Cambridge A Level Mathematics - Further 9231 — 2017 May/June Paper 2 · Variant 3

*2032562688* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/23 Paper 2 May/June 2017 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. 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. JC17 06_9231_23/RP © UCLES 2017 [Turn over

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3 1 3m O a A uniform disc with centre O, mass m and radius a is free to rotate without resistance in a vertical plane about a horizontal axis through O. One end of a light inextensible string is attached to the rim of the disc and wrapped around the rim. The other end of the string is attached to a block of mass 3m (see diagram). The system is released from rest with the block hanging vertically. While the block is in motion, it experiences a constant vertical resisting force of magnitude 0.9mg. Find the tension in the string in terms of m and g. [5] … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17 [Turn over

4 2 A particle P moves on a straight line in simple harmonic motion. The centre of the motion is O, and the amplitude of the motion is 2.5 m. The points L and M are on the line, on opposite sides of O, with OL = 1.5 m. The magnitudes of the accelerations of P at L and at M are in the ratio 3 : 4. (i) Find the distance OM. [2] … … … … The time taken by P to travel directly from L to M is 2 s. (ii) Find the period of the motion. [5] … … … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17

5 … … … … … … … … … … … … … … … (iii) Find the speed of P when it passes through L. [2] … … … … … … … … … © UCLES 2017 9231/23/M/J/17 [Turn over

6 3 Two uniform small smooth spheres A and B have equal radii and each has mass m. 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 2 3. Sphere B is initially at a distance d from a fixed smooth vertical wall which is perpendicular to the direction of motion of A. The coefficient of restitution between B and the wall is 1 3. (i) Show that the speed of B after its collision with the wall is 5 18u. [4] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17

7 (ii) Find the distance of B from the wall when it collides with A for the second time. [6] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17 [Turn over

8 4 1 21 O A B 5a 3a A uniform rod AB of length 3a and weight W is freely hinged to a fixed point at the end A. The end B is below the level of A and is attached to one end of a light elastic string of natural length 4a. The other end of the string is attached to a point O on a vertical wall. The horizontal distance between A and the wall is 5a. The string and the rod make angles 1 and 21 respectively with the horizontal (see diagram). The system is in equilibrium with the rod and the string in the same vertical plane. It is given that sin 1 = 3 5 and you may use the fact that cos 21 = 7 25. (i) Find the tension in the string in terms of W. [3] … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17

9 (ii) Find the modulus of elasticity of the string in terms of W. [4] … … … … … … … … … … … … (iii) Find the angle that the force acting on the rod at A makes with the horizontal. [3] … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17 [Turn over

10 5 P Q ! a u v O A particle 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 moving in complete vertical circles with the string taut. When the particle is at the point P, where OP makes an angle ! with the upward vertical through O, its speed is u. When the particle is at the point Q, where angle QOP = 90Å, its speed is v (see diagram). It is given that cos ! = 4 5. (i) Show that v2 = u2 + 14 5 ag. [2] … … … … … … The tension in the string when the particle is at Q is twice the tension in the string when the particle is at P. (ii) Obtain another equation relating u2, v2, a and g, and hence find u in terms of a and g. [5] … … … … … … … © UCLES 2017 9231/23/M/J/17

11 … … … … … … … … … … … … (iii) Find the least tension in the string during the motion. [3] … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17 [Turn over

12 6 The independent variables X and Y have distributions with the same variance 32. Random samples of N observations of X and 2N observations of Y are taken, and the results are summarised by Σx = 4, Σ x2 = 10, Σ y = 8, Σy2 = 102. These data give a pooled estimate of 10 for 32. Find N. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17

13 7 A random sample of twelve pairs of values of x and y is taken from a bivariate distribution. The equations of the regression lines of y on x and of x on y are respectively y = 0.46x + 1.62 and x = 0.93y + 8.24. (i) Find the value of the product moment correlation coefficient for this sample. [2] … … … … … … … (ii) Using a 5% significance level, test whether there is non-zero correlation between the variables. [4] … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17 [Turn over

