow through from their angle 21.3° in part (c). Award (M0)(A0) for halving 110° and/or assuming E is the midpoint of BD in any method seen.
\n\n
OR
\n(M1)
\n(A1)(ft)(G2)
\n\n
Note: Follow through from parts (a) and (c). Award (M0)(A0) for halving 110° and/or assuming E is the midpoint of BD in any method seen.
\n\n
[2 marks]
\n(M1)(M1)
\n\n
Note: Award (M1) for correct use of Pythagoras to find DE (or correct trigonometric equation, , to find DE), (M1) for the sum of 100, their DE and their CE.
\n\n
(A1)(ft)(G2)
\n\n
Note: Follow through from part (d). Use of 3 sf values gives an answer of .
\n\n
[3 marks]
\nAll answers in this question should be given to four significant figures.
\n
In a local weekly lottery, tickets cost each.
In the first week of the lottery, a player will receive for each ticket, with the probability distribution shown in the following table. For example, the probability of a player receiving is . The grand prize in the first week of the lottery is .
\nIf nobody wins the grand prize in the first week, the probabilities will remain the same, but the value of the grand prize will be in the second week, and the value of the grand prize will continue to double each week until it is won. All other prize amounts will remain the same.
\nFind the value of .
\nDetermine whether this lottery is a fair game in the first week. Justify your answer.
\nGiven that the grand prize is not won and the grand prize continues to double, write an expression in terms of for the value of the grand prize in the week of the lottery.
\nThe week is the first week in which the player is expected to make a profit. Ryan knows that if he buys a lottery ticket in the week, his expected profit is .
\nFind the value of .
\nconsidering that sum of probabilities is (M1)
\n\nA1
\n\n
[2 marks]
\nvalid attempt to find (M1)
\n\nA1
\nNo, not a fair game A1
\nfor a fair game, would be OR players expected winnings are R1
\n\n
[4 marks]
\nrecognition of GP with (M1)
\nOR A1
\n\n
[2 marks]
\nrecognizing (M1)
\ncorrect expression for week (or week) (A1)
\n\ncorrect inequality (accept equation) (A1)
\nOR
\n\n
EITHER
\nOR (A1)
\n
OR
in week or in week (A1)
\n
THEN
A1
\nexpected profit per ticket (M1)
\nA1
\n\n
[7 marks]
\nA function is defined by .
\nA function is defined by .
\nShow that is an even function.
\nBy considering limits, show that the graph of has a horizontal asymptote and state its equation.
\nShow that for .
\nBy using the expression for and the result , show that is decreasing for .
\n\n
Find an expression for , justifying your answer.
\nState the domain of .
\nSketch the graph of , clearly indicating any asymptotes with their equations and stating the values of any axes intercepts.
\nEITHER
\nR1
\n
OR
a sketch graph of with line symmetry in the -axis indicated R1
\n
THEN
so is an even function. AG
\n\n
[1 mark]
\nas A1
\nso the horizontal asymptote is A1
\n\n
[2 marks]
\nattempting to use the quotient rule to find M1
\nA1
\nattempting to use the chain rule to find M1
\nlet and so and
\nM1
\nA1
\nA1
\nAG
\n\n
[6 marks]
\n
EITHER
for (A1)
\nso A1
\n
OR
and A1
\nA1
\n
THEN
R1
\n
Note: Award R1 for stating that in , the numerator is negative, and the denominator is positive.
so is decreasing for AG
Note: Do not accept a graphical solution
\n
[3 marks]
\nM1
\nA1
\nA1
\ndomain of is and so the range of must be
\nhence the positive root is taken (or the negative root is rejected) R1
\n
Note: The R1 is dependent on the above A1.
so A1
Note: The final A1 is not dependent on R1 mark.
\n
[5 marks]
\ndomain is A1
\n
Note: Accept correct alternative notations, for example, or .
Accept if correct to s.f.
\n
[1 mark]
\n A1A1A1
Note: A1 for correct domain and correct range and -intercept at
A1 for asymptotic behaviour
A1 for
Coordinates are not required.
Do not accept or other inexact values.
\n
[3 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 find the mean age;
\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\"/)
(M1)
\n\n
Note: Award (M1) for correct substitutions into mean formula.
\n\n
(A1) (C2)
\n[2 marks]
\n16.5 (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]
\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]
\nThree airport runways intersect to form a triangle, ABC. The length of AB is 3.1 km, AC is 2.6 km, and BC is 2.4 km.
\nA company is hired to cut the grass that grows in triangle ABC, but they need to know the area.
\nFind the size, in degrees, of angle BÂC.
\nFind the area, in km2, of triangle ABC.
\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 cosine rule formula, (A1) for correct substitutions.
\n48.8° (48.8381…°) (A1) (C3)
\n[3 marks]
\n(M1)(A1)(ft)
\nNote: Award (M1) for substituted area of a triangle formula, (A1) for correct substitution.
\n3.03 (km2) (3.033997…(km2)) (A1)(ft) (C3)
\nNote: Follow through from part (a).
\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]
\nThis question asks you to explore the behaviour and some key features of the function , where and .
\nIn parts (a) and (b), only consider the case where .
\nConsider .
\nConsider , where .
\nNow consider where and .
\nBy using the result from part (f) and considering the sign of , show that the point on the graph of is
\nSketch the graph of , stating the values of any axes intercepts and the coordinates of any local maximum or minimum points.
\nUse your graphic display calculator to explore the graph of for
\n• the odd values and ;
\n• the even values and .
\nHence, copy and complete the following table.
\nShow that .
\nState the three solutions to the equation .
\nShow that the point on the graph of is always above the horizontal axis.
\nHence, or otherwise, show that , for .
\na local minimum point for even values of , where and .
\na point of inflexion with zero gradient for odd values of , where and .
\nConsider the graph of , where , and .
\nState the conditions on and such that the equation has four solutions for .
\ninverted parabola extended below the -axis A1
\n-axis intercept values A1
Note: Accept a graph passing through the origin as an indication of .
local maximum at A1
\n
Note: Coordinates must be stated to gain the final A1.
Do not accept decimal approximations.
[3 marks]
A1A1A1A1A1A1
Note: Award A1 for each correct value.
For a table not sufficiently or clearly labelled, assume that their values are in the same order as the table in the question paper and award marks accordingly.
\n
[6 marks]
METHOD 1
\nattempts to use the product rule (M1)
\nA1A1
\n
Note: Award A1 for a correct and A1 for a correct .
EITHER
attempts to factorise (involving at least one of or ) (M1)
\nA1
\n
OR
attempts to express as the difference of two products with each product containing at least one of or (M1)
\nA1
\n
THEN
AG
\n
Note: Award the final (M1)A1 for obtaining any of the following forms:
\n
\n
\n
\n
METHOD 2
\n(M1)
\nA1
\nattempts to use the chain rule (M1)
\nA1A1
\n
Note: Award A1 for and A1 for .
AG
\n
[5 marks]
\nA2
\nNote: Award A1 for either two correct solutions or for obtaining
Award A0 otherwise.
\n
[2 marks]
\nattempts to find an expression for (M1)
\n\nA1
\n
EITHER
since (for and so ) R1
\n
Note: Accept any logically equivalent conditions/statements on and .
Award R0 if any conditions/statements specified involving , or both are incorrect.
\n
OR
\n(since ), raised to an even power () (or equivalent reasoning) is always positive (and so ) R1
\n
Note: The condition is given in the question. Hence some candidates will assume and not state it. In these instances, award R1 for a convincing argument.
Accept any logically equivalent conditions/statements on on and .
Award R0 if any conditions/statements specified involving , or both are incorrect.
THEN
so is always above the horizontal axis AG
\n
Note: Do not award (M1)A0R1.
\n
[3 marks]
\nMETHOD 1
\nA1
\n
EITHER
as and R1
\n
OR
and are all R1
\n\n
Note: Do not award A0R1.
Accept equivalent reasoning on correct alternative expressions for and accept any logically equivalent conditions/statements on and .
Exceptions to the above are condone and condone .
\nAn alternative form for is .
\n
THEN
hence AG
\n\n
METHOD 2
\nand A1
\n(since is continuous and there are no stationary points between and )
\nthe gradient (of the curve) must be positive between and R1
\n
Note: Do not award A0R1.
hence AG
\n
[2 marks]
\nfor even:
\n(and are both ) R1
\nA1
\nand (seen anywhere) A1
\n\n
Note: Candidates can give arguments based on the sign of to obtain the R mark.
For example, award R1 for the following:
If is even, then is odd and hence .
Do not award R0A1.
The second A1 is independent of the other two marks.
The A marks can be awarded for correct descriptions expressed in words.
Candidates can state as a point of zero gradient from part (d) or show, state or explain (words or diagram) that . The last A mark can be awarded for a clearly labelled diagram showing changes in the sign of the gradient.
The last A1 can be awarded for use of a specific case (e.g. ).
hence is a local minimum point AG
\n
[3 marks]
\nfor odd:
\n, (and are both ) so R1
\n
Note: Candidates can give arguments based on the sign of to obtain the R mark.
For example, award R1 for the following:
If is odd, then is even and hence .
and (seen anywhere) A1
Note: The A1 is independent of the R1.
Candidates can state as a point of zero gradient from part (d) or show, state or explain (words or diagram) that . The last A mark can be awarded for a clearly labelled diagram showing changes in the sign of the gradient.
The last A1 can be awarded for use of a specific case (e.g. ).
\n
hence is a point of inflexion with zero gradient AG
\n\n
[2 marks]
\nconsiders the parity of (M1)
\n
Note: Award M1 for stating at least one specific even value of .
must be even (for four solutions) A1
Note: The above 2 marks are independent of the 3 marks below.
\n
A1A1A1
\n\n
Note: Award A1 for the correct lower endpoint, A1 for the correct upper endpoint and A1 for strict inequality signs.
\n The third A1 (strict inequality signs) can only be awarded if A1A1 has been awarded.
For example, award A1A1A0 for . Award A1A0A0 for .
Award A1A0A0 for .
\n\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]
\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]
\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 function is defined by , where .
\nWrite down the equation of
\nFind the coordinates where the graph of crosses
\nthe vertical asymptote of the graph of .
\nthe horizontal asymptote of the graph of .
\nthe -axis.
\nthe -axis.
\nSketch the graph of on the axes below.
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\n(accept ) A1
\n\n
[1 mark]
\n(accept and ) A1
\n\n
[1 mark]
\n A1
Note: Award A1 for completely correct shape: two branches in correct quadrants with asymptotic behaviour.
\n
[1 mark]
\nUse l’Hôpital’s rule to determine the value of
\n.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
using l’Hôpital’s rule,
\nM1A1
\n(M1)A1
\nA1
\nA1
\n
[6 marks]
Assuming the Maclaurin series for and , show that the Maclaurin series for is
\n\nBy differentiating the series in part (a), show that the Maclaurin series for is .
\nHence determine the Maclaurin series for as far as the term in .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nMETHOD 1
\nattempts to substitute into
\nM1
\nA1
\nattempts to expand the up to and including the term M1
\nA1
\nAG
\n\n
METHOD 2
\nattempts to substitute into M1
\n\nattempts to find the Maclaurin series for up to and including the term M1
\nA1
\n\nA1
\nAG
\n
[4 marks]
A1A1
\n\nattempts to expand the up to and including the term M1
\nA1
\nAG
\n
[4 marks]
METHOD 1
\nlet
\nuses to form M1
\nA1
\n(A1)
\nattempts to equate coefficients,
\nM1
\nA1
\nso
\n\n
METHOD 2
\nuses to form M1
\nA1
\n(A1)
\nattempts to expand the up to and including the term M1
\n\n\nA1
\n
Note: Accept use of long division.
[5 marks]
Box 1 contains 5 red balls and 2 white balls.
\nBox 2 contains 4 red balls and 3 white balls.
\nA box is chosen at random and a ball is drawn. Find the probability that the ball is red.
\nLet be the event that “box 1 is chosen” and let be the event that “a red ball is drawn”.
\nDetermine whether events and are independent.
\nvalid approach to find (M1)
\ntree diagram (must include probabilty of picking box) with correct required probabilities
\nOR OR
\n(A1)
\nA1
\n\n
[3 marks]
\nevents and are not independent, since OR OR
\nOR an explanation e.g. different number of red balls in each box A2
\n\n
Note: Both conclusion and reasoning are required. Do not split the A2.
\n\n
[2 marks]
\nThe curve has a gradient function given by
\n.
\nThe curve passes through the point .
\nOn the same set of axes, sketch and label isoclines for and , and clearly indicate the value of each -intercept.
\nHence or otherwise, explain why the point is a local minimum.
\nFind the solution of the differential equation , which passes through the point . Give your answer in the form .
\nExplain why the graph of does not intersect the isocline .
\nSketch the graph of on the same set of axes as part (a)(i).
\nattempt to find equation of isoclines by setting M1
\nparallel lines with positive gradient A1
\n-intercept for A1
\n
Note: To award A1, each -intercept should be clear, but condone a missing label (eg. ).
If candidates represent the lines using slope fields, but omit the lines, award maximum of M1A0A1.
\n
[3 marks]
at point A1
\n
EITHER
to the left of , the gradient is negative R1
\nto the right of , the gradient is positive R1
\n
Note: Accept any correct reasoning using gradient, isoclines or slope field.
If a candidate uses left/right or without explicitly referring to the point or a correct region on the diagram, award R0R1.
\n
OR
A1
\nA1
\n
Note: accept correct reasoning that is increasing as increases.
THEN
hence is a local minimum AG
\n
[3 marks]
integrating factor (M1)
\n(A1)
\n(M1)
\nA1
\n(M1)
\nA1
\n
Note: Award A1 for the correct RHS.
substituting gives
M1
\n\nA1
\n
[8 marks]
METHOD 1
\nEITHER
\nattempt to solve for the intersection (M1)
\n
OR
attempt to find the difference (M1)
\n
THEN
for all R1
\n
Note: Accept or equivalent reasoning.
therefore the curve does not intersect the isocline AG
\n
METHOD 2
\nis an (oblique) asymptote to the curve R1
\n
Note: Do not accept “the curve is parallel to \"
is the isocline for R1
therefore the curve does not intersect the isocline AG
\n\n
METHOD 3
\nThe initial point is above , so R1
\n\nR1
\ntherefore the curve does not intersect the isocline AG
\n\n
[2 marks]
\n\n
concave up curve with minimum at approximately A1
\nasymptote of curve is isocline A1
\n
Note: Only award FT from (b) if the above conditions are satisfied.
[2 marks]
A triangular postage stamp, ABC, is shown in the diagram below, such that and .
\n![\"M17/5/MATSD/SP1/ENG/TZ1/13\"](\"../assets/uploads/tinymce_asset/asset/3571/Schermafbeelding_2017-08-15_om_11.34.31.png\"/)
Find the length of BC.
\nFind the area of the postage stamp.
\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 sine rule formula, (A1) for correct substitutions.
\n\n
(A1) (C3)
\n[3 marks]
\n(M1)(A1)(ft)
\n\n
Note: Award (M1) for substituted area of a triangle formula, (A1) for correct substitutions.
\n\n
(A1)(ft) (C3)
\n\n
Note: Follow through from part (a).
\n[3 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.
\nFind the value of a and of b.
\nWrite down the correlation coefficient.
\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.
\nevidence of set up (M1)
\neg correct value for a or b or r (seen in (ii)) or r2 (= 0.973)
\n9.91044, −31.3194
\na = 9.91, b = −31.3, y = 9.91x − 31.3 A1A1 N3
\n\n
[3 marks]
\n0.986417
\nr = 0.986 A1 N1
\n\n
[1 mark]
\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]
\nThis question asks you to investigate and prove a geometric property involving the roots of the equation where for integers , where .
\n
The roots of the equation where are , where . Each root can be represented by a point , respectively, on an Argand diagram.
For example, the roots of the equation where are and . On an Argand diagram, the root can be represented by a point and the root can be represented by a point .
\nConsider the case where .
\nThe roots of the equation where are , and . On the following Argand diagram, the points and lie on a circle of radius unit with centre .
\nLine segments and are added to the Argand diagram in part (a) and are shown on the following Argand diagram.
\nis the length of and is the length of .
\nConsider the case where .
\nThe roots of the equation where are and .
\nOn the following Argand diagram, the points and lie on a circle of radius unit with centre . , and are line segments.
\nFor the case where , the equation where has roots and .
\nIt can be shown that .
\nNow consider the general case for integer values of , where .
\nThe roots of the equation where are . On an Argand diagram, these roots can be represented by the points respectively where are line segments. The roots lie on a circle of radius unit with centre .
\ncan be expressed as .
\nConsider where .
\nShow that .
\nHence, deduce that .
\nShow that .
\nBy factorizing , or otherwise, deduce that .
\nShow that .
\nSuggest a value for .
\nWrite down expressions for and in terms of .
\nHence, write down an expression for in terms of and .
\nExpress as a product of linear factors over the set .
\nHence, using the part (g)(i) and part (f) results, or otherwise, prove your suggested result to part (e).
\nMETHOD 1
\nattempts to expand (M1)
\nA1
\nAG
\n\n
METHOD 2
\nattempts polynomial division on M1
\nA1
\nso AG
\n\n
Note: In part (a), award marks as appropriate where has been converted into Cartesian, modulus-argument (polar) or Euler form.
\n\n
[2 marks]
\n(since is a root of ) R1
\nand R1
\nAG
\n\n
Note: In part (a), award marks as appropriate where has been converted into Cartesian, modulus-argument (polar) or Euler form.
\n\n
[2 marks]
\nMETHOD 1
\nattempts to find either or (M1)
\naccept any valid method
\ne.g. from either or
\ne.g. use of Pythagoras’ theorem
\ne.g. by calculating the distance between points
\nA1
\nA1
\n
Note: Award a maximum of M1A1A0 for any decimal approximation seen in the calculation of either or or both.
so AG
\n
METHOD 2
\nattempts to find (M1)
\nA1
\nand since R1
\nso AG
\n\n
[3 marks]
\nMETHOD 1
\nA1
\n( is a root hence) and R1
\nAG
\n
Note: Condone the use of throughout.
\n
METHOD 2
\nconsiders the sum of roots of (M1)
\nthe sum of roots is zero (there is no term) A1
\nAG
\n\n
METHOD 3
\nsubstitutes for (M1)
\ne.g.
\nA1
\n
Note: This can be demonstrated geometrically or by using vectors. Accept Cartesian or modulus-argument (polar) form.
AG
\n
METHOD 4
\nA1
\nas R1
\nAG
\n\n
[2 marks]
\nMETHOD 1
\nA1
\nattempts to find either or (M1)
\n\n
Note: For example and .
Various geometric and trigonometric approaches can be used by candidates.
\n
A1A1
\n
Note: Award a maximum of A1M1A1A0 if labels such as are not clearly shown.
Award full marks if the lengths are shown on a clearly labelled diagram.
Award a maximum of A1M1A1A0 for any decimal approximation seen in the calculation of either or or both.
AG
\n
METHOD 2
\nattempts to find M1
\nA1
\nsince and A1
\nand since R1
\nso AG
\n\n
METHOD 3
\nA1
\nattempts to find M1
\nA1
\nsince and R1
\nso AG
\n\n
[4 marks]
\nA1
\n\n
[1 mark]
\nA1A1
\n\n
[2 marks]
\nA1A1
\n
Note: Accept from symmetry.
[1 mark]
considers the equation (M1)
\nthe roots are (A1)
\nso A1
\n
[3 marks]
METHOD 1
\nsubstitutes into M1
\n(A1)
\ntakes modulus of both sides M1
\n\nA1
\nso AG
\n\n
Note: Award a maximum of M1A1FTM1A0 from part (e).
\n\n
METHOD 2
\nare the roots of M1
\ncoefficient of is and the coefficient of is A1
\nproduct of the roots is A1
\nA1
\nso AG
\n
[4 marks]
Consider the differential equation , where .
\nIt is given that when .
\nSolve the differential equation, giving your answer in the form .
\nThe graph of against has a local maximum between and . Determine the coordinates of this local maximum.
\nShow that there are no points of inflexion on the graph of against .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nputs so that M1
\nA1
\nattempts to express as a single rational fraction in
\nM1
\nattempts to separate variables M1
\n\nA1A1
\nsubstitutes and attempts to find the value of M1
\nA1
\nthe solution is
\nA1
\n
[9 marks]
at a maximum, M1
\nattempts to substitute into their solution M1
\n\nattempts to solve for (M1)
\nA1
\n
Note: Accept all answers that round to the correct answer.
Accept .
[4 marks]
METHOD 1
\nattempts (quotient rule) implicit differentiation M1
\n\ncorrectly substitutes into
\nA1
\nA1
\nthis expression can never be zero therefore no points of inflexion R1
\n\n
METHOD 2
\nattempts implicit differentiation on M1
\nA1
\n\nA1
\nand therefore no points of inflexion R1
\n
Note: Accept putting and obtaining contradiction.
[4 marks]
A jigsaw puzzle consists of many differently shaped pieces that fit together to form a picture.
