Cambridge A Level Mathematics - Further 9231 — 2012 May/June Paper 2 · Variant 3

*8267572846* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 May/June 2012 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. JC12 06_9231_23/RP © UCLES 2012 [Turn over

2 1 Two smooth spheres A and B, of equal radii and of masses 3m and 6m respectively, are at rest on a smooth horizontal surface. Sphere A is projected directly towards B with speed u. The coefficient of restitution between A and B is e. Show that the kinetic energy lost in the collision between A and B is mu2(1 −e2). [7] 2 A B P x O 4a Two light elastic strings, each of natural length a and modulus of elasticity 2mg, are attached to a particle P of mass m. The strings join the particle to the points A and B which are fixed and at a distance 4a apart on a smooth horizontal surface. The particle is at rest at the mid-point O of AB. The particle is now displaced a small distance in a direction perpendicular to AB, on the surface, and released from rest. At time t, the displacement of P from O is x (see diagram). Show that ¨x = −4gx a 1 −1 21 + x2 4a2  −1 2 !. [5] Given that x a is so small that  x a 2 and higher powers may be neglected, show that the motion of P is approximately simple harmonic and state the period of the motion. [2] 3 The point O is on the fixed line l. Points A and B on l are such that OA = 6 m and OB = 8 m, with O between A and B. A particle P oscillates on l in simple harmonic motion with centre O. When P is at A its speed is 4 m s−1, and when P is at B its speed is 3 m s−1. Show that the amplitude of the motion is 10 m and find the period of the motion. [4] Find the time taken by P to travel directly from A to B, through O. [4] 4 A smooth sphere, with centre O and radius a, has its lowest point fixed on a horizontal plane. A particle P of mass m is projected horizontally with speed u from the highest point on the outer surface of the sphere. In the subsequent motion, OP makes an angle θ with the upward vertical through O. Show that, while P remains in contact with the sphere, the magnitude of the reaction of the sphere on P is mg(3 cos θ −2) −mu2 a . [4] The particle loses contact with the surface of the sphere when θ = α. Given that u = 1 2 √(ga), find (i) cos α, [2] (ii) the vertical component of the velocity of P as it strikes the horizontal plane. [5] © UCLES 2012 9231/23/M/J/12

3 5 A B O 6a 2a A uniform rod AB, of mass m and length 6a, is rigidly attached at B to a point on the circumference of a uniform circular lamina of mass m, radius 2a and centre O. The lamina and the rod are in the same vertical plane, and ABO is a straight line (see diagram). Show that the moment of inertia of the system about an axis l through A perpendicular to the plane of the lamina is 78ma2. [4] A particle of mass 2m is now attached at B and the system is free to rotate in a vertical plane about the fixed axis l which is horizontal. Initially AB is horizontal, with O moving downwards and the system having angular velocity 3 5 q g a. At time t, AB makes an angle θ with the downward vertical through A. (i) Find, in terms of a, g and θ, an expression for d2θ dt2 . [4] (ii) Find the angular velocity of the system when B is vertically below A. [3] 6 A random sample of 10 observations of a normal random variable X has mean x, where x = 8.254, Σ(x −x)2 = 0.912. Using a 5% significance level, test whether the mean of X is greater than 8.05. [7] 7 The waiting time, T minutes, before a customer is served in a restaurant has distribution function F given by F(t) =  1 −e−λt t ≥0, 0 t < 0, where λ is a positive constant. The standard deviation of T is 8. Find (i) the value of λ, [2] (ii) the probability that a customer has to wait between 5 and 10 minutes before being served, [2] (iii) the median value of T. [3] © UCLES 2012 9231/23/M/J/12 [Turn over

4 8 Residents of three towns A, B and C were asked to grade the reliability of their digital television signal as good, satisfactory or poor. A random sample of responses from each town is taken and the numbers in each category are given in the following table. Good Satisfactory Poor Town A 24 34 14 Town B 58 60 26 Town C 20 34 30 Test, at the 2.5% significance level, whether grade of reliability is independent of town. [7] Identify which town makes the greatest contribution to the test statistic and relate your answer to the context of the question. [2] 9 The continuous random variable X has probability density function f given by f(x) =   1 2a −a ≤x ≤a, 0 otherwise, where a is a positive constant. Find the distribution function of X. [3] The random variable Y is defined by Y = eX. Find the distribution function of Y. [3] Given that a = 4, find the value of k for which P(Y ≥k) = 0.25. [3] 10 Engineers are investigating the speed of the internet connection received by households in two towns P and Q. The speeds, in suitable units, in P and Q are denoted by x and y respectively. For a random sample of 50 houses in town P and a random sample of 40 houses in town Q the results are summarised as follows. Σx = 240 Σ x2 = 1224 Σy = 168 Σy2 = 754 Calculate a 95% confidence interval for µP −µQ, where µP and µQ are the population mean speeds for P and Q. [6] Test, at the 1% significance level, whether µP is greater than µQ. [6] © UCLES 2012 9231/23/M/J/12