14 8 The number, x, of beech trees was counted in each of 50 randomly chosen regions of equal size in beech forests in country A. The number, y, of beech trees was counted in each of 40 randomly chosen regions of the same equal size in beech forests in country B. The results are summarised as follows. Σ x = 1416 Σx2 = 41 100 Σy = 888 Σ y2 = 20 140 Find a 95% confidence interval for the difference between the mean number of beech trees in regions of this size in country A and in country B. [9] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17

15 … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17 [Turn over

16 9 The continuous random variable X has probability density function f given by fx = T 0 x < 0, ae−x ln2 x ≥0, where a is a positive constant. (i) Find the value of a. [2] … … … … (ii) State the value of EX. [1] … … (iii) Find the interquartile range of X. [4] … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17

17 The variable Y is related to X by Y = 2X. (iv) Find the probability density function of Y. [5] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17 [Turn over

18 10 Roberto owns a small hotel and offers accommodation to guests. Over a period of 100 nights, the numbers of rooms, x, that are occupied each night at Roberto’s hotel and the corresponding frequencies are shown in the following table. Number of rooms occupied (x) 0 1 2 3 4 5 6 ≥7 Number of nights 4 9 18 26 20 16 7 0 (i) Show that the mean number of rooms that are occupied each night is 3.25. [1] … … … The following table shows most of the corresponding expected frequencies, correct to 2 decimal places, using a Poisson distribution with mean 3.25. Number of rooms occupied (x) 0 1 2 3 4 5 6 ≥7 Observed frequency 4 9 18 26 20 16 7 0 Expected frequency 3.88 12.60 20.48 22.18 18.02 11.72 (ii) Show how the expected value of 22.18, for x = 3, is obtained and find the expected values for x = 6 and for x ≥7. [4] … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17

19 (iii) Use a goodness-of-fit test at the 5% significance level to determine whether the Poisson distribution is a suitable model for the number of rooms occupied each night at Roberto’s hotel. [7] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17 [Turn over

20 11 Answer only one of the following two alternatives. EITHER O A B a 3a The diagram shows a uniform thin rod AB of length 3a and mass 8m. The end A is rigidly attached to the surface of a sphere with centre O and radius a. The rod is perpendicular to the surface of the sphere. The sphere consists of two parts: an inner uniform solid sphere of mass 3 2m and radius a surrounded by a thin uniform spherical shell of mass m and also of radius a. The horizontal axis l is perpendicular to the rod and passes through the point C on the rod where AC = a. (i) Show that the moment of inertia of the object, consisting of rod, shell and inner sphere, about the axis l is 289 15 ma2. [6] … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17

21 The object is free to rotate about the axis l. The object is held so that CA makes an angle ! with the downward vertical and is released from rest. (ii) Given that cos ! = 1 6, find the greatest speed achieved by the centre of the sphere in the subsequent motion. [6] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17 [Turn over

22 OR The times taken to run 200 metres at the beginning of the year and at the end of the year are recorded for each member of a large athletics club. The time taken, in seconds, at the beginning of the year is denoted by x and the time taken, in seconds, at the end of the year is denoted by y. For a random sample of 8 members, the results are shown in the following table. Member A B C D E F G H x 24.2 23.8 22.8 25.1 24.5 24.0 23.8 22.8 y 23.9 23.6 22.8 24.5 24.2 23.5 23.6 22.7 [Σ x = 191, Σx2 = 4564.46, Σy = 188.8, Σy2 = 4458.4, Σ xy = 4510.99.] (i) Find, showing all necessary working, the equation of the regression line of y on x. [4] … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17

23 The athletics coach believes that, on average, the time taken by an athlete to run 200 metres decreases between the beginning and the end of the year by more than 0.2 seconds. (ii) Stating suitable hypotheses and assuming a normal distribution, test the coach’s belief at the 10% significance level. [8] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/23/M/J/17

