Cambridge A Level Mathematics - Further 9231 — 2013 Oct/Nov Paper 2 · Variant 3

*4500527322* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 October/November 2013 3 hours Additional Materials: Answer Booklet/Paper Graph Paper List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. 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 5 printed pages and 3 blank pages. JC13 11_9231_23/RP © UCLES 2013 [Turn over

2 1 A particle P is moving in a fixed circle of radius 0.5 m. At time t s its speed is 2 −2t + t2 m s−1. Find the magnitude of each of the radial and transverse components of the acceleration of P when t = 3. [3] 2 Three uniform small smooth spheres A, B and C, of equal radii and of masses 4m, ,m and m respectively, are at rest in a straight line on a smooth horizontal plane, with B between A and C. Sphere A is projected directly towards B with speed u. The coefficient of restitution between A and B, and between B and C, is 1 2. Show that the speed of B after it is struck by A is 6u , + 4. [4] Given that the speed of C after it is struck by B is u, find the value of ,. [5] 3 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 path of the particle is a complete vertical circle with centre O. When P is at its lowest point, its speed is u. When P is at the point A, the tension in the string is T and the string makes an angle 1 with the downward vertical, where cos 1 = 3 5. When P is at the point B, above the level of O, the tension in the string is 1 8T and angle BOA = 90Å. Find u in terms of a and g. [9] 4 A particle P of mass m is attached to one end of a light elastic string of natural length 4a. The other end of the string is attached to a fixed point O. The particle rests in equilibrium at the point E, vertically below O, where OE = 5a. The particle is pulled down a vertical distance 1 2a from E and released from rest. Show that the motion of P is simple harmonic and state the period of the motion. [6] Find the two possible values of the distance OP when the speed of P is equal to one half of its maximum speed. [4] © UCLES 2013 9231/23/O/N/13

3 5 P F D B C E A a 2a 60° Q Two parallel vertical smooth walls EF and CD meet a horizontal plane at E and C respectively. A uniform smooth rod AB, of weight 2W and length 3a, is freely hinged to the horizontal plane at the point A, between E and C. The end B rests against CD. A uniform smooth circular disc of weight W is in contact with the wall EF at the point P and with the rod at the point Q. It is given that angle BAC is 60Å and that AQ = a (see diagram). The rod and the disc are in equilibrium in the same vertical plane, which is perpendicular to both walls. Show that (i) the magnitude of the reaction at P is ï3W, [3] (ii) the magnitude of the reaction at B is 7ï3 9 W. [4] Find, in the form kW, the magnitude of the reaction on AB at A, giving the value of k correct to 3 significant figures. [5] 6 The random variable T is the time, in suitable units, between two successive arrivals in a hospital casualty department. The probability density function of T is f, where ft = T 0.2e−0.2t t ≥0, 0 otherwise. State the expected value of T. [1] Write down the distribution function of T and find PT > 10. [4] 7 Two independent random 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 Σ x = 10.0, Σ x2 = 25.0, Σ y = 15.0, Σ y2 = 43.5. Given that the pooled estimate of 32 is 2, find the value of n. [7] © UCLES 2013 9231/23/O/N/13 [Turn over

4 8 A factory produces china mugs. Random samples of size 6 are selected at regular intervals, and the mugs are inspected for defects. During one week, 100 samples are selected and the numbers of defective mugs found are summarised in the following table. Number of defective mugs 0 1 2 3 4 5 6 Number of samples 11 43 35 8 2 1 0 Fit a binomial distribution to the data and carry out a goodness of fit test at the 5% significance level. [10] 9 A random sample of 9 observations of a normally distributed random variable X gave the following summarised data. Σ x = 94.5 Σ x2 = 993.6 Test, at the 5% significance level, whether the population mean of X is 10.2. [7] Calculate a 90% confidence interval for the population mean of X. [3] 10 The lengths, x m, and masses, y kg, of 12 randomly chosen babies born at a particular hospital last year are summarised as follows. Σ x = 7.50 Σ x2 = 4.73 Σ y = 38.6 Σ y2 = 124.84 Σ xy = 24.25 Find the value of the product moment correlation coefficient for this sample. [3] Obtain an estimate for the mass of a baby, born last year at the hospital, whose length is 0.64 m. [4] Test, at the 2% significance level, whether there is non-zero correlation between the two variables. [4] © UCLES 2013 9231/23/O/N/13

