Cambridge A Level Mathematics - Further 9231 — 2012 May/June Paper 2 · Variant 1

*9498306157* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/21 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 4 printed pages. JC12 06_9231_21/RP © UCLES 2012 [Turn over

2 1 A circular flywheel of radius 0.3 m, with moment of inertia about its axis 18 kg m2, is rotating freely with angular speed 6 rad s−1. A tangential force of constant magnitude 48 N is applied to the rim of the flywheel, in order to slow the flywheel down. Find the time taken for the angular speed of the flywheel to be reduced to 2 rad s−1. [4] 2 Two particles, of masses 3m and m, are moving in the same straight line towards each other with speeds 2u and u respectively. When they collide, the impulse acting on each particle has magnitude 4mu. Show that the total loss in kinetic energy is 4 3mu2. [6] 3 A particle P of mass m is projected horizontally with speed p7 2ga from the lowest point of the inside of a fixed hollow smooth sphere of internal radius a and centre O. The angle between OP and the downward vertical at O is denoted by θ. Show that, as long as P remains in contact with the inner surface of the sphere, the magnitude of the reaction between the sphere and the particle is 3 2mg(1 + 2 cos θ). [4] Find the speed of P (i) when it loses contact with the sphere, [3] (ii) when, in the subsequent motion, it passes through the horizontal plane containing O. (You may assume that this happens before P comes into contact with the sphere again.) [3] 4 AB is a diameter of a uniform circular disc D of mass 9m, radius 3a and centre O. A lamina is formed by removing a circular disc, with centre O and radius a, from D. Show that the moment of inertia of the lamina, about a fixed horizontal axis l through A and perpendicular to the plane of the lamina, is 112ma2. [5] A particle of mass 3m is now attached to the lamina at B. The system is free to rotate about the axis l. The system is held with B vertically above A and is then slightly displaced and released from rest. The greatest speed of B in the subsequent motion is k√(ga). Find the value of k, correct to 3 significant figures. [7] 5 A B C 3a 5a 6a Two uniform rods AB and BC are smoothly jointed at B and rest in equilibrium with C on a rough horizontal floor and with A against a rough vertical wall. The rod AB is horizontal and the rods are in a vertical plane perpendicular to the wall. The rod AB has mass 3m and length 3a, the rod BC has mass 5m and length 5a, and C is at a distance 6a from the wall (see diagram). Show that the normal reaction exerted by the floor on the rod BC at C has magnitude 13 2 mg. [5] The coefficient of friction at both A and C is µ. Find the least possible value of µ for which the rods do not slip at either A or C. [7] © UCLES 2012 9231/21/M/J/12

3 6 The probability that a particular type of light bulb is defective is 0.01. A large number of these bulbs are tested, one by one. Assuming independence, find the probability that the tenth bulb tested is the first to be found defective. [2] The first defective bulb is the Nth to be tested. Write down the value of E(N). [1] Find the least value of n such that P(N ≤n) is greater than 0.9. [3] 7 A random sample of 8 swimmers from a swimming club were timed over a distance of 100 metres, once in an outdoor pool and once in an indoor pool. Their times, in seconds, are given in the following table. Swimmer A B C D E F G H Outdoor time 66.2 62.4 60.8 65.4 68.8 64.3 65.2 67.2 Indoor time 66.1 60.3 60.9 65.2 66.4 63.8 62.4 69.8 Assuming a normal distribution, test, at the 5% significance level, whether there is a non-zero difference between mean time in the