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 2017 9231/23/M/J/17

® IGCSE is a registered trademark. This document consists of 14 printed pages. © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 May/June 2017 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 will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 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 2017 © UCLES 2017 Page 2 of 14 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 2017 © UCLES 2017 Page 3 of 14 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 2017 © UCLES 2017 Page 4 of 14 Question Answer Marks Guidance 1 EITHER: T × a = (½ ma2) d2θ / dt2or (½ ma) d2x / dt2 (B1 Find eqn of motion for disc 3mg – 0⋅9mg – T = 3 m × a d2θ / dt2 or 3m d2x / dt2 M1 A1) Find eqn of motion for block OR: Taθ = ½ (½ ma2) (dθ / dt)2 (B1 Find eqn. of energy for disc (in terms of θ or x) (3mg – 0⋅9mg – T) aθ = ½ 3 m (a dθ / dt)2 M1 A1) Find eqn of energy for block (in terms of θ or x) 6T = 2⋅1mg – T, T = 0⋅3 mg M1 A1 Combine two eqns to find T Total: 5 2(i) OL/OM = ¾, OM = 2 [m] M1 A1 Find OM by equating ratios of distances and acceln. Total: 2 2(ii) EITHER: ω tLM = sin–1 (OL/2⋅5) + sin–1(OM/2⋅5) or ω tOL = sin–1 (OL/2⋅5), ω (2 – tOL) = sin–1 OM/2⋅5) so 2ω = sin–1 0⋅6 + sin–1 0⋅8 [= 0⋅6435 + 0⋅9273] (*M1 A1) Find eqn. (AEF) for ω, for example using x = a sin ωt OR: ω tLM = cos–1 (–OM/2⋅5) – cos–1(OL/2⋅5) or ω tAL = cos–1 (OL/2⋅5), ω (2 + tAL) = cos–1 (–OM/2⋅5) so 2ω = cos–1 (–0⋅8) – cos–1 0⋅6 [= 2⋅498 – 0⋅927] or π – cos–1 0⋅8 – cos–1 0⋅6 (*M1 A1) or x = a cos ωt (A is at 2⋅5 from O, near L) ω = π / 4 or 0⋅785 (M1 dep *M1) DM1 A1 Simplify to find ω (may be implied by T) T = 2π / ω = 8 [s] B1 FT Find period T (FT on ω) Total: 5

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 5 of 14 Question Answer Marks Guidance 2(iii) vL = ω√(2⋅52 – 1⋅52) = 2ω = π / 2 or 1⋅57 [m s–1] M1 A1 Find speed vL at L Total: 2 3(i) mvA + mvB = mu (AEF) *M1 Use conservation of momentum (allow vA + vB = u) vB – vA = ⅔ u *M1 Use Newton’s restitution law (consistent LHS signs) vB = 5u/6 A1 Combine to find vB wB = ⅓ vB = 5u/18 AG B1 Verify speed wB of B after collision with wall (ignore sign) Total: 4

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 6 of 14 Question Answer Marks Guidance 3(ii) vA = u / 6 DA1 Find vA (dependent on above *M1 *M1) EITHER: (d – x) / vA = d / vB + x / wB (AEF) (M1 A1 EITHER: Equate times in terms of reqd. distance x 6(d – x) = 1⋅2 d + 3⋅6 x M1 A1) Substitute for speeds to formulate an eqn. in x OR: xA = (d/vB) vA = (6d/5u) u/6 = 0⋅2 d (M1 OR: Find dist. xA moved by A when B reaches wall t2 = (0⋅8 d) / (vA + wB) = 9d/5u M1 A1 Find remaining time t2 yA = vA t2 = 0⋅3 d or yB = wB t2 = 0⋅5 d A1) Find remaining distance moved by A or B OR2: xA = (d/vB) vA = (6d/5u) u/6 = 0⋅2 d (M1 OR2: Find dist. xA moved by A when B reaches wall (0⋅8 d – x) / vA = x / wB or 0⋅8 d/(vA + wB) = x/wB M1 A1 Equate remaining times to formulate an eqn. in x 4⋅8 d – 6 x = 3⋅6 x or 1⋅8 d = 3⋅6 x A1) x = ½ d A1 Find x Total: 6 4(i) T × 3a sin θ = W × 1⋅5a cos 2θ M1 A1 Take moments for rod about A T = 7W/30 or 0⋅233W A1 Find tension T Total: 3