\nJill is doing a 1000-piece jigsaw puzzle. She started by sorting the edge pieces from the interior pieces. Six times she stopped and counted how many of each type she had found. The following table indicates this information.
\nJill models the relationship between these variables using the regression equation .
\nWrite down the value of and of .
\nUse the model to predict how many edge pieces she had found when she had sorted a total of 750 pieces.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\neg correct value for or (ignore incorrect labels)
\n,
\n, (accept ) A1A1 N3
\n[3 marks]
\nvalid approach (M1)
\neg 750 = , edge + interior = 750
\ncorrect working (A1)
\neg 750 − = 6.9298 + 8.807 , 93.4684
\n93 (pieces) (accept 94) A1 N3
\n[3 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]
\nThe function is defined by where .
\nThe seventh derivative of is given by .
\nUse the Maclaurin series for to write down the first three non-zero terms of the Maclaurin series for .
\nHence find the first three non-zero terms of the Maclaurin series for .
\nUse your answer to part (a)(i) to write down an estimate for .
\nUse the Lagrange form of the error term to find an upper bound for the absolute value of the error in calculating , using the first three non-zero terms of the Maclaurin series for .
\nWith reference to the Lagrange form of the error term, explain whether your answer to part (b) is an overestimate or an underestimate for .
\nsubstitution of in (M1)
\nA1
\n
[2 marks]
(M1)
\n
Note: Award (M1) if this is seen in part (a)(i).
attempt to differentiate their answer in part (a) (M1)
M1
Note: Award M1 for equating their derivatives.
A1
[4 marks]
A1
\n
Note: Accept an answer that rounds correct to or better.
[1 mark]
attempt to find the maximum of for (M1)
\nmaximum of occurs at (A1)
\n(for all ) (A1)
\nuse of (M1)
\nsubstitution of and and their value of and their value of into Lagrange error term (M1)
\n
Note: Award (M1) for substitution of and and their value of and their value of into Lagrange error term.
upper bound A1
\n
Note: Accept an answer that rounds correct to or better.
[6 marks]
(for all ) R1
\n
Note: Accept or “the error term is negative”.
the answer in (b) is an overestimate A1
Note: The A1 is dependent on the R1.
[2 marks]
A discrete random variable has the probability distribution given by the following table.
\nGiven that , determine the value of and 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\nA1
\n(M1)
\n\nA1
\n\n
[4 marks]
\nConsider the function , where and .
\nConsider the function , where .
\nThe graph of may be obtained by transforming the graph of using a sequence of three transformations.
\nWrite down an expression for .
\nHence, given that does not exist, show that .
\nShow that exists.
\ncan be written in the form , where .
\nFind the value of and the value of .
\nHence find .
\nState each of the transformations in the order in which they are applied.
\nSketch the graphs of and on the same set of axes, indicating the points where each graph crosses the coordinate axes.
\nA1
\n
[1 mark]
since does not exist, there must be two turning points R1
\n( has more than one solution)
\nusing the discriminant M1
\nA1
\nAG
\n
[4 marks]
METHOD 1
\nM1
\n\nA1
\nhence exists AG
\n\n
METHOD 2
\nM1
\n\nthere is (only) one point with gradient of and this must be a point of inflexion (since is a cubic.) R1
\nhence exists AG
\n
[2 marks]
A1
\n(M1)
\n\nA1
\n
[3 marks]
(M1)
\n
Note: Interchanging and can be done at any stage.
(M1)
A1
\n
Note: must be seen for the final A mark.
[3 marks]
translation through , A1
\n
Note: This can be seen anywhere.
EITHER
a stretch scale factor parallel to the -axis then a translation through A2
OR
a translation through then a stretch scale factor parallel to the -axis A2
Note: Accept ‘shift’ for translation, but do not accept ‘move’. Accept ‘scaling’ for ‘stretch’.
[3 marks]
A1A1A1M1A1
Note: Award A1 for correct ‘shape’ of (allow non-stationary point of inflexion)
Award A1 for each correct intercept of
Award M1 for attempt to reflect their graph in , A1 for completely correct including intercepts
[5 marks]
A cylindrical container with a radius of 8 cm is placed on a flat surface. The container is filled with water to a height of 12 cm, as shown in the following diagram.
\n![\"M17/5/MATSD/SP1/ENG/TZ2/12\"](\"../assets/uploads/tinymce_asset/asset/3585/Schermafbeelding_2017-08-16_om_06.17.11.png\"/)
A heavy ball with a radius of 2.9 cm is dropped into the container. As a result, the height of the water increases to cm, as shown in the following diagram.
\n![\"M17/5/MATSD/SP1/ENG/TZ2/12.b\"](\"../assets/uploads/tinymce_asset/asset/3586/Schermafbeelding_2017-08-16_om_06.18.54.png\"/)
Find the volume of water in the container.
\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 the volume of a cylinder formula.
\n\n
(A1) (C2)
\n[2 marks]
\n(M1)(M1)(M1)
\n\n
Note: Award (M1) for correct substitution into the volume of a sphere formula (this may be implied by seeing 102.160…), (M1) for adding their volume of the ball to their part (a), (M1) for equating a volume to the volume of a cylinder with a height of .
\n\n
OR
\n(M1)(M1)(M1)
\n\n
Note: Award (M1) for correct substitution into the volume of a sphere formula (this may be implied by seeing 102.160…), (M1) for equating to the volume of a cylinder, (M1) for the height of the water level increase, . Accept for if adding 12 is implied by their answer.
\n\n
(A1)(ft) (C4)
\n\n
Note: If 3 sf answer used, answer is 12.5 (12.4944…). Follow through from part (a) if first method is used.
\n\n
[4 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]
A 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]
\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 null.
\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. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.
OR 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]
A 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 , for . The following diagram shows part of the graph of .
\n\n
For the graph of
\nfind the -coordinates of the -intercepts.
\nfind the coordinates of the vertex.
\nThe function can be written in the form .
\nWrite down the value of and the value of .
\nsetting (M1)
\n(accept ) A1
\n\n
[2 marks]
\nMETHOD 1
\nA1
\nsubstituting their -coordinate into (M1)
\nA1
\n\n\n
METHOD 2
\nattempt to complete the square (M1)
\n(M1)
\nA1A1
\n\n
\n
[3 marks]
\nA1
\nA1
\n\n
[2 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).
\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.
(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]
\nConsider the curve defined by .
\nShow that .
\nProve that, when .
\nHence find the coordinates of all points on , for , where .
\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
\n
Note: Award A1 for LHS, M1 for attempt at chain rule, A1 for RHS.
M1
\n
Note: Award M1 for collecting derivatives and factorising.
AG
[5 marks]
setting
\n(M1)
\nA1
\nOR OR A1
\n
Note: If they offer values for , award A1 for at least two correct values in two different ‘quadrants’ and no incorrect values.
R1
A1
\nAG
\n
[5 marks]
OR (M1)
\nA1A1
\nA1A1
\n
Note: Allow ‘coordinates’ expressed as for example.
Note: Each of the A marks may be awarded independently and are not dependent on (M1) being awarded.
Note: Mark only the candidate’s first two attempts for each case of .
[5 marks]
\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. It appeared in a paper that permitted the use of a calculator, and so might not be suitable for all forms of practice.
(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]
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 .
\nShow that the stationary points of the curve are at x = 1 and x = 2.
\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.
\n(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]
\n6x2 − 18x + 12 = 0 (M1)
\nNote: Award (M1) for equating their derivative to 0. If the derivative is not explicitly equated to 0, but a subsequent solving of their correct equation is seen, award (M1).
\n6(x − 1)(x − 2) = 0 (or equivalent) (M1)
\nNote: Award (M1) for correct factorization. The final (M1) is awarded only if answers are clearly stated.
\nAward (M0)(M0) for substitution of 1 and of 2 in their derivative.
\nx = 1, x = 2 (AG)
\n[2 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]
\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]
Jashanti 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.
\n400 (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]
\nGiven that and when , find in terms of .
\nMETHOD 1
\nrecognition that (M1)
\n(A1)
\nsubstitute both and values into their integrated expression including (M1)
\n\n\nA1
\n\n
METHOD 2
\n(M1)(A1)
\nA1
\nA1
\n\n
[4 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\"/)
Draw the line on the axes.
\nWrite down the number of solutions to .
\nFind the range of values of for which has no solution.
\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]
\nConsider the function defined by , where and .
\nConsider the case where .
\nState the equation of the vertical asymptote on the graph of .
\nState the equation of the horizontal asymptote on the graph of .
\nUse an algebraic method to determine whether is a self-inverse function.
\nSketch the graph of , stating clearly the equations of any asymptotes and the coordinates of any points of intersections with the coordinate axes.
\nThe region bounded by the -axis, the curve , and the lines and is rotated through about the -axis. Find the volume of the solid generated, 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.
A1
\n
[1 mark]
A1
\n
[1 mark]
METHOD 1
\nM1
\nA1
\n\nA1
\n\n\n, (hence is self-inverse) R1
\n
Note: The statement could be seen anywhere in the candidate’s working to award R1.
\n
METHOD 2
\n\nM1
\n
Note: Interchanging and can be done at any stage.
A1
A1
\n(hence is self-inverse) R1
\n
[4 marks]
attempt to draw both branches of a rectangular hyperbola M1
\nand A1
\nand A1
\n
[3 marks]
METHOD 1
\n(M1)
\nEITHER
\nattempt to express in the form M1
\nA1
\nOR
\nattempt to expand or and divide out M1
\nA1
\nTHEN
\nA1
\n\nA1
\n\nA1
\n\n
METHOD 2
\n(M1)
\nsubstituting A1
\n\nM1
\nA1
\nA1
\n
Note: Ignore absence of or incorrect limits seen up to this point.
A1
[6 marks]
Find the equation of the tangent to the curve at the point where .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(A1)
\nappreciate the need to find (M1)
\nA1
\nA1
\nA1
\n
[5 marks]
Consider the equation , where and .
\nFind the value of and the value of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
substituting and M1
\n\n\n
equate real and imaginary: M1
\nAND A1
\n
Note: If they multiply top and bottom by the conjugate, the equations and may be seen. Allow for A1.
solving simultaneously:
A1A1
\n
[5 marks]
The first term in an arithmetic sequence is and the fifth term is .
\nFind the common difference of the sequence, expressing 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.
(A1)
\n\nattempt to write an integer (eg or ) in terms of M1
\n\nattempt to combine two logs into one M1
\n\n\nattempt to use power rule for logs M1
\n\nA1
\n
[5 marks]
Note: Award method marks in any order.
The plane has equation and the line has vector equation
\n.
\nThe plane contains the point and the line .
\nGiven that meets at the point , find the coordinates of .
\nFind the shortest distance from the point to .
\nFind the equation of , giving your answer in the form .
\nDetermine the acute angle between and .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(M1)
\n(A1)
\n(M1)
\nso A1
\n
Note: Do not award the final A1 if a vector given instead of coordinates
[4 marks]
METHOD 1
\n\nsubstituting into equation of the plane M1
\n\nA1
\ndistance (M1)
\nA1
\n
METHOD 2
choice of any point on the plane, eg to use in distance formula (M1)
\nso distance A1A1
\n
Note: Award A1 for numerator, A1 for denominator.
A1
\n
[4 marks]
EITHER
\nidentify two vectors (A1)
\neg, and
\n(M1)
\n
OR
identify three points in the plane (A1)
eg gives and
\nsolving system of equations (M1)
\n
THEN
A1
Note: Accept .
[3 marks]
vector normal to is eg
\nvector normal to is eg (A1)
\nrequired angle is , where M1A1
\n(A1)
\n\nA1
\nNote: Award the penultimate (A1) but not the final A1 for the obtuse angle or .
\n
[5 marks]
Using 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 particle moves in a straight line such that after time seconds, its velocity, in , is given by , where .
\nAt time , has displacement ; at time , .
\nAt successive times when the acceleration of is, the velocities of form a geometric sequence. The acceleration of is zero at times where and the respective velocities are .
\nFind the times when comes to instantaneous rest.
\nFind an expression for in terms of .
\nFind the maximum displacement of , in metres, from its initial position.
\nFind the total distance travelled by in the first seconds of its motion.
\nShow that, at these times, .
\nHence show that .
\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]
attempt to use integration by parts M1
\n\n
EITHER
A1
A1
\n\nM1
\n
OR
A1
A1
\n\nM1
\n
THEN
A1
at M1
\nA1
\n\n
[7 marks]
EITHER
\nsubstituting into their equation for (M1)
\n\n
OR
using GDC to find maximum value (M1)
OR
\nevaluating (M1)
\n
THEN
A1
[2 marks]
METHOD 1
\n
EITHER
distance required (M1)
\n
OR
distance required (M1)
\n\n
THEN
A1
\n
METHOD 2
\n
using successive minimum and maximum values on the displacement graph (M1)
A1
\n
[2 marks]
valid attempt to find using product rule and set M1
\nA1
\nAG
\n
[2 marks]
attempt to evaluate in exact form M1
\n\n\nA1
\n
Note: The A1 is for any two consecutive correct, or showing that or .
showing that
eg M1A1
\nshowing that M1
\neg
\n
Note: Award the A1 for any two consecutive terms.
AG
[5 marks]
Hyungmin 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 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]
There 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]
\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]
\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 diagram shows part of the graph of . The graph has a local maximum point at and a local minimum point at .
\nDetermine the values of , and .
\nHence find the area of the shaded region.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
the principal axis is
\nso A1
\nthe amplitude is
\nso A1
\n
EITHER
one period is (M1)
\n\n\n
OR
Substituting a point eg
\n\nChoice of correct solution (M1)
\n
THEN
A1
\n\n
Note: and can be both given as negatives for full marks
[4 marks]
roots are (A1)
\n(M1)
\n(A1)
\nso area A1
\n
[4 marks]
At 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]
\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]
In a group of 35 students, some take art class (A) and some take music class (M). 5 of these students do not take either class. This information is shown in the following Venn diagram.
\nOne student from the group is chosen at random. Find the probability that
\nWrite down the number of students in the group who take art class.
\nthe student does not take art class.
\nthe student takes either art class or music class, but not both.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid approach (M1)
\n, , 11 + 6
\nnumber of students taking art class = 17 A1 N2
\n\n
[2 marks]
\nvalid approach (M1)
\n13 + 5, 35 − 17, 18, 1 − P(A)
\n0.514285
\nP(A') = (exact), 0.514 A1 N2
\n\n
[2 marks]
\nvalid approach (M1)
\n11 + 13, 35 − 6 − 5, 24
\n0.685714
\nP(A or M but not both) = (exact), 0.686 A1 N2
\n\n
[2 marks]
\nFind the term independent of in the expansion of .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
use of Binomial expansion to find a term in either or (M1)(A1)
\n
Note: Award M1 for a product of three terms including a binomial coefficient and powers of the two terms, and A1 for a correct expression of a term in the expansion.
finding the powers required to be and (M1)(A1)
constant term is (M1)
\n
Note: Ignore all ’s in student’s expression.
therefore term independent of is A1
[6 marks]
Use mathematical induction to prove that for .
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
\nso true for A1
\n
Note: Award A1 if is proved.
assume proposition true for , i.e. M1
Notes: Do not award M1 if using instead of .
Assumption of truth must be present.
Subsequent marks are not dependent on this M1 mark.
(M1)
M1
\n\nA1
\n
Note: Award A1 for correct derivative.
A1
Note: The final A1 can be awarded for either of the two lines above.
hence true for and true true R1
therefore true for all
\n
Note: Only award the final R1 if the three method marks have been awarded.
[7 marks]
At a gathering of teachers, seven are male and five are female. A group of five of these teachers go out for a meal together. Determine the possible number of groups in each of the following situations:
\nThere are more males than females in the group.
\nTwo of the teachers, Gary and Gerwyn, refuse to go out for a meal together.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
identifying two or three possible cases (M1)
\ntotal number of possible groups is (A1)(A1)
\n
Note: Award A1 for any two correct cases, A1 for the other one.
A1
\n
[4 marks]
METHOD 1
\nidentifying at least two of the three possible cases- Gary goes, Gerwyn goes or neither goes (M1)
\ntotal number of possible groups is (A1)
\n\nA1
\n\n
METHOD 2
\nidentifying the overall number of groups and no. of cases where both Gary and Gerwyn go. (M1)
\ntotal number of possible groups is (A1)
\n\nA1
\n
[3 marks]
The 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.
\nvalid 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]
\nA small bead is free to move along a smooth wire in the shape of the curve .
\nFind an expression for .
\nAt the point on the curve where , it is given that
\nFind the value of at this exact same instant.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid attempt to use chain rule or quotient rule (M1)
\nOR A1A1
\n
[3 marks]
Note: Award A1 for numerator and A1 for denominator, or A1 for each part if the second alternative given.
valid attempt to use chain rule (M1)
\nor equivalent (A1)
\n\nA1
\n
[3 marks]
Events 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.
\nvalid 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]
\nThe th term of an arithmetic sequence is given by .
\nState the value of the first term, .
\nGiven that the th term of this sequence is , find the value of .
\nFind the common difference, .
\nA1
\n\n
[1 mark]
\n(A1)
\nA1
\n\n
[2 marks]
\nvalid approach to find (M1)
\nOR recognize gradient is OR attempts to solve
\nA1
\n\n
[2 marks]
\nA large majority of candidates earned full marks for this question. In part (a), a surprising number of candidates did not substitute into the given expression, erroneously stating . Many of these candidates were able to earn follow-through marks in later parts of the question. In part (b), algebraic errors led a few candidates to find inappropriate values for , such as .
\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]
A function is defined by , where .
\nThe graph of has a vertical asymptote and a horizontal asymptote.
\nWrite down the equation of the vertical asymptote.
\nWrite down the equation of the horizontal asymptote.
\nOn the set of axes below, sketch the graph of .
\nOn your sketch, clearly indicate the asymptotes and the position of any points of intersection with the axes.
\nHence, solve the inequality .
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\n\n
rational function shape with two branches in opposite quadrants, with two correctly positioned asymptotes and asymptotic behaviour shown A1
\n\n
Note: The equations of the asymptotes are not required on the graph provided there is a clear indication of asymptotic behaviour at and (or at their FT asymptotes from part (a)).
\n\n
axes intercepts clearly shown at and A1A1
\n\n
[3 marks]
\nA1
\n\n
Note: Accept correct alternative correct notation, such as and .
\n\n
[1 mark]
\nIt is pleasing to note that many candidates were familiar with the shape of the graph of a rational function of the form , and a large number of them were able to sketch an appropriate graph. Part (c) was a struggle for the majority of candidates, with only a few answering correctly. Despite the word \"hence\" and the single mark available in this part, most candidates who attempted part (c) did so by trying to solve the inequality algebraically, rather than seeing the connection to the values in their graph.
\nConsider any three consecutive integers, , and .
\nProve that the sum of these three integers is always divisible by .
\nProve that the sum of the squares of these three integers is never divisible by .
\n(A1)
\nA1
\nwhich is always divisible by AG
\n\n
[2 marks]
\nA1
\nattempts to expand either or (do not accept or ) (M1)
\nA1
\ndemonstrating recognition that is not divisible by or seen after correct expression divided by R1
\n\n
is divisible by and so is never divisible by
\nOR the first term is divisible by , the second is not
\nOR OR
\nhence the sum of the squares is never divisible by AG
\n\n
[4 marks]
\nMost candidates were able to earn full marks in part (a), though some were not able to provide the required reasoning to earn full marks in part (b). In many cases, candidates did not seem to understand the nature of a general deductive proof, and instead substituted different consecutive integers (such as 1, 2,3 ), showing the desired result for these specific values, rather than an algebraic generalization for any three consecutive integers.
\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]
\nFind the least positive value of for which .
\ndetermines (or ) as the first quadrant (reference) angle (A1)
\nattempts to solve (M1)
\n\n
Note: Award M1 for attempting to solve
\n\n
and so is rejected (R1)
\nA1
\n(must be in radians) A1
\n\n
[5 marks]
\nThis question proved to be a struggle for many candidates, and some candidates made no attempt here. While a good number of candidates recognized the reference angle of , this led to a final answer of , which many left as their final answer. In other cases, some candidates heeded the requirement that must be a positive value, however they gave an incorrect final answer of . Few candidates correctly rejected their initial reference angle of and correctly solved an equation using .
\nThe following diagram shows part of the graph of a quadratic function .
\nThe graph of has its vertex at , and it passes through point as shown.
\nThe function can be written in the form .
\nThe line is tangent to the graph of at .
\nNow consider another function . The derivative of is given by , where .
\nWrite down the equation of the axis of symmetry.
\nWrite down the values of and .
\nPoint has coordinates . Find the value of .
\nFind the equation of .
\nFind the values of for which is an increasing function.
\nFind the values of for which the graph of is concave-up.
\nA1
\n\n
Note: Must be an equation in the form “ ”. Do not accept or .
\n\n
[1 mark]
\n(accept ) A1A1
\n\n
[2 marks]
\nattempt to substitute coordinates of (M1)
\n\nA1
\n\n
[2 marks]
\nrecognize need to find derivative of (M1)
\nor A1
\n(may be seen as gradient in their equation) (A1)
\nor A1
\n\n
Note: Award A0 for .