outdoor pool and mean time in the indoor pool. [8] 8 The number of flaws in a randomly chosen 100 metre length of ribbon is modelled by a Poisson distribution with mean 1.6. The random variable X metres is the distance between two successive flaws. Show that the distribution function of X is given by F(x) = 1 −e−0.016x x ≥0, 0 x < 0, and deduce that X has a negative exponential distribution, stating its mean. [4] Find (i) the median distance between two successive flaws, [3] (ii) the probability that there is a distance of at least 50 metres between two successive flaws. [2] 9 A random sample of 8 observations of a normal random variable X gave the following summarised data, where x denotes the sample mean. Σx = 42.5 Σ(x −x)2 = 15.519 Test, at the 5% significance level, whether the population mean of X is greater than 4.5. [7] Calculate a 95% confidence interval for the population mean of X. [3] 10 Random samples of employees are taken from two companies, A and B. Each employee is asked which of three types of coffee (Cappuccino, Latte, Ground) they prefer. The results are shown in the following table. Cappuccino Latte Ground Company A 60 52 32 Company B 35 40 31 Test, at the 5% significance level, whether coffee preferences of employees are independent of their company. [7] Larger random samples, consisting of N times as many employees from each company, are taken. In each company, the proportions of employees preferring the three types of coffee remain unchanged. Find the least possible value of N that would lead to the conclusion, at the 1% significance level, that coffee preferences of employees are not independent of their company. [4] © UCLES 2012 9231/21/M/J/12 [Turn over

4 11 Answer only one of the following two alternatives. EITHER A particle P of mass m is attached to one end of a light elastic string of modulus of elasticity 4mg and natural length l. 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. The particle is pulled down a vertical distance 1 8l from E and released from rest. Show that the motion of P is simple harmonic with period π r l g. [4] At an instant when P is moving vertically downwards through E, the string is cut. When P has descended a further distance 7 16l under gravity, it strikes a fixed smooth plane which is inclined at 30◦ to the horizontal. The coefficient of restitution between P and the plane is 1 3. Show that the speed of P immediately after the impact is 1 4 √(5gl). [8] OR A new restaurant S has recently opened in a particular town. In order to investigate any effect of S on an existing restaurant R, the daily takings, x and y in thousands of dollars, at R and S respectively are recorded for a random sample of 8 days during a six-month period. The results are shown in the following table. Day 1 2 3 4 5 6 7 8 x 1.2 1.4 0.9 1.1 0.8 1.0 0.6 1.5 y 0.3 0.4 0.6 0.6 0.25 0.75 0.6 0.35 (i) Calculate the product moment correlation coefficient for this sample. [4] (ii) Stating your hypotheses, test, at the 2.5% significance level, whether there is negative correlation between daily takings at the two restaurants and comment on your result in the context of the question. [5] Another sample is taken over N randomly chosen days and the product moment correlation coefficient is found to be −0.431. A test, at the 5% significance level, shows that there is evidence of negative correlation between daily takings in the two restaurants. (iii) Find the range of possible values of N. [3] 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 2012 9231/21/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 9231/21 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 21 © 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 