Cambridge A Level Mathematics - Further 9231 — 2018 May/June Paper 2 · Variant 3

*4173383933* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/23 Paper 2 May/June 2018 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 22 printed pages and 2 blank pages. JC18 06_9231_23/2R © UCLES 2018 [Turn over

2 1 A particle P is moving in a fixed circle of radius 0.8 m. At time t s its velocity is t2 −t + 2 m s−1. Find the magnitudes of the radial and the transverse components of the acceleration of P when t = 2. [3] Radial component … … … … … Transverse component … … … … … 2 Two uniform small spheres A and B have equal radii and masses 4m and m respectively. Sphere A is moving with speed u on a smooth horizontal surface when it collides directly with sphere B which is at rest. The coefficient of restitution between the spheres is e. (i) Show that after the collision A moves with speed 1 5u4 −e and find the speed of B. [4] … … … … … … … … … … © UCLES 2018 9231/23/M/J/18

3 Sphere B continues to move until it collides with a fixed smooth vertical barrier which is perpendicular to the direction of motion of B. The coefficient of restitution between B and the barrier is 3 4e. After this collision, the speeds of A and B are equal. (ii) Find the value of e. [3] … … … … … … … … … … … … … The spheres A and B now collide directly again. (iii) Determine whether sphere B collides with the barrier for a second time. [2] … … … … … … … … © UCLES 2018 9231/23/M/J/18 [Turn over

4 3 A particle P moves on the positive x-axis in simple harmonic motion. The centre of the motion is a distance d m from the origin O, where 0 < d < 6.5. The points A and B are on the positive x-axis, with OA = 6.5 m and OB = 7.5 m. The magnitude of the acceleration of P when it is at B is twice the magnitude of the acceleration of P when it is at A. (i) Find d. [2] … … … … … … … … The period of the motion is 0 s and the maximum acceleration of P during the motion is 10 m s−2. (ii) Find the speed of P when it is 7 m from O. [5] … … … … … … … … … … … … © UCLES 2018 9231/23/M/J/18

5 … … … … … … … … … (iii) Find the time taken by P to travel directly from A to B. [3] … … … … … … … … … … … … … … … © UCLES 2018 9231/23/M/J/18 [Turn over

6 4 A uniform rod AB has length 2a and weight W. The end A rests on rough horizontal ground and the end B rests against a smooth vertical wall. The angle between the rod and the horizontal is 1, where tan 1 = 4 3. One end of a light inextensible rope is attached to a point C on the rod. The other end is attached to a point where the vertical wall and the horizontal ground meet. The rope is taut and perpendicular to the rod. The rope and rod are in a vertical plane perpendicular to the wall. (i) Show that AC = 18 25a. [2] … … … … … … … … … … The magnitude of the frictional force at A is equal to one quarter of the magnitude of the normal reaction force at A. (ii) Show that the tension in the rope is 1 4W. [6] … … … … … … … … … © UCLES 2018 9231/23/M/J/18

7 … … … … … … … … … … … … … … … (iii) Find expressions, in terms of W, for the magnitudes of the normal reaction forces at A and B. [2] … … … … … … … … … © UCLES 2018 9231/23/M/J/18 [Turn over

8 5 P O Q 12ag a A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle is held with the string taut and horizontal. It is projected downwards with speed 12ag. At the lowest point of its motion, P collides directly with a particle Q of mass km which is at rest (see diagram). In the collision, P and Q coalesce. The tension in the string immediately after the collision is half of its value immediately before the collision. Find the possible values of k. [11] … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/23/M/J/18

9 … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/23/M/J/18 [Turn over

10 6 A random sample of 15 observations of pairs of values of two variables gives a product moment correlation coefficient of 0.430. (i) Test at the 10% significance level whether there is evidence of non-zero correlation between the variables. [4] … … … … … … … … A second random sample of N observations gives a product moment correlation coefficient of 0.615. Using a 5% significance level, there is evidence of positive correlation between the variables. (ii) Find the least possible value of N, justifying your answer. [2] … … … … … 7 The probability that a driver passes an advanced driving test has a fixed value p for each attempt. A driver keeps taking the test until he passes. The random variable X denotes the number of attempts required for the driver to pass. The variance of X is 3.75. (i) Show that 15p2 + 4p −4 = 0 and hence find the value of p. [4] … … … … © UCLES 2018 9231/23/M/J/18

