Cambridge A Level Mathematics - Further 9231 — 2019 May/June Paper 2 · Variant 1

*8015586296* Cambridge Assessment International Education Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/21 Paper 2 May/June 2019 3 hours Candidates answer on the Question Paper. Additional Materials: List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST Write your centre number, candidate number and name in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all the questions in the space provided. If additional space is required, you should use the lined page at the end of this booklet. The question number(s) must be clearly shown. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. Where a numerical value is necessary, take the acceleration due to gravity to be 10 m s−2. The use of a calculator is expected, where appropriate. Results obtained solely from a graphic calculator, without supporting working or reasoning, will not receive credit. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 25 printed pages and 3 blank pages. JC19 06_9231_21/FP © UCLES 2019 [Turn over

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3 1 A particle P moves along an arc of a circle with centre O and radius 2 m. At time t seconds, the angle POA is 1, where 1 = 1 −cos 2t, and A is a fixed point on the arc of the circle. (i) Show that the magnitude of the radial component of the acceleration of P when t = 1 60 is 6 m s−2. [2] … … … … … … … … … … (ii) Find the magnitude of the transverse component of the acceleration of P when t = 1 60. [2] … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19 [Turn over

4 2 A particle P moves on a straight line in simple harmonic motion. The centre of the motion is O. The points A and B are on the line on opposite sides of O such that OA = 3.5 m and OB = 1 m. The speed of P when it is at B is twice its speed when it is at A. The maximum acceleration of P is 1 m s−2. (i) Find the speed of P when it is at O. [4] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19

5 (ii) Find the time taken by P to travel directly from A to B. [4] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19 [Turn over

6 3 Three uniform small spheres A, B and C have equal radii and masses 2m, 4m and m respectively. The spheres are moving in a straight line on a smooth horizontal surface, with B between A and C. The coefficient of restitution between each pair of spheres is e. Spheres A and B are moving towards each other with speeds 2u and u respectively. The first collision is between A and B. (i) Find the velocities of A and B after this collision. [3] … … … … … … … … … … … … Sphere C is moving towards B with speed 4 3u and now collides with it. As a result of this collision, B is brought to rest. (ii) Find the value of e. [4] … … … … … … … … © UCLES 2019 9231/21/M/J/19

7 … … … … … … … … … … … (iii) Find the total kinetic energy lost by the three spheres as a result of the two collisions. [3] … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19 [Turn over

8 4 1 A B C D 5 2a 3 2a 5 2a A uniform rod AB of length 4a and weight W rests with the end A in contact with a rough vertical wall. A light inextensible string of length 5 2a has one end attached to the point C on the rod, where AC = 5 2a. The other end of the string is attached to a point D on the wall, vertically above A. The vertical plane containing the rod AB is perpendicular to the wall. The angle between the rod and the wall is 1, where tan 1 = 2 (see diagram). The end A of the rod is on the point of slipping down the wall and the coefficient of friction between the rod and the wall is -. Find, in either order, the tension in the string and the value of -. [10] … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19

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10 5 ring hollow sphere L 2a a O A B C a a A thin uniform rod AB has mass kM and length 2a. The end A of the rod is rigidly attached to the surface of a uniform hollow sphere with centre O, mass kM and radius 2a. The end B of the rod is rigidly attached to the circumference of a uniform ring with centre C, mass M and radius a. The points C, B, A, O lie in a straight line. The horizontal axis L passes through the mid-point of the rod and is perpendicular to the rod and in the plane of the ring (see diagram). The object consisting of the rod, the ring and the hollow sphere can rotate freely about L. (i) Show that the moment of inertia of the object about L is 3 28k + 3Ma2. [6] … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19

11 The object performs small oscillations about L, with the ring above the sphere as shown in the diagram. (ii) Find the set of possible values of k and the period of these oscillations in terms of k. [6] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19 [Turn over

12 6 A fair six-sided die is thrown until a 3 or a 4 is obtained. The number of throws taken is denoted by the random variable X. (i) State the mean value of X. [1] … … … … (ii) Find the probability that obtaining a 3 or a 4 takes exactly 6 throws. [1] … … … … … (iii) Find the probability that obtaining a 3 or a 4 takes more than 4 throws. [2] … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19