21 © 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 21 © University of Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 1 EITHER: Use C = Iα to find angular acceleration α: (omitting minus sign loses this A1 only) Integrate or use ω1 = ω0 + αt to find time t: OR: Use energy to find angle θ : Use θ = ½(ω0 +ω1)t to find time t: α = – 48 × 0⋅3 / 18 [= – 0⋅8] M1 A1 t = (2 – 6)/α = 5 [s] M1 A1 θ = ½ 18(62 – 22)/(48 × 0⋅3) [= 20] (M1 A1) t = θ / ½(6 + 2) = 5 [s] (M1 A1) 4 [4] 2 Find two momentum eqns, e.g.: Solve to find both speeds after colln.: Find total loss in KE: 3mv3m = 3m × 2u – 4mu or mvm = – mu + 4mu or 3mv3m + mvm = 3m × 2u – mu M1 M1 v3m = 2u/3 and vm = 3u A1 ½m {3(2u)2 + u2 – 3v3m 2 – vm 2} M1 A1 = ½m (12 + 1 – 4/3 – 9) u2 = (13/2 – 31/6 or 16/3 – 4) mu2 = (4/3) mu2 A.G. A1 6 [6] 3 (i) (iii) Use conservation of energy: Equate radial forces: Eliminate v to find R: Find cos θ1 when R = 0: Find speed at this point: EITHER: Use energy to find reqd speed v2: Simplify: OR: Find horiz. comp. of v2: Find vertical comp. of v2: Combine comps. to find v2: ½mv2 = ½m(7ga/2) – mga (1 – cos θ) B1 [v2 = 3ga/2 + 2ga cos θ] R = mv2/a + mg cos θ B1 R = (3mg/2) (1 + 2 cos θ) A.G. M1 A1 cos θ1 = – ½ [θ1 = 2π/3] B1 v1 2 = ½ ga, v1 = √(½ ga) M1 A1 ½mv2 2 = ½mv1 2 – mga cos θ1 or ½m(7ga/2) – mga M1 ½v2 2 = ¼ga + ½ga, v2 = √(3ga/2) M1 A1 v1 cos (π–θ) [= ½v1 = √(ga/8)] (M1) √{v1 2 sin2 (π–θ) + ga} [= √(11ga/8)] (M1) v2 = √(ga/8 + 11ga/8) = √(3ga/2) (A1) 4 3 3 [10]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 21 © University of Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 4 Find MI of disc D about O: Find MI of removed disc D′ about O: Find MI of circular lamina about O: Find MI of circular lamina about A: Find MI of lamina plus particle about A: State or imply that speed is max when AB vertical Use energy when AB vertical (or at general point): Substitute for IA′ and find max ang. speed ω (or ω2): Equate 6aω to k√(ga) to find k: ID = ½ (9m)(3a)2 = (81/2)ma2 B1 ID′ = ½ ma2 B1 IO = ID – ID′ [= 40 ma2] M1 IA = IO +8m(3a)2 or ½(243 – 19)ma2 = 112ma2 A.G. M1 A1 IA′ = IA + (3m)(6a)2 = 220ma2 B1 M1 ½ IA′ω2 = 8mg × 6a + 3mg × 12a or 2 × 11mg × 42a/11 [= 84mga] M1 A1 ω2 = 84mga/110ma2 = 42g/55a A1 k = √(36 × 42/55) = 5⋅24 M1 A1 5 7 [12] 5 Resolve forces vertically: Take moments about B for AB: Combine to find RC: Substitute for RC in above resln. eqn to find FA: Take moments about B for BC: Substitute for RC to find FC: Find limiting value µC for µ at C [or A] (A.E.F.) : Relate RA , FC by e.g. horizontal resolution: Deduce least possible value of µ for system: RC + FA = 3mg+ 5mg B1 3aFA = (3a/2)3mg [FA = 3mg/2] M1 A1 RC = 8mg – 3mg/2 = 13mg/2 A.G. M1 A1 FA = 8mg – 13mg/2 = 3mg/2 B1 4aFC = 3aRC – (3a/2)5mg M1 FC = 3mg A1 µC = 6/13 [= 0⋅462 or µA = 0⋅5] M1 A1 RA = FC [= 3mg] B1 µ min = max[µA, µC] = 0⋅5 B1 5 7 [12] 6 Find prob. that 10th bulb is first defective one: State or find E(N): Formulate condition for n: (equality throughout loses this M1 only) Take logs (any base) to give inequality for n: Find nmin: (1 – 0⋅01)9 × 0⋅01 = 0⋅00914 (allow 0⋅00913) M1 A1 E(N) = 1/0⋅01 = 100 B1 P(N ≤ n) = 1 – P(N > n) = 1 – 0⋅99n > 0⋅9, 0⋅99n < 0⋅1 M1 n > log 0⋅1 / log 0⋅99 M1 n > 229⋅1, nmin = 230 A1 2 1 3 [6]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 21 © University of Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 7 Consider differences e.g. outdoor – indoor times: Calculate sample mean: Estimate population variance: (allow biased here: 2⋅679 or 1⋅6372) State hypotheses (A.E.F.): Calculate value of t (to 3 sf): State or use correct tabular t value: (or can compared with 1⋅46[3]) Correct conclusion (AEF, dep *A1, *B1): 0⋅1 2⋅1 -0⋅1 0⋅2 2⋅4 0⋅5 2⋅8 –2⋅6 M1 d = 5⋅4 / 8 = 0⋅675 M1 s2 = (25⋅08 – 5⋅42/8) / 7 = 3⋅062 or 1⋅7502 M1 H0: µo – µi = 0, H1: µo – µi ≠ 0 B1 t = d/(s/√8) = 1⋅09 M1 *A1 t7, 0.975 = 2⋅36[5] *B1 No difference between mean times B1 8 [8] 8 (i) (ii) Relate F(x) to Poisson distribution (ignore x < 0): Equate F(x) to 1 – e–λx or f(x) to e–λx to find mean λ: Formulate eqn for median m of X: Find value of m: Find P(X ≥ 50) (or > 50): F(x) = 1 – P(X > x) = 1 – P(no flaws in length x) M1 = 1 – e–(x/100)1⋅6 = 1 – e–0⋅016x A.G. M1 A1 λ = 1/0⋅016 = 62⋅5 B1 1 – e–0⋅016m = ½ M1 m = – ln ½ / 0⋅016 = 43⋅3 M1 A1 1 – F(50) = e–0⋅8 = 0⋅449 M1 A1 4 3 2 [9] 9 Calculate sample mean: Estimate population variance: (allow biased here: 1⋅940 or 1⋅3932) State hypotheses (A.E.F.): Calculate value of t (to 3 sf): State or use correct tabular t value: (or can compared with 4⋅5 + 0⋅998 = 5⋅49[8]) 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.: d = 42⋅5 / 8 = 5⋅3125 M1 s2 = 15⋅519 / 7 = 2⋅217 or 1⋅4892 M1 H0: µ = 4⋅5, H1: µ > 4⋅5 B1 t = (d – 4⋅5)/(s/√8) = 1⋅54 M1 *A1 t7, 0.95 = 1⋅89[5] *B1 Mean is not greater than 4⋅5 B1 5⋅3125 ± t √{15⋅519/(7 × 8)} M1 t7, 0.975 = 2⋅36[5] A1 5⋅31 ± 1⋅24[5] or [4⋅07, 6⋅56] A1 7 3 [10]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 21 © University of Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 10 Find expected values (to 1 d.p.): (lose A1 if rounded to integers) State (at least) null hypothesis (A.E.F.): Calculate value of χ2: State or use correct tabular χ2 value (to 3 sf): Conclusion consistent with values (A.E.F): Calculate new value χnew 2 of χ2: State or use correct tabular χ2 value: Find Nmin: 54⋅72 52⋅992 36⋅288 40⋅28 39⋅008 26⋅712 M1 A1 H0: Preferences are independent B1 χ2 = 0⋅5095 + 0⋅0186 + 0⋅5067 + 0⋅6921 + 0⋅0252 + 0⋅6883 = 2⋅44 or 2⋅45 M1 A1 χ2, 0.95 2 = 5⋅99[1] B1 Preferences are independent A1 √ χnew 2 = N ×χ2 M1 χ2, 0.99 2 = 9⋅21 B1 N > 9⋅21/2⋅45, Nmin = 4 M1 A1 7 4 [11] 11 (a) Resolve vertically at equilibrium with extn. e: Use Newton’s Law at general point: Simplify to give standard SHM eqn: S.R.: Stating this without derivation (max 3/4): Find period T using SHM with ω = √(4g/l): Find speed vE at E using v2 = ω2 (A2 – x2) with x = 0: Find speed vP before striking plane (A.E.F.): Find comps. of speed V after striking plane: Combine to find V: 4mge / l = mg [e = ¼l] B1 m d2x/dt2 = mg – 4mg(e+x)/l [ or – mg + 4mg(e–x)/l ] M1 d2x/dt2 = – (4g/l) x A1 (B1) T = 2π/√(4g/l) = π√(l/g) A.G. B1 vE = ωl/8 = ¼√(gl) M1 A1 vP = √(gl/16 + 14gl/16) = ¼√(15gl) M1 A1 Parallel to plane: vP sin 30° or ½ vP or √(15gl/64) B1 Normal to plane: ⅓ vP cos 30° or ⅓(√3/2) vP or √(5gl/64) B1 V2 = 15gl/64 + 5gl/64 = 5gl/16 M1 V = ¼√(5gl) A.G. A1 4 8 [12]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2012 9231 21 © University of Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total (b) (i) (ii) (iii) Find correlation coefficient r: (A0 if only 3 s.f. used) State both hypotheses: State or use correct tabular one-tail r value: Valid method for reaching conclusion: Correct conclusion (AEF, dep *A1, *B1): Valid comment, consistent with values Find critical tabular one-tail r value: Deduce range of possible values of N: Σx = 8⋅5, Σx2 = 9⋅67, Σxy = 3⋅955, Σy = 3⋅85, Σy2 = 2⋅0775 r = (3⋅955 – 8⋅5 × 3⋅85/8) / √{(9⋅67 – 8⋅52/8) (2⋅0775 – 3⋅852/8)} M1 A1 = –0⋅1356 / √(0⋅6387 × 0⋅2247) A1 = –0⋅1356 / (0⋅7992 × 0⋅4740) [or –0⋅01695 / √(0⋅07984 × 0⋅02809) = –0⋅01695 / (0⋅2826 × 0⋅1676) ] = – 0⋅358 *A1 H0: ρ = 0, H1: ρ < 0 B1 r8, 2.5% = 0⋅707 *B1 Accept H0 if |r| < tabular value M1 There is no negative correlation A1 No effect of S on R (A.E.F.) B1√ r16, 5% = 0⋅426 or r15, 5% = 0⋅441 M1 A1 N ≥ 16 A1 4 5 3 [12]