\n\n
[4 marks]
\nMETHOD 1
\nRecognizing that for to be increasing, , or (M1)
\nThe vertex must be above the -axis, (R1)
\nA1
\n\n
METHOD 2
\nattempting to find discriminant of (M1)
\n\nrecognizing discriminant must be negative (R1)
\nOR
\nA1
\n\n
[3 marks]
\nrecognizing that for to be concave up, (M1)
\nwhen (R1)
\nA1
\n\n
[3 marks]
\nIn parts (a) and (b) of this question, a majority of candidates recognized the connection between the coordinates of the vertex and the axis of symmetry and the values of and , and most candidates were able to successfully substitute the coordinates of point Q to find the value of . In part (c), the candidates who recognized the need to use the derivative to find the gradient of the tangent were generally successful in finding the equation of the line, although many did not give their equation in the proper form in terms of and , and instead wrote , thus losing the final mark. Parts (d) and (e) were much more challenging for candidates. Although a good number of candidates recognized that in part (d), and in part (e), very few were able to proceed beyond this point and find the correct inequalities for their final answers.
\nConsider the binomial expansion where and .
\nShow that .
\nThe third term in the expansion is the mean of the second term and the fourth term in the expansion.
\nFind the possible values of .
\nEITHER
\nrecognises the required term (or coefficient) in the expansion (M1)
\nOR OR
\n\ncorrect working A1
\nOR OR
\n
OR
lists terms from row of Pascal’s triangle (M1)
\nA1
\n
THEN
AG
\n\n
[2 marks]
\n(A1)
\ncorrect equation A1
\nOR
\ncorrect quadratic equation A1
\nOR (or equivalent)
\nvalid attempt to solve their quadratic (M1)
\nOR
\nA1
\n\n
Note: Award final A0 for obtaining .
\n\n
[5 marks]
\nThe majority of candidates answered part (a) correctly, either by using the formula or Pascal's Triangle. In part (b) of the question, most candidates were able to correctly find the value of and set up a correct equation showing the mean of the second and fourth terms. While some struggled to complete the required algebra to solve the equation, the majority of candidates who found a correct quadratic equation were able to solve it correctly. A few candidates included in their final answer, thus not earning the final mark.
\nA biased four-sided die with faces labelled and is rolled and the result recorded. Let be the result obtained when the die is rolled. The probability distribution for is given in the following table where and are constants.
\nFor this probability distribution, it is known that .
\nNicky plays a game with this four-sided die. In this game she is allowed a maximum of five rolls. Her score is calculated by adding the results of each roll. Nicky wins the game if her score is at least ten.
\nAfter three rolls of the die, Nicky has a score of four.
\nDavid has two pairs of unbiased four-sided dice, a yellow pair and a red pair.
\nBoth yellow dice have faces labelled and . Let represent the sum obtained by rolling the two yellow dice. The probability distribution for is shown below.
\nThe first red die has faces labelled and . The second red die has faces labelled and , where and . The probability distribution for the sum obtained by rolling the red pair is the same as the distribution for the sum obtained by rolling the yellow pair.
\nShow that and .
\nFind .
\nAssuming that rolls of the die are independent, find the probability that Nicky wins the game.
\nDetermine the value of .
\nFind the value of , providing evidence for your answer.
\nuses to form a linear equation in and (M1)
\ncorrect equation in terms of and from summing to A1
\nOR (or equivalent)
\nuses to form a linear equation in and (M1)
\ncorrect equation in terms of and from A1
\nOR (or equivalent)
\n\n
Note: The marks for using and the marks for using may be awarded independently of each other.
\n\n
evidence of correctly solving these equations simultaneously A1
\nfor example, or
\nso and AG
\n\n
[5 marks]
\nvalid approach (M1)
\nOR
\nA1
\n\n
[2 marks]
\nrecognises at least one of the valid scores (, or ) required to win the game (M1)
\n\n
Note: Award M0 if candidate also considers scores other than , or (such as ).
\n\n
let represent the score on the last two rolls
\na score of is obtained by rolling or
\nA1
\na score of is obtained by rolling or
\nA1
\na score of is obtained by rolling
\nA1
\n\n
Note: The above 3 A1 marks are independent of each other.
\n\n\n
A1
\n\n
[5 marks]
\n(M1)
\nA1
\n\n
[2 marks]
\nMETHOD 1
\nEITHER
\n\nA1
\n\n
OR
and A1
\n\n
OR
and A1
\n\n
THEN
A1
\n\n
Note: Award A0A0 for obtained without working/reasoning/justification.
\n\n
METHOD 2
\nEITHER
\ncorrectly lists a relevant part of the sample space A1
\nfor example, or
\nor
\n\n
OR
eliminates possibilities (exhaustion) for
\nconvincingly shows that A1
\n, for example, from and so
\n\n\n
THEN
\nA1
\n\n
[2 marks]
\nA majority of candidates knew to set up equations using the sum of the probabilities in the distribution equal to 1 and/or the expected value equal to 2, however some candidates simply substituted the given values of and into their equations, which is considered working backwards and not doing what is required by the command term \"show that\". For the candidates who did set up both equations in terms of and , nearly all were successful in solving the resulting system of equations. Many candidates answered part (b) correctly using the given values from part (a). In part (c), most candidates recognized a sum of 6 (or more) was required in the final two rolls, but very few were able to find all the different outcomes to make this happen, especially for sums that can happen in more than one way, such as and . While some candidates were able to correctly answer parts (d) and (e), some did not attempt
these questions parts, and many did not justify their final answer in part (e).
Use l’Hôpital’s rule to find
\n.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to use l’Hôpital’s rule M1
\nA1A1
\n
Note: Award A1 for the numerator and A1 for the denominator.
substitution of into their expression (M1)
hence use l’Hôpital’s rule again
\n
Note: If the first use of l’Hôpital’s rule results in an expression which is not in indeterminate form, do not award any further marks.
attempt to use product rule in numerator M1
A1
\nA1
\n
[7 marks]
The 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]
\nConsider the functions , for , and for .
\nThe following diagram shows the graphs of and .
\nThe graphs of and intersect at points and . The coordinates of are .
\nIn the following diagram, the shaded region is enclosed by the graph of , the graph of , the -axis, and the line , where .
\nThe area of the shaded region can be written as , where .
\nFind the coordinates of .
\nFind the value of and the value of .
\n(M1)
\nOR (A1)
\nvalid attempt to solve their quadratic (M1)
\nOR OR
\n(may be seen in answer) A1
\n(accept ) A1
\n\n
[5 marks]
\nrecognizing two correct regions from to and from to (R1)
\ntriangle OR OR
\narea of triangle is OR OR (A1)
\ncorrect integration (A1)(A1)
\n\n\n
Note: Award A1 for and A1 for .
Note: The first three A marks may be awarded independently of the R mark.
\n
substitution of their limits (for ) into their integrated function (in terms of ) (M1)
\n\nA1
\nadding their two areas (in terms of ) and equating to (M1)
\n\nequating their non-log terms to (equation must be in terms of ) (M1)
\n\nA1
\n\nA1
\n\n
[10 marks]
\nNearly all candidates knew to set up an equation with in order to find the intersection of the two graphs, and most were able to solve the resulting quadratic equation. Candidates were not as successful in part (b), however. While some candidates recognized that there were two regions to be added together, very few were able to determine the correct boundaries of these regions, with many candidates integrating one or both functions from to . While a good number of candidates were able to correctly integrate the function(s), without the correct bounds the values of and were unattainable.
\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]
\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]
\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]
\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 person is selected at random from Group X. Find the probability that their reaction time is greater than 0.65 seconds.
\nThe probability that the reaction time of a person in Group Y is greater than 0.65 seconds is 0.396. Find the value of σ.
\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.
\n* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
0.010724
\n0.0107 A2 N2
\n\n
[2 marks]
\ncorrect z-value (A1)
\n0.263714…
\nevidence of appropriate approach (M1)
\neg ,
\ncorrect substitution (A1)
\neg ,
\n0.219934
\nσ = 0.220 A1 N3
\n\n
[4 marks]
\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]
\nWrite down the remainder when is divided by .
\nUse Fermat’s little theorem to find the remainder when is divided by .
\nProve that a number in base is divisible by if, and only if, the sum of its digits is divisible by .
\nThe base number is divisible by . Find the possible values of the digit .
\nthe remainder is A1
\n
[1 mark]
(from Fermat’s little theorem) (A1)
\n(M1)
\n
Note: Award M1 for a exponent consistent with the correct use of Fermat’s little theorem.
A1
the remainder is A1
\n
[4 marks]
METHOD 1
\nlet M1
\n
Note: The above M1 is independent of the A marks below.
A1
EITHER
(for all ) A1
\n
OR
A1
\n
THEN
so if and only if R1
\nso if and only if AG
\n\n
METHOD 2
\nlet (M1)
\nM1A1
\n
Note: Award M1 for attempting to express in the form .
as R1
so if and only if AG
\n
[4 marks]
METHOD 1
\nthe sum of the digits is (A1)
\nuses (or equivalent) to attempt to find a valid value of (M1)
\nA1A1
\n
Note: Award A1 for and A1 for .
\n
METHOD 2
\n(A1)
\n\nattempts to find a valid value of such that
\n(M1)
\nA1A1
\n
Note: Award A1 for and A1 for .
[4 marks]
A 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]
\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]
\nThe function is defined for all . The line with equation is the tangent to the graph of at .
\nThe function is defined for all where and .
\nWrite down the value of .
\nFind .
\nFind .
\nHence find the equation of the tangent to the graph of at .
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\n(M1)
\n\nA1
\n\n
[2 marks]
\nattempt to use chain rule to find (M1)
\nOR
\nA1
\n\n
OR A1
\n\n
[3 marks]
\nA particle moves along the -axis. The velocity of is at time seconds, where for . When is at the origin .
\nFind the value of when reaches its maximum velocity.
\nShow that the distance of from at this time is metres.
\nSketch a graph of against , clearly showing any points of intersection with the axes.
\nFind the total distance travelled by .
\nvalid approach to find turning point (, average of roots) (M1)
\nOR OR
\n(s) A1
\n\n
[2 marks]
\nattempt to integrate (M1)
\nA1A1
\n
Note: Award A1 for , A1 for .
attempt to substitute their into their solution for the integral (M1)
distance
\n(or equivalent) A1
\n(m) AG
\n\n
[5 marks]
\nvalid approach to solve (may be seen in part (a)) (M1)
\nOR
\ncorrect - intercept on the graph at A1
\n
Note: The following two A marks may only be awarded if the shape is a concave down parabola. These two marks are independent of each other and the (M1).
correct domain from to starting at A1
Note: The must be clearly indicated.
vertex in approximately correct place for and A1
\n
[4 marks]
\nrecognising to integrate between and , or and OR (M1)
\n\nA1
\n\nA1
\nvalid approach to sum the two areas (seen anywhere) (M1)
\nOR
\ntotal distance travelled (m) A1
\n\n
[5 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]
\nConsider the function where and .
\nThe graph of contains the point .
\nConsider the arithmetic sequence where and .
\nShow that .
\nWrite down an expression for .
\nFind the value of .
\nShow that and are four consecutive terms in a geometric sequence.
\nFind the value of and the value of .
\nOR (M1)
\nOR OR OR A1
\nAG
\n\n
[2 marks]
\nA1
\n
Note: Accept .
Accept any equivalent expression for e.g. .
\n
[1 mark]
\ncorrect substitution (A1)
\nOR
\ncorrect working involving log/index law (A1)
\nOR OR OR OR OR OR
\nA1
\n\n
[3 marks]
\nMETHOD 1
\nequating a pair of differences (M1)
\n\n\n\nA1A1
\nand A1
\nand are in geometric sequence AG
\n
Note: If candidate assumes the sequence is geometric, award no marks for part (i). If has been found, this will be awarded marks in part (ii).
\n
METHOD 2
\nexpressing a pair of consecutive terms, in terms of (M1)
\nand OR and
\ntwo correct pairs of consecutive terms, in terms of A1
\n(must include 3 ratios) A1
\nall simplify to A1
\nand are in geometric sequence AG
\n\n
[4 marks]
\nMETHOD 1 (geometric, finding )
\nOR (M1)
\n(seen anywhere) A1
\nOR (M1)
\nA1A1
\n\n
METHOD 2 (arithmetic)
\nOR (M1)
\n(seen anywhere) A1
\nOR (M1)
\nA1A1
\n\n
METHOD 3 (geometric using proportion)
\nrecognizing proportion (M1)
\nOR OR
\ntwo correct proportion equations A1
\nattempt to eliminate either or (M1)
\nOR
\nA1A1
\n\n
[5 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.
\nevidence 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 following table shows the probability distribution of a discrete random variable .
\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.
evidence of using (M1)
\ncorrect working (A1)
\neg ,
\nA1 N2
\n[3 marks]
\n\n
valid approach to find (M1)
\neg ,
\ncorrect working (A1)
\neg
\nA1 N2
\n[3 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 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 .
\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]
\nShow that .
\nHence or otherwise, solve the equation for .
\nMETHOD 1
\nattempt to write all LHS terms with a common denominator of (M1)
\nOR
\nOR A1
\nAG
\n\n
METHOD 2
\nattempt to use algebraic division on RHS (M1)
\ncorrectly obtains quotient of and remainder A1
\nas required. AG
\n\n
[2 marks]
\nconsider the equation (M1)
\n\n
EITHER
attempt to factorise in the form (M1)
\n
Note: Accept any variable in place of .
OR
attempt to substitute into quadratic formula (M1)
\n\n
THEN
or (A1)
\n
Note: Award A1 for only.
one of OR (accept or ) (A1)
(must be in radians) A1
\n
Note: Award A0 if additional answers given.
\n
[5 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.
\nrecognizing 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 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]
\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 .
\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 .
\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]
\nConsider the function , .
\nThe graph of has a horizontal tangent line at and at . Find .
\nvalid method (M1)
\neg , ,
(accept ) A1 N2
\n[2 marks]
\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]
\nConsider a function with domain . The following diagram shows the graph of , the derivative of .
\nThe graph of , the derivative of , has -intercepts at and . There are local maximum points at and and a local minimum point at .
\nFind all the values of where the graph of is increasing. Justify your answer.
\nFind the value of where the graph of has a local maximum.
\nFind the value of where the graph of has a local minimum. Justify your answer.
\nFind the values of where the graph of has points of inflexion. Justify your answer.
\nThe total area of the region enclosed by the graph of , the derivative of , and the -axis is .
\nGiven that , find the value of .
\nSpecial note: In this question if candidates use the word 'gradient' in their reasoning. e.g. gradient is positive, it must be clear whether this is the gradient of or the gradient of to earn the R mark.
\n\n
increases when A1
\nincreases when OR is above the -axis R1
\n
Note: Do not award A0R1.
\n
[2 marks]
\nSpecial note: In this question if candidates use the word 'gradient' in their reasoning. e.g. gradient is positive, it must be clear whether this is the gradient of or the gradient of to earn the R mark.
\n\n
A1
\n\n
[1 mark]
\nSpecial note: In this question if candidates use the word 'gradient' in their reasoning. e.g. gradient is positive, it must be clear whether this is the gradient of or the gradient of to earn the R mark.
\n\n
is minimum when A1
\nbecause when and when
\n(may be seen in a sign diagram clearly labelled as )
\nOR because changes from negative to positive at
\nOR and slope of is positive at R1
\n\n
Note: Do not award A0 R1
\n\n
[2 marks]
\nSpecial note: In this question if candidates use the word 'gradient' in their reasoning. e.g. gradient is positive, it must be clear whether this is the gradient of or the gradient of to earn the R mark.
\n\n
has points of inflexion when and A2
\n
has turning points at and
OR
\nand and changes from increasing to decreasing or vice versa at each of these -values (may be seen in a sign diagram clearly labelled as and ) R1
\n\n
Note: Award A0 if any incorrect answers are given. Do not award A0R1
\n\n
[3 marks]
\nSpecial note: In this question if candidates use the word 'gradient' in their reasoning. e.g. gradient is positive, it must be clear whether this is the gradient of or the gradient of to earn the R mark.
\n\n
recognizing area from to (seen anywhere) M1
\n\nrecognizing to negate integral for area below -axis (M1)
\nOR
\n(for any integral) (M1)
\nOR (A1)
\n(A1)
\nA1
\n\n
[6 marks]
\nThe function is defined by , where .
\nWrite down the equation of
\nFind the coordinates where the graph of crosses
\nthe vertical asymptote of the graph of .
\nthe horizontal asymptote of the graph of .
\nthe -axis.
\nthe -axis.
\nSketch the graph of on the axes below.
\nThe function is defined by , where and .
\nGiven that , determine the value of .
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\n(accept ) A1
\n\n
[1 mark]
\n(accept and ) A1
\n\n
[1 mark]
\n A1
Note: Award A1 for completely correct shape: two branches in correct quadrants with asymptotic behaviour.
\n
[1 mark]
\nMETHOD 1
\n\nattempt to find in terms of (M1)
\nOR exchange and and attempt to find in terms of
\nA1
\n\n\n\nA1
\n
Note: Condone use of
A1
\n\n
METHOD 2
\n\nattempt to find an expression for and equate to (M1)
\nA1
\n\n\nA1
\nequating coefficients of (or similar)
\nA1
\n\n
[4 marks]
\nThe equation has two real, distinct roots.
\nFind the possible values for .
\nConsider the case when . The roots of the equation can be expressed in the form , where . Find the value of .
\nattempt to use discriminant M1
\n\n(A1)
\n\nattempt to find critical values M1
\nrecognition that discriminant (M1)
\nor A1
\n
Note: Condone ‘or’ replaced with ‘and’, a comma, or no separator
\n
[5 marks]
\nvalid attempt to use (or equivalent) M1
\n\n\nA1
\n\n
[2 marks]
\nProve by mathematical induction that for .
\nHence or otherwise, determine the Maclaurin series of in ascending powers of , up to and including the term in .
\nHence or otherwise, determine the value of .
\nFor
\nLHS: A1
\nRHS: A1
\nso true for
\nnow assume true for ; i.e. M1
\n
Note: Do not award M1 for statements such as \"let \". Subsequent marks can still be awarded.
attempt to differentiate the RHS M1
A1
\nA1
\nso true for implies true for
\ntherefore true and true true
\ntherefore, true for all R1
\n
Note: Award R1 only if three of the previous four marks have been awarded
\n
[7 marks]
\nMETHOD 1
\nattempt to use (M1)
\n
Note: For , may be seen.
use of (M1)
\nA1
\n\n
METHOD 2
\n' Maclaurin series of ' (M1)
\n(A1)
\nA1
\n\n
[3 marks]
\nMETHOD 1
\nattempt to substitute into M1
\n(A1)
\n
EITHER
A1
\n\n
OR
A1
\n\n
THEN
so A1
\n\n
METHOD 2
\nM1
\n(A1)
\nattempt to use L'Hôpital's rule M1
\n\n\nA1
\n\n
[4 marks]
\nConsider the equation . The roots of this equation are , and , where and .
\nThe roots , and are represented by the points , and respectively on an Argand diagram.
\nConsider the equation .
\nVerify that is a root of this equation.
\nFind and , expressing these in the form , where and .
\nPlot the points , and on an Argand diagram.
\nFind .
\nBy using de Moivre’s theorem, show that is a root of this equation.
\nDetermine the value of .
\nA1
\nA1
\n\nAG
\n\n
Note: Candidates who solve the equation correctly can be awarded the above two marks. The working for part (i) may be seen in part (ii).
\n\n
[2 marks]
\n(M1)
\n(M1)
\nA1
\nA1
\n\n
[4 marks]
\nEITHER
\nattempt to express , , in Cartesian form and translate 1 unit in the positive direction of the real axis (M1)
\n
OR
attempt to express , and in Cartesian form (M1)
\n
THEN
Note: To award A marks, it is not necessary to see , or , the , or the solid lines
\nA1A1A1
\n\n
[4 marks]
\nvalid attempt to find M1
\nOR
\nvalid attempt to find M1
\n\nA1
\n\n
[3 marks]
\nMETHOD 1
\nM1
\nA1
\nA1
\n
Note: This step to change from to may occur at any point in MS.
AG
\n\n
METHOD 2
\nM1
\nA1
\nA1
\n
Note: This step to change from to may occur at any point in MS.
AG
\n\n
METHOD 3
\nLHS
\n\nM1A1
\n
Note: Award M1 for applying de Moivre’s theorem (may be seen in modulus- argument form.)
RHS
A1
\nAG
\n\n
METHOD 4
\n\n\n(M1)
\n\n(A1)
\nA1
\nAG
\n
Note: If the candidate does not interpret their conclusion, award (M1)(A1)A0 as appropriate.