11 … … … … … … … … … (ii) Find PX = 5. [1] … … … … … (iii) Find P3 ≤X ≤7. [2] … … … … … … … … … © UCLES 2018 9231/23/M/J/18 [Turn over

12 8 For a random sample of 6 observations of pairs of values x, y, the equation of the regression line of y on x is y = bx + 1.306, where b is a constant. The corresponding equation of the regression line of x on y is x = 0.6331y + d, where d is a constant. The values of x from the sample are 2.3 2.8 3.7 p 6.1 6.4 and the sum of the values of y is 46.5. The product moment correlation coefficient is 0.9797. (i) Find the value of b correct to 3 decimal places. [2] … … … … … … … … … (ii) Find the value of p. [4] … … … … … … … … … … … © UCLES 2018 9231/23/M/J/18

13 … … … … … … … … … … (iii) Use the equation of the regression line of x on y to estimate the value of x when y = 8.5. [3] … … … … … … … … … … … … … … © UCLES 2018 9231/23/M/J/18 [Turn over

14 9 The continuous random variable X has probability density function given by fx = t 1 20 @ 3 −1 x A 1 ≤x ≤9, 0 otherwise. The random variable Y is defined by Y = ïX. (i) Show that the probability density function of Y is given by gy = T 1 103y −1 1 ≤y ≤3, 0 otherwise. [7] … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/23/M/J/18

15 … … … … … … … … … … … … … … … (ii) Find the mean value of Y. [2] … … … … … … … … … © UCLES 2018 9231/23/M/J/18 [Turn over

16 10 During the summer months, all members of a large swimming club take part in intensive training. The times taken to swim 50 metres at the beginning of the summer and at the end of the summer are recorded for each member of the club. The time taken, in seconds, at the beginning of the summer is denoted by x and the time taken at the end of the summer is denoted by y. For a random sample of 9 members the results are shown in the following table. Member A B C D E F G H I x 38.5 40.2 32.3 35.1 36.2 41.4 32.0 38.2 38.2 y 37.4 38.1 31.6 34.7 34.2 38.6 31.8 36.3 36.8 The swimming coach believes that, on average, the time taken by a swimmer to swim 50 metres will decrease by more than one second as a result of the intensive training. (i) Stating suitable hypotheses and assuming a normal distribution, test the coach’s belief at the 10% significance level. [8] … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/23/M/J/18

17 … … … … … … … … … … … … (ii) Find a 95% confidence interval for the population mean time taken to swim 50 metres after the intensive training, assuming a normal distribution. [4] … … … … … … … … … … … … © UCLES 2018 9231/23/M/J/18 [Turn over

18 11 Answer only one of the following two alternatives. EITHER A B C D a 2a An object is formed from a square frame ABCD with a square lamina attached in one corner of the frame. The frame consists of four identical thin rods, each of mass M and length 2a. The lamina has mass kM and edges of length a. It has one vertex at C and adjacent sides in contact with CB and CD (see diagram). (i) Show that the moment of inertia of the object about an axis l through A perpendicular to the plane of the object is 2 3Ma27k + 20. [8] … … … … … … … … … … … … … © UCLES 2018 9231/23/M/J/18

19 The object is released from rest with the edge AB horizontal and D vertically above A. The object rotates freely about the fixed axis l. The angular speed of the object is 1 2 O@5g a A when D is first vertically below A. (ii) Find the value of k. [6] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/23/M/J/18 [Turn over