13 (iv) Find the greatest integer n such that the probability of obtaining a 3 or a 4 in fewer than n throws is less than 0.95. [3] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19 [Turn over

14 7 The continuous random variable X has probability density function f given by fx = d 3 4x2 + 1 4 1 ≤x ≤3, 0 otherwise. (i) Find the distribution function of X. [3] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19

15 (ii) Find the exact value of the interquartile range of X. [5] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19 [Turn over

16 8 A large number of runners are attending a summer training camp. A random sample of 6 runners is chosen and their times to run 1500 m at the beginning of the camp and at the end of the camp are recorded. Their times, in minutes, are shown in the following table. Runner A B C D E F Time at beginning of camp 3.82 3.62 3.55 3.71 3.75 3.92 Time at end of camp 3.72 3.55 3.52 3.68 3.54 3.73 The organiser of the training camp claims that a runner’s time will improve by more than 0.05 minutes between the beginning and end of the camp. Assuming that differences in time over the two runs are normally distributed, test at the 10% significance level whether the organiser’s claim is justified. [8] … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19

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18 9 A random sample of 50 observations of the continuous random variable X was taken and the values are summarised in the following table. Interval 0 ≤x < 0.8 0.8 ≤x < 1.6 1.6 ≤x < 2.4 2.4 ≤x < 3.2 3.2 ≤x < 4 Observed frequency 18 16 8 6 2 It is required to test the goodness of fit of the distribution with probability density function f given by fx = d 3 164 −x 1 2 0 ≤x < 4, 0 otherwise. The relevant expected frequencies, correct to 2 decimal places, are given in the following table. Interval 0 ≤x < 0.8 0.8 ≤x < 1.6 1.6 ≤x < 2.4 2.4 ≤x < 3.2 3.2 ≤x < 4 Expected frequency 14.22 12.54 10.59 8.18 4.47 (i) Show how the expected frequency for 1.6 ≤x < 2.4 is obtained. [3] … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19

19 (ii) Carry out a goodness of fit test at the 5% significance level. [7] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19 [Turn over

20 10 The values from a random sample of five pairs x, y taken from a bivariate distribution are shown below. x 3 4 4 6 8 y 5 7 q 6 7 The equation of the regression line of x on y is given by x = 5 4y + c. (i) Given that q is an integer, find its value. [5] … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19

21 (ii) Find the value of c. [3] … … … … … … … … … … … … (iii) Find the value of the product moment correlation coefficient. [3] … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19 [Turn over

22 11 Answer only one of the following two alternatives. EITHER A particle P, of mass m, is able to move in a vertical circle on the smooth inner surface of a sphere with centre O and radius a. Points A and B are on the inner surface of the sphere and AOB is a horizontal diameter. Initially, P is projected vertically downwards with speed / 21 2 ag ! from A and begins to move in a vertical circle. At the lowest point of its path, vertically below O, the particle P collides with a stationary particle Q, of mass 4m, and rebounds. The speed acquired by Q, as a result of the collision, is just sufficient for it to reach the point B. (i) Find the speed of P and the speed of Q immediately after their collision. [7] … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19

23 In its subsequent motion, P loses contact with the inner surface of the sphere at the point D, where the angle between OD and the upward vertical through O is 1. (ii) Find cos 1. [5] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19 [Turn over

24 OR A farmer grows two different types of cherries, Type A and Type B. He assumes that the masses of each type are normally distributed. He chooses a random sample of 8 cherries of Type A. He finds that the sample mean mass is 15.1 g and that a 95% confidence interval for the population mean mass, - g, is 13.5 ≤- ≤16.7. (i) Find an unbiased estimate for the population variance of the masses of cherries of Type A. [3] … … … … … … … … … The farmer now chooses a random sample of 6 cherries of Type B and records their masses as follows. 12.2 13.3 16.4 14.0 13.9 15.4 (ii) Test at the 5% significance level whether the mean mass of cherries of Type B is less than the mean mass of cherries of Type A. You should assume that the population variances for the two types of cherry are equal. [9] … … … … … … … … © UCLES 2019 9231/21/M/J/19

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26 Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/M/J/19

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28 BLANK PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge. © UCLES 2019 9231/21/M/J/19