\n
[3 marks]
\nMETHOD 1
\nM1
\nA1
\nattempt to use conjugate to rationalise M1
\nA1
\nA1
\n\nA1
\n
Note: Their final imaginary part does not have to be correct in order for the final three A marks to be awarded
\n
METHOD 2
\nM1
\nattempt to use conjugate to rationalise M1
\nA1
\nA1
\n\nA1
\n\nA1
\n
Note: Their final imaginary part does not have to be correct in order for the final three A marks to be awarded
\n
METHOD 3
\nattempt to multiply through by M1
\nA1
\nattempting to re-write in r-cis form M1
\nA1
\nA1
\n\nA1
\n\n
METHOD 4
\nattempt to multiply through by M1
\nA1
\nattempting to re-write in r-cis form M1
\nA1
\nattempt to re-write in Cartesian form M1
\n\nA1
\n
Note: Their final imaginary part does not have to be correct in order for the final A mark to be awarded
\n
[6 marks]
\nSolve the differential equation , given that at .
\nGive your answer in the form .
\n(M1)
\nattempt to find integrating factor (M1)
\n(A1)
\n\n\n\nattempt to use integration by parts (M1)
\nA1
\n\nsubstituting into an integrated equation involving M1
\n\n\nA1
\n\n
[7 marks]
\nSolve the equation , where .
\nattempt to use change the base (M1)
\n\nattempt to use the power rule (M1)
\n\nattempt to use product or quotient rule for logs, (M1)
\n\n
Note: The M marks are for attempting to use the relevant log rule and may be applied in any order and at any time during the attempt seen.
(A1)
\nA1
\n\n
[5 marks]
\nConsider the expression where .
\nThe binomial expansion of this expression, in ascending powers of , as far as the term in is , where .
\nFind the value of and the value of .
\nState the restriction which must be placed on for this expansion to be valid.
\nattempt to expand binomial with negative fractional power (M1)
\nA1
\nA1
\n\nattempt to equate coefficients of or (M1)
\n\nattempt to solve simultaneously (M1)
\nA1
\n\n
[6 marks]
\nA1
\n\n
[1 mark]
\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]
\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]
\nIn Lucy’s music academy, eight students took their piano diploma examination and achieved scores out of 150. For her records, Lucy decided to record the average number of hours per week each student reported practising in the weeks prior to their examination. These results are summarized in the table below.
\nFind Pearson’s product-moment correlation coefficient, , for these data.
\nThe relationship between the variables can be modelled by the regression equation . Write down the value of and the value of .
\nOne of these eight students was disappointed with her result and wished she had practised more. Based on the given data, determine how her score could have been expected to alter had she practised an extra five hours per week.
\nuse of GDC to give (M1)
\n\nA1
\n
Note: Award the (M1) for any correct value of , , or seen in part (a) or part (b).
[2 marks]
A1
\n
[1 mark]
attempt to find their difference (M1)
\nOR
\n\n\n
the student could have expected her score to increase by marks. A1
\n
Note: Accept an increase of or .
[2 marks]
Consider the function , for .
\nFind the values of for which .
\nSketch the graph of on the following grid.
\nA1A1
\n
[2 marks]
A1A1A1
Note: Award A1 for approximately correct shape. Only if this mark is awarded, award A1 for approximately correct roots and maximum point and A1 for approximately correct endpoints.
Allow for roots, for maximum and for endpoints.
[3 marks]
The 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]
\nConsider a triangle , where and .
\nFind the smallest possible perimeter of triangle .
\nEITHER
\nattempt to use cosine rule (M1)
\nOR (A1)
\nat least one correct value for (A1)
\nOR
\nusing their smaller value for to find minimum perimeter (M1)
\n\n
OR
attempt to use sine rule (M1)
\nOR OR OR (A1)
\nat least one correct value for (A1)
\nOR
\nusing their acute value for to find minimum perimeter (M1)
\nOR
\n
THEN
minimum perimeter . A1
\n\n
[5 marks]
\nA factory manufactures lamps. It is known that the probability that a lamp is found to be defective is . A random sample of lamps is tested.
\nFind the probability that there is at least one defective lamp in the sample.
\nGiven that there is at least one defective lamp in the sample, find the probability that there are at most two defective lamps.
\nrecognize that the variable has a Binomial distribution (M1)
\n\nattempt to find (M1)
\nOR OR OR
\n
Note: The two M marks are independent of each other.
A1
\n
[3 marks]
\nrecognition of conditional probability (M1)
\nOR
\n
Note: Recognition must be shown in context either in words or symbols but not just .
OR (A1)
OR OR (A1)
\n\nA1
\n\n
[4 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]
\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]
\nThe following diagram shows a semicircle with centre and radius . Points and lie on the circumference of the circle, such that and , where .
\nGiven that the areas of the two shaded regions are equal, show that .
\nHence determine the value of .
\nattempt to find the area of either shaded region in terms of and (M1)
\n
Note: Do not award M1 if they have only copied from the booklet and not applied to the shaded area.
Area of segment A1
Area of triangle A1
\ncorrect equation in terms of only (A1)
\n\nA1
\nAG
\n
Note: Award a maximum of M1A1A0A0A0 if a candidate uses degrees (i.e., ), even if later work is correct.
Note: If a candidate directly states that the area of the triangle is , award a maximum of M1A1A0A1A1.
\n
[5 marks]
A1
\n
Note: Award A0 if there is more than one solution. Award A0 for an answer in degrees.
[1 mark]
Let . 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]
\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 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]
\nPoints and lie on opposite banks of a river, such that is the shortest distance across the river. Point represents the centre of a city which is located on the riverbank. , and .
\nThe following diagram shows this information.
\nA boat travels at an average speed of . A bus travels along the straight road between and at an average speed of .
\nFind the travel time, in hours, from to given that
\nThere is a point , which lies on the road from to , such that . The boat travels from to , and the bus travels from to .
\nAn excursion involves renting the boat and the bus. The cost to rent the boat is per hour, and the cost to rent the bus is per hour.
\nthe boat is taken from to , and the bus from to .
\nthe boat travels directly to .
\nFind an expression, in terms of for the travel time , from to , passing through .
\nFind the value of so that is a minimum.
\nWrite down the minimum value of .
\nFind the new value of so that the total cost to travel from to via is a minimum.
\nWrite down the minimum total cost for this journey.
\nOR OR (M1)
\ntime (hours)
\ntime (hours) A1
\n\n
[2 marks]
\n(A1)
\ntime (hours)
\ntime (hours) A1
\n\n
[2 marks]
\n(A1)
\n(A1)
\nA1
\n\n
[3 marks]
\nvalid approach to find the minimum for (may be seen in (iii)) (M1)
\ngraph of OR OR graph of
\n\nA1
\n\n
[2 marks]
\n(hours) A1
\n
Note: Only allow FT in (b)(ii) and (iii) for and a function that has a minimum in that interval.
\n
[1 mark]
\n(A1)
\nvalid approach to find the minimum for (may be seen in (ii)) (M1)
\ngraph of OR OR graph of
\n\nA1
\n
Note: Only allow FT from (b) if the function has a minimum in .
\n
[3 marks]
\nA1
\n\n
Note: Only allow FT from (c)(i) if the function has a minimum in .
\n\n
[1 mark]
\nThe sum of the first terms of a geometric sequence is given by .
\nFind the first term of the sequence, .
\nFind .
\nFind the least value of such that .
\n(M1)
\nA1
\n
[2 marks]
(A1)
\nsubstituting their values for and into (M1)
\nA1
\n
[3 marks]
attempt to substitute their values into the inequality or formula for (M1)
\nOR
\nattempt to solve their inequality using a table, graph or logarithms
\n(must be exponential) (M1)
\n
Note: Award (M0) if the candidate attempts to solve .
correct critical value or at least one correct crossover value (A1)
OR OR
\nOR OR
\nleast value is A1
\n\n
[4 marks]
\nA function is defined by , where .
\nThe graph of has a vertical asymptote and a horizontal asymptote.
\nWrite down the equation of the vertical asymptote.
\nWrite down the equation of the horizontal asymptote.
\nOn the set of axes below, sketch the graph of .
\nOn your sketch, clearly indicate the asymptotes and the position of any points of intersection with the axes.
\nHence, solve the inequality .
\nSolve the inequality .
\nA1
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\nrational function shape with two branches in opposite quadrants, with two correctly positioned asymptotes and asymptotic behaviour shown A1
\naxes intercepts clearly shown at and A1A1
\n\n
[3 marks]
\nA1
\n
Note: Accept correct alternative correct notation, such as and .
\n
[1 mark]
\nEITHER
\nattempts to sketch (M1)
\n
OR
attempts to solve (M1)
\n\n
Note: Award the (M1) if and are identified.
\n\n
THEN
\nor A1
\n\n
Note: Accept the use of a comma. Condone the use of ‘and’. Accept correct alternative notation.
\n\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 height of water, in metres, in Dungeness harbour is modelled by the function , where is the number of hours after midnight, and and are constants, where and .
\nThe following graph shows the height of the water for hours, starting at midnight.
\nThe first high tide occurs at and the next high tide occurs hours later. Throughout the day, the height of the water fluctuates between and .
\nAll heights are given correct to one decimal place.
\nShow that .
\nFind the value of .
\nFind the value of .
\nFind the smallest possible value of .
\nFind the height of the water at .
\nDetermine the number of hours, over a 24-hour period, for which the tide is higher than metres.
\nOR A1
\nAG
\n\n
[1 mark]
\nOR (M1)
\nA1
\n\n
[2 marks]
\nOR (M1)
\nA1
\n\n
[2 marks]
\nMETHOD 1
\nsubstituting and for example into their equation for (A1)
\n\nattempt to solve their equation (M1)
\nA1
\n\n
METHOD 2
\nusing horizontal translation of (M1)
\n(A1)
\nA1
\n\n
METHOD 3
\n(A1)
\nattempts to solve their for (M1)
\n\nA1
\n\n
[3 marks]
\nattempt to find when or , graphically or algebraically (M1)
\n\nA1
\n\n
[2 marks]
\nattempt to solve (M1)
\ntimes are and (A1)
\ntotal time is
\n\n(hours) A1
\n
Note: Accept .
\n
[3 marks]
\nA function is defined by where .
\nThe graph of is shown below.
\nShow that is an odd function.
\nThe range of is , where .
\nFind the value of and the value of .
\nattempts to replace with M1
\n\nA1
\n\n
Note: Award M1A1 for an attempt to calculate both and independently, showing that they are equal.
Note: Award M1A0 for a graphical approach including evidence that either the graph is invariant after rotation by about the origin or the graph is invariant after a reflection in the -axis and then in the -axis (or vice versa).
\n
so is an odd function AG
\n\n
[2 marks]
\nattempts both product rule and chain rule differentiation to find M1
\nA1
\n\nsets their M1
\nA1
\nattempts to find at least one of (M1)
\n\n
Note: Award M1 for an attempt to evaluate at least at one of their roots.
\n\n
and A1
\n\n
Note: Award A1 for .
\n\n
[6 marks]
\nProve by contradiction that the equation has no integer roots.
\nMETHOD 1 (rearranging the equation)
\nassume there exists some such that M1
\n
Note: Award M1 for equivalent statements such as ‘assume that is an integer root of ’. Condone the use of throughout the proof.
Award M1 for an assumption involving .
\nNote: Award M0 for statements such as “let’s consider the equation has integer roots…” ,“let be a root of …”
\nNote: Subsequent marks after this M1 are independent of this M1 and can be awarded.
\n\n
attempts to rearrange their equation into a suitable form M1
\n
EITHER
A1
\nis even R1
\nwhich is not even and so cannot be an integer R1
\n
Note: Accept ‘ which gives a contradiction’.
OR
A1
\nR1
\nis even which is not true and so cannot be an integer R1
\n
Note: Accept ‘ is even which gives a contradiction’.
OR
A1
\nR1
\nis is not an integer and so cannot be an integer R1
\n
Note: Accept ‘ is not an integer which gives a contradiction’.
OR
A1
\nR1
\nis not an integer and so cannot be an integer R1
\n
Note: Accept is not an integer which gives a contradiction’.
THEN
so the equation has no integer roots AG
\n\n
METHOD 2
\nassume there exists some such that M1
\n
Note: Award M1 for equivalent statements such as ‘assume that is an integer root of ’. Condone the use of throughout the proof. Award M1 for an assumption involving and award subsequent marks based on this.
Note: Award M0 for statements such as “let’s consider the equation has integer roots…” ,“let be a root of …”
\nNote: Subsequent marks after this M1 are independent of this M1 and can be awarded.
\n\n
let (and )
\nfor all is a (strictly) increasing function M1A1
\nand R1
\nthus has only one real root between and , which gives a contradiction
\n(or therefore, contradicting the assumption that for some ), R1
\nso the equation has no integer roots AG
\n\n
[5 marks]
\nA function is defined by , where .
\nA function is defined by , where .
\nThe inverse of is .
\nA function is defined by , where .
\nSketch the curve , clearly indicating any asymptotes with their equations. State the coordinates of any local maximum or minimum points and any points of intersection with the coordinate axes.
\nShow that .
\nState the domain of .
\nGiven that , find the value of .
\nGive your answer in the form , where .
\n\n
-intercept A1
\n
Note: Accept an indication of on the -axis.
vertical asymptotes and A1
horizontal asymptote A1
\nuses a valid method to find the -coordinate of the local maximum point (M1)
\n
Note: For example, uses the axis of symmetry or attempts to solve .
local maximum point A1
Note: Award (M1)A0 for a local maximum point at and coordinates not given.
three correct branches with correct asymptotic behaviour and the key features in approximately correct relative positions to each other A1
\n
[6 marks]
\nM1
\n
Note: Award M1 for interchanging and (this can be done at a later stage).
\n
EITHER
\nattempts to complete the square M1
\nA1
\n\nA1
\n\n\n
OR
\nattempts to solve for M1
\nA1
\n
Note: Award A1 even if (in ) is missing
A1
\n
THEN
\nA1
\nand hence is rejected R1
\n\n
Note: Award R1 for concluding that the expression for must have the ‘’ sign.
The R1 may be awarded earlier for using the condition .
\n\n
AG
\n\n
[6 marks]
\ndomain of is A1
\n\n
[1 mark]
\nattempts to find (M1)
\n(A1)
\n\nattempts to solve for M1
\n\n\n
EITHER
\nA1
\nattempts to find their M1
\nA1
\n\n
Note: Award all available marks to this stage if is used instead of .
\n
OR
A1
\nattempts to solve their quadratic equation M1
\nA1
\n
Note: Award all available marks to this stage if is used instead of .
THEN
(as ) A1
\n\n\n
Note: Award A1 for
\n\n
[7 marks]
\nPart (a) was generally well done. It was pleasing to see how often candidates presented complete sketches here. Several decided to sketch using the reciprocal function. Occasionally, candidates omitted the upper branches or forgot to calculate the y-coordinate of the maximum.
\nPart (b): The majority of candidates knew how to start finding the inverse, and those who attempted completing the square or using the quadratic formula to solve for y made good progress (both methods equally seen). Otherwise, they got lost in the algebra. Very few explicitly justified the rejection of the negative root.
\nPart (c) was well done in general, with some algebraic errors seen in occasions.
\nBy using the substitution or otherwise, find an expression for in terms of , where is a non-zero real number.
\nMETHOD 1
\n(A1)
\nattempts to express the integral in terms of M1
\nA1
\nA1
\n\n
Note: Condone the absence of or incorrect limits up to this point.
\n\n
M1
\nA1
\n\n
Note: Award M1 for correct substitution of their limits for into their antiderivative for (or given limits for into their antiderivative for ).
\n\n
METHOD 2
\n(A1)
\napplies integration by inspection (M1)
\nA2
\n\n
Note: Award A2 if the limits are not stated.
\n\n
M1
\n\n
Note: Award M1 for correct substitution into their antiderivative.
\n\n
A1
\n\n
[6 marks]
\nA continuous random variable has the probability density function
\n.
\nThe following diagram shows the graph of for .
\nGiven that , find an expression for the median of in terms of and .
\nlet be the median
\n
EITHER
attempts to find the area of the required triangle M1
\nbase is (A1)
\nand height is
\narea A1
\n\n
OR
\nattempts to integrate the correct function M1
\n\nOR A1A1
\n\n
Note: Award A1 for correct integration and A1 for correct limits.
\n\n
THEN
\nsets up (their) or area M1
\n\n
Note: Award M0A0A0M1A0A0 if candidates conclude that and set up their area or sum of integrals .
\n\n\n
(A1)
\n\n
as , rejects
\nso A1
\n\n
[6 marks]
\nIn the following Argand diagram, the points , and are the vertices of triangle described anticlockwise.
\nThe point represents the complex number , where . The point represents the complex number , where .
\nAngles are measured anticlockwise from the positive direction of the real axis such that and .
\nIn parts (c), (d) and (e), consider the case where is an equilateral triangle.
\nLet and be the distinct roots of the equation where and .
\nShow that where is the complex conjugate of .
\nGiven that , show that is a right-angled triangle.
\nExpress in terms of .
\nHence show that .
\nUse the result from part (c)(ii) to show that .
\nConsider the equation , where and .
\nGiven that , deduce that only one equilateral triangle can be formed from the point and the roots of this equation.
\n(A1)
\nA1
\nAG
\n
Note: Accept working in modulus-argument form
\n
[2 marks]
\nA1
\n\n(as ) A1
\nso is a right-angled triangle AG
\n\n
[2 marks]
\nEITHER
\n(since ) (M1)
\n
OR
(M1)
\n
THEN
A1
\n\n
Note: Accept working in either modulus-argument form to obtain or in Cartesian form to obtain .
\n\n
[2 marks]
\nsubstitutes into M1
\nA1
\n\n
EITHER
\nA1
\n
OR
A1
\n\n
THEN
\n\nand A1
\nso AG
\n\n
Note: For candidates who work on the LHS and RHS separately to show equality, award M1A1 for , A1 for and A1 for . Accept working in either modulus-argument form or in Cartesian form.
\n\n
[4 marks]
\nMETHOD 1
\nand (A1)
\nA1
\nA1
\nsubstitutes into their expression M1
\nOR A1
\n
Note: If is not clearly recognized, award maximum (A0)A1A1M1A0.
\n
so AG
\n\n
METHOD 2
\nand (A1)
\nA1
\nA1
\nsubstitutes and into their expression M1
\nOR A1
\n
Note: If is not clearly recognized, award maximum (A0)A1A1M1A0.
so AG
\n
[5 marks]
\nA1
\nfor
\nand which does not satisfy R1
\nfor
\nand A1
\nso (for ), only one equilateral triangle can be formed from point and the two roots of this equation AG
\n\n
[3 marks]
\nThe vast majority of candidates scored full marks in parts (a) and (b). If they did not, it was normally due to the lack of rigour in setting out of the answer to a \"show that\" question. Part (c) was, though, more often than not poorly done. Many candidates could not use the given condition (equilateral triangle) to find in terms of . Part (d) was well answered by a rather high number of candidates.
\nOnly a handful of students made good progress in (e), not even finding the possible values for .
\nIn this question, give all answers correct to two decimal places.
\nSam invests in a savings account that pays a nominal annual rate of interest of , compounded half-yearly. Sam makes no further payments to, or withdrawals from, this account.
\nDavid also invests in a savings account that pays an annual rate of interest of , compounded yearly. David makes no further payments or withdrawals from this account.
\nFind the amount that Sam will have in his account after years.
\nFind the value of required so that the amount in David’s account after years will be equal to the amount in Sam’s account.
\nFind the interest David will earn over the years.
\nNote: The first time an answer is not given to two decimal places, the final A1 in that part is not awarded.
\n\n
EITHER
\nOR
\n\n
\n
\n
(M1)(A1)
\n\n
Note: Award (M1) for an attempt to use a financial app in their technology with at least two entries seen, and award (A1) for all entries correct. Accept a positive or negative value for .
\n
OR
(M1)(A1)
\n\n
Note: Award (M1) for substitution into compound interest formula.
Award (A1) for correct substitution.
THEN
A1
\n\n
[3 marks]
\nNote: The first time an answer is not given to two decimal places, the final A1 in that part is not awarded.
\n\n
EITHER
\n\n\n\n\n(M1)
\n\n
Note: Award (M1) for an attempt to use a financial app in their technology with at least two entries seen.
\n
OR
(M1)
\n\n
THEN
\n\nA1
\n\n
Note: Ignore omission of opposite signs for and if is obtained.
\n\n
[2 marks]
\nNote: The first time an answer is not given to two decimal places, the final A1 in that part is not awarded.
\n\n
A1
\n\n
[1 mark]
\nThis was the only short question which was not common to both HL and SL papers.
\nCandidates found this question a straightforward start, with the majority being able to attempt all three parts. While most candidates initially gave answers to two decimal places as required, many subsequently rounded their answers to three significant figures. In parts (a) and (b), most used the compound interest formula rather than the finance app on their GDC. Difficulty using the formula arose in determining the value of r (0.0274 was often seen, rather than 2.74) and the value of k. Many candidates struggled to correctly interpret 'half-yearly', and attempted to use either or in the formula.