5 11 Answer only one of the following two alternatives. EITHER A C B q 3a a The diagram shows a uniform rod AB, of length 4a and weight W, resting in equilibrium with its end A on rough horizontal ground. The rod rests at C on the surface of a smooth cylinder whose axis is horizontal. The cylinder rests on the ground and is fixed to it. The rod is in a vertical plane perpendicular to the axis of the cylinder and is inclined at an angle θ to the horizontal, where cos θ = 3 5. A particle of weight kW is attached to the rod at B. Given that AC = 3a, show that the least possible value of the coefficient of friction µ between the rod and the ground is 8(2k + 1) 13k + 19 . [9] Given that µ = 9 10, find the set of values of k for which equilibrium is possible. [3] OR For a random sample of 5 pairs of values of x and y, the equations of the regression lines of y on x and x on y are respectively y = −0.5x + 5 and x = −1.2y + 7.6. Find the value of the product moment correlation coefficient for this sample. [3] Test, at the 5% significance level, whether the population product moment correlation coefficient differs from zero. [4] The following table shows the sample data. x 1 2 5 5 p y 5 3 4 2 q Find the values of p and q. [5] © UCLES 2012 9231/23/M/J/12

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. 9231/23/M/J/12

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2012 question paper for the guidance of teachers 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 must be read in conjunction with the question papers and the report on the examination. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 23 © University of Cambridge International Examinations 2012 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: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 23 © University of Cambridge International Examinations 2012 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: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 23 © University of Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 1 Use conservation of momentum, e.g.: 3mvA + 6mvB = 3mu M1 Use Newton’s law of restitution (A1 if both correct): vA – vB = – eu M1 A1 Solve for vA and vB (A.E.F.; A0 if any dirn. error): vA = ⅓(1 – 2e)u, vB = ⅓(1+e)u M1 A1 Find loss of kinetic energy by substituting for vA, vB: ½m(3u2 – 3vA 2 – 6vB 2) = mu2{3/2 – (1 – 4e + 4e2)/6 – (1 + 2e + e2)/3} (some working needed to earn A1; = mu2{3/2 – ½(1 + 2e2)} ignore any earlier dirn. error) = mu2(1 – e2) A.G. M1 A1 7 [7] 2 Find AP (or BP) in terms of x and a: AP = √(4a2 + x2) B1 Find T by Hooke’s Law for one string [or both]: T = 2mg(AP– a) / a B1 Find eqn of motion at general point: md2x/dt2 = – 2Tx/AP M1 Combine to find d2x/dt2 : d2x/dt2 = – (4gx/a)(1 – a/AP) M1 A.G. = – (4gx/a){1 – ½(1 + x2/4a2)–½} A1 M.R. Vertical motion loses only this A1 Motion along AB loses all marks Neglect (x/a)2: d2x/dt2 = – (4gx/a) ½ = – 2gx/a B1 State period of motion using 2π/ω (A.E.F.): 2π √(a/2g) or π √(2a/g) B1 5 2 [7] 3 Apply v2 = ω2 (A2 – x2) at both points: 42 = ω2(A2 – 62), 32 = ω2(A2 – 82) M1 Combine to find A: 7A2 = 322 – 182, A = 10 A.G. M1 A1 Find period T using ω: ω = ½, T = 2π/ω = 4π or 12⋅6 [s] B1 EITHER: Use x = A sin ωt to find time from O to A: tOA = ω–1 sin–1 (6/10) [= 1⋅287] M1 Use x = A sin ωt to find time from O to B: tOB = ω–1 sin–1 (8/10) [= 