This document consists of 17 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/21 Paper 2 May/June 2019 MARK SCHEME Maximum Mark: 100 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the May/June 2019 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 2 of 17 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 3 of 17 GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 4 of 17 Mark Scheme Notes The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Mark for a correct result or statement independent of method marks. Abbreviations AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) CAO Correct Answer Only (emphasising that no “follow through” from a previous error is allowed) CWO Correct Working Only – often written by a ‘fortuitous’ answer ISW Ignore Subsequent Working SOI Seen or implied SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance)

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 5 of 17 Question Answer Marks Guidance 1(i) aR = 2 (dθ /dt)2 = 2 (2 sin 2t)2 = 2 (2 sin π/3) 2 M1 Verify radial acceleration aR at t = π/6 from rω2 = 2 (√3) 2 = 6 [m s-2] AG A1 2 1(ii) aT = 2 d2θ /dt2 = 2 (4 cos 2t) = 2 (4 cos π/3) M1 Find transverse acceleration aT at t = π/6 by differentiation = 4 [m s-2] A1 2

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 6 of 17 Question Answer Marks Guidance 2(i) vB = 2vA , vB2 = 4 vA2 , ω2 (a2 – 12) = 4ω2 (a2 – 3⋅52) M1 Find amplitude a: allow M1 for vA = 2 vB 3 a2 = 48, a = [±] 4 [m] A1 1 = aω2, ω = 1/√a [ = ½] B1 Find ω from given maximum acceleration vO = aω = √a = 2 [m s-1] B1 Find speed vO at O 4 2(ii) ω tAB = sin-1 (3.5/a) + sin-1(1/a) = sin-1 0⋅875 + sin-1 0⋅25 M1 A1 Find equation (AEF) for ω tAB, combining tA and tB, using for example x = a sin ωt allow sign errors for the M1 = 1⋅065 + 0⋅253 (or tAB = 2⋅131 + 0⋅505) A1 or x = a cos ωt tAB = 2 × 1⋅318 = 2⋅64 [s] A1 Hence find tAB Alternative method for question 2(ii) ω tAB = cos-1 (–1/a) – cos-1(3.5/a) = cos-1 (–0⋅25) – cos-1 0⋅875 or π – cos-1 0⋅25 – cos-1 0⋅875 (AEF) M1 A1 or x = a cos ωt = 1⋅823 – 0⋅505 (or tAB = 3⋅647 – 1⋅011) A1 tAB = 2 × 1⋅318 = 2⋅64 [s] A1 Hence find tAB 4

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 7 of 17 Question Answer Marks Guidance 3(i) 2mvA + 4mvB = 4mu – 4mu [vA + 2vB = 0] (AEF) M1 Use conservation of momentum for A and B (m may be omitted) vB – vA = e (2u + u) [vB – vA = 3eu] M1 Use Newton’s restitution law with consistent LHS signs vA = – 2eu and vB = eu A1 Combine to find vA and vB (A0 if directions unclear) 3 3(ii) [4mvB′] + mvC = 4mvB – (4/3) mu (AEF) M1 Use conservation of momentum for B & C (m may be omitted) vC [– vB′] = e (vB + 4u/3) M1 Use Newton’s restitution law 4vB – (4/3) u = evB + 4eu/3, 4e – 4/3 = e2 + 4e/3 M1 Combine to find quadratic equation for e using vB′ = 0 3e2 – 8e + 4 = 0, e = 2/3 [vC = 4u/3] A1 Find value of e, (implicitly) rejecting e = 2 4