\nIn part (b), a correct equation was usually seen, but lengthy analytical methods to solving it were favoured, with varying degrees of success, over an approach using the GDC. A common error was an attempt to use logarithms rather than the tenth root. It was not uncommon to see a final answer of either 2.76% or 0.0276, rather than 2.76. Many did not know what was meant by 'interest' in part (c). Those that did often recalculated David's amount using their value of r, rather than subtract $1700 from Sam's amount.
\nThe candidates who attempted to use a finance app on their GDC in this question were generally able to give accurate and concise answers. In the case where they obtained an incorrect answer, most gave sufficient detail of the values they were using in the app, to be awarded the method mark. The most common error seen was the use of an incorrect value for the number of payments per year or the number of compounding periods per year.
\nThe number of hours spent exercising each week by a group of students is shown in the following table.
\nThe median is hours.
\nFind the value of .
\nFind the standard deviation.
\nEITHER
\nrecognising that half the total frequency is (may be seen in an ordered list or indicated on the frequency table) (A1)
\n
OR
(A1)
\n
OR
(A1)
\n
THEN
A1
\n\n
[2 marks]
\nMETHOD 1
\n\nA2
\n\n
METHOD 2
\nEITHER
\n(A1)
\n
OR
(A1)
\n
THEN
A1
\n\n
[2 marks]
\nMost candidates attempted both parts, with varying levels of success, particularly in part (b).
\nIn part (a), the most successful approach seen was from candidates who made an ordered list to visualize the given data set, which enabled them to recognise either the number of sixes required for the median to lie at 4.5, or the total frequency. The most common error was to mistake the median for the mean, which led to a non-integer value of .
\nPart (b) proved to be more challenging, with many candidates either not taking into account the frequency of the exercise time when generating the summary statistics or treating frequency as an additional variable and using two-variable statistics on their GDC. With both, this led to being the most common wrong answer seen. A few candidates gave the sample standard deviation rather than the population standard deviation. A number of candidates attempted to use the standard deviation formula but were usually not successful. This formula is not in the course, although it can be obtained in the HL section of the formula booklet.
\nA company is designing a new logo. The logo is created by removing two equal segments from a rectangle, as shown in the following diagram.
\nThe rectangle measures by . The points and lie on a circle, with centre and radius , such that , where . This information is shown in the following diagram.
\nFind the area of one of the shaded segments in terms of .
\nGiven that the area of the logo is , find the value of .
\nvalid approach to find area of segment by finding area of sector – area of triangle (M1)
\n\n(A1)
\narea A1
\n\n
[3 marks]
\nEITHER
\narea of logo = area of rectangle – area of segments (M1)
\n(A1)
\n
OR
area of one segment (M1)
\n(A1)
\n
THEN
(do not accept an answer in degrees) A1
\n\n
Note: Award (M1)(A1)A0 if there is more than one solution.
Award (M1)(A1FT)A0 if the candidate works in degrees and obtains a final answer of
\n
[3 marks]
\nThe first part of this question proved particularly challenging for many candidates. Most were able to obtain the area of the sector, but then did not recognise that the segment was formed by subtracting the area of the triangle. Those that did, often struggled to find the appropriate triangle area formula to do so.
\nThe second part of the question was generally answered more successfully, although sometimes only one of the segments was subtracted from the whole rectangle. Many candidates who had a correct equation lacked the requisite calculator skills to obtain a solution. Those that had an incorrect expression for the area from part (a), often obtained . This was a significantly easier equation to solve, and consequently these candidates were not awarded the final A mark for a final answer of .
\nA discrete random variable, , has the following probability distribution:
\nShow that .
\nFind the value of , giving a reason for your answer.
\nHence, find .
\nOR (or equivalent) A1
\nAG
\n\n
[1 mark]
\none of OR (M1)
\nA1
\nreasoning to reject eg therefore R1
\n\n
[3 marks]
\nattempting to use the expected value formula (M1)
\n\nA1
\n
Note: Award M1A0 if additional values are given.
\n
[2 marks]
\nPart (a) was well done in this question, with most candidates recognising that the probabilities needed to sum to 1. Many candidates also approached part (b) appropriately. While many did so by graphing the quadratic on the GDC and identifying the zeros, most solved the equation analytically. Those that used the GDC, often assumed there was only one x-intercept and did not investigate the relevant area of the graph in more detail. While some who found the two required values of k recognised that k = 0.2 should be rejected by referring to the original probabilities, most had lost sight of the context of the question, and were unable to give a valid reason using P(X = 1) to reject this solution. Those that obtained one solution in part (b), were generally able to find the expected value successfully in part (c).
\nA particle moves along a straight line so that its velocity, , after seconds is given by for .
\nFind the value of when the particle is at rest.
\nFind the acceleration of the particle when it changes direction.
\nFind the total distance travelled by the particle.
\nrecognizing at rest (M1)
\n\n(seconds) A1
\n\n
Note: Award (M1)A0 for additional solutions to eg or .
\n\n
[2 marks]
\nrecognizing particle changes direction when OR when (M1)
\n\nA2
\n\n
[3 marks]
\ndistance travelled OR
\n(A1)
\n\n(metres) A1
\n\n
[2 marks]
\nThe majority of candidates found this question challenging but were often able to gain some of the marks in each part. However, it was not uncommon to see candidates manage either all of this question, or none of it, which unfortunately suggested that not all candidates had covered this content.
\nIn part (a), while many recognized when the particle is at rest, a common error was to assume that . It was pleasing to see the majority of those who had the correct equation, manage to progress to the correct value of t, having recognised the domain and the angle measure. Inevitably, a few candidates ignored the domain and obtained , or found .
\nPart (b) was not well done. The most successful approach was to use the GDC to find the gradient of the curve at the value of t obtained in part (a). This was well communicated, concise and generally accurate, although some either rounded incorrectly, or obtained rather than . Of those that did not use the GDC, many were aware that and made an attempt to find an expression for . However, the majority appeared not to recognise when the particle would change direction and went on to substitute an incorrect value of , not realising that they had obtained the required value earlier.
\nThose that had been successful in parts (a) and (b), particularly if they had been using their GDC, were generally able to complete part (c). However, it was disappointing that many candidates who understood that an integral was required, did not refer to the formula booklet and omitted the absolute value from the integrand.
\nLet and be two independent events such that and .
\nGiven that , find the value of .
\nFind .
\nMETHOD 1
\nEITHER
\none of OR A1
\n
OR
A1
\n
THEN
\nattempt to equate their with their expression for M1
\nA1
\nA1
\n\n
METHOD 2
\nattempt to form at least one equation in and using independence M1
\nOR
\n\nand A1
\n(A1)
\nA1
\n\n
[4 marks]
\nMETHOD 1
\nrecognising (M1)
\n\nA1
\n\n
METHOD 2
\n\n(A1)
\nA1
\n\n
[2 marks]
\nThis question asks you to explore properties of a family of curves of the type for various values of and , where .
\nOn the same set of axes, sketch the following curves for and , clearly indicating any points of intersection with the coordinate axes.
\nNow, consider curves of the form , for , where .
\nNext, consider the curve .
\nThe curve has two points of inflexion. Due to the symmetry of the curve these points have the same -coordinate.
\nis defined to be a rational point on a curve if and are rational numbers.
\nThe tangent to the curve at a rational point intersects the curve at another rational point .
\nLet be the curve , for . The rational point lies on .
\nWrite down the coordinates of the two points of inflexion on the curve .
\nBy considering each curve from part (a), identify two key features that would distinguish one curve from the other.
\nBy varying the value of , suggest two key features common to these curves.
\nShow that , for .
\nHence deduce that the curve has no local minimum or maximum points.
\nFind the value of this -coordinate, giving your answer in the form , where .
\nFind the equation of the tangent to at .
\nHence, find the coordinates of the rational point where this tangent intersects , expressing each coordinate as a fraction.
\nThe point also lies on . The line intersects at a further point. Determine the coordinates of this point.
\napproximately symmetric about the -axis graph of A1
\nincluding cusp/sharp point at A1
\n\n
[2 marks]
\n\n
Note: Final A1 can be awarded if intersections are in approximate correct place with respect to the axes shown. Award A1A1A1A0 if graphs ‘merge’ or ‘cross’ or are discontinuous at -axis but are otherwise correct. Award A1A0A0A0 if only one correct branch of both curves are seen.
\nNote: If they sketch graphs on separate axes, award a maximum of 2 marks for the ‘best’ response seen. This is likely to be A1A1A0A0.
\napproximately symmetric about the -axis graph of with approximately correct gradient at axes intercepts A1
some indication of position of intersections at , A1
[2 marks]
\n\n
Note: Final A1 can be awarded if intersections are in approximate correct place with respect to the axes shown. Award A1A1A1A0 if graphs ‘merge’ or ‘cross’ or are discontinuous at -axis but are otherwise correct. Award A1A0A0A0 if only one correct branch of both curves are seen.
\nNote: If they sketch graphs on separate axes, award a maximum of 2 marks for the ‘best’ response seen. This is likely to be A1A1A0A0.
\nand A1
\n\n
[1 mark]
\nAny two from:
\nhas a cusp/sharp point, (the other does not)
\ngraphs have different domains
\nhas points of inflexion, (the other does not)
\ngraphs have different -axis intercepts (one goes through the origin, and the other does not)
\ngraphs have different -axis intercepts A1
\n\n
Note: Follow through from their sketch in part (a)(i). In accordance with marking rules, mark their first two responses and ignore any subsequent.
\n\n
[1 mark]
\nAny two from:
\nas ,
\nas is approximated by (or similar)
\nthey have intercepts at
\nthey have intercepts at
\nthey all have the same range
\n(or -axis) is a line of symmetry
\nthey all have the same line of symmetry
\nthey have one -axis intercept
\nthey have two -axis intercepts
\nthey have two points of inflexion
\nat -axis intercepts, curve is vertical/infinite gradient
\nthere is no cusp/sharp point at -axis intercepts A1A1
\n\n
Note: The last example is the only valid answer for things “not” present. Do not credit an answer of “they are all symmetrical” without some reference to the line of symmetry.
\nNote: Do not allow same/ similar shape or equivalent.
\nNote: In accordance with marking rules, mark their first two responses and ignore any subsequent.
\n\n
[2 marks]
\nMETHOD 1
\nattempt to differentiate implicitly M1
\nA1
\nOR A1
\nAG
\n\n
METHOD 2
\nattempt to use chain rule M1
\nA1A1
\n\n
Note: Award A1 for , A1 for
\n\n
AG
\n\n
[3 marks]
\nEITHER
\nlocal minima/maxima occur when
has no (real) solutions (or equivalent) R1
OR
, so R1
\n
THEN
so, no local minima/maxima exist AG
\n\n
[1 mark]
\nEITHER
\nattempt to use quotient rule to find M1
\nA1A1
\n
Note: Award A1 for correct and correct denominator, A1 for correct .
Note: Future A marks may be awarded if the denominator is missing or incorrect.
\n
stating or using (may be seen anywhere) (M1)
OR
attempt to use product rule to find M1
\nA1A1
\n
Note: Award A1 for correct first term, A1 for correct second term.
setting (M1)
OR
attempts implicit differentiation on M1
\nA1
\nrecognizes that (M1)
\n\n(A1)
\n
THEN
A1
\nattempt to use quadratic formula or equivalent (M1)
\n\nA1
\n\n
Note: Accept any integer multiple of and (e.g. and ).
\n\n
[7 marks]
\nattempt to find tangent line through (M1)
\nOR A1
\n\n
[2 marks]
\nattempt to solve simultaneously with (M1)
\n
Note: The M1 mark can be awarded for an unsupported correct answer in an incorrect format (e.g. ).
obtain A1
\n
[2 marks]
\nattempt to find equation of (M1)
\n(A1)
\nsolve simultaneously with (M1)
\nA1
\nA1
\n\n\n
OR
\nattempt to find vector equation of (M1)
\n(A1)
\n\n\nattempt to solve (M1)
\n\nA1
\nA1
\n\n\n
[5 marks]
\nThis was a relatively straightforward start, though it was disappointing to see so many candidates sketch their graphs on two separate axes, despite the question stating they should be sketched on the same axes.
\nOf those candidates producing clear sketches, the vast majority were able to recognise the points of inflexion and write down their coordinates. A small number embarked on a mostly fruitless algebraic approach rather than use their graphs as intended. The distinguishing features between curves tended to focus on points of intersection with the axes, which was accepted. Only a small number offered ideas such as on both curves. A number of (incorrect) suggestions were seen, stating that both curves tended towards a linear asymptote.
\nA majority of candidates' suggestions related to the number of intersection points with the coordinate axes, while the idea of the x-axis acting as a line of symmetry was also often seen.
\nThe required differentiation was straightforward for the majority of candidates.
\nThe majority employed the quotient rule here, often doing so successfully to find a correct expression for . Despite realising that , the resulting algebra to find the required solution proved a step too far for most. A number of slips were seen in candidates' working, though better candidates were able to answer the question confidently.
\nMistakes proved to be increasingly common by this stage of the paper. Various equations of lines were suggested, with the incorrect appearing more than once. Only the better candidates were able to tackle the final part of the question with any success; it was pleasing to see a number of clear algebraic (only) approaches, though this was not necessary to obtain full marks.
\nSignificant work on this question part was rarely seen, and it may have been the case that many candidates chose to spend their remaining time on the second question, especially if they felt they were making little progress with part f. Having said that, correct final answers were seen from better candidates, though these were few and far between.
\nThis question asks you to investigate conditions for the existence of complex roots of polynomial equations of degree and .
\n
The cubic equation , where , has roots and .
Consider the equation , where .
\nNoah believes that if then and are all real.
\nNow consider polynomial equations of degree .
\nThe equation , where , has roots and .
\nIn a similar way to the cubic equation, it can be shown that:
\n\n\n\n.
\nThe equation , has one integer root.
\nBy expanding show that:
\n\n\n.
\nShow that .
\nHence show that .
\nGiven that , deduce that and cannot all be real.
\nUsing the result from part (c), show that when , this equation has at least one complex root.
\nBy varying the value of in the equation , determine the smallest positive integer value of required to show that Noah is incorrect.
\nExplain why the equation will have at least one real root for all values of .
\nFind an expression for in terms of and .
\nHence state a condition in terms of and that would imply has at least one complex root.
\nUse your result from part (f)(ii) to show that the equation has at least one complex root.
\nState what the result in part (f)(ii) tells us when considering this equation .
\nWrite down the integer root of this equation.
\nBy writing as a product of one linear and one cubic factor, prove that the equation has at least one complex root.
\nattempt to expand M1
\nOR A1
\nA1
\ncomparing coefficients:
\nAG
\nAG
\nAG
\n\n
Note: For candidates who do not include the AG lines award full marks.
\n\n
[3 marks]
\n(A1)
\nattempt to expand (M1)
\nor equivalent A1
\nAG
\n\n
Note: Accept equivalent working from RHS to LHS.
\n\n
[3 marks]
\nEITHER
\nattempt to expand (M1)
\nA1
\n\nor equivalent A1
\nAG
\n
OR
attempt to write in terms of (M1)
\n\nA1
\nA1
\nAG
\n\n
Note: Accept equivalent working where LHS and RHS are expanded to identical expressions.
\n\n
[3 marks]
\nA1
\nif all roots were real R1
\n
Note: Condone strict inequality in the R1 line.
Note: Do not award A0R1.
roots cannot all be real AG
\n
[2 marks]
\nand A1
\nso the equation has at least one complex root R1
\n\n
Note: Allow equivalent comparisons; e.g. checking
\n\n
[2 marks]
\nuse of GDC (eg graphs or tables) (M1)
\nA1
\n\n
[2 marks]
\ncomplex roots appear in conjugate pairs (so if complex roots occur the other root will be real OR all 3 roots will be real).
\nOR
\na cubic curve always crosses the -axis at at least one point. R1
\n\n
[1 mark]
\nattempt to expand (M1)
\n(A1)
\n\nA1
\n\n
[3 marks]
\nOR A1
\n\n
Note: Allow FT on their result from part (f)(i).
\n\n
[1 mark]
\nOR R1
\nhence there is at least one complex root. AG
\n\n
Note: Allow FT from part (f)(ii) for the R mark provided numerical reasoning is seen.
\n\n
[1 mark]
\n(so) nothing can be deduced R1
\n\n
Note: Do not allow FT for the R mark.
\n\n
[1 mark]
\nA1
\n\n
[1 mark]
\nattempt to express as a product of a linear and cubic factor M1
\nA1A1
\n\n
Note: Award A1 for each factor. Award at most A1A0 if not written as a product.
\n\n
since for the cubic, R1
\nthere is at least one complex root AG
\n\n
[4 marks]
\nThe first part of this question proved to be very accessible, with the majority of candidates expanding their brackets as required, to find the coefficients and .
\nThe first part of this question was usually answered well, though presentation in the second part sometimes left a lot to be desired. The expression was expected to be seen more often, as a 'pivot' to reaching the required result. Algebra was often lengthy, but untidily so, sometimes leaving examiners to do some mental tidying up on behalf of the candidate.
\nA good number of candidates recognised the reasoning required in this part of the question and were able to score both marks.
\nMost candidates found applying this specific case to be very straightforward.
\nMost candidates offered incorrect answers in the first part; despite their working suggested utilisation of the GDC, it was clear that many did not appreciate what the question was asking. The second part was usually answered well, with the idea of complex roots occurring in conjugate pairs being put to good use.
\nSome very dubious algebra was seen here, and often no algebra at all. Despite this, a good number of candidates seemed to make the 'leap' to the correct expression , perhaps fortuitously so in a number of cases.
\nOf those finding in part f, a surprising number of answers seen employed the test of checking whether .
\nPart i was usually not answered successfully, which may have been due to shortage of time. However, it was pleasing to see a number of candidates reach the end of the paper and successfully factorise the given quartic using a variety of methods. The final part required the test. Though correct reasoning was sometimes seen, it was rare for this final mark to be gained.
\nAll lengths in this question are in centimetres.
\nA solid metal ornament is in the shape of a right pyramid, with vertex and square base . The centre of the base is . Point has coordinates and point has coordinates .
\nThe volume of the pyramid is , correct to three significant figures.
\nFind .
\nGiven that , find .
\nFind the height of the pyramid, .
\nA second ornament is in the shape of a cuboid with a rectangular base of length , width and height . The cuboid has the same volume as the pyramid.
\nThe cuboid has a minimum surface area of . Find the value of .
\nattempt to use the distance formula to find (M1)
\n\n\nA1
\n\n
[2 marks]
\nMETHOD 1
\nattempt to apply cosine rule OR sine rule to find (M1)
\nOR (A1)
\n\nA1
\n\n
METHOD 2
\nLet be the midpoint of
\nattempt to apply right-angled trigonometry on triangle (M1)
\n(A1)
\n\n
A1
\n\n
[3 marks]
\nMETHOD 1
\nequating volume of pyramid formula to (M1)
\n(A1)
\n\nA1
\n\n
METHOD 2
\nLet be the midpoint of
\n(M1)
\n(A1)
\n\nA1
\n\n
[3 marks]
\n(A1)
\nA1
\n\n
Note: Condone use of .
\n
attempt to substitute into their expression for surface area (M1)
EITHER
attempt to find minimum turning point on graph of area function (M1)
\n
OR
OR (M1)
\n
THEN
minimum surface area A1
\n\n
[5 marks]
\nParts (a), (b) and (c) were completed well by many candidates, but few were able to make any significant progress in part (d).
\nIn part (a), many candidates were able to apply the distance formula and successfully find AV. However, a common error was to work in two-dimensions and to apply Pythagoras' Theorem once, neglecting completely the z-coordinates. Many recognised the need to use either the sine or cosine rule in part (b) to find the length AB. Common errors in this part included: the GDC being set incorrectly in radians; applying right-angled trigonometry on a 40°, 40°, 70° triangle; or using as the length of AX in triangle AVX.
\nDespite the formula for the volume of a pyramid being in the formula booklet, a common error in part (c) was to omit the factor of from the volume formula, or not to recognise that the area of the base of the pyramid was .
\nThe most challenging part of this question proved to be the optimization of the surface area of the cuboid in part (d). Although some candidates were able to form an equation involving the volume of the cuboid, an expression for the surface area eluded most. A common error was to gain a surface area which involved eight sides rather than six. It was surprising that few who were able to find both the equation and an expression were able to progress any further. Of those that did, few used their GDC to find the minimum surface area directly, with most preferring the more time consuming analytical approach.
\nA function, , has its derivative given by , where . The following diagram shows part of the graph of .
\nThe graph of has an axis of symmetry .
\nThe vertex of the graph of lies on the -axis.
\nThe graph of has a point of inflexion at .
\nFind the value of .
\nWrite down the value of the discriminant of .
\nHence or otherwise, find the value of .
\nFind the value of the gradient of the graph of at .
\nSketch the graph of , the second derivative of . Indicate clearly the -intercept and the -intercept.
\nWrite down the value of .
\nFind the values of for which the graph of is concave-down. Justify your answer.