20 OR A scientist carries out an experiment to investigate the quantity X, which takes the values 0, 1, 2, 3, 4, 5 or 6. He believes that the values taken by X follow a binomial distribution. He conducts 250 trials. His results are summarised in the following table. x 0 1 2 3 4 5 6 Observed frequency 22 83 72 53 17 3 0 (i) Show that unbiased estimates of the mean and variance for these results are 1.876 and 1.266 respectively, correct to 3 decimal places. By evaluating the mean and variance of the distribution B6, 0.313, explain why X could have this distribution. [4] … … … … … … … … The expected frequencies corresponding to the distribution B6, 0.313 are shown in the following table. x 0 1 2 3 4 5 6 Observed frequency 22 83 72 53 17 3 0 Expected frequency 26.3 71.9 81.8 49.7 17.0 3.1 0.2 (ii) Show how the expected frequency for x = 4 is calculated. [2] … … … … … … © UCLES 2018 9231/23/M/J/18

21 (iii) Test at the 5% significance level whether the scientist’s belief is correct. [8] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/23/M/J/18 [Turn over

22 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 2018 9231/23/M/J/18

23 BLANK PAGE © UCLES 2018 9231/23/M/J/18

24 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 International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. © UCLES 2018 9231/23/M/J/18

IGCSE™ is a registered trademark. This document consists of 16 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 May/June 2018 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 2018 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 2 of 16 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.

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 3 of 16 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.

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 4 of 16 Mark Scheme Notes Marks are of the following three types: 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. • When a part of a question has two or more ‘method’ steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. • The symbol FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously ‘correct’ answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. • Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 5 of 16 The following abbreviations may be used in a mark scheme or used on the scripts: 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 SR Special Ruling (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) Penalties MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become ‘follow through’ marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR –2 penalty may be applied in particular cases if agreed at the coordination meeting. PA –1 This is deducted from A or B marks in the case of premature approximation. The PA –1 penalty is usually discussed at the meeting.

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 6 of 16 Question Answer Marks Guidance 1 aR = (22 – 2 + 2) 2 / 0⋅8 = 42 / 0⋅8 = 20 [m s–2] M1 A1 Find radial acceleration aR at t = 2 from v2/r aT = 2t – 1 = 3 [m s–2] B1 Find transverse acceleration aT at t = 2 by differentiation 3 Question Answer Marks Guidance 2(i) 4mvA + mvB = 4mu (AEF) M1 Use momentum (allow m omitted) vB – vA = eu M1 Use Newton’s law (M0 if LHS signs inconsistent) vA = (1/5) (4 – e) u AG A1 Combine to verify/find speeds of A and B after collision vB = (4/5) (1 + e) u A1 4 2(ii) vB′ = [–] ¾ e vB [= – (3/5) e (1 + e) u] M1 Relate vel. vB′ of B after colln. with wall to vB (1/5) (4 – e) u = (3/5) e (1 + e) u M1 Equate speeds of A and B (ignore sign of vB′ for both M1s) 3e2 + 4e – 4 = 0, e = 2/3 A1 Solve resulting quadratic for e, [implicitly] rejecting root –2 3 2(iii) EITHER: 4mwA + mwB = 3mvA [= 2mu] (AEF) wB – wA = e × 2vA = (4/3)vA or (8/9) u (AEF) [wA = ⅓ vA = (2/9) u], wB = (5/3)vA or (10/9) u OR: Momentum before colln. is 3mvA = 2mu > 0, so after colln. momentum or speed of B > 0 wB > 0 so B collides again with barrier M1 A1 Use momentum (allow m omitted) with vA = –vB′ [= ⅔ u] Use Newton’s law (M0 if LHS signs inconsistent) Combine to find velocity of B after final collision Allow any similar valid argument Allow vA > 0 by [implicit] inspection 2

