Cambridge A Level Mathematics - Further 9231 — 2019 Oct/Nov Paper 2 · Variant 1

*5777159437* Cambridge Assessment International Education Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/21 Paper 2 October/November 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 23 printed pages and 1 blank page. JC19 11_9231_21/RP R © UCLES 2019 [Turn over

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3 1 A particle P is moving in a circle of radius 2 m. At time t seconds, its velocity is t −12 m s−1. At a particular time T seconds, where T > 0, the magnitude of the radial component of the acceleration of P is 8 m s−2. Find the magnitude of the transverse component of the acceleration of P at this instant. [5] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19 [Turn over

4 2 E A B C D 4a 3a 1 A uniform square lamina ABCD of side 4a and weight W rests in a vertical plane with the edge AB inclined at an angle 1 to the horizontal, where tan 1 = 1 3. The vertex B is in contact with a rough horizontal surface for which the coefficient of friction is -. The lamina is supported by a smooth peg at the point E on AB, where BE = 3a (see diagram). (i) Find expressions in terms of W for the normal reaction forces at E and B. [5] … … … … … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19

5 … … … … … … … … … (ii) Given that the lamina is about to slip, find the value of -. [3] … … … … … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19 [Turn over

6 3 Three uniform small spheres A, B and C have equal radii and masses 5m, 5m and 3m respectively. The spheres are at rest on a smooth horizontal surface, in a straight line, with B between A and C. The coefficient of restitution between each pair of spheres is e. Sphere A is projected directly towards B with speed u. (i) Show that the speed of A after its collision with B is 1 2u1 −e and find the speed of B. [3] … … … … … … … … … … … … Sphere B now collides with sphere C. Subsequently there are no further collisions between any of the spheres. (ii) Find the set of possible values of e. [6] … … … … … … … … © UCLES 2019 9231/21/O/N/19

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8 4 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 and P is held with the string taut and horizontal. The particle P is projected vertically downwards with speed 2ag so that it begins to move along a circular path. The string becomes slack when OP makes an angle 1 with the upward vertical through O. (i) Show that cos 1 = 2 3. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19

9 (ii) Find the greatest height, above the horizontal through O, reached by P in its subsequent motion. [4] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19 [Turn over

10 5 L O A B C a a 3 2a 1 2a A thin uniform rod AB has mass ,M and length 2a. The end A of the rod is rigidly attached to the surface of a uniform hollow sphere (spherical shell) with centre O, mass 3M and radius a. The end B of the rod is rigidly attached to the surface of a uniform solid sphere with centre C, mass 5M and radius a. The rod lies along the line joining the centres of the spheres, so that CBAO is a straight line. The horizontal axis L is perpendicular to the rod and passes through the point of the rod that is a distance 1 2a from B (see diagram). The object consisting of the rod and the two spheres can rotate freely about L. (i) Show that the moment of inertia of the object about L is @408 + 7, 12 A Ma2. [6] … … … … … … … … … © UCLES 2019 9231/21/O/N/19

11 … … … … … … … The period of small oscillations of the object about L is 50 O@2a g A . (ii) Find the value of ,. [6] … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19 [Turn over

12 6 A random sample of 9 members is taken from the large number of members of a sports club, and their heights are measured. The heights of all the members of the club are assumed to be normally distributed. A 95% confidence interval for the population mean height, - metres, is calculated from the data as 1.65 ≤- ≤1.85. (i) Find an unbiased estimate for the population variance. [3] … … … … … … … … (ii) Denoting the height of a member of the club by x metres, find Σx2 for this sample of 9 members. [4] … … … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19

13 7 The time, T days, before an electrical component develops a fault has distribution function F given by Ft = T 1 −e−at t ≥0, 0 otherwise, where a is a positive constant. The mean value of T is 200. (i) Write down the value of a. [1] … … (ii) Find the probability that an electrical component of this type develops a fault in less than 150 days. [2] … … … … … A piece of equipment contains n of these components, which develop faults independently of each other. The probability that, after 150 days, at least one of the n components has not developed a fault is greater than 0.99. (iii) Find the smallest possible value of n. [4] … … … … … … … … … … © UCLES 2019 9231/21/O/N/19 [Turn over