\nEITHER
\nattempt to use (M1)
\n\n
OR
attempt to complete the square (M1)
\n\n
OR
attempt to differentiate and equate to (M1)
\n\n
THEN
A1
\n\n
[2 marks]
\ndiscriminant A1
\n\n
[1 mark]
\nEITHER
\nattempt to substitute into (M1)
\nA1
\n
OR
(M1)
\nA1
\n
THEN
A1
\n\n
[3 marks]
\nA1
\nattempt to find (M1)
\n\ngradient A1
\n\n
[3 marks]
\n A1A1
Note: Award A1 for line with positive gradient, A1 for correct intercepts.
\n
[2 marks]
\nA1
\n\n
[1 mark]
\nA1
\n(for ) OR the is below the -axis (for )
\nOR (sign diagram must be labelled ) R1
\n
[2 marks]
\nCandidates did score well on this question. As always, candidates are encouraged to read the questions carefully for key words such as 'value' as opposed to 'expression'. So, if asked for the value of the discriminant, their answer should be a number and not an expression found from . As such the value of the discriminant in (b)(i) was often seen in (b)(ii). Please ask students to use a straight edge when sketching a straight line! Overall, the reasoning mark for determining where the graph of f is concave-down, was an improvement on previous years. Sign diagrams were typically well labelled, and the description contained clarity regarding which function was being referred to.
\nThe function is defined by , where .
\nFor the graph of
\nThe graphs of and intersect at and , where .
\nwrite down the equation of the vertical asymptote.
\nfind the equation of the horizontal asymptote.
\nFind .
\nUsing an algebraic approach, show that the graph of is obtained by a reflection of the graph of in the -axis followed by a reflection in the -axis.
\nFind the value of and the value of .
\nHence, find the area enclosed by the graph of and the graph of .
\nA1
\n\n
[1 mark]
\nattempt to substitute into OR table with large values of OR sketch of showing asymptotic behaviour (M1)
\nA1
\n\n
[2 marks]
\nattempt to interchange and (seen anywhere) M1
\nOR (A1)
\nOR (A1)
\n(accept ) A1
\n\n
[4 marks]
\nreflection in -axis given by (M1)
\n(A1)
\nreflection of their in -axis given by accept \"now \" M1
\n\nOR A1
\nAG
\n\n
Note: If the candidate attempts to show the result using a particular coordinate on the graph of rather than a general coordinate on the graph of , where appropriate, award marks as follows:
M0A0 for eg
M0A0 for
\n
[4 marks]
\nattempt to solve using graph or algebraically (M1)
\nAND A1
\n\n
Note: Award (M1)A0 if only one correct value seen.
\n\n
[2 marks]
\nattempt to set up an integral to find area between and (M1)
\n(A1)
\n\nA1
\n\n
[3 marks]
\nCandidates mostly found the first part of this question accessible, with many knowing how to find the equation of both asymptotes. Common errors included transposing the asymptotes, or finding where an asymptote occurred but not giving it as an equation.
\nCandidates knew how to start part (b)(i), with most attempting to find the inverse function by firstly interchanging and . However, many struggled with the algebra required to change the subject, and were not awarded all the marks. A common error in this part was for candidates to attempt to find an expression for , rather than one for . Few candidates were able to answer part (b)(ii). Many appeared not to know that a reflection in the -axis is given by , or that a reflection in the -axis is given by . Many of those that did, multiplied both the numerator and denominator by when taking the negative of their , i.e. was often simplified as . However, the majority of candidates either did not attempt this question part or attempted to describe a graphical approach often involving a specific point, rather than an algebraic approach.
\nThose that attempted part (c), and had the correct expression for , were usually able to gain all the marks. However, those that had an incorrect expression, or had found , often proceeded to find an area, even when there was not an area enclosed by their two curves.
\nConsider the series , where and .
\nConsider the case where the series is geometric.
\nNow consider the case where the series is arithmetic with common difference .
\nShow that .
\nGiven that and , find the value of .
\nShow that .
\nWrite down in the form , where .
\nThe sum of the first terms of the series is .
\nFind the value of .
\nEITHER
\nattempt to use a ratio from consecutive terms M1
\nOR OR
\n\n
Note: Candidates may use and consider the powers of in geometric sequence
\nAward M1 for .
\n
OR
and M1
\n
THEN
OR A1
\nAG
\n\n
Note: Award M0A0 for or with no other working seen.
\n\n
[2 marks]
\n(A1)
\nOR A1
\nA1
\n\n
[3 marks]
\nMETHOD 1
\nattempt to find a difference from consecutive terms or from M1
\ncorrect equation A1
\nOR
\n
Note: Candidates may use and consider the powers of in arithmetic sequence.
Award M1A1 for
\n\n
A1
\nAG
\n\n
METHOD 2
\nattempt to use arithmetic mean M1
\nA1
\nA1
\nAG
\n\n
METHOD 3
\nattempt to find difference using M1
\n\n\n
OR A1
\nA1
\nAG
\n\n
[3 marks]
\nA1
\n\n
[1 mark]
\nMETHOD 1
\n\nattempt to substitute into and equate to (M1)
\n\ncorrect working with (seen anywhere) (A1)
\nOR OR
\ncorrect equation without A1
\nOR or equivalent
\n
Note: Award as above if the series is considered leading to .
attempt to form a quadratic (M1)
attempt to solve their quadratic (M1)
\n\nA1
\n\n
METHOD 2
\nlisting the first terms of the sequence (A1)
\n\nrecognizing first terms sum to M1
\nth term is (A1)
\nth term is (A1)
\nsum of th and th term (A1)
\nA1
\n\n
[6 marks]
\nMany candidates were able to identify the key relationship between consecutive terms for both geometric and arithmetic sequences. Substitution into the infinity sum formula was good with solving involving the natural logarithm done quite well. The complexity of the equation formed using 𝑆𝑛 was a stumbling block for some candidates. Those who factored out and cancelled the ln𝑥 expression were typically successful in solving the resulting quadratic.
\nConsider .
\nExpand and simplify in ascending powers of .
\nBy using a suitable substitution for , show that .
\nShow that , where is a positive real constant.
\nIt is given that , where . Find the value of .
\nEITHER
\nattempt to use binomial expansion (M1)
\n\n
OR
(M1)
\n
THEN
A1
\n\n
[2 marks]
\n(A1)
\nSo,
\nA1
\nattempt to substitute any double angle rule for into (M1)
\nA1
\nAG
\n
Note: Allow working RHS to LHS.
\n
[4 marks]
\nrecognizing to integrate (M1)
\n
EITHER
applies integration by inspection (M1)
\n\nA1
\nA1
\n
OR
(M1)
\n\nA1
\nOR A1
\n
THEN
AG
\n\n
[4 marks]
\nEITHER
\nM1
\nOR (M1)
\n
OR
M1
\n(M1)
\n
THEN
(A1)
\n(A1)
\nA1
\n\n
[5 marks]
\nMany candidates successfully expanded the binomial, with the most common error being to omit the negative sign with a. The connection between (a)(i) and (ii) was often noted but not fully utilised with candidates embarking on unnecessary complex algebraic expansions of expressions involving double angle rules. Candidates often struggled to apply inspection or substitution when integrating. As a 'show that' question, b(i) provided a useful result to be utilised in (ii). So even without successfully completing (i) candidates could apply it in part (ii). Not many managed to do so.
\nConsider the points , and . The line passes through the point and is perpendicular to .
\nFind the equation of .
\nThe line passes through the point .
\nFind the value of .
\n(A1)
\nfinding using their (M1)
\n\nA1
\n\n
Note: Do not accept
\n\n
[3 marks]
\nsubstituting into their (M1)
\nOR
\nA1
\n\n
[2 marks]
\nFinding the gradient of a line was well understood and many candidates also correctly found the perpendicular slope. Even with an error in their part (a), follow through marks in part (b) allowed many candidates to earn full marks for finding k despite their incorrect equation resulting in arithmetic of greater complexity.
\nA bakery makes two types of muffins: chocolate muffins and banana muffins.
\nThe weights, grams, of the chocolate muffins are normally distributed with a mean of and standard deviation of .
\nThe weights, grams, of the banana muffins are normally distributed with a mean of and standard deviation of .
\nEach day of the muffins made are chocolate.
\nOn a particular day, a muffin is randomly selected from all those made at the bakery.
\nThe machine that makes the chocolate muffins is adjusted so that the mean weight of the chocolate muffins remains the same but their standard deviation changes to . The machine that makes the banana muffins is not adjusted. The probability that the weight of a randomly selected muffin from these machines is less than is now .
\nFind the probability that a randomly selected chocolate muffin weighs less than .
\nIn a random selection of chocolate muffins, find the probability that exactly weigh less than .
\nFind the probability that the randomly selected muffin weighs less than .
\nGiven that a randomly selected muffin weighs less than , find the probability that it is chocolate.
\nFind the value of .
\n(M1)
\n\nA1
\n\n
[2 marks]
\nrecognition of binomial eg (M1)
\n\nA1
\n\n
[2 marks]
\nLet represent ‘chocolate muffin’ and represent ‘banana muffin’
\n(A1)
\n
EITHER
(or equivalent in words) (M1)
\n
OR
tree diagram showing two ways to have a muffin weigh (M1)
\n
THEN
(A1)
\n\nA1
\n\n
[4 marks]
\nrecognizing conditional probability (M1)
\n\n
Note: Recognition must be shown in context either in words or symbols, not just
\n\n
(A1)
\n\nA1
\n\n
[3 marks]
\nMETHOD 1
\n(M1)
\n\n(A1)
\nattempt to solve for using GDC (M1)
\n
Note: Award (M1) for a graph or table of values to show their with a variable standard deviation.
A2
\n\n
METHOD 2
\n(M1)
\n\n(A1)
\nuse of inverse normal to find score of their (M1)
\n\ncorrect substitution (A1)
\n\n\nA1
\n\n
[5 marks]
\nThis question was common to both HL and SL papers.
\nThe first two parts of this question were generally well done, with many candidates demonstrating an understanding of how to find, using their GDC, the required probability from a normal distribution in part (a), and recognising the binomial probability in part (b).
\nParts (c) and (d) were not done well, although many that were able to make progress in part (d) were often able to give concise solutions. Most that attempted part (c) did very poorly, while few attempted part (d). Both parts proved challenging, principally due to difficulties in determining the different possible outcomes with combined events. In part (c)(i), tree diagrams were unfortunately rarely seen, as were attempts to set out the ways of selecting a muffin weighing less than 61 g, either in words, or using appropriate notation involving probabilities. Those who did understand these concepts on the other hand were much more likely to be able to find the conditional probability in part (c)(ii) and be successful in part (d). Common errors included not considering both types of muffin, and in part (d) using a probability instead of a z-value.
\nFind the value of .
\n(A1)
\nA1A1
\nsubstituting limits into their integrated function and subtracting (M1)
\nOR
\nA1
\n\n
[5 marks]
\nA mixed response was noted for this question. Candidates who simplified the algebraic fraction before integrating were far more successful in gaining full marks in this question. Many candidates used other valid approaches such as integration by substitution and integration by parts with varying degrees of success. A small number of candidates substituted the limits without integrating.
\nConsider the expansion of where . Determine all possible values of for which the expansion has a non-zero constant term.
\nEITHER
\nattempt to obtain the general term of the expansion
\nOR (M1)
\n
OR
recognize power of starts at and goes down by each time (M1)
\n
THEN
recognizing the constant term when the power of is zero (or equivalent) (M1)
\nor or OR (or equivalent) A1
\nis a multiple of or one correct value of (seen anywhere) (A1)
\nA1
\n
Note: Accept is a (positive) multiple of or
Do not accept
Note: Award full marks for a correct answer using trial and error approach
showing and for recognizing that this pattern continues.
\n
[5 marks]
\nThere was a mixed response to this question. Candidates who used a trial and error approach were more successful in obtaining completely correct answers than those who tried to solve algebraically by finding the general term to form an equation relating n and r . Poor explanations were often noted in the trial and error approach. Candidates often failed to make progress after obtaining in the algebraic approach. Some candidates did not attempt this question.
\nConsider the vectors and such that and .
\nConsider the vector such that .
\nConsider the vector such that , where .
\nFind the possible range of values for .
\nGiven that is a minimum, find .
\nFind such that and is perpendicular to .
\n(A1)
\n(accept min and max ) A1
\n\n
Note: Award (A1)A0 for and seen with no indication that they are the endpoints of an interval.
\n\n
[2 marks]
\nrecognition that or is a negative multiple of (M1)
\nOR
\nA1
\n\n
\n
[2 marks]
\nMETHOD 1
\nis perpendicular to
\nis in the direction (M1)
\n(A1)
\n(M1)
\n(A1)
\nA1
\n\n
METHOD 2
\nis perpendicular to
\nattempt to set scalar product OR product of gradients (M1)
\n(A1)
\n\nattempt to solve simultaneously to find a quadratic in or (M1)
\nOR
\nA1A1
\nNote: Award A1 independently for each value. Accept values given as and or equivalent.
\n\n
[5 marks]
\nThe expression can be written as . Write down the value of .
\nHence, find the value of .
\nA1
\n\n\n
[1 mark]
\nA1A1
\nsubstituting limits into their integrated function and subtracting (M1)
\nOR
\nA1
\n\n
[4 marks]
\nMany candidates could give the value of p correctly. However, many did struggle with the integration, including substituting limits into the integrand, without integrating at all. An incorrect value of p often resulted in arithmetic of greater complexity.
\nMary, three female friends, and her brother, Peter, attend the theatre. In the theatre there is a row of empty seats. For the first half of the show, they decide to sit next to each other in this row.
\nFor the second half of the show, they return to the same row of empty seats. The four girls decide to sit at least one seat apart from Peter. The four girls do not have to sit next to each other.
\nFind the number of ways these five people can be seated in this row.
\nFind the number of ways these five people can now be seated in this row.
\n(A1)(A1)
\n(accept ) A1
\n\n
[3 marks]
\nMETHOD 1
\n(Peter apart from girls, in an end seat) OR
\n(Peter apart from girls, not in end seat) (A1)
\ncase 1: Peter at either end
\nOR (A1)
\ncase 2: Peter not at the end
\nOR (A1)
\nTotal number of ways
\nA1
\n\n
METHOD 2
\n(Peter next to girl, in an end seat) OR
\n(Peter next to one girl, not in end seat) OR
\n(Peter next to two girls, not in end seat) (A1)
\ncase 1: Peter at either end
\n(A1)
\ncase 2: Peter not at the end
\n(A1)
\nTotal number of ways
\nA1
\n\n
[4 marks]
\nConsider the function , where .
\nThe curve is rotated about the -axis to form a solid of revolution that is used to model a water container.
\nAt , the container is empty. Water is then added to the container at a constant rate of .
\nSketch the curve , clearly indicating the coordinates of the endpoints.
\nShow that the inverse function of is given by .
\nState the domain and range of .
\nShow that the volume, , of water in the container when it is filled to a height of metres is given by .
\nHence, determine the maximum volume of the container.
\nFind the time it takes to fill the container to its maximum volume.
\nFind the rate of change of the height of the water when the container is filled to half its maximum volume.
\ncorrect shape (concave down) within the given domain A1
\nand A1
\n\n
Note: The coordinates of endpoints may be seen on the graph or marked on the axes.
\n\n
[2 marks]
\ninterchanging and (seen anywhere) M1
\n\nA1
\nA1
\nAG
\n\n
[3 marks]
\nOR domain A1
\nOR OR range A1
\n\n
[2 marks]
\nattempt to substitute into the correct volume formula (M1)
\nA1
\nA1
\nAG
\n
Note: Award marks as appropriate for correct work using a different variable e.g.
[3 marks]
attempt to substitute into (M1)
\n\nA1
\n\n
[2 marks]
\ntime (M1)
\n\nA1
\n\n
[2 marks]
\nattempt to find the height of the tank when (M1)
\n\n(A1)
\nattempt to use the chain rule or differentiate with respect to (M1)
\nOR (A1)
\nattempt to substitute their and (M1)
\n\nA1
\n\n
[6 marks]
\nPart a) was generally well done, with the most common errors being to use an incorrect domain or not to give the coordinates of the endpoints. Some graphs appeared to be straight lines; some candidates drew sketches which were too small which made it more difficult for them to show the curvature.
\nMost candidates were able to show the steps to find an inverse function in part b), although occasionally a candidate did not explicitly swop the x and y variables before writing down the inverse function, which was given in the question. Many candidates struggled to identify the domain and range of the inverse, despite having a correct graph.
\nPart c) required a rotation around the y-axis, but a number of candidates attempted to rotate around the x-axis or failed to include limits. In the same vein, many substituted 2 into the formula instead of the square root of 3 when answering the second part. Many subsequently gained follow through marks on part d).
\nThere were a number of good attempts at related rates in part e), with the majority differentiating V with respect to t, using implicit differentiation. However, many did not find the value of h which corresponded to halving the volume, and a number did not differentiate with respect to t, only with respect to h.
\nThe continuous random variable has probability density function
\n\nFind the value of .
\nFind .
\nattempt to integrate (M1)
\nA1
\nNote: Award (M1)A0 for .
Condone absence of up to this stage.
\n
equating their integrand to M1
\n\nA1
\n\n
[4 marks]
\nA1
\n
Note: Condone absence of limits if seen at a later stage.
EITHER
attempt to integrate by inspection (M1)
\n\nA1
\n
Note: Condone the use of up to this stage.
OR
for example,
\n
Note: Other substitutions may be used. For example .
M1
\n
Note: Condone absence of limits up to this stage.
A1
\n
Note: Condone the use of up to this stage.
THEN
A1
\n
Note: Award A0M1A1A0 for their or for working with incorrect or no limits.
\n
[4 marks]
\nMost candidates who attempted part (a) knew that the integrand must be equated to 1 and only a small proportion of these managed to recognize the standard integral involved here. The effect of 3 in 3x2 was missed by many resulting in very few completely correct answers for this part. Part (b) proved to be challenging for vast majority of the candidates and was poorly done in general. Stronger candidates who made good progress in part (a) were often successful in part (b) as well. Most candidates used a substitution, however many struggled to make progress using this approach. Often when using a substitution, the limits were unchanged. If the function was re-written in terms of x, this did not result in an error in the final answer.
\nConsider the equation , where .
\nWrite down an expression for the product of the roots, in terms of .
\nHence or otherwise, determine the values of such that the equation has one positive and one negative real root.
\nproduct of roots A1
\n\n
[1 mark]
\nrecognition that the product of the roots will be negative (M1)
\n\ncritical values seen (A1)
\nA1
\n\n
[3 marks]
\nConsider the function .
\nThe function is given by , where .
\nShow that is an odd function.
\nSolve the inequality .
\nattempt to replace with M1
\n\n
EITHER
A1
\n
OR
A1
\n
Note: Award M1A0 for a graphical approach including evidence that either the graph is invariant after rotation by about the origin or the graph is invariant after a reflection in the -axis and then in the -axis (or vice versa).
so is an odd function AG
\n\n
[2 marks]
\nattempt to find at least one intersection point (M1)
\n\n\nA1
\nA1
\nA1
\n\n
[4 marks]
\nA survey at a swimming pool is given to one adult in each family. The age of the adult, years old, and of their eldest child, years old, are recorded.
\nThe ages of the eldest child are summarized in the following box and whisker diagram.
\nThe regression line of on is . The regression line of on is .
\nFind the largest value of that would not be considered an outlier.
\nOne of the adults surveyed is years old. Estimate the age of their eldest child.
\nFind the mean age of all the adults surveyed.
\n(A1)
\nattempt to find (M1)
\n\nA1
\n\n
[3 marks]
\nchoosing (M1)
\n\n(years old) A1
\n\n
[2 marks]
\nattempt to solve system by substitution or elimination (M1)
\n(years old) A1
\n\n
[2 marks]
\nMany candidates correctly found the value of 16. Some then incorrectly went on to state that 15 was therefore the minimum value that was not an outlier. For part (b) students needed to choose the appropriate rule to use to estimate the child's age. It was clear that many did not know there was a choice to be made and used both equations. As the mean point (𝑐̅,𝑎̅ ) lies on both regression lines, in part (c) candidates needed to solve the system of equations to find the mean adult age, 𝑎̅. Few candidates seemed to be aware of this.
\nConsider the functions where and where .
\nFind .
\nSolve the equation where .
\n(A1)
\nA1
\n\n
[2 marks]
\nrecognising to use or M1
\nOR (values may be seen in right triangle) (A1)
\n(seen anywhere) (accept degrees) (A1)
\n\nA1A1
\n\n
Note: Do not award the final A1 if any additional solutions are seen.
Award A1A0 for correct answers in degrees.
Award A0A0 for correct answers in degrees with additional values.
\n
[5 marks]
\nDetermining the composite function was very well done. In part (b) very few candidates showed any recognition that tan (or cot) were required to solve this trigonometric equation. Many saw the 2x and simply employed one of the double angle rules but could not then progress to an answer.
\nConsider the curve with equation , where and .
\nThe tangent to the curve at the point where is parallel to the line .
\nFind the value of .