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 7 of 14 Question Answer Marks Guidance 4(ii) OB = (5a – 3a cos 2θ) / cos θ = 26a/5 or 5⋅2a M1 A1 Find length OB of string T = λ (OB – 4a)/4a M1 Find modulus λ using Hooke’s Law = λ (6a/5)/4a = 3λ /10, λ = 7W/9 or 0⋅778W A1 Total: 4 4(iii) X = T cos θ [= 14W/75 or 0⋅187W] M1 Find horizontal component X of force at A Y = T sin θ + W [= 57W/50 or 1⋅14W] φ = tan–1 (Y/X) = tan–1 (171/28) M1 Find vertical component Y of force at A Find angle φ which force at A makes with horizontal = 80⋅7° or 1⋅41 radians B1 Total: 3

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 8 of 14 Question Answer Marks Guidance 5(i) ½mv2 = ½mu2 + mga (cos α + sin α) v2 = u2 + 2 ag (4/5 + 3/5) = u2 + 14 ag/5 AG M1 A1 Verify v by conservation of energy (A0 if no m) Total: 2 5(ii) TP = mu2/a – mg cos α B1 Find tension TP at P by using F = ma TQ = mv2/a + mg sin α B1 Find tension TQ at Q by using F = ma v2 = 2u2 – ag(2 cos α + sin α) = 2u 2 – 11ag/5 (AEF) M1 A1 Relate u2, v2 using TQ = 2TP (A0 if reqd. eqn omitted) u = √(5ag) or 2⋅24√(ag) [v2 = 39ag/5] A1 Eliminate v2 to find u Total: 5 5(iii) Tmin + mg = mV 2/a B1 Find tension Tmin at top from F = ma radially ½mV2 = ½mu 2 – mga (1 – cos α) or ½mv2 – mga (1 + sin α) M1 Find V 2 at top by conservation of energy (A0 if no m) [V2 = (5 – 2/5)ag = (39/5 – 16/5)ag = 23ag/5] Tmin = 23mg/5 – mg = 18mg/5 or 3⋅6mg A1 Combine to find Tmin Total: 3 6 (10 – 42/N + 102 – 82/2N) / (N + 2N – 2) (AEF) M1 A1 State or find expression for pooled estimate of σ2 (confusing biased/unbiased estimates may still earn M1) 112 – 48/N = 10 (3N – 2), 15N 2 – 66N + 24 = 0 M1 A1 Equate to 10 and rearrange as quadratic N = (66 ± 54) / 30 = 4 A1 Solve quadratic for N, rejecting root 0⋅4 Total: 5

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 9 of 14 Question Answer Marks Guidance 7(i) r = √(0⋅46 × 0⋅93) = 0⋅654 M1 A1 Find correlation coefficient r Total: 2 7(ii) H0: ρ = 0, H1: ρ ≠ 0 B1 State both hypotheses (B0 for r …) r12, 5% = 0⋅576 B1 State or use correct tabular two-tail r-value Accept H1 if |r| > tab. value (AEF) M1 State or imply valid method for conclusion (M0 if r or tab. value has magnitude > 1) 0⋅654 [or 0⋅65] > 0⋅576 so there is non-zero correlation A1 Correct conclusion (AEF) Total: 4 8 x = 28⋅32 and y = 22⋅2 B1 Find both sample means sA 2 = (41 100 – 14162/50) / 49 and sB 2 = (20 140 – 8882/40) / 39 M1 Estimate both population variances sA 2 = 20⋅39 and sB 2 = 164/15 or 10⋅93 (to 3 s.f.) A1 (allow biased here: 19⋅98 and 10⋅66) EITHER: s2 = sA 2/50 + sB 2/40 = 0⋅681 or 0⋅8252 (19⋅98/50 + 10⋅66/40 = 0⋅666 is M1 A0, max 8/9) (M1 A1 EITHER: Estimate combined variance (if biased values used wrongly here, giving zs = 1⋅60, only this A1 is lost) x – y ± z s M1 Find confidence interval for difference z0.975 = 1⋅96 A1 Use appropriate tabular value z s = 1⋅62 A1 Evaluate semi-interval length 6⋅12 ± 1⋅62 or [4⋅50, 7⋅74] A1) State confidence interval (in either form)