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 7 of 16 Question Answer Marks Guidance 3(i) ω 2 (7⋅5 – d ) = 2 ω 2 (6⋅5 – d ), d = 5⋅5 M1 A1 Find d by relating accelns. at A and B (ω 2 may be omitted) 2 3(ii) ω = 2π/T = 2 B1 Find ω from period T (may be implied) a = 10/ω 2 = 2⋅5 [m] (ignore sign; FT on ω ) B1√ Find amplitude a from max. acceleration 10 v 2 = ω 2 (a2 – x2), x = 7 – d [= 1⋅5] M1, M1 Find speed v when OP = 7 (2nd M1 dep. on 1st M1) = 22 (2⋅52 – 1⋅52) = 16, v = 4 [m s–1] A1 5 3(iii) ω –1 sin–1 (xB / a) – ω –1 sin–1 (xA / a) or ω –1 cos–1 (xA / a) – ω –1 cos–1 (xB / a) M1 Find time from A to B from x = a sin ωt or a cos ωt (all terms must be correct ( FT on a, ω ) for M1) = ½ sin–1 (2/2⋅5) – ½ sin–1 (1/2⋅5) or ½ cos–1 (1/2⋅5) – ½ cos–1 (2/2⋅5) = ½ (0⋅9273 – 0⋅4115) or 0⋅4636 – 0⋅2058 or ½ (1⋅1593 – 0⋅6435) or 0⋅5796 – 0⋅3218 A1 To 3 d.p., AEF throughout = 0⋅258 [s] A1 3

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 8 of 16 Question Answer Marks Guidance 4(i) AC = AD cos θ and AD = 2 a cos θ [= 6a/5] so AC = 2 a cos2 θ = 2 a (3/5) 2 = 18a/25 AG M1 A1 Find AC (D denotes other end of rope from C) 2 4(ii) A: RB × 2a sin θ – W × a cos θ – T × AC = 0 [ RB × 8a/5 – W × 3a/5 – T × 18a/25 = 0 so 40 RB – 15 W – 18 T = 0 ] B: FA × 2a sin θ – RA × 2a cos θ + W × a cos θ + T × (2a – AC) = 0 [ FA × 8a/5 – RA × 6a/5 + W × 3a/5 + T × 32a/25 = 0 so 15 W + 32 T = 30 RA – 40 FA = 20 RA] C: FA × AC sin θ – RA × AC cos θ + RB × (2a – AC) sin θ – W × (a – AC) cos θ = 0 [ FA × 72a/125 – RA × 54a/125 + RB × 128a/125 – W × 21a/125 = 0 so 128 RB – 21 W = 54 RA – 72 FA = 36 RA ] D: RA × 2a cos θ – RB × 2a sin θ – W × a cos θ = 0 [ RA × 6a/5 – RB × 8a/5 – W × 3a/5 = 0 so 6 RA – 8 RB – 3W = 0 ] G: FA × a sin θ – RA × a cos θ + RB × a sin θ + T × (a – AC) = 0 [FA × 4a/5 – RA × 3a/5 + RB × 4a/5 + T × 7a/25 = 0 so 20 FA – 15 RA + 20 RB + 7 T = 0 ] M1 A1 Take moments for rod about one chosen point (Note that a vertical resolution will then give T, earning 6/6) (G is mid-point of AB) Horizontally: RB – FA = T sin θ [= 4T/5] B1 Find two more indep. eqns, e.g. resolution of forces on rod Vertically: RA – W = T cos θ [= 3T/5] B1 (a second moment eqn. may be used) T = W/4 AG M1 A1 Find or verify T using FA = ¼ RA , sin θ = 4/5 , cos θ = 3/5 6