\nevidence of using product rule (M1)
\nA1
\ncorrect working for one of (seen anywhere) A1
\nat
\n
OR
slope of tangent is
\n
their at equals the slope of (seen anywhere) (M1)
A1
\n\n
[5 marks]
\nThe product rule was well recognised and used with 𝑥=1 properly substituted into this expression. Although the majority of the candidates tried equating the derivative to the slope of the tangent line, the slope of the tangent line was not correctly identified; many candidates incorrectly substituted 𝑥=1 into the tangent equation, thus finding the y-coordinate instead of the slope.
\nConsider and , where and .
\nThe graph of is obtained by two transformations of the graph of .
\nDescribe these two transformations.
\nThe -intercept of the graph of is at .
\nGiven that , find the smallest value of .
\ntranslation (shift) by to the right/positive horizontal direction A1
\ntranslation (shift) by upwards/positive vertical direction A1
\n\n
Note: accept translation by
\nDo not accept ‘move’ for translation/shift.
\n\n
[2 marks]
\nMETHOD 1
\nminimum of is (may be seen in sketch) (M1)
\n\n(accept ) A1
\nsubstituting and their to find (M1)
\n\n(A1)
\nsmallest value of is A1
\n\n
METHOD 2
\nsubstituting to find an expression (for ) in terms of (M1)
\n\nA1
\nminimum of is (M1)
\n\n(accept ) (A1)
\nsmallest value of is A1
\n\n
METHOD 3
\nA1
\n-intercept of is a maximum (M1)
\namplitude of is (A1)
\nattempt to find least maximum (M1)
\n\nsmallest value of is A1
\n\n
[5 marks]
\nCandidates knew aspects of the transformations performed but some were unable to correctly describe them fully, e.g., omitting direction (right/up/positive) or using 'move' instead of translate/shift. Each description requires three parts: transformation type, size and direction. e.g., translation of q units up. For part (b) few candidates were able to fully navigate the reasoning required in this question. A common error was to evaluate , instead of 1. Those who used sketches to assist in their thinking were typically more successful.
\nConsider the differential equation for and . It is given that when .
\nUse Euler’s method, with a step length of , to find an approximate value of when .
\nUse the substitution to show that .
\nBy solving the differential equation, show that .
\nFind the actual value of when .
\nUsing the graph of , suggest a reason why the approximation given by Euler’s method in part (a) is not a good estimate to the actual value of at .
\nattempt to use Euler’s method (M1)
\n, where
\ncorrect intermediate -values (A1)(A1)
\n\n\n
Note: A1 for any two correct -values seen
\n\n\n
A1
\n\n
Note: For the final A1, the value must be the last value in a table or a list, or be given as a final answer, not just embedded in a table which has further lines.
\n\n
[4 marks]
\n(A1)
\nreplacing with and with M1
\nA1
\n(since )
\nAG
\n\n
[3 marks]
\nattempt to separate variables and (M1)
\n\n(A1)
\nattempt to express in partial fraction form M1
\n\nA1
\n\nA1
\n\n
Note: Condone absence of modulus signs throughout.
\n
EITHER
attempt to find using M1
\n\n\nexpressing both sides as a single logarithm (M1)
\n\n
OR
expressing both sides as a single logarithm (M1)
\n\nattempt to find using M1
\n\n
THEN
(since )
\nsubstitute (seen anywhere) M1
\n(since )
\n\nattempt to make the subject M1
\nA1
\nAG
\n\n
[10 marks]
\nactual value at A1
\n\n
[1 mark]
\ngradient changes rapidly (during the interval considered) OR
\nthe curve has a vertical asymptote at R1
\n\n
[1 mark]
\nMost candidates showed evidence of an attempt to use Euler's method in part a), although very few explicitly wrote down the formulae, they used in order to calculate successive y-value. In addition, many seemed to take a step-by-step approach rather than using the recursive capabilities of the graphical display calculator.
\nThere were many good attempts at part b), but not all candidates recognised that this would help them to solve part c).
\nPart c) was done very well by many candidates, although there were a significant number who failed to recognise the need for partial fractions and could not make further progress. A common error was to integrate without a constant of integration, which meant that the initial condition could not be used. The reasoning given for the estimate being poor was often too vague and did not address the specific nature of the function given clearly enough.
\nThe following diagram shows a circle with centre and radius metres.
\nPoints and lie on the circle and radians.
\nFind the length of the chord .
\nFind the area of the shaded sector.
\nEITHER
\nuses the cosine rule (M1)
\n(A1)
\n
OR
uses right-angled trigonometry (M1)
\n(A1)
\n
OR
uses the sine rule (M1)
\n\n(A1)
\n
THEN
A1
\n\n
[3 marks]
\nlet the shaded area be
\n
METHOD 1
attempt at finding reflex angle (M1)
\n\nsubstitution into area formula (A1)
\nOR
\n\nA1
\n\n
METHOD 2
\nlet the area of the circle be and the area of the unshaded sector be
\n(M1)
\n(A1)
\n\nA1
\n\n
[3 marks]
\nMost students used the cosine rule to correctly find AB in part (a), although many found the arc length instead of the chord.
\nPart (b) was generally correctly solved. Some candidates found the area of the unshaded region rather than the shaded one.
\nThe derivative of a function is given by , where . The graph of passes through the point . Find .
\nMETHOD 1
\nrecognises that (M1)
\n(A1)(A1)
\n
Note: Award A1 for each integrated term.
\n
substitutes and into their integrated function (must involve ) (M1)
\n\nA1
\n\n
METHOD 2
\nattempts to write both sides in the form of a definite integral (M1)
\n(A1)
\n(A1)(A1)
\n
Note: Award A1 for and A1 for .
A1
\n
[5 marks]
\nWhile many students were successful in solving this question, some did not consider the constant of integration or struggled to integrate the exponential term. A few students lost the final mark for stopping at and not giving the formula for .
\nGemma and Kaia started working for different companies on January 1st 2011.
\nGemma’s starting annual salary was , and her annual salary increases on January 1st each year after 2011.
\nKaia’s annual salary is based on a yearly performance review. Her salary for the years 2011, 2013, 2014, 2018, and 2022 is shown in the following table.
\nFind Gemma’s annual salary for the year 2021, to the nearest dollar.
\nAssuming Kaia’s annual salary can be approximately modelled by the equation , show that Kaia had a higher salary than Gemma in the year 2021, according to the model.
\nMETHOD 1
\nusing geometric sequence with (M1)
\ncorrect expression or listing terms correctly (A1)
\nOR OR listing terms
\nGemma’s salary is (must be to the nearest dollar) A1
\n\n
METHOD 2
\n\n\n\n\n\n(M1)(A1)
\nGemma’s salary is (must be to the nearest dollar) A1
\n\n
[3 marks]
\nfinds and (accept ) (A1)(A1)
\n
Note: Award (A1)(A1) for , or , or
Kaia’s salary in 2021 is (accept from ) A1
Kaia had a higher salary than Gemma in 2021 AG
\n\n
[3 marks]
\nMany errors were seen in part (a). Some candidates used the incorrect formula or used an incorrect value for the exponent e.g. 9 was often seen. Others lost the final mark for not answering to the nearest dollar.
Very few tried to make a table of values.
In part (b) students often let represent the number of years since a given year, rather than the year itself. Despite this, most were able to find the correct amount with their equation and were awarded marks as appropriate. Some students did not realise regression on GDC was expected and tried to work with a few given data points, others had difficulty dealing with the constant in the regression equation if it was reported using scientific notation.
\nEvents and are independent and .
\nGiven that , find .
\nsubstitution of for in (M1)
\n\nsubstitution of for (M1)
\n(or equivalent) (A1)
\n\n
Note: The first two M marks are independent of each other.
\n\n
attempts to solve their quadratic equation (M1)
\n\nA2
\n\n
Note: Award A1 if both answers are given as final answers for .
\n\n
[6 marks]
\nThis question proved difficult for many students. One common error was to use , which simplified the problem greatly, resulting in a linear, not a quadratic equation.
\nA random sample of nine adults were selected to see whether sleeping well affected their reaction times to a visual stimulus. Each adult’s reaction time was measured twice.
\nThe first measurement for reaction time was taken on a morning after the adult had slept well. The second measurement was taken on a morning after the same adult had not slept well.
\nThe box and whisker diagrams for the reaction times, measured in seconds, are shown below.
\nConsider the box and whisker diagram representing the reaction times after sleeping well.
\nState the median reaction time after sleeping well.
\nVerify that the measurement of seconds is not an outlier.
\nState why it appears that the mean reaction time is greater than the median reaction time.
\nNow consider the two box and whisker diagrams.
\nComment on whether these box and whisker diagrams provide any evidence that might suggest that not sleeping well causes an increase in reaction time.
\n(s) A1
\n\n
[1 mark]
\n(s) (A1)
\nsubstituting their into correct expression for upper fence (A1)
\n(s)
\nR1
\nso (s) is not an outlier AG
\n\n
[3 marks]
\nEITHER
\nthe median is closer to the lower quartile (positively skewed) R1
\n
OR
The distribution is positively skewed R1
\n
OR
the range of reaction times below the median is smaller than the range of reaction times above the median R1
\n\n
Note: These are sample answers from a range of acceptable correct answers. Award R1 for any correct statement that explains this.
Do not award R1 if there is also an incorrect statement, even if another statement in the answer is correct. Accept a correctly and clearly labelled diagram.
\n
[1 mark]
\nEITHER
\nthe distribution for ‘not sleeping well’ is centred at a higher reaction time R1
\n
OR
The median reaction time after not sleeping well is equal to the upper quartile reaction time after sleeping well R1
\n
OR
of reaction times are seconds after sleeping well, compared with after not sleeping well R1
\n
OR
the sample size of is too small to draw any conclusions R1
\n\n
Note: These are sample answers from a range of acceptable correct answers. Accept any relevant correct statement that relates to the median and/or quartiles shown in the box plots. Do not accept a comparison of means. Do not award R1 if there is also an incorrect statement, even if another statement in the answer is correct.
\nAward R0 to “correlation does not imply causation”.
\n\n
[1 mark]
\nParts (a) and (b) were generally known, but answers to parts (c) and (d) showed poor understanding of interpreting data. Many students thought they could find the mean by considering only the end points. Others assumed it would be halfway between the quartiles. When it came to evidence, many were far too quick to say the diagrams 'proved' something. Most compared only the medians and thought that was sufficient evidence, completely ignoring the fact the median only represented one data point. Others just compared the maximum and minimum. A few commented correctly that 9 subjects was too small a sample to prove anything.
\nA particle moves in a straight line such that its velocity, , at time seconds is given by .
\nDetermine when the particle changes its direction of motion.
\nFind the times when the particle’s acceleration is .
\nFind the particle’s acceleration when its speed is at its greatest.
\nrecognises the need to find the value of when (M1)
\n\n(s) A1
\n\n
[2 marks]
\nrecognises that (M1)
\n\n(s) A1A1
\n\n
Note: Award M1A1A0 if the two correct answers are given with additional values outside .
\n\n
[3 marks]
\nspeed is greatest at (A1)
\n\nA1
\n\n
[2 marks]
\nIn part (a) many did not realize the change of motion occurred when . A common error was finding or thinking that it was at the maximum of .
\nIn part (b), most candidates knew to differentiate but some tried to substitute in for , while others struggled to differentiate the function by hand rather than using the GDC. Many candidates tried to solve the equation analytically and did not use their technology. Of those who did, many had their calculators in degree mode.
\nAlmost all candidates who attempted part (c) thought the greatest speed was the same as the maximum of .
\nA farmer is placing posts at points , , and in the ground to mark the boundaries of a triangular piece of land on his property.
\nFrom point , he walks due west metres to point .
From point , he walks metres on a bearing of to reach point .
This is shown in the following diagram.
\nThe farmer wants to divide the piece of land into two sections. He will put a post at point , which is between and . He wants the boundary to divide the piece of land such that the sections have equal area. This is shown in the following diagram.
\nFind the distance from point to point .
\nFind the area of this piece of land.
\nFind .
\nFind the distance from point to point .
\n(A1)
\nattempt to substitute into cosine rule (M1)
\n(A1)
\n\nA1
\n\n
[4 marks]
\ncorrect substitution into area formula (A1)
\n\n\narea A1
\n\n
[2 marks]
\nattempt to substitute into sine rule or cosine rule (M1)
\nOR (A1)
\n\nA1
\n\n
[3 marks]
\nMETHOD 1
\nrecognizing that for areas to be equal, (M1)
\nA1
\nattempt to substitute into cosine rule to find (M1)
\ncorrect substitution into cosine rule (A1)
\n\n\nA1
\n\n
METHOD 2
\ncorrect expressions for areas of triangle and triangle using A1
\nand OR
\nand
\ncorrect equation in terms of (A1)
\nor
\nor (A1)
\nsubstituting their value of into equation to solve for (M1)
\nor
\n\nA1
\n\n
[5 marks]
\nStudents performed well on parts (a)-(c), correctly applying the cosine rule, the sine formula for area and the sine rule. Part (d) proved challenging. A common error was to falsely assume that segment BD bisected angle ABC.
\nA significant number of candidates did not have their calculator in degree mode or started in radians and changed to degrees part way through but used answers they had obtained when they were in radian mode. They got answers which were clearly impossible from the diagram, but most did not notice this.
\nAccuracy was a great problem throughout this question: premature rounding, incorrect rounding, or quoting more figures for the answer than they had used in the calculation.
\nA scientist conducted a nine-week experiment on two plants, and , of the same species. He wanted to determine the effect of using a new plant fertilizer. Plant was given fertilizer regularly, while Plant was not.
\nThe scientist found that the height of Plant , at time weeks can be modelled by the function , where .
\nThe scientist found that the height of Plant , at time weeks can be modelled by the function , where .
\nUse the scientist’s models to find the initial height of
\nPlant .
\nPlant correct to three significant figures.
\nFind the values of when .
\nFor , find the total amount of time when the rate of growth of Plant was greater than the rate of growth of Plant .
\n(cm) A1
\n\n
[1 mark]
\n(M1)
\n\n(cm) A1
\n\n
[2 marks]
\nattempts to solve for (M1)
\n\n(weeks) A2
\n\n
[3 marks]
\nrecognises that and are required (M1)
\nattempts to solve for (M1)
\nand OR and OR and (A1)
\n\n
Note: Award full marks for .
\nAward subsequent marks for correct use of these exact values.
\n\n
OR OR
\n(A1)
\nattempts to calculate the total amount of time (M1)
\n\n(weeks) A1
\n\n
[6 marks]
\nMany students did not change their calculators back to radian mode. This meant they had no chance of correctly answering parts (c) and (d), since even if follow through was given, there were not enough intersections on the graphs.
\nMost managed part (a) and some attempted to equate the functions in part b) but few recognised that 'rate of growth' was the derivatives of the given functions, and of those who did, most were unable to find them.
\nAlmost all the candidates who did solve part (c) gave the answer , when working with more significant figures would have given them 3.14. They lost the last mark.
\nThe time it takes Suzi to drive from home to work each morning is normally distributed with a mean of minutes and a standard deviation of minutes.
\nOn of days, it takes Suzi longer than minutes to drive to work.
\nSuzi will be late to work if it takes her longer than minutes to drive to work. The time it takes to drive to work each day is independent of any other day.
\nSuzi will work five days next week.
\nSuzi will work days this month. She will receive a bonus if she is on time at least of those days.
\nSo far this month, she has worked days and been on time of those days.
\nFind the value of .
\nOn a randomly selected day, find the probability that Suzi’s drive to work will take longer than minutes.
\nFind the probability that she will be late to work at least one day next week.
\nGiven that Suzi will be late to work at least one day next week, find the probability that she will be late less than three times.
\nFind the probability that Suzi will receive a bonus.
\nMETHOD 1
\n\nor (M1)
\nattempt to solve for graphically or numerically using the GDC (M1)
\ngraph of normal curve for and OR and
OR table of values for or
(min) A2
\n\n
METHOD 2
\n\nor (M1)
\n(A1)
\nvalid equation using their -score (clearly identified as -score and not a probability) (M1)
\nOR
\n\n(min) A1
\n\n
[4 marks]
\n(M1)
\n\nA1
\n\n
[2 marks]
\nrecognizing binomial probability (M1)
\n\nOR
\n(M1)
\n\nA1
\n\n
[3 marks]
\nrecognizing conditional probability in context (M1)
\nfinding (may be seen in conditional probability) (A1)
\n(may be seen in conditional probability) (A1)
\n(A1)
\n\nA1
\n\n
[5 marks]
\nMETHOD 1
\nrecognizing that Suzi can be late no more than once (in the remaining six days) (M1)
\n, where is the number of days late (A1)
\n(M1)
\n\nA1
\n
Note: The first two marks may be awarded independently.
\n
METHOD 2
\nrecognizing that Suzi must be on time at least five times (of the remaining six days) (M1)
\n, where is the number of days on time (A1)
\nOR OR OR (M1)
\n\nA1
\n
Note: The first two marks may be awarded independently.
\n
[4 marks]
\nIn part (a) many candidates did not know to use inverse normal to find a value. Some did find , then rounded it to 3 sf and got an incorrect value for sigma.
\nPart (b) was mostly well done.
\nIn (c) most recognised the binomial and handled 'at least one' correctly.
\nIn (d) many recognised conditional probability, but most candidates were not able to find the intersection of the events as P(1) + P(2).
\nIn part (e), those candidates who did understand what to do often misunderstood that they needed to look at 1 or no more lates and just considered one more late. Something similar happened to those who approached the question by considering the times Suzi was on time.
\nThis question was only correctly answered by a few, and students tended to perform either very well or very poorly.
\nConsider integers and such that is exactly divisible by . Prove by contradiction that and cannot both be odd.
\nAssume that and are both odd. M1
\n
Note: Award M0 for statements such as “let and be both odd”.
Note: Subsequent marks after this M1 are independent of this mark and can be awarded.
Then and A1
A1
\n(A1)
\n( is always divisible by ) but is not divisible by . (or equivalent) R1
\nis not divisible by , a contradiction. (or equivalent) R1
\nhence and cannot both be odd. AG
\n
Note: Award a maximum of M1A0A0(A0)R1R1 for considering identical or two consecutive odd numbers for and .
\n
[6 marks]
\nMost candidates did not present their proof in a formal manner and merely relied on an algebraic approach rendering the proof incomplete. Very few candidates earned the first mark for making a clear assumption that a and b are both odd. A significant number of candidates only considered consecutive or identical odd numbers. The required reasoning to complete the proof were often poorly expressed or missing altogether. Only a small number of candidates were awarded all the available marks for this question.
\nConsider the complex numbers and , where .
\nFind an expression for in terms of .
\nHence, given that , find the value of .
\nM1
\nA1A1
\n
Note: Award A1 for and A1 for .
\n
[3 marks]
\n(M1)
\n
EITHER
(since , for ) A1
OR
(or equivalent) A1
\n
THEN
A1
\n\n
[3 marks]
\nPart (a) was generally well done with many completely correct answers seen. Part (b) proved to be challenging with many candidates incorrectly equating the ratio of their imaginary and real parts to instead of . Stronger candidates realized that when , it forms an isosceles right-angled triangle and equated the real and imaginary parts to obtain the value of b .
\nThe following diagram shows the curve , where .
\nThe curve from point to point is rotated about the -axis to form the interior surface of a bowl. The rectangle , of height , is rotated about the -axis to form a solid base.
\nThe bowl is assumed to have negligible thickness.
\nGiven that the interior volume of the bowl is to be , determine the height of the base.
\nattempts to express in terms of (M1)
\nA1
\n
Note: Correct limits are required.
\n
Attempts to solve for (M1)
\nNote: Award M1 for attempting to solve or equivalent for .
\n(cm) A2
\n\n
[5 marks]
\nThis question was a struggle for many candidates. To start with, many candidates had difficulty understanding the diagram. Some candidates tried to include the base in their equation.
\nBecause of this confusion, the question was poorly attempted. Some only received one mark for rearranging the equation to make the subject but were unable to set the correct definite integral with correct terminals. Again, many candidates tried to solve by hand instead of using their GDC. The correct answer was not seen that often.
\nThose candidates who recognised that the volume was around the y -axis and used their GDC to solve, usually achieved full marks for this question.
\nConsider the series , where and .
\nConsider the case where the series is geometric.
\nNow consider the case where the series is arithmetic with common difference .
\nShow that .
\nHence or otherwise, show that the series is convergent.
\nGiven that and , find the value of .
\nShow that .
\nWrite down in the form , where .
\nThe sum of the first terms of the series is .
\nFind the value of .
\nEITHER
\nattempt to use a ratio from consecutive terms M1
\nOR OR
\n\n
Note: Candidates may use and consider the powers of in geometric sequence
\nAward M1 for .
\n
OR
and M1
\n
THEN
OR A1
\nAG
\n\n
Note: Award M0A0 for or with no other working seen.
\n\n
[2 marks]
\nEITHER
\nsince, and R1
\n
OR
since, and R1
\n
THEN
the geometric series converges. AG
\n
Note: Accept instead of .