14 8 A random sample of 8 elephants from region A is taken and their weights, x tonnes, are recorded. (1 tonne = 1000 kg.) The results are summarised as follows. Σx = 32.4 Σx2 = 131.82 A random sample of 10 elephants from region B is taken. Their weights give a sample mean of 3.78 tonnes and an unbiased variance estimate of 0.1555 tonnes2. The distributions of the weights of elephants in regions A and B are both assumed to be normal with the same population variance. Test at the 10% significance level whether the mean weight of elephants in region A is the same as the mean weight of elephants in region B. [9] … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19

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16 9 A random sample of five pairs of values of x and y is taken from a bivariate distribution. The values are shown in the following table, where p and q are constants. x 1 2 3 4 5 y 4 p q 2 1 The equation of the regression line of y on x is y = −0.5x + 3.5. (i) Find the values of p and q. [7] … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19

17 … … … … … … … … … … (ii) Find the value of the product moment correlation coefficient. [3] … … … … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19 [Turn over

18 10 The random variable X has probability density function f given by fx = d 1 30 @ 8 x2 + 3x2 −14 A 2 ≤x ≤4, 0 otherwise. (i) Find the distribution function of X. [3] … … … … … … … … … … … … The random variable Y is defined by Y = X2. (ii) Find the probability density function of Y. [4] … … … … … … … … © UCLES 2019 9231/21/O/N/19

19 … … … … … … … … … … (iii) Find the value of y such that PY < y = 0.8. [3] … … … … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19 [Turn over

20 11 Answer only one of the following two alternatives. EITHER The points A and B are a distance 1.2 m apart on a smooth horizontal surface. A particle P of mass 2 3 kg is attached to one end of a light spring of natural length 0.6 m and modulus of elasticity 10 N. The other end of the spring is attached to the point A. A second light spring, of natural length 0.4 m and modulus of elasticity 20 N, has one end attached to P and the other end attached to B. (i) Show that when P is in equilibrium AP = 0.75 m. [3] … … … … … … … … The particle P is displaced by 0.05 m from the equilibrium position towards A and then released from rest. (ii) Show that P performs simple harmonic motion and state the period of the motion. [6] … … … … … … … … … … © UCLES 2019 9231/21/O/N/19

21 … … … … … … … (iii) Find the speed of P when it passes through the equilibrium position. [2] … … … … … (iv) Find the speed of P when its acceleration is equal to half of its maximum value. [3] … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19 [Turn over

22 OR The number of puncture repairs carried out each week by a small repair shop is recorded over a period of 40 weeks. The results are shown in the following table. Number of repairs in a week 0 1 2 3 4 5 ≥6 Number of weeks 6 15 9 6 3 1 0 (i) Calculate the mean and variance for the number of repairs in a week and comment on the possible suitability of a Poisson distribution to model the data. [3] … … … … … … … Records over a longer period of time indicate that the mean number of repairs in a week is 1.6. The following table shows some of the expected frequencies, correct to 3 decimal places, for a period of 40 weeks using a Poisson distribution with mean 1.6. Number of repairs in a week 0 1 2 3 4 5 ≥6 Expected frequency 8.076 12.921 10.337 5.513 2.205 a b (ii) Show that a = 0.706 and find the value of the constant b. [3] … … … … … … … … © UCLES 2019 9231/21/O/N/19

23 … … … … … … (iii) Carry out a goodness of fit test of a Poisson distribution with mean 1.6, using a 10% significance level. [8] … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/21/O/N/19

24 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. … … … … … … … … … … … … … … … … … … … … … 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/O/N/19

This document consists of 16 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/21 Paper 2 October/November 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 October/November 2019 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 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. 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 October/November 2019 © UCLES 2019 Page 3 of 16 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. DM or DB 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 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. 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.