5 11 Answer only one of the following two alternatives. EITHER A 6a 2a B O C A rigid body consists of a thin uniform rod AB, of mass 4m and length 6a, joined at B to a point on the circumference of a uniform circular disc, with centre O, mass 8m and radius 2a. The point C on the circumference of the disc is such that BC is a diameter and ABC is a straight line (see diagram). The body rotates about a smooth fixed horizontal axis through C, perpendicular to the plane of the disc. The angle between CA and the downward vertical at time t is denoted by 1. (i) Given that the body is performing small oscillations about the downward vertical, show that the period of these oscillations is approximately 160 O@ a 11g A . [9] (ii) Given instead that the body is released from rest in the position given by cos 1 = 0.6, find the maximum speed of A. [5] OR Guided tours of a museum begin every 60 minutes. A randomly chosen tourist arrives X minutes after the start of a tour. The continuous random variable X has probability density function f given by fx = t x −202 24 000 0 < x < 60, 0 otherwise. The random variable T is the time that the tourist has to wait for the next tour to begin. Show that the distribution function G of T is given by Gt = ” 0 t ≤0, 8 9 −40 −t3 72 000 0 < t < 60, 1 t ≥60. [5] Find the median and the mean of T. [9] © UCLES 2013 9231/23/O/N/13

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8 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. University of 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 2013 9231/23/O/N/13

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2013 series 9231 FURTHER MATHEMATICS 9231/23 Paper 2, maximum raw mark 100 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 October/November 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 23 © Cambridge International Examinations 2013 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 √ 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.

Page 3 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 23 © Cambridge International Examinations 2013 The following abbreviations may be used in a mark scheme or used on the scripts: AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) 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 MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) 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.

Page 4 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 23 © Cambridge International Examinations 2013 Question Number Mark Scheme Details Part Mark Total 1 Find radial component of acceleration: Find transverse component of acceleration: (2 – 2 × 3 + 32)2 / 0⋅5 = 50 [m s-2] M1 A1 – 2 + 2 × 3 = 4 [m s-2] B1 3 3 2 Use conservation of momentum, e.g.: Use restitution (must be consistent with prev. eqn.): Solve for vB : (or verify eqns are satisfied by this vB) Use conservation of momentum, e.g.: Use restitution (must be consistent with prev. eqn.): Eliminate wB: Put wC = u, substitute for vB and solve for λ: 4mvA + λmvB = 4mu B1 vA – vB = – ½ u B1 4(vB – ½ u) + λvB = 4u vB = 6u / (λ + 4) A.G. M1 A1 λmwB + mwC = λmvB B1 wB – wC = – ½ vB B1 (1 + λ)wC = (1 + ½)λvB M1 (1 + λ) = 9λ /(λ + 4) λ2 – 4λ + 4 = 0, λ = 2 M1 A1 4 5 9

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 23 © Cambridge International Examinations 2013 3 Equate radial forces at A and B (M1 for either): Find 2 energy eqns (M1 for either), e.g.: Find u by e.g, first eliminating T: and then finding one of vA 2 or vB 2: Hence u : T – mg cos θ = mvA 2/a M1 A1 T/8 + mg sin θ = mvB 2/a A1 ½mvA 2 = ½mu2 – mga(1 – cos θ) M1 A1 ½mvB 2 = ½mu2 – mga(1 + sin θ) A1 mvA 2/a + mg cos θ = 8 mvB 2/a – 8 mg sin θ vA 2 = 8vB 2 – 8ga(4/5) – ga(3/5) vA 2 = 8vB 2 – 7ga M1 ½mvA 2 = ½mvB 2 + mga(cos θ + sin θ) vA 2 = vB 2 + (14/5) ga vA 2 = (21/5)ga or vB 2 = (7/5)ga M1 u2 = vA 2 + (4/5)ga or vB 2 + (18/5)ga = 5ga, u = √(5ga) A1 9 9