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 9 of 16 Question Answer Marks Guidance 4(iii) RA = 23W / 20 or 1⋅15 W [FA = 23W / 80 = 0⋅2875 W ] B1 Find RA, RB (can assume T = W/4) RB = 39W / 80 or 0⋅487[5] W B1 2 Question Answer Marks Guidance 5 ½mv1 2 = ½mu2 + mga M1 Find v1 2 at lowest point from consvn. of energy (M0 if no m) v1 2 = (12 + 2) ag = 14 ag A1 Mv2 = mv1 with M = (1 + k) m M1 Find new v2 from consvn. of momentum v2 = v1 / (1 + k) [= √(14 ag) / (1 + k) ] A1 T1 = mv1 2/a + mg = (14 + 1) mg = 15mg M1 A1 Find tension T1 just before collision by using F = ma radially T2 = Mv2 2/a + Mg = (1 + k) {14 / (1 + k)2 + 1} mg M1 Find tension T2 just after collision by using F = ma radially (M1 needs M, not m, throughout) or {14 / (1 + k) + (1 + k)} mg (AEF) A1 14 + (1 + k)2 = 15 (1 + k)/2 M1 Equate T2 and ½ T1 to give any eqn in k 2k2 – 11 k + 15 = 0, k = 2⋅5 or 3 (M1 dep) M1 A1 Solve resulting quadratic for k (M1 dep on all previous M1s, and requires quadratic eqn) 11

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 10 of 16 Question Answer Marks Guidance 6(i) H0: ρ = 0, H1: ρ ≠ 0 B1 State both hypotheses (B0 for r …) EITHER: r15, 10% = 0⋅441 *B1 State or use correct tabular two-tail r-value Accept H0 if 0⋅430 < tab. r-value (AEF) M1 State or imply valid method for conclusion OR: tr = r√((n–2) / (1 – r2)) = 1⋅72, t13,0⋅95 = 1⋅771 (*B1) (Rarely seen) Accept H0 if rt < tab. T-value (AEF) (M1) No [non-zero] correlation (AEF) A1 Correct conclusion (dep *B1) 4 6(ii) r8, 5% = 0⋅621, r9, 5% = 0⋅582 so Nmin = 9 (or equivalent argument) M1 A1 Find Nmin from relevant [one-tail] tabular values SC: Award B1 for stating 9 with inadequate justification 2

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 11 of 16 Question Answer Marks Guidance 7(i) (1 – p)/p2 = 3⋅75, 15p2 + 4p – 4 = 0 AG M1 A1 Find given eqn. for p using Var(X) = (1 – p)/p2 (5p – 2)(3p + 2) = 0, p = 2/5 or 0⋅4 M1 A1 Solve quadratic for p (A0 if p = – ⅔ not [implicitly] rejected) 4 7(ii) P(X = 5) = (1 – p)4 p = 0⋅64 × 0⋅4 = 0⋅0518 or 162/3125 B1 Find P(X = 5) 1 7(iii) EITHER:P(3 ⩽ X ⩽ 7) = (1 – p)2 – (1 – p)7 M1 Find P(3 ⩽ X ⩽ 7) = 0⋅62 – 0⋅67 = 0⋅36 – 0⋅028 = 0⋅332 A1 M0 for P(X ⩽ 7) – P(X ⩽ 3) [= 0⋅188] or similar error OR: P(3 ⩽ X ⩽7) = ∑i=3 7 (1 – p) i –1 p (M1) = (0⋅62 + 0⋅63 + 0⋅64 + 0⋅65 + 0⋅66) × 0⋅4 = 0⋅830016 × 0⋅4 = 0⋅332 (A1) 2

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 12 of 16 Question Answer Marks Guidance 8(i) b × 0⋅6331 = 0⋅97972, b = 1⋅516 (3 d.p.) M1 A1 Find b from given gradients and coefficient 2 8(ii) 46⋅5/6 [= 7⋅75] = b × x + 1⋅306 or b × (Σ x) /6 + 1⋅306 M1 Find p from means and regression line of y on x x = 4⋅25 or Σ x = 25⋅5 (3 s.f.) A1 (intermediate values may be implied) 6 x or Σ x = 21⋅3 + p, p = 4⋅2 (1 d.p.) M1 A1 4 8(iii) EITHER: (Σ x) /6 = 0⋅6331 (Σ y) /6 + d [d = –0⋅656] M1 Find d from means and regression line of x on y x = 0⋅6331 × 8⋅5 + d = 4⋅725[3] or 4⋅73 M1 A1 Estimate x when y = 8.5 using regression line of x on y OR: x – (Σ x) /6 = 0⋅6331 {8⋅5 – (Σ y) /6}, (M2) Combine above into single step x = 4⋅725[3] or 4⋅73 (A1) SC: x = (8⋅5 – 1⋅306) / b [= 4⋅7452] = 4⋅75 (B1) SC: Estimate x when y = 8.5 using regression line of y on x 3