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 4 of 16 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 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) WWW Without Wrong Working AWRT Answer Which Rounds To

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 5 of 16 Question Answer Marks Guidance 1 (T – 1)4 / 2 = 8 M1 A1 Equate radial acceln. to 8 at t = T from v2/r T = 3 (or T – 1 = 2) A1 Hence find positive value of T (or of T – 1) aT = 2 (T – 1) = 4 [m s-2] M1 A1 Find magnitude of transverse acceleration at t = T 5 Question Answer Marks Guidance 2(i) RE × 3a = W cos θ × 2a – W sin θ × 2a or RE × 3a = W × 2a (1 – tan θ) cos θ or RE × 3a = 2 2 sin 4 W π θ   −     M1 A1 Take moments about B RE = 4W / 3√10 A1 Find normal reaction at E. AEF RB = W – RE cos θ = 3W/5 M1 A1 Find normal reaction at B by resolving forces vertically 5 2(ii) FB = RE sin θ = 2W/15 M1 A1 Find friction at B by resolving forces horizontally µ = (2/15) / (3/5) = 2/9 A1 Find µ from FB = µRB 3

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 6 of 16 Question Answer Marks Guidance 3(i) 5mvA + 5mvB = 5mu [vA + vB = u] and vB – vA = e u M1 Use consvn. of momentum for A and B and use Newton’s restitution law with consistent LHS signs. AEF vA = ½ (1 – e) u A1 Combine to verify speed of A. AG vB = ½ (1 + e) u A1 Find speed of B 3 3(ii) 5mvB′ + 3mvC = 5mvB [5vB′ + 3vC = 5vB] vC – vB′ = evB M1 Use consvn. of momentum for B and C and use Newton’s restitution law with consistent LHS signs. AEF vB′ = (1/8) (5 – 3e) vB [vC = (1/8) (5 + 5e) vB ] A1 Combine to find vB′ (vC not reqd as B, C cannot collide again) 1 2 (1 – e) u ⩽ (1/8) (5 – 3e) × 1 2 (1 + e) u M1 Find condition on e using vA ⩽ vB′ 3e2 – 10e + 3 ⩽ 0 A1 Simplify to a quadratic inequality 1 3 ⩽ e A1 Solve to give a lower bound on e 1 3 ⩽ e ⩽ 1 A1 Non-strict inequality 6

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 7 of 16 Question Answer Marks Guidance 4(i) 1 2 mv2 = 1 2 mu2 – mga cos θ M1 Use conservation of energy to slack point P1 mv2/a – mg cos θ = 0 M1 A1 Equate tension at P1 to 0 by using F = ma A1 if both eqns correct, with m included. AG v2 = 2ag – 2ag cos θ = ag cos θ M1 Combine to verify cos θ using u = √(2ag) cos θ = 2/3 A1 5 4(ii) vV = v sin θ = √(2ag/3) (√5/3) or vV2 = (10/27) ag M1 Find vertical speed vV at P1 h = vV2/2g = (5/27) a or 0⋅185 a M1 A1 Find height risen above P1 by considering vertical motion h + a cos θ = (23/27) a or 0⋅852 a A1 Find total height risen above level of O 4

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 8 of 16 Question Answer Marks Guidance 5(i) Irod = 1 3 λMa2 + λM(a/2)2 [= (7/12) λMa2] B1 Find or state MI of rod AB about axis L IO = 2 3 3M a2 + 3M (5a/2)2 [= (83/4) Ma2] M1 A1 Find MI of hollow sphere centre O about axis L IC = (2/5) 5Ma2 + 5M (3a/2)2 [= (53/4) Ma2] M1 A1 Find MI of solid sphere centre C about axis L I = (7λ/12 + 83/4 + 53/4) Ma2 I = ((7λ + 408) / 12) Ma2 A1 Verify MI of object about axis L. AG 6 5(ii) [–] I d2θ/dt2 = [– 3Mg × (5a/2) sin θ + 5Mg × (3a/2) sin θ] – λMg × (a/2) sin θ M1 A1 Use eqn of circular motion to find d2θ/dt2 where θ is angle of rod with vertical. AEF d2θ/dt2 = – {6gλ / (7λ + 408)a} θ M1* Approximate sin θ by θ to give standard form of SHM eqn T = 2π √{(7λ + 408)a / 6gλ} = 5π√(2a/g) DM1A1 Find possible values of λ by equating period T to 5π√(2a/g).AEF λ = 6 A1 6