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 23 © Cambridge International Examinations 2013 4 Resolve vertically at equilibrium with modulus λ: Use Newton’s Law at general point: Simplify to give standard SHM eqn: S.R.: Stating this without derivation (max 4/6): Find period T = 2π/ω with ω = √(g/a): Equate speed at P to one-half maximum speed: Find x2: Find OP: λa / 4a = mg [λ = 4mg] M1 A1 m d2x/dt2 = mg – λ (a+x)/4a [ or – mg + λ (a–x)/4a ] M1 A1 d2x/dt2 = – (g/a) x A1 (B1) T = 2π/√(g/a) or 2π√(a/g) B1 ω2(A2 – x2) = ¼ ω2A2, M1 A1 x2 = ¾ A2 = ¾ (½ a)2 [= 3a2/16] A1 OP = (5 ± ¼√3) a (A.E.F.) A1 6 4 10 5 (i) Find RP by e.g. moments about Q for disc: RP r cos 60° = W r sin 60° M1 A1 RP = W tan 60° = √3 W A.G. A1 3 (ii) Find RQ by resolving vertically for disc: Find RB by e.g. moments about A for AB: Resolve horizontally and vertically for rod (M1 for either) (or for rod and disc): Find magnitude R of reaction at A: RQ cos 60° = W, RQ = 2W B1 RB 3a sin 60° RB = 2W (3a/2) cos 60° + RQ a M1 A1 RB = W(3/2 + 2) / (3√3/2) RB = (7√3 / 9) W A.G. A1 XA = RB – RQ sin 60° M1 = – (2√3/9) W A1 YA = 2W + RQ cos 60° = 3W A1 R = √(4 × 3/81 + 9) W M1 R = √(247/27) W = 3⋅02 W A1 3 4 5 12

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 23 © Cambridge International Examinations 2013 6 State or find E(T): State or find distribution function of T: Find P(T > 10): E(T) = 1/0⋅2 = 5 B1 F(t) = 1 – exp (-0⋅2t) (t ù 0) B1 F(t) = 0 (otherwise or t < 0) B1 P(T > 10) = 1 – F(10) = 1 – (1 – e-2) M1 P(T > 10) = e-2 or 0⋅135 A1 1 4 5 7 Find Σ (x –x)2 : Find Σ (y –y)2 : Equate pooled estimate of σ2 to 2: Rearrange to give quadratic eqn for n: Find n (integer value): Σ (x –x)2 = 25 – 102/n B1 Σ (y –y)2 = 43⋅5 – 152/2n B1 (25 – 102/n + 43⋅5 – 152/2n) / (3n – 2) = 2 M1 A1 12n2 – 145n + 425 = 0 M1 n = (145 ± 25)/24 = 5 (≠ 7⋅08) M1 A1 7 7 8 Find value of p for binomial dist.: State (at least) null hypothesis: Find expected binomial values (to 1 d.p.): Combine last four cells since exp. value < 5: Calculate value of χ2 (to 2 d.p.; A1 dep *M1): State or use consistent tabular value (to 2 d.p.): Correct conclusion (A.E.F., dep *A1, *B1): mean = 150/100, p = 1⋅5/6 = ¼ M1 A1 H0: B(6, p) fits data (A.E.F.) B1 17⋅80 35⋅60 29⋅66 13⋅18 3⋅30 0.44 0.02 M1 A1 O: 11 43 35 11 E: 17⋅80 35⋅60 29⋅66 16⋅94 *M1 χ2 = 2⋅60 + 1⋅54 + 0⋅96 + 2⋅08 χ2 = 7⋅18 (or 7⋅14 if 1 d.p.) M1 *A1 χ2, 0.95 2 = 5⋅991 (cells combined) *B1 [χ3, 0.95 2 = 7⋅815, χ4, 0.95 2 = 9⋅488] χ2 > 5⋅99 so distn. does not fit B1 10 10

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 23 © Cambridge International Examinations 2013 9 Calculate sample mean: Estimate population variance: (allow biased here: 0⋅15 or 0⋅38732) State hypotheses (A.E.F.): Calculate value of t (to 3 s.f.): State or use correct tabular t value (to 3 s.f.): (or can comparex with 10⋅2 + 0⋅316 = 10⋅52 ) Correct conclusion (AEF, dep *A1, *B1): Find confidence interval (allow z in place of t) e.g.: Use of correct tabular value: Evaluate C.I. correct to 3 s.f.: x = 94⋅5 / 9 = 10⋅5 M1 s2 = (993⋅6 – 94⋅52 /9) / 8 s2 = 0⋅16875 or 0⋅41082 M1 H0: µ = 10⋅2, H1: µ ≠ 10⋅2 B1 t = (x – 10⋅2)/(s/√9) = 2⋅19 M1 *A1 t8, 0.975 = 2⋅306 *B1 Population mean is 10⋅2 B1 10⋅5 ± t √{1⋅35/(8 × 9)} M1 t8, 0.95 = 1⋅86[0] A1 10⋅5 ± 0⋅255 or [10⋅2, 10⋅8] A1 7 3 10