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 13 of 16 Question Answer Marks Guidance 9(i) EITHER: F(x) = ∫ f(x) dx = (1/20) (3x – 2√x [+ c]) c = –1 so F(x) = (1/20) (3x – 2√x – 1) M1 Find or state distribution function F(x) for 1 ⩽ x ⩽ 9 (may be implied by G(y)) or (3/20) x – (1/10) √x – 1/20 G(y) [= P(Y < y) = P(√X < y) = P(X < y 2) ] A1 Find or state G(y) from Y = √X for 1 ⩽ x ⩽ 9 or 1 ⩽ y ⩽ 3 = F(y 2) = (1/20) (3 y 2 – 2 y – 1) M1 Allow A1√ as FT on expression found for F(x) or (3/20) y 2 – (1/10) y – 1/20 A2 Verify g(y) (differentiation may be implied) g(y) = G′(y) = (1/10) (3y – 1) AG [ for 1 ⩽ y ⩽ 3, g(y) = 0 otherwise ] M1 A1 SC Missing/incorrect c can earn M1 M1 A1√ M1 (max 4/7) 7 OR: Use of g(y) = f(x) × d d x / y (*M1) Reference to standard result required (not in syllabus) f(x) = (1/20) (3 – 1/y) (dep *M1) (M1 A1) Find f(x) using x = y 2 dx/dy = 2y (dep *M1) (M1 A1) Find dx/dy using x = y 2 g(y) = f(x) × dx/dy = (1/10) (3y – 1) AG [ for 1 ⩽ y ⩽ 3, g(y) = 0 otherwise ] (M1 A1) (7) 9(ii) E(Y) = (1/10) ∫(3y 2 – y) dy M1 Find mean of Y from ∫ y g( y) dy = (1/10) [ y 3 – ½ y 2]1 3 = 11/5 or 2⋅2 A1 2

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 14 of 16 Question Answer Marks Guidance 10(i) H0: µx – µy = 1, H1: µx – µy > 1 (AEF) B1 State both hypotheses (B0 for x …) di: 1⋅1 2⋅1 0⋅7 0⋅4 2⋅0 2⋅8 0⋅2 1⋅9 1⋅4 M1 Consider differences di, e.g. xi – yi d = 12⋅6 / 9 = 1⋅4 B1 Find sample mean s2 = (23⋅72 – 12⋅62/9) / 8 [ = 0⋅76 or 0⋅87182 ] M1 Estimate population variance (allow biased here: [ 0⋅6756 or 0⋅82192 ]) t8, 0.9 = 1⋅397 or 1⋅40 B1 State or use correct tabular t-value t = ( d – 1) / (s/√9) = 1⋅38 [Accept H0:] No evidence for coach’s belief M1 A1 Find t (or compare d – 1 = 0⋅4 with t8, 0.9 s /√9 = 0⋅406) Consistent conclusion (FT on both t-values) or time will not decrease by more than 1 s (AEF) B1√ SC Wrong (hypothesis) test can earn only B1 for hypotheses B1√ for conclusion (max 2/8) 8 10(ii) sy 2 = (11 395⋅79 – 319⋅52/9) / 8 = 6⋅693 or 2⋅5872 (to 3 sf) B1 Estimate population variance using y’s sample (allow use of biased here: 5⋅950 or 2⋅4392) 319⋅5/9 ± t √(sy 2 /9) M1 Find confidence interval (M0 if s2 not sy 2) t8, 0.975 = 2⋅306 or 2⋅31 A1 State or use correct tabular value of t 35⋅5 ± 1⋅99 or [33⋅5, 37⋅5] (allow 35⋅5 ± 2⋅0) A1 Evaluate C.I. (either form) SC B1 (max 1/4) for 1⋅4 ± 0⋅67 or [0⋅73, 2⋅07] 4