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 8 of 17 Question Answer Marks Guidance 3(iii) For A: Loss = ½ 2m(2u)2 - ½ 2m(4/3u)2 = (20/9) mu2 For B: Loss = ½ 4m u2 = ½ mu2 For C: Loss = 0 M1 vA = – (4/3) u , [vB = (2/3) u], vC = (4/3) u One correct M1 Other two correct Einitial – Efinal or L1 + L2 = (38/9) mu2 A1 Hence find loss in KE Alternative method for question 3(iii) Einitial = ½ 2m(2u)2 + ½ 4mu2 + ½ m(4u/3)2 = 4mu2 + 2mu2 + (8/9) mu2 = (62/9) mu2 M1 Find initial KE of 3 particles in terms of m and u Efinal = ½ 2mvA2 [+ ½ 4mvB′ 2 ] + ½ mvC 2 = (16/9) mu2 + (8/9) mu2 = (24/9) mu2 M1 Find final KE of 3 particles in terms of m and u Einitial – Efinal or L1 + L2 = (38/9) mu2 A1 Hence find loss in KE Alternative method for question 3(iii) L1 = ½ 2m(2u)2 + ½ 4mu2 – ½ 2mvA2 – ½ 4mvB 2 = 4mu2 + 2mu2 – (16/9) mu2 – (8/9) mu2 = (30/9) mu2 M1 Find losses in KE in both collisions in terms of m and u L2 = ½ 4mvB 2 + ½ m(4u/3)2 – [½ 4mvB′ 2 ] – ½ mvC 2 = (8/9) mu2 + (8/9) mu2 – (8/9) mu2 = (8/9) mu2 M1 Einitial – Efinal or L1 + L2 = (38/9) mu2 A1 Hence find loss in KE 3

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 9 of 17 Question Answer Marks Guidance 4 A: T × 5a/2 sin (π – 2θ) – W × 2a sin θ = 0 C: FA × 5a/2 sin θ – RA × 5a/2 cos θ – W × ½ a sin θ = 0 B: FA × 4a sin θ – RA × 4a cos θ – W × 2a sin θ + T × 3a/2 sin (π – 2θ) = 0 G: FA × 2a sin θ – RA × 2a cos θ – T × ½ a sin (π – 2θ) = 0 (G is mid-point of AB) D: RA × 5a cos θ – W × 2a sin θ = 0 M1 A1 Take moments for rod about one chosen point [sin θ = 2/√5, cos θ = 1/√5, sin (π – 2θ) = sin 2θ = 4/5] RA = T sin θ FA = W – T cos θ B1 Find two more independent equations B1 e.g. resolution of forces on rod (a second moment equation may be used) FA = µRA B1 Relate FA and RA (may be implied) T = (2W sin θ ) / (5/2 sin 2θ ) = 2W / (5 cos θ ) M1 Find T by any method (e.g. from moments about A) = 2W/√5 or (2√5/5) W or 0⋅894 W A1 FA = (3/5) W and RA = (4/5) W µ = ¾ or 0⋅75 M1 A1 Find or imply FA and RA by any method (e.g. from resolutions) and hence µ A1 10

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 10 of 17 Question Answer Marks Guidance 5(i) Irod = ⅓ kMa2 B1 Find or state MI of rod AB about axis L Isphere = ⅔ kM (2a)2 + kM (3a)2 [= (35k/3) Ma2] M1 A1 M1 for one term correct, A1 for both terms correct Iring = ½ × Ma2 + M (2a)2 [= (9/2) Ma2] M1 A1 M1 for one term correct, A1 for both terms correct I = (k/3 + 35k/3 + 9/2) Ma2 = (3/2) (8k + 3) Ma2 AG A1 MI of object about axis L 6 5(ii) [–] I d2θ/dt2 = – kMg × 3a sin θ + Mg × 2a sin θ [ = – (3k – 2) Mga sin θ ] M1 A1 Use equation of circular motion to find d2θ/dt2 where θ is angle of rod with vertical d2θ/dt2 = – {2g (3k – 2) / 3a(8k + 3)} θ (AEF) M1 Approximate sin θ by θ to give standard form of SHM equation (M0 if wrong sign or cos θ ≈ θ used) SHM if 3k – 2 > 0, k > 2/3 M1 A1 Find possible values of k T = 2π √{3a(8k + 3) / 2g (3k – 2)} or π √{6a(8k + 3) / g (3k – 2)} A1 Find period T 6

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 11 of 17 Question Answer Marks Guidance 6(i) Mean = 3 B1 State mean of X 1 6(ii) P(X = 6) = q5 p with p = 1/3, q = 2/3 (AEF) = 32/729 or 0⋅0439 B1 Find probability of score of 3 or 4 on exactly 6 throws 1 6(iii) P(X > 4) = q4 = 16/81 or 0⋅1975 or 0⋅198 M1 A1 Find probability of score of 3 or 4 on more than 4 throws 2 6(iv) 1 – qn – 1 < 0⋅95 (AEF) M1 Formulate condition for n (1 – qn is M0) 0⋅05 < (2/3) n – 1 , n – 1 < log 0⋅05 / log 2/3 M1 Set q = 2/3, rearrange and take logs (any base) to give bound n – 1 < 7⋅39, nmax = 8 A1 Find nmax (> or = can earn M1 M1 A0, max 2/3) 3