Award R0 if both values of not considered.
\n
[1 mark]
\n(A1)
\nOR A1
\nA1
\n\n
[3 marks]
\nMETHOD 1
\nattempt to find a difference from consecutive terms or from M1
\ncorrect equation A1
\nOR
\n
Note: Candidates may use and consider the powers of in arithmetic sequence.
Award M1A1 for
\n\n
A1
\nAG
\n\n
METHOD 2
\nattempt to use arithmetic mean M1
\nA1
\nA1
\nAG
\n\n
METHOD 3
\nattempt to find difference using M1
\n\n\n
OR A1
\nA1
\nAG
\n\n
[3 marks]
\nA1
\n\n
[1 mark]
\nMETHOD 1
\n\nattempt to substitute into and equate to (M1)
\n\n(A1)
\n(A1)
\ncorrect working with (seen anywhere) (A1)
\nOR OR
\ncorrect equation without A1
\nOR or equivalent
\n
Note: Award as above if the series is considered leading to .
attempt to form a quadratic (M1)
attempt to solve their quadratic (M1)
\n\nA1
\n\n
METHOD 2
\n(A1)
\n(A1)
\nlisting the first terms of the sequence (A1)
\n\nrecognizing first terms sum to M1
\nth term is (A1)
\nth term is (A1)
\nsum of th and th term (A1)
\nA1
\n\n
[8 marks]
\nPart (a)(i) was well done with few candidates incorrectly using the value of p to verify rather than to 'show' the given result. In part (a)(ii) most did not consider both values of r and some did know the condition for convergence of a geometric series. Part (a)(iii) was generally well done but some had difficulty in simplifying the surd. Part (b) (i) and (ii) was generally well done. Although many completely correct answers to part b (iii) were noted, weaker candidates often made errors in properties of logarithms or algebraic manipulation leading to an incorrect quadratic equation.
\nConsider , where .
\nShow that a finite limit only exists for .
\nUsing l’Hôpital’s rule, show algebraically that the value of the limit is .
\n(as , the indeterminate form is required for the limit to exist)
\nM1
\nA1
\nso AG
\n
Note: Award M1A0 for using to show the limit is .
\n
[2 marks]
\nA1A1
\n
Note: Award A1 for a correct numerator and A1 for a correct denominator.
recognises to apply l’Hôpital’s rule again (M1)
Note: Award M0 if their limit is not the indeterminate form .
EITHER
A1A1
\n
Note: Award A1 for a correct first term in the numerator and A1 for a correct second term in the numerator.
OR
A1A1
\n
Note: Award A1 for a correct numerator and A1 for a correct denominator.
THEN
substitutes into the correct expression to evaluate the limit A1
\n
Note: The final A1 is dependent on all previous marks.
AG
\n
[6 marks]
\nPart (a) Many candidates recognised the indeterminate form and provided a nice algebraic proof. Some verified by substituting the given value. Therefore, there is a need to teach the candidates the difference between proof and verification. Only a few candidates were able to give a complete 'show that' proof.
\nPart (b) Many candidates realised that they needed to apply the L'Hôpital's rule twice. There were many mistakes in differentiation using the chain rule. Not all candidates clearly showed the final substitution.
\nRachel and Sophia are competing in a javelin-throwing competition.
\nThe distances, metres, thrown by Rachel can be modelled by a normal distribution with mean and standard deviation .
\nThe distances, metres, thrown by Sophia can be modelled by a normal distribution with mean and standard deviation .
\nIn the first round of competition, each competitor must have five throws. To qualify for the next round of competition, a competitor must record at least one throw of metres or greater in the first round.
\nFind the probability that only one of Rachel or Sophia qualifies for the next round of competition.
\nRachel:
\n(A1)
\nSophia:
\n(A1)
\nrecognises binomial distribution with (M1)
\nlet represent the number of Rachel’s throws that are longer than metres
\n\neither or (A1)
\nlet represent the number of Sophia’s throws that are longer than metres
\n\neither or (A1)
\n
EITHER
uses (M1)
\n\n
OR
uses (M1)
\n\n\n
THEN
\n\nA1
\n
Note: M marks are not dependent on the previous A marks.
\n
[7 marks]
\nIn general, well answered with many candidates getting full marks. On the other end of the spectrum, some candidates could only get the first two marks for normal distribution without recognising the binomial distribution and often just stopping there. Some candidates lost the last two marks for not being able to combine the probabilities correctly.
\nConsider the set of six-digit positive integers that can be formed from the digits and .
\nFind the total number of six-digit positive integers that can be formed such that
\nthe digits are distinct.
\nthe digits are distinct and are in increasing order.
\n(M1)
\nA1
\n
Note: Award M1A0 for .
Note: Award M1A0 for
\n\n
[2 marks]
\nMETHOD 1
\nEITHER
\nevery unordered subset of digits from the set of non-zero digits can be arranged in exactly one way into a -digit number with the digits in increasing order. A1
\n
OR
A1
\n
THEN
A1
\n\n
METHOD 2
\nEITHER
\nremoves digits from the set of non-zero digits and these remaining digits can be arranged in exactly one way into a -digit number with the digits in increasing order. A1
\n
OR
A1
\n\n
THEN
\nA1
\n\n
[2 marks]
\nPart (a) A number of candidates got the correct answer here with the valid approach and recognising that zero could not occupy the first position. Some lost a mark by including zero. Many candidates used an incorrect method with combinations or simplified permutations.
\nPart (b) Only very few candidates got the correct answer. Many left it blank or provided unreasonably enormous numbers as their answers.
\nSome candidates had the answer to part (b) showing in part (a).
\nA small number of candidates tried to list all possibilities but mostly unsuccessfully.
\nConsider the three planes
\n\n\n\nShow that the three planes do not intersect.
\nVerify that the point lies on both and .
\nFind a vector equation of , the line of intersection of and .
\nFind the distance between and .
\nMETHOD 1
\nattempt to eliminate a variable M1
\nobtain a pair of equations in two variables
\n
EITHER
and A1
\nA1
\n
OR
and A1
\nA1
\n
OR
and A1
\nA1
\n
THEN
the two lines are parallel ( or or ) R1
\n\n
Note: There are other possible pairs of equations in two variables.
To obtain the final R1, at least the initial M1 must have been awarded.
\n
hence the three planes do not intersect AG
\n\n
METHOD 2
\nvector product of the two normals (or equivalent) A1
\n(or equivalent) A1
\n\n
Note: Award A0 if “” is missing. Subsequent marks may still be awarded.
\n\n
Attempt to substitute in M1
\n\n, a contradiction R1
\nhence the three planes do not intersect AG
\n\n
METHOD 3
\nattempt to eliminate a variable M1
\nA1
\nA1
\n, a contradiction R1
\n\n
Note: Accept other equivalent alternatives. Accept other valid methods.
To obtain the final R1, at least the initial M1 must have been awarded.
\n
hence the three planes do not intersect AG
\n\n
[4 marks]
\nand A1
\n\n
[1 mark]
\nMETHOD 1
\nattempt to find the vector product of the two normals M1
\n\nA1
\nA1A1
\n\n
Note: Award A1A0 if “” is missing.
Accept any multiple of the direction vector.
Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize lack of “” only once.
\n
METHOD 2
\nattempt to eliminate a variable from and M1
\nOR OR
\nLet
\nsubstituting in to obtain
\nand (for all three variables in parametric form) A1
\nA1A1
\n\n
Note: Award A1A0 if “” is missing.
Accept any multiple of the direction vector. Accept other position vectors which satisfy both the planes and .
\n
[4 marks]
\nMETHOD 1
\nthe line connecting and is given by
\nattempt to substitute position and direction vector to form (M1)
\nA1
\nsubstitute in M1
\n\nA1
\nattempt to find distance between and their point (M1)
\n\nA1
\n\n
METHOD 2
\nunit normal vector equation of is given by (M1)
\nA1
\nlet be the plane parallel to and passing through ,
then the normal vector equation of is given by
M1
\n\n
unit normal vector equation of is given by
\nA1
\ndistance between the planes is (M1)
\nA1
\n\n
[6 marks]
\nPart (a) was well attempted using a variety of approaches. Most candidates were able to gain marks for part (a) through attempts to eliminate a variable with many subsequently making algebraic errors. Part (b)(i) was well done. For part (b)(ii) few successful attempts were noted, many candidates failed to use an appropriate notation \"r =\" while giving the vector equation of a line. Part (c) proved to be challenging for most candidates with very few correct answers seen. Many candidates did not attempt part (c).
\nA scientist conducted a nine-week experiment on two plants, and , of the same species. He wanted to determine the effect of using a new plant fertilizer. Plant was given fertilizer regularly, while Plant was not.
\nThe scientist found that the height of Plant , at time weeks can be modelled by the function , where .
\nThe scientist found that the height of Plant , at time weeks can be modelled by the function , where .
\nUse the scientist’s models to find the initial height of
\nPlant .
\nPlant correct to three significant figures.
\nFind the values of when .
\nFor , prove that Plant was always taller than Plant .
\nFor , find the total amount of time when the rate of growth of Plant was greater than the rate of growth of Plant .
\n(cm) A1
\n\n
[1 mark]
\n(M1)
\n\n(cm) A1
\n\n
[2 marks]
\nattempts to solve for (M1)
\n\n(weeks) A2
\n\n
[3 marks]
\nA1
\n
EITHER
for A1
\nand as R1
\n
OR
the minimum value of R1
\nso for A1
\n
THEN
hence for , Plant was always taller than Plant AG
\n\n
[3 marks]
\nrecognises that and are required (M1)
\nattempts to solve for (M1)
\nand OR and OR and (A1)
\n\n
Note: Award full marks for .
\nAward subsequent marks for correct use of these exact values.
\n\n
OR OR (A1)
\nattempts to calculate the total amount of time (M1)
\n\n(weeks) A1
\n\n
[6 marks]
\nPart (a) In general, very well done, most students scored full marks. Some though had an incorrect answer for part(a)(ii) because they had their GDC in degrees.
\nPart (b) Well attempted. Some accuracy errors and not all candidates listed all three values.
\nPart (c) Most students tried a graphical approach (but this would only get them one out of three marks) and only some provided a convincing algebraic justification. Many candidates tried to explain in words without a convincing mathematical justification or used numerical calculations with specific time values. Some arrived at the correct simplified equation for the difference in heights but could not do much with it. Then only a few provided a correct mathematical proof.
\nPart (d) In general, well attempted by many candidates. The common error was giving the answer as 3.15 due to the pre-mature rounding. Some candidates only provided the values of time when the rates are equal, some intervals rather than the total time.
\nThe function is defined by , where .
\nThe function is defined by , where .
\nFind the Maclaurin series for up to and including the term.
\nHence, find an approximate value for .
\nShow that satisfies the equation .
\nHence, deduce that .
\nUsing the result from part (c), find the Maclaurin series for up to and including the term.
\nHence, or otherwise, determine the value of .
\nMETHOD 1
\nrecognition of both known series (M1)
\nand
\nattempt to multiply the two series up to and including term (M1)
\n\n(A1)
\nA1
\n\n
METHOD 2
\n\nA1
\n\n\nand A1
\nsubstitute into or its derivatives to obtain Maclaurin series (M1)
\n\nA1
\n\n
[4 marks]
\n(A1)
\nsubstituting their expression and attempt to integrate M1
\n\n\n
Note: Condone absence of limits up to this stage.
\n\n
A1
\nA1
\n\n
[4 marks]
\nattempt to use product rule at least once M1
\nA1
\nA1
\n
EITHER
A1
\n
OR
A1
\n
THEN
AG
\n\n
Note: Accept working with each side separately to obtain .
\n\n
[4 marks]
\nA1
\nAG
\n\n
Note: Accept working with each side separately to obtain .
\n\n
[1 mark]
\nattempt to substitute into a derivative (M1)
\nA1
\n(A1)
\nattempt to substitute into Maclaurin formula (M1)
\nA1
\n\n
Note: Do not award any marks for approaches that do not use the part (c) result.
\n\n
[5 marks]
\nMETHOD 1
\nM1
\n(A1)
\nA1
\n\n
Note: Condone the omission of in their working.
\n\n
METHOD 2
\nindeterminate form, attempt to apply l'Hôpital's rule M1
\n\n, using l'Hôpital's rule again
\n\n, using l'Hôpital's rule again
\nA1
\nA1
\n\n
[3 marks]
\nPart (a) was well answered using both methods. A number failed to see the connection between parts (a) and (b). Far too often, a candidate could not add three fractions together in part (b). There were many good responses to part (c) with candidates showing results on both sides are equal. A number of candidates failed to use the result from (c) in part (d). There were some good responses to part (e), with candidates working successfully with the series from (d) or applying l'Hôpital's rule. In particular, some responses were missing the appropriate limit notation and candidates following method 2 did not always show that the initial expression was of an indeterminate form before applying l'Hôpital's rule. Many candidates did not attempt parts (d) and (e).
\nThe following table shows values of and for different values of .
\nBoth and are one-to-one functions.
\nFind .
\nFind .
\nFind the value of such that .
\nA1
\n\n
[1 mark]
\nevidence of using composite function (M1)
\nOR
\nA1
\n\n
[2 marks]
\nA2
\n\n
[2 marks]
\nThis question was completed successfully by most of the candidates. In part (b) of the question, a few candidates did not recognize the notation for a composite function and instead incorrectly thought they were supposed to multiply values for and .
\nTwo airplanes, and , have position vectors with respect to an origin given respectively by
\n\n\nwhere represents the time in minutes and .
\nEntries in each column vector give the displacement east of , the displacement north of and the distance above sea level, all measured in kilometres.
\nThe two airplanes’ lines of flight cross at point .
\nFind the three-figure bearing on which airplane is travelling.
\nShow that airplane travels at a greater speed than airplane .
\nFind the acute angle between the two airplanes’ lines of flight. Give your answer in degrees.
\nFind the coordinates of .
\nDetermine the length of time between the first airplane arriving at and the second airplane arriving at .
\nLet represent the distance between airplane and airplane for .
\nFind the minimum value of .
\nlet be the required angle (bearing)
\n
EITHER
(M1)
\n
Note: Award M1 for a labelled sketch.
OR
(M1)
\n\n
THEN
A1
\n
Note: Do not accept or or .
\n
[2 marks]
\nMETHOD 1
\nlet be the speed of and let be the speed of
\nattempts to find the speed of one of or (M1)
\nor
\n
Note: Award M0 for and .
(km min-1) and (km min-1) A1
so travels at a greater speed than AG
\n\n
METHOD 2
\nattempts to use
\nand (M1)
\nfor example:
\nand
\nand
\nand A1
\nso travels at a greater speed than AG
\n\n
[2 marks]
\nattempts to use the angle between two direction vectors formula (M1)
\n(A1)
\nor
\nattempts to find the acute angle using their value of (M1)
\nA1
\n\n
[4 marks]
\nfor example, sets and forms at least two equations (M1)
\n\n\n\n
Note: Award M0 for equations involving only.
EITHER
attempts to solve the system of equations for one of or (M1)
\nor A1
\n
OR
attempts to solve the system of equations for and (M1)
\nor A1
\n
THEN
substitutes their or value into the corresponding or (M1)
\nA1
\n
Note: Accept . Accept km east of , km north of and km above sea level.
\n
[5 marks]
\nattempts to find the value of (M1)
\n\nminutes ( seconds) A1
\n\n
[2 marks]
\nEITHER
\nattempts to find (M1)
\n\nattempts to find their (M1)
\nA1
\n
OR
attempts to find (M1)
\n\nattempts to find their (M1)
\nA1
\n\n
Note: Award M0M0A0 for expressions using two different time parameters.
\n
THEN
either attempts to find the local minimum point of or attempts to find the value of such that (or equivalent) (M1)
\n\n\nminimum value of is (km) A1
\n
Note: Award M0 for attempts at the shortest distance between two lines.
\n
[5 marks]
\nGeneral comment about this question: many candidates were not exposed to this setting of vectors question and were rather lost.
\nPart (a) Probably the least answered question on the whole paper. Many candidates left it blank, others tried using 3D vectors. Out of those who calculated the angle correctly, only a small percentage were able to provide the correct true bearing as a 3-digit figure.
\nPart (b) Well done by many candidates who used the direction vectors to calculate and compare the speeds. A number of candidates tried to use the average rate of change but mostly unsuccessfully.
\nPart (c) Most candidates used the correct vectors and the formula to obtain the obtuse angle. Then only some read the question properly to give the acute angle in degrees, as requested.
\nPart (d) Well done by many candidates who used two different parameters. They were able to solve and obtain two values for time, the difference in minutes and the correct point of intersection. A number of candidates only had one parameter, thus scoring no marks in part (d) (i). The frequent error in part (d)(ii) was providing incorrect units.
\nPart (e) Many correct answers were seen with an efficient way of setting the question and using their GDC to obtain the answer, graphically or numerically. Some gave time only instead of actually giving the minimal distance. A number of candidates tried to find the distance between two skew lines ignoring the fact that the lines intersect.
\nThe population, , of a particular species of marsupial on a small remote island can be modelled by the logistic differential equation
\n\nwhere is the time measured in years and are positive constants.
\nThe constant represents the maximum population of this species of marsupial that the island can sustain indefinitely.
\nLet be the initial population of marsupials.
\nIn the context of the population model, interpret the meaning of .
\nShow that .
\nHence show that the population of marsupials will increase at its maximum rate when . Justify your answer.
\nHence determine the maximum value of in terms of and .
\nBy solving the logistic differential equation, show that its solution can be expressed in the form
\n.
\nAfter years, the population of marsupials is . It is known that .
\nFind the value of for this population model.
\nrate of growth (change) of the (marsupial) population (with respect to time) A1
\n\n
[1 mark]
\n
Note: Do not accept growth (change) in the (marsupials) population per year.
METHOD 1
\nattempts implicit differentiation on be expanding (M1)
\nA1A1
\nA1
\nand so AG
\n\n
METHOD 2
\nattempts implicit differentiation (product rule) on M1
\nA1
\nsubstitutes into their M1
\n\n\nA1
\nso AG
\n\n
[4 marks]
\n(M1)
\nA2
\nNote: Award A1 for only.
\nuses the second derivative to show that concavity changes at or the first derivative to show a local maximum at M1
EITHER
a clearly labelled correct sketch of versus showing corresponding to a local maximum point for R1
\n
OR
a correct and clearly labelled sign diagram (table) showing corresponding to a local maximum point for R1
\n
OR
for example, with and with showing corresponds to a local maximum point for R1
\nso the population is increasing at its maximum rate when AG
\n\n
[5 marks]
\nsubstitutes into (M1)
\n\nthe maximum value of is A1
\n\n
[2 marks]
\nMETHOD 1
\nattempts to separate variables M1
\n\nattempts to write in partial fractions form M1
\n\nA1
\n\n\nA1A1
\n
Note: Award A1 for and A1 for and . Absolute value signs are not required.
\n
attempts to find in terms of and M1
\nwhen and so
\nA1
\nso AG
\n\n
METHOD 2
\nattempts to separate variables M1
\n\nattempts to write in partial fractions form M1
\n\n
A1
\n\n
\n
A1A1
\n
Note: Award A1 for and A1 for and . Absolute value signs are not required.
attempts to find in terms of and M1
\nwhen and so
\nA1
\nAG
\n\n
METHOD 3
\nlets and forms M1
\nmultiplies both sides of the differential equation by and makes the above substitutions M1
\n\n(linear first-order DE) A1
\n(M1)
\n\nA1
\nattempts to find in terms of and M1
\nwhen and so
\n\nA1
\nAG
\n\n
[7 marks]
\nsubstitutes and into M1
\n\nA1
\n\n
[2 marks]
\nAn extremely tricky question even for the strong candidates. Many struggled to understand what was expected in parts (b) and (c). As the question was set all with pronumerals instead of numbers many candidates found it challenging, thrown at deep water for parts (b), (c) and (e). It definitely was the question to show their skills for the Level 7 candidates provided that they did not run out of time.
\nPart (a) Very well answered, mostly correctly referring to the rate of change. Some candidates did not gain this mark because their sentence did not include the reference to the rate of change. Worded explanations continue being problematic to many candidates.
\nPart (b) Many candidates were confused how to approach this question and did not realise that they
needed to differentiate implicitly. Some tried but with errors, some did not fully show what was required.
Part (c) Most candidates started with equating the second derivative to zero. Most gave the answer omitting the other two possibilities. Most stopped here. Only a small number of candidates provided the correct mathematical argument to show it is a local maximum.
\nPart (d) Well done by those candidates who got that far. Most got the correct answer, sometimes not fully simplified.
\nPart (e) Most candidates separated the variables, but some were not able to do much more. Some candidates knew to resolve into partial fractions and attempted to do so, mainly successfully. Then they integrated, again, mainly successfully and continued to substitute the initial condition and manipulate the equation accordingly.
\nPart (f) Algebraic manipulation of the logarithmic expression was too much for some candidates with a common error of 0.33 given as the answer. The strong candidates provided the correct exact or rounded value.
\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 .
\nMETHOD 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]
\n