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 9 of 16 Question Answer Marks Guidance 6(i) t √(s2/9) = ½ (1⋅85 – 1⋅65) [= 0⋅1] M1 Find estimate s2 of population variance (must be t) t8, 0.975 = 2⋅306 (to 3 s.f.) A1 Use of correct tabular t-value s2 = 9 × 0⋅043372 = 0⋅0169 or 0⋅130[1]2 A1 3 6(ii) x = 1 2 (1⋅65 + 1⋅85) = 1⋅75 or Σ x = 9 × 1⋅75 = 15⋅75 M1 A1 Find sample meanx s2 = (Σ x2 – 9 ×x2) / 8 or {Σ x2 – (Σ x)2 / 9} / 8 M1 or Σ x Σ x2 = 8 × 0⋅0169 + 15⋅752/9 = 27⋅7 A1 Find Σ x2 from s2 4

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 10 of 16 Question Answer Marks Guidance 7(i) a = 1/200 or 0⋅005 B1 State a or find a by equating mean value to 1/a 1 7(ii) p = P(T < 150) = F(150) = 1 – e - 150a M1 Find P(T < 150) p = 1 – e - 0⋅75 = 0⋅528 A1 2 7(iii) 1 – p n > 0⋅99 M1 Formulate condition for n 0⋅01 > (1 – e - 0⋅75) n or 0⋅01 > 0⋅528 n A1 n > log 0⋅01 / log 0⋅528 M1 Rearrange and take logs to give bound n > 7⋅20 [or 7⋅21] so nmin = 8 A1 Find nmin 4

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 11 of 16 Question Answer Marks Guidance 8 A x = 32⋅4 / 8 = 4⋅05 B1 Find sample mean for A sA2 = (131⋅82 – 32⋅42/8) / 7 sA2 = 3/35 (or 0⋅08571 or 0⋅29282 both to 3 s.f.) M1 Estimate or imply popln. variance for A H0: µA = µB , H1: µA ≠ µB B1 State hypotheses. AEF s2 = (7 sA2 + 9 sB2) / 16 = 0⋅12497 or 0⋅35352 or 3999 32000 M1 A1 Estimate (pooled) common variance t16, 0.95 = 1⋅746 B1* State or use correct tabular t value [–] t = (xA – B x ) / s √(1/8 + 1/10) M1 = 0⋅27 / 0⋅1677 = 1⋅61 A1 Find value of t (or can comparexA –xB = 0⋅27 with 0⋅293) t < 1⋅75 so [accept H0] mean masses are the same DB1 Correct conclusion (FT on t, dep B1*). AEF 9

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 12 of 16 Question Answer Marks Guidance 9(i) Σ x = 15, Σ y = 7 + p + q, Σ xy = 17 + 2p + 3q Σ x2 = 55, [Σ y2 = 21 + p2 + q2] M1 Find required summations Sxx = 55 – 152 / 5 = 10 and Sxy = 17 + 2p + 3q – 15 × (7 + p + q) / 5 = – 4 – p M1 A1 – 0⋅5 = Sxy / Sxx = (– 4 – p) / 10 p = 1 M1 A1 Find p from gradient in eqn. of regression line (7 + p + q) / 5 = – 0⋅5 × 15/5 + 3⋅5 q = 2 M1 A1 Find q from means and regression line 7 9(ii) Σ y = 10, Σ y2 = 26, Syy = 26 – 102/5 = 6 M1 Find Syy r = Sxy / √(Sxx Syy) = – 5 / √(10 × 6) M1 Find correlation coefficient r r = – 0⋅645[5] [allow – 0⋅646] A1 3

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 13 of 16 Question Answer Marks Guidance 10(i) F(x) = ∫ f(x) dx = (1/30) (– 8/x + x3 – 14x) [+ c] M1 Find or state distribution function F(x) for 2 ⩽ x ⩽ 4 F(x) = (1/30) (– 8/x + x3 – 14x + 24) M1 Using F(2) = 0 or F(4) = 1 to find c if necessary. AEF F(x) = 0 (x < or ⩽ 2), F(x) = 1 ( x > or ⩾ 4) A1 State F(x) for other values of x 3 10(ii) G(y) = P(Y < y) = P(X2 < y) G(y) = P(X < √y) = F(√y) G(y) = ( )( ) 3 1 1 2 2 2 1/30 8 / 14 24 y y y − + − + M1 A1 Find or state G(y) for 2 ⩽ x ⩽ 4 from Y = X2 (allow < or ⩽ throughout) Alternative method for question 10(ii) Use 1 2 x y = to find ( ) ( )( ) 1 2 d 1 f 1/30 8 / 3 14 , d 2 x x y y y y − = + − = − (M1 A1) Find f(x) and d d x y for use in ( ) ( ) d g f d y y x x = × ( ) ( ) ( ) ( ) ( ) 3 1 1 2 2 2 g [ G ] 1/30 4 / 3/2 7 / y y y y y ′ = = + − for 4 ⩽ y ⩽ 16 [g(y) = 0 otherwise] A1 A1 Find g(y). AEF State corresponding range of y for G(y) or g(y) 4 10(iii) ( )( ) 3 1 1 2 2 2 1/30 8 / 14 24 0.8 y y y − + − + = M1 Set G(y) = 0⋅8 – 8 + y2 – 14 y = 0, y = 7 + √57 or 14⋅5[5] [rejecting 7 – √57; allow 14⋅6] M1 A1 Rearrange to give quadratic in y and solve to find value of y 3