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 12 of 17 Question Answer Marks Guidance 7(i) F(x) = ∫ f(x) dx = – 3/(4x) + x/4 [+ c] M1 Find or state distribution function F(x) for 1 ≤x≤3 = – 3/(4x) + x/4 + ½ or ¼ (– 3/x + x + 2) A1 using F(1) = 0 or F(3) = 1 to find c if necessary F(x) = 0 (x < or ≤1), F(x) = 1 (x > or≥3) A1 State F(x) for other values of x 3 7(ii) ∫1Q f(x) dx = – 3/4Q + Q/4 + ½ = ¼ [or ¾] (AEF) M1 Formulate equation for either quartile value Q Q2 + [or – ] Q – 3 = 0 A1 Q1 = ½ (– 1 + √13), Q3 = ½ (1 + √13) A1 A1 Find lower quartile Q1 and upper quartile Q3 Q3 – Q1 = 1 A1 Find interquartile range 5

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 13 of 17 Question Answer Marks Guidance 8 H0: µb – µe = 0⋅05, H1: µb – µe > 0⋅05 (AEF) B1 State both hypotheses (B0 for x … ) di: 0⋅10 0⋅07 0⋅03 0⋅03 0⋅21 0⋅19 (or in sec) M1 Consider differences di from e.g. xb – ye d = 0⋅63 / 6 = 0⋅105 (or 6⋅3 sec) B1 Find sample mean s2 = (0⋅0969 – 0⋅632/6) / 5 [ = 123/20 000 or 0⋅00615 or 0⋅07842 ] (or 22⋅14) M1 Estimate population variance (allow biased here: [41/8000 or 0⋅005125 or 0⋅07162 ]) t5, 0.9 = 1⋅476 (to 3 sf) B1 State or use correct tabular t-value t = ( d – 0⋅05) / (s/√6) = 1⋅72 M1 A1 Find value of t (or compare − d 0⋅05 = 0⋅055 with (t5, 0.9)s/√6 = 0⋅0473) [Reject H0:] Evidence for organiser’s belief or times improve by more than 0⋅05 min (AEF) B1 FT on both t-values Consistent conclusion SC Wrong type of hypothesis test can earn only B1 for hypotheses B1FT for conclusion (max 2/8) 8

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 14 of 17 Question Answer Marks Guidance 9 (i) E3 = (3/16) ∫1.62.4 (4 – x)1/2 dx = (3/16) [ – (2/3) (4 – x)3/2 ] 1.62.4 M1 State or imply expression for required expected value E3 of X = (2⋅43/2 – 1⋅63/2) / 8 = 1⋅694/8 or 0⋅2118 A1 Find expected value E3 (may be implied in finding 50 E3) (M1 A1 requires adequate explicit working) 50 E3 = 10⋅59 AG A1 Hence verify corresponding expected frequency 3 9(ii) H0: Distribution fits data (AEF) B1 State (at least) null hypothesis in full Oi 18 16 8 8 Ei: 14⋅22 12⋅54 10⋅59 12⋅65 X2 = 1⋅005 + 0⋅955 + 0⋅633 + 1⋅709 = 4⋅30 M1 Combine values consistent with all exp. values ≥ 5 M1 A1 Find value of X2 from Σ (Ei – Oi)2 / Ei [or Σ Oi2/Ei – n ] No. n of cells: 5 4 3 χn-1, 0.95 2: 9⋅488 7⋅815 5⋅991 (to 3 s.f.) B1 FT on number, n, of cells used to find X2 State or use consistent tabular value χn-1, 0.95 2 Accept H0 if X2 < tabular value (AEF) M1 State or imply valid method for conclusion 4⋅30 [± 0⋅01] < 7⋅81[5] so distribution fits [data] or distribution is a suitable model (AEF) A1 Conclusion (requires both values approx. correct) 7