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 15 of 16 Question Answer Marks Guidance 11A(i) Ilamina = ⅓ kM ((½ a)2 + (½ a)2) [= (k/6) Ma2] B1 Find or state MI of lamina about axis at its centre I ′ lamina = Ilamina + kM ((3a /2)2 + (3a/2)2) [= (14k/3) Ma2] M1 A1 Find MI of lamina about axis at A EITHER: IAB = ⅓ Ma2 + Ma2 [= (4/3) Ma2 ] B1 Find or state MI of AB (or AD) about axis at A IBC = ⅓ Ma2 + M((a2 + (2a)2) [= (16/3) Ma2] M1 A1 Find or state MI of BC (or DC) about axis at A OR: Irod = ⅓ Ma2 + Ma2 [= (4/3) Ma2 ] (B1) Find or state MI of any rod about axis at centre of frame Iframe = 4 × Irod + 4M (a 2 + a 2) [= (40/3) Ma2] (M1 A1) Find or state MI of frame about axis at A I = (14k /3 + 2 × 4/3 + 2 × 16/3) Ma2 or (14k /3 + 40/3) Ma2 M1 Verify MI of system about axis at A = ⅔ Ma2 (7k + 20) AG A1 A0 if inadequate explanation 8 11A(ii) ½ I ω 2 = kMg × 3a + 2Mg × 2a + Mg × 4a or kMg × 3a + 4Mg × 2a M1 A2 Find ω 2 or angular speed ω when D below B by energy Award A1 if error in only term ω 2 = {3(3k + 8) / (7k + 20)} g/a 4 × 3(3k + 8) = 5(7k + 20), k = 4 A1 Find k by equating ω 2 to {½√(5g /a)}2 M1 A1 6

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 16 of 16 Question Answer Marks Guidance 11B(i) x = 469/250 = 1⋅876 and AG σ2 = (1195 – 4692/250) / 249 = 1⋅266 AG B1 Verify mean and unbiased variance of data, showing method B0 if biased variance, but allow other marks (max 3/4) 6 × 0⋅313 = 1⋅878 and 6 × 0⋅313 × 0⋅687 = 1⋅29[0] M1 A1 Find mean np and variance npq of binomial distribution Means and variances are similar for X and B(6, 0⋅313) A1 Valid explanation (AEF; needs 1⋅878, 1⋅290 correct to 0⋅002) 4 11B(ii) 250 × 6C4 × 0⋅3134 × 0⋅6872 [= 250 × 0⋅06795 = 16⋅99] Allow 49⋅7 × ( 6C4 × 0⋅313) / ( 6C3 × 0⋅687) M1 A1 Show how expected frequency for x = 4 is found (AEF for 6C4 = 15) 2 11B(iii) H0: [Binomial] distribution fits data (AEF) B1 State (at least) null hypothesis Oi : 22 83 72 53 20 Ei : 26⋅3 71⋅9 81⋅8 49⋅7 20⋅3 *M1 A1 Combine last 3 cells so that all exp. value ≥ 5 X 2 = 0⋅703 + 1⋅714 + 1⋅174 + 0⋅219 + 0⋅004 = 3⋅81 (to 3 s.f.) M1 A1 Find value of χ 2 from Σ (Ei – Oi)2 / Ei [or Σ Oi 2/Ei – n ] (A1 dep *M1) No. n of cells: 7 6 5 χn–2, 0.95 2: 11⋅07 9⋅488 7⋅815 B1√ State or use consistent tabular value χn–2, 0.95 2 (to 3 s.f.) [FT on number, n, of cells used to find X 2] Accept H0 if X 2 < tabular value M1 State or imply valid method for conclusion 3⋅81 [± 0⋅01] < 7⋅81[5] so distribution fits or scientist’s belief is correct A1 Conclusion (AEF; requires both values approx. correct) 8