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 14 of 16 Question Answer Marks Guidance 11A(i) 10 (AP – 0⋅6) / 0⋅6 = 20 (1⋅2 – AP – 0⋅4) / 0⋅4 M1 A1 Verify AP by equating equilibrium tensions. AEF 4 AP – 2⋅4 = 9⋅6 – 12 AP AP = 0⋅75 [m] A1 AG 3 11A(ii) m d2x/dt2 = – 10 (0⋅15 + x) / 0⋅6 + 20 (0⋅05 – x) / 0⋅4 or m d2x/dt2 = + 10 (0⋅15 – x) / 0⋅6 – 20 (0⋅05 + x) / 0⋅4 M1 A1 A1 Apply Newton’s law at 0⋅75 + x or 0⋅75 – x from A (M1 requires LHS and 2 tensions: A1 for each correct tension) ⅔ d2x/dt2 = – (80 / 1⋅2) x , d2x/dt2 = – 100 x M1 A1 Simplify to give SHM eqn. in standard form T = 2π/ω = 2π/10 = π/5 or 0⋅628 [s] DB1 State the period T with FT on ω from SHM eqn. 6 11A(iii) (a = 0⋅75 – 0⋅7 = 0⋅05) vmax = ω × a M1 Find speed at equilibrium position from ωa vmax = 0⋅5 [m s–1] A1 2 11A(iv) x = a/2 = 0⋅025 M1 Find value of x giving half max. acceln. v = ω √ (a2 – x2) = 10 √ (0⋅052 – 0⋅0252) M1 v = 0⋅433 [m s-1] A1 Find corresponding speed 3

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 15 of 16 Question Answer Marks Guidance 11B(i) x = (1/40) Σ x f(x) = 68/40 = 1⋅7 B1 Find mean of sample (1/40) Σ x2 f(x) = 178/40 = 4⋅45, Var = 4⋅45 – 1⋅72 = 1⋅56 B1 Find variance of sample Mean and variance are similar so Poisson may be suitable B1 State valid comment 3 11B(ii) a = 40 × 1⋅65 e–1⋅6 /5! = 40 × 0⋅01764 M1 AG a = 0⋅706 A1 Verify a from Poisson term b = 40 – 39⋅758 = 0⋅242 B1* Find b 3 11B(iii) H0: Distribution fits/models data B1 State (at least) null hypothesis in full Oi: 6 15 9 10 Ei: 8⋅076 12⋅921 10⋅337 8⋅666 DM1 A1 Combine values consistent with all exp. values ⩾ 5 (FT on b, dep B1*) X2 = 0⋅5337 + 0⋅3345 + 0⋅1729 + 0⋅2053 = 1⋅25 M1 Find value of X 2 from Σ (Ei – Oi)2 / Ei [or Σ Oi2/Ei – n ] X2 = 1⋅25 A1 No. n of cells: 7 6 5 4 3 χn-1, 0.9 2: 10⋅64 9⋅236 7.779 6⋅251 4⋅605 DB1 State or use consistent tabular value χn–1, 0.9 2 (to 3 s.f.) [FT on number, n, of cells used to find X2] Accept H0 if X 2 < tabular value (using their values) M1 AEF 1⋅25 [± 0⋅1] < 6⋅25 so distn. fits [data] or distn. is a suitable model A1 Conclusion (requires both values approx. correct). AEF 8

9231/21 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 16 of 16