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 10 of 14 Question Answer Marks Guidance OR: Assume equal [population] variances s2 = (49 sA 2 + 39 sB 2) / 88 or (41 100 – 14162/50 + 20 140 – 8882/40) / 88 (B1 OR: State assumption Find pooled estimate of common variance (M1 A1 for sA 2 and sB 2 may be implied here) = 16⋅2 or 4⋅022 B1 x – y ± z s √(1/50 + 1/40) M1 Find confidence interval for difference z0.975 = 1⋅96 A1 Use appropriate tabular z-value (or appropriate t-value from calculator or interpolation) (t88, 0.975 = 1⋅9873 or 1⋅99) 1⋅67 (1⋅70) A1 Evaluate semi-interval length 6⋅12 ± 1⋅67 or [4⋅45, 7⋅79] (6⋅12 ± 1⋅70 or [4⋅42, 7⋅82]) A1) Evaluate confidence interval (in either form) Total: 9 9(i) (a / ln 2) [– e–x ln 2]0 ∞ = a / ln 2 so a = ln 2 or 0⋅693 M1 A1 State a or find a by equating ∫0 ∞ f(x) dx to 1 Total: 2 9(ii) E(X) = 1 / ln 2 or 1⋅44 B1 State or find E(X) Total: 1

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 11 of 14 Question Answer Marks Guidance 9(iii) F(Q) = 1 – e–Q ln 2 = ¼ or ¾ M1 Formulate equation for either quartile value Q Q1 = (ln 4/3) / (ln 2) [= 0⋅415 ] (AEF) A1 Find one [lower] quartile Q1 Q3 = (ln 4) / (ln 2) [= 2] (AEF) A1 Find other [upper] quartile Q3 Q3 – Q1 [= (ln 3) / (ln 2)] = 1⋅58 [or 1⋅59] A1 Find interquartile range (allow Q1 – Q3) Total: 4 9(iv) EITHER: G(y) = P(Y < y) = P(2X < y) = P(X < (ln y) / (ln 2)) = F((ln y) / (ln 2)) or F(log2 y) (AEF) (M1 A1 Find or state G(y) for x ⩾ 0 from Y = 2X (allow < or ≤ throughout) = 1 – e– ln y or 1 – 1/y A1) OR: Use x = (ln y) / (ln 2) to find both (M1 Find f(x) and dx/dy for use in g(y) = f(x) × dx/dy f(x) = (ln 2) e–x ln 2 = (ln 2) e –ln y = (1/y) ln 2 A1 and dx/dy = 1 / (y ln 2) A1) g(y) [= G′(y)] = 1/y 2 A1 Find g(y) in simplest form for y ⩾ 1 [g(y) = 0 otherwise] A1 State corresponding range of y for G(y) or g(y) Total: 5 10(i) x = (1/100) Σ x f(x) = 325/100 = 3⋅25 AG B1 Verify given mean (B0 forx = 325/100 = 3⋅25) Total: 1

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 12 of 14 Question Answer Marks Guidance 10(ii) E3 = 100 λ3 e–λ /3! with λ = 3⋅25 M1 A1 State expression for reqd. expected value E3 (M1 for E3 or E6) E6 = 100 λ6 e–λ /6! = 6⋅35 A1 Find exp. value E6 E ⩾7 = 100 – Σ0 6 Ei = 4⋅77 A1 Find exp. value E ⩾7 Total: 4 10(iii) H0: Distribution fits data (AEF) B1 State (at least) null hypothesis in full Oi: 13 18 26 20 16 7 Ei: 16⋅48 20⋅48 22⋅18 18⋅02 11⋅72 11⋅12 M1FT Combine values consistent with all exp. values ⩾ 5 (FT on E6 and E⩾7) χ2 = 0⋅735 + 0⋅300 + 0⋅658 + 0⋅218 + 1⋅563 + 1⋅527 M1 Find χ 2 = 5⋅00 A1 No. n of cells: 8 7 6 5 4 χn–2, 0.95 2: 12⋅59 11⋅07 9⋅488 7⋅815 5⋅991 B1FT State or use consistent tabular value χn-2, 0.95 2 (to 3 s.f.) [FT on number, n, of cells used to find χ2] Accept H0 if χ 2 < tabular value (AEF) 5⋅00 [± 0⋅1] < 9⋅49 so distn. fits [data] M1 State or imply valid method for conclusion Conclusion (requires both values correct) or distn. is a suitable model (AEF) A1 Total: 7