Page 9 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 23 © Cambridge International Examinations 2013 10 Find correlation coefficient r: (A0 if only 3 s.f. used) Calculate gradient b in y –y = b(x –x) : Find regression line of y on x (A.E.F.): (allow use of x on y) Find y when x = 0⋅64: State both hypotheses: State or use correct tabular two-tail r value: Valid method for reaching conclusion: Correct conclusion (AEF, dep *A1, *B1): r = (24⋅25 – 7⋅5 × 38⋅6/12) / √{(4⋅73 – 7⋅52/12) (124⋅84 – 38⋅62/12)} M1 A1 = 0⋅125 / √(0⋅0425 × 0⋅6767) = 0⋅125 / (0⋅2062 × 0⋅8226) [or 0⋅01042 / √(0⋅003542 × 0⋅05639) = 0⋅01042 / (0⋅05951 × 0⋅2375) ] = 0⋅737 *A1 b = 0⋅125 / 0⋅0425 = 2⋅94[1] B1 y = 38⋅6/12 + 2⋅94 (x – 7⋅5/12) M1 = 3⋅21[7] + 2⋅94 (x – 0⋅625) or 1⋅38 + 2⋅94x A1 y = 3⋅26 [kg] B1 H0: ρ = 0, H1: ρ ≠ 0 B1 r12, 2% = 0⋅658 *B1 Reject H0 if |r| > tabular value M1 There is non-zero correlation A1 3 4 4 11

Page 10 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 23 © Cambridge International Examinations 2013 11 (a) (i) (ii) State or find MI of rod AB about C: Find MI of disc about C: Find MI of body about C: Use eqn of circular motion to find d2θ/dt2 : Approximate sin θ by θ and substitute for I: Find period T = 2π/ω with ω = √(11g/64a): Use energy to find max. ang. vel. ω: Substitute for I and cos θ and simplify: Find maximum speed of A (A.E.F.): IAB = ⅓ 4m(3a)2 + 4m(7a)2 M1 A1 [ = 208 ma2 ] Idisc = ½ 8m(2a)2 + 8m(2a)2 M1 A1 [ = 48ma2 ] I = IAB + Idisc = 256ma2 A1 I d2θ/dt2 = [–] (4 × 7 + 8 × 2)mga sin θ M1 A1 d2θ/dt2 = – (11g/64a) θ A1 T = 16π√(a/11g) A.G. B1 ½Iω2 = 4mg 7a (1 – cos θ) + 8mg 2a (1 – cos θ) M1 A1 ω2 = 2 (11mga) 0⋅4 / (256ma2) ω2 = 11g / 80a A1 vA = 10aω = √(55ga/4) M1 A1 9 5 14

Page 11 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 23 © Cambridge International Examinations 2013 11 (b) Find F(x) for 0 < x < 60 by integration: State F(x) for other values of x: Find G(t) for 0 < t < 60 from X + T = 60: Formulate eqn for median m of T: Find value of m: Find g(t) for 0 < t < 60: Find E(T) from ∫ t g(t) dt: F(x) = ⅓ (x – 20)3/24000 + ⅓ 203/24000 M1 F(x) = (x – 20)3 / 72000 + 1/9 A1 F(x) = 0 (x ≤ 0), 1 (x ≥ 60) B1 G(t) = P(T < t) G(t) = P(60 – X < t) G(t) = P(X > 60 – t) G(t) = 1 – F(60 – t) = 8/9 – (40 – t)3 / 72000 A.G. M1 A1 8/9 – (40 – m)3 / 72000 = ½ M1 (40 – m)3 = (8/9 – ½) 72000 (40 – m)3 = 28000 M1 m = 40 – 280001/3 = 9⋅63 A1 g(t) = (40 – t)2 / 24000 M1 A1 E(T) = ∫(402t – 80t2 + t3) dt /24000 M1 = [½ 402t2 – ⅓ 80t3 + ¼t4]0 60/24000 A1 = 120 – 240 + 135 = 15 A1 5 9 14