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 15 of 17 Question Answer Marks Guidance 10(i) Σ x = 25, Σ y = 25 + q, Σ xy = 135 + 4q, Σ x2 = 141, Σ y2 = 159 + q2 B1 Find required values Sxy = 135 + 4q – 25(25 + q)/5 = [10 – q] (AEF) [ Sxx = 141 – 252/5 = 16 ] Syy = (159 + q2) – (25 + q)2/5 [= 34 – 10q + 4q2/5 ] M1 A1 (Sxy , Sxx , Syy may be scaled by the same constant) 40 – 4q = 170 – 50q + 4q2, 2 q2 – 23q + 65 = 0 (2q – 13)(q – 5) = 0, q = 5 M1 A1 Equate gradient 5/4 in line of x on y to Sxy / Syy and solve quadratic to find integer value of q 5 10(ii) c = 25/5 – (5/4)(25 + q)/5 = – (5 + q) / 4 M1 A1 Find c from x – (5/4) y = – 5/2 or – 2⋅5 A1 3 10(iii) r = Sxy / √(Sxx Syy) = 5 / √(16 × 4) M1 A1 Find correlation coefficient r = 5/8 or 0⋅625 A1 3

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 16 of 17 Question Answer Marks Guidance 11E(i) ½muP2 = ½m (21ag/2) + mga [uP2 = (25/2) ag] M1 A1 uP is speed of P at lowest point, vQ is speed of Q immediately after collision. Apply conservation of energy at lowest point (A0 if no m) ½ 4mvQ2 = 4mag M1 Find speed vQ at lowest point by conservation of energy (A0 if no m) vQ = √(2ag) or 1⋅41√(ag) or 4⋅47√a A1 muP = [±] mvP + 4mvQ M1 Find vP using conservation of momentum (m may be omitted) vP = [±] (– 5/√2 + 4√2) √(ag) A1 vP = (3/√2) √(ag) or 2⋅12 √(ag) or 0⋅671√a (AEF) A1 Hence find speed of P 7 11E(ii) VP is speed of P when it loses contact ½mVP2 = ½mvP2 – mga (1 + cos α) [VP2 = (9/2)ag – 2ga (1 + cos α) = (5/2 – 2 cos α) ag] M1 A1 Apply conservation of energy at D (A0 if no m) [RD =] mVP2/a – mg cos α = 0 [VP2 = ag cos α] M1 A1 Apply F = ma radially at D with reaction = 0 (5/2 – 2 cos α) ag = ag cos α , cos α = 5/6 or 0⋅833 A1 Combine to find cos α 5

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 17 of 17 Question Answer Marks Guidance 11O(i) tsA / √8 = ½ (16⋅7 – 13⋅5) [= 1⋅6] M1 Relate sA to semi-width of confidence interval t7, 0.975 = 2⋅365 (to 3 s.f.) A1 State or use correct tabular t value [sA = √8 × 1⋅6 / 2⋅365 = 1⋅9135], sA 2 = 3⋅66[16] A1 Hence find unbiased estimate of A’s population variance 3 11O(ii) H0: µA = µB , H1: µA >µB (AEF) B1 State hypotheses (B0 for x …) [ 15.1], 85.2 / 6 14.2 = = = A B x x B1 Find sample mean for B sB2 = (1221⋅06 – 85⋅22/6) / 5 = 561/250 or 2⋅244 or 1⋅4982 (all to 3 s.f.) M1 Estimate or imply population variance for B (allow biased here: 1⋅87 or 1⋅3672) s2 = (7 sA2 + 5 sB2) / 12 = 3⋅0709 or 1⋅7522 M1 A1 Estimate (pooled) common variance (sB2 not needed explicitly) t12, 0.95 = 1⋅782 B1 State or use correct tabular t value [-] t = ( A x – B x ) / (s √(1/8 + 1/6)) = 0⋅951 t < 1⋅78 so [accept H0] M1 A1 Find value of t (or can compare A x – B x = 0⋅9 with 1⋅69) Correct conclusion mean mass of B not less than mean mass of A (AEF) B1 9 SC1: Implicitly taking sA2, sB2 as unequal population variances (may also earn first B1 B1 M1) z = ( A x – B x ) / √(sA2/8 + sB2/6) = 0⋅9 / √(0⋅8317) = 0⋅987 z < 1⋅645 so DepSC1: mean mass of B not less than mean mass of A (AEF) Comparison with z0.95 and conclusion (FT on z) (can earn at most 5/9)