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 13 of 14 Question Answer Marks Guidance 11(a)(i) IAB = ⅓ 8m (3a/2)2 + 8m (a/2)2 or 8m (3a)2 /12 + 8m (a/2)2 [= 8 ma2] M1 A1 Find or state MI of rod AB about axis l Ishell = ⅔ ma2 + m (2a)2 [= (14/3) ma2] M1 Find MI of shell about axis l Isphere = (2/5) (3m/2)a2 + (3m/2) (2a)2 [= (33/5) ma2] M1 A1 Find MI of sphere about axis l I = (8 + 14/3 + 33/5) ma2 = (289/15) ma2 AG A1 Verify MI of object about axis l Total: 6 11(a)(ii) ½ I ω2 = (5mg/2) 2a (1 – cos α ) – 8mg (a/2) (1 – cos α ) or ((21/2)mg × 2a/21) (1 – cos α) M1 A1 Find ω2 or angular speed ω when CA vertical by energy = mga (1 – cos α ) = 5mga/6 A1 ω2 = 25g/289a or 0⋅0865 g/a or ω = (5/17) √(g/a) or 0⋅294 √(g/a) A1 vmax = 2aω = 2a √(25g/289a) = √(100ag/289) or (10/17)√(ag) M1 Find maximum speed vmax of O from rω or 0⋅588√(ag) or 1⋅86√a (AEF) A1 (A1 requires some simplification) Total: 6 11(b)(i) e.g. Sxy = 4510⋅99 – 191 × 188⋅8/8 = 3⋅39 or 0⋅424 Sxx = 4564⋅46 – 1912/8 = 4⋅335 or 0⋅542 b = Sxy / Sxx = 3⋅39/4⋅335 = 0⋅782 M1 A1 Find reqd. values (y – 188⋅8/8) = b (x – 191/8) (y – 23⋅6) = 0⋅782 (x – 23⋅875), y = 0⋅782x + 4⋅93 M1 A1 Find gradient b in y –y = b (x –x) and hence eqn. of regression line (may be implied by writing y = a + bx and finding a, b) Total: 4

9231/23 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 14 of 14 Question Answer Marks Guidance 11(b)(ii) H0: µx – µy = 0⋅2, H1: µx – µy > 0⋅2 (AEF) B1 State both hypotheses (B0 forx … ) di: 0⋅3 0⋅2 0 0⋅6 0⋅3 0⋅5 0⋅2 0⋅1 M1 Consider differences di, e.g. xi – yi d = 2⋅2 / 8 = 0⋅275 B1 Find sample mean s2 = (0⋅88 – 2⋅22/8) / 7 M1 Estimate population variance (allow biased here: [11/320 or 0⋅0344 or 0⋅1852 ]) [ = 11/280 or 0⋅0393 or 0⋅1982] t7, 0.9 = 1⋅41[5] B1 State or use correct tabular t-value t = ( d – 0⋅2) / (s/√8) = 1⋅07 M1 A1 Find value of t (or compare d – 0⋅2 = 0⋅075 with t7, 0.9 s/√8 = 0⋅099) [Accept H0:] No evidence for coach’s belief (AEF) B1 FT Consistent conclusion (FT on both t-values) SR Wrong (hypothesis) test can earn only B1 for hypotheses B1FT for conclusion (max 2/8) Total: 8