1⋅855] M1 Combine to find time from A to B: tOA + tOB = 3⋅14 or π [s] M1 A1 OR: Use x = A cos ωt to find time to A: tA = ω–1 cos–1 (6/10) [= 1⋅855] (M1) Use x = A cos ωt to find time to B: tB = ω–1 cos–1 (-8/10) [= 4⋅996] (M1) Combine to find time from A to B: tB – tA = 3⋅14 or π [s] (M1 A1) 4 4 [8] 4 (i) Use conservation of energy (B0 for v2 = …): ½mv2 = ½mu2 + mga(1 – cos θ) B1 Equate radial forces: R = mg cos θ – mv2/a B1 Eliminate v to find R: R = mg(3 cos θ – 2) – mu2/a A.G. M1 A1 Find cos α when R = 0: cos α = ⅓(¼ + 2) = ¾ M1 A1 4 2 (ii) Find v2 at this point: v1 2 = ¼ ga + ½ ga = ¾ ga B1 Find vertical comp. of v1 at plane: v2 2 = v1 2 sin2 α + 2ga(1 + cos α) M1 A1 (using u in place of v1 can earn M1 A0) = (21/64 + 7/2) ga A1 (A.E.F.) v2 = √(245ga/64) or (7/8)√(5ga) or 1⋅96√(ga) or 6⋅19√a A1 5 [11]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 23 © University of Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 5 (i) Find MI of rod about A: Irod = (4/3)m(3a)2 = 12ma2 B1 Find MI of circular lamina about A: Ilamina = 2ma2 + 64ma2 = 66ma2 M1 A1 Find MI of rod plus lamina about A: I = Irod + Ilamina = 78ma2 A.G. A1 Find new MI including particle about A: I′ = I + (2m)(6a)2 = 150ma2 B1 Use eqn of circular motion to find d2θ/dt2 : I′d2θ/dt2 = [–] mg(3 + 8 + 12)a sin θ M1 A1 (using I can earn B0 M1 A1 A0) d2θ/dt2 = – (23g/150a) sin θ A1 4 4 (ii) Use energy when AB vertical to find ω: ½I′ω2 = ½I′ω0 2 + (3 + 8 + 12)mga M1 A1 (or by integration) ω2 = 9g/25a + 46mga/150ma2 ω = √(2g/3a) or 0⋅816√(g/a) (using I can earn M1 A1 A0) or 2⋅58/√a (A.E.F.) A1 3 [11] 6 Estimate population variance: s2 = 0⋅912 / 9 (allow biased here: 0⋅0912 or 0⋅30202) = 0⋅1013 or 38/375 or 0⋅31832 M1 State hypotheses (B0 forx …): H0: µ = 8⋅05, H1: µ > 8⋅05 B1 Calculate value of t (to 3 sf): t = ( x – 8⋅05)/(s/√10) = 2⋅03 M1 A1 State or use correct tabular t value: t9, 0.95 = 1⋅83[3] B1 (or can comparex with 8⋅05 + 0⋅184[5] = 8⋅23) Valid method for reaching conclusion: Reject H0 if t > tabular value M1 Conclusion (A.E.F., needs correct values): 2⋅03 > 1⋅83 so mean is greater A1 7 [7] 7 (i) State or find λ: λ = 1/(standard deviation) = 1/8 M1 A1 2 (ii) Find F(10) – F(5): e–5λ – e–10λ = e–0⋅625 – e–1⋅25 = 0⋅249 M1 A1 2 (iii) Formulate eqn for median m of T: 1 – e–λm = ½ M1 Find value of m: m = – 8 ln ½ = 5⋅55 (or 5⋅54) M1 A1 3 [7] 8 Find expected values (to 1 d.p.): 24⋅48 30⋅72 16⋅8 (lose A1 if rounded to integers) 48⋅96 61⋅44 33⋅6 28⋅56 35⋅84 19⋅6 M1 A1 State (at least) null hypothesis (A.E.F.): H0: Grade independent of town B1 Calculate value of χ2 to 1 d.p.: χ2 = 0⋅009 + 0⋅350 + 0⋅467 + 1⋅669 + 0⋅034 + 1⋅719 + 2⋅567 + 0⋅094 + 5⋅518 = 12⋅4[3] M1 A1 Compare with consistent tabular value (to 1 d.p.): χ4, 0.975 2 = 11⋅1[4] B1 Conclusion consistent with values (A.E.F): Grade not independent of town A1 State town with max. contribn. ( on exp. values): Town C B1 State valid comment ( on exp. values), e.g.: Signal poorer than expected in C B1 7 2 [9] 9 Integrate f(x) to find F(x) for –a ≤ x ≤ a (A.E.F.): F(x) = x/2a + c = (x + a)/2a M1 A1 (Finding F(x) = x/2a earns M1 A0) State F(x) for other intervals: F(x) = 0 (x < –a), F(x) = 1 (x > a) B1 Relate dist. fn. G(y) of Y to X for central interval: G(y) = P(Y < y) = P(eX < y) (working may be omitted; ignore other intervals) = P(X < ln y) = F(ln y) = (ln y + a)/2a; (e–a ≤ y ≤ ea) M1 A1; A1 Find k from P(Y ≥ k) = ¼ with a = 4: 1 – G(k) = ¼, (ln k)/2a + ½ = ¾ M1 A1 (Using G(k) = ¼ can earn M1) ln k = 2, k = e2 or 7⋅39 A1 3 3 3 [9]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 23 © University of Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 10 Estimate population variance using P’s sample: sP 2 = (1224 – 2402/50) / 49 (allow use of biased: σ P,50 2 = 1⋅44 or 1⋅22) = 1⋅469 or 72/49 or 1⋅2122 M1 A1 Estimate population variance using Q’s sample: sQ 2 = (754 – 1682/40) / 39 (allow use of biased: σ Q,40 2 = 1⋅21 or 1⋅12) = 1⋅241 or 242/195 or 1⋅1142 A1 Estimate population variance for combined sample: s2 = sP 2 /50 + sQ 2/40 (allow use of σ P,50 2, σQ,40 2: 0⋅05905 or 0⋅2432) = 0⋅0604 or 0⋅2462 M1 Find confidence interval: (240/50 – 168/40) ± zs M1 Use z = 1⋅96 and evaluate correct to 3 d.p.: 0⋅6 ± 0⋅482 or [0⋅118, 1⋅082] A1 (allow use of 0⋅05905 instead of 0⋅0604: 0⋅6 ± 0⋅476 or [0⋅124, 1⋅076]) State hypotheses: H0: µP = µQ , H1: µP > µQ B1 Calculate value of z (to 2 d.p.): z = (240/50 – 168/40)/ s = 2⋅44 M1 *A1 (allow 0⋅6/0⋅243 = 2⋅47) State or use correct tabular z value: z0.99 = 2⋅326 [or 2⋅33] *B1 (or can compare 0⋅6 with 0⋅572 or 0⋅56[5]) Valid method for reaching conclusion: Reject H0 if z > tabular value M1 Correct conclusion (AEF, dep *A1, *B1): µP is greater than µQ A1 S.R. Allow (implicit) assumption of equal variances, but deduct A1 if not explicit: Pooled estimate of common variance s2 : (50σ P,50 2 + 40σ Q,40 2)/88 = 1⋅368 Confidence interval: 0⋅6 ± 1⋅96s√(1/50+1/40) = 0⋅6 ± 0⋅486 or [0⋅114, 1⋅086] Value of z (to 2 dp): 0⋅6/s√(1/50+1/40) = 2⋅42 6 6 [12] 11 (a) EITHER: Take moments about A for rod to find RC: RC 3a = kW 4a cos θ + W 2a cos θ M1 A1 RC = (2k + 1) (2W/5) A1 Resolve horizontal forces on rod to find FA: FA = RC sin θ = (2k + 1) (8W/25) M1 A1 Resolve vertical forces on rod to find RA: RA = W + kW – RC cos θ = (13k + 19) (W/25) M1 A1 OR: Take moments about C for rod: FA 3a sin θ – RA 3a cos θ = kWa cos θ – Wa cos θ (M1 A1) 4FA – 3RA = (k – 1)W (A1) Resolve forces parallel to rod: FA cos θ + RA sin θ = kW sin θ + Wsin θ (M1) 3FA + 4RA = 4(k + 1)W (A1) Combine to find FA, RA: FA = (2k + 1) (8W/25) and RA = (13k + 19) (W/25) (M1 A1) Use FA ≤ µRA to find µmin: µmin = 8(2k + 1)/(13k + 19) A.G. M1 A1 (No use of inequality loses A1) Find bound on k (allow use of equality): 160k + 80 ≤ (or =) 117k + 171 M1 Bound = 91/43 or 2⋅12 A1 Correct use of inequality k ≤ 91/43 or 2⋅12 A1 9 3 [12]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 23 © University of Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 11 (b) Find correlation coefficient r: r2 = (–0⋅5)(–1⋅2) = 0⋅6 or 0⋅7752 M1 A1 r = – 0⋅775 *A1 State both hypotheses (B0 for r …): H0: ρ = 0, H1: ρ ≠ 0 B1 State or use correct tabular one-tail r value: r5, 2.5% = 0⋅878 *B1 Valid method for reaching conclusion: Accept H0 if |r| < tabular value M1 Correct conclusion (AEF, dep *A1, *B1): Coeff. does not differ from zero A1 EITHER: Find two indep. eqns for p, q, e.g.: 14 + q = (–0⋅5)(13 + p) + 5 × 5 (M1 for either) or (14 + q)/5 = (–0⋅5)(13 + p)/5 + 5 M1 A1 13 + p = (–1⋅2)(14 + q) + 5 × 7⋅6 or (13 + p)/5 = (–1⋅2)(14 + q)/5 + 7⋅6 A1 Combine to find p, q: p = 7, q = 1 A1, A1 OR: Combine regression lines to findx,y: x = 4; y = 3 (M1 A1; A1) Use sample data to find p, q: p = 7, q = 1 (A1, A1) 3 4 5 [12]