Cambridge A Level Mathematics - Further 9231 — 2011 May/June Paper 2 · Variant 3

*3104481736* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 May/June 2011 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. JC11 06_9231_23/2R © UCLES 2011 [Turn over

2 1 Three small spheres, A, B and C, of masses m, km and 6m respectively, have the same radius. They are at rest on a smooth horizontal surface, in a straight line with B between A and C. The coefficient of restitution between A and B is 1 2 and the coefficient of restitution between B and C is e. Sphere A is projected towards B with speed u and is brought to rest by the subsequent collision. Show that k = 2. [3] Given that there are no further collisions after B has collided with C, show that e ≤1 3. [5] 2 O B A P a 3a A uniform circular disc with centre A has mass M and radius 3a. A second uniform circular disc with centre B has mass 1 9M and radius a. The two discs are rigidly joined together so that they lie in the same plane with their circumferences touching. The line of centres meets the circumference of the larger disc at P and the circumference of the smaller disc at O. A particle of mass 1 3M is attached at P (see diagram). Show that the moment of inertia of the system about an axis through O, perpendicular to the plane of the discs, is 51Ma2. [6] The system is free to rotate about a fixed horizontal axis through O, perpendicular to the plane of the discs. The system is held with OP horizontal and is then released from rest. Given that a = 0.5 m, find the greatest speed of P in the subsequent motion, giving your answer correct to 2 significant figures. [5] 3 A B C W 2W 4a 4a 3a 3a The diagram shows two uniform rods BA and AC, smoothly hinged at A. The rod BA has length 8a and weight W; the rod AC has length 6a and weight 2W. The rods are in equilibrium in a vertical plane with B and C resting on a rough horizontal floor and angle CAB equal to 90◦. Show that the normal contact force at B is 26 25W. [4] The coefficient of friction between each rod and the floor is µ. Find the least possible value of µ. [8] © UCLES 2011 9231/23/M/J/11

3 4 A particle P of mass m is suspended from a fixed point O by a light inextensible string of length a. When hanging at rest under gravity, P is given a horizontal velocity of magnitude √(3ag) and subsequently moves freely in a vertical circle. Show that the tension T in the string when OP makes an angle θ with the downward vertical is given by T = mg(1 + 3 cos θ). [4] When the string is horizontal, it comes into contact with a small smooth peg Q which is at the same horizontal level as O and at a distance x from O, where x < a. Given that P completes a vertical circle about Q, find the least possible value of x. [8] 5 The continuous random variable X has probability density function f given by f(x) = 0.01e−0.01x x ≥0, 0 x < 0. (i) State the value of E(X). [1] (ii) Find the median value of X. [3] (iii) Find the probability that X lies between the median and the mean. [2] 6 The independent random variables X and Y have distributions with the same variance σ2. Random samples of 5 observations of X and n observations of Y are made and the results are summarised by Σx = 5.5, Σx2 = 15.05, Σy = 8.0, Σy2 = 36.4. Given that the pooled estimate of σ2 is 3, find the value of n. [7] 7 A fair die is thrown until a 6 appears for the first time. Assuming that the throws are independent, find (i) the probability that exactly 5 throws are needed, [3] (ii) the probability that fewer than 8 throws are needed, [2] (iii) the least integer n such that the probability of obtaining a 6 before the nth throw is at least 0.99. [3] 8 A company decides that its employees should follow an exercise programme for 30 minutes each day, with the aim that they lose weight and increase productivity. The weights, in kg, of a random sample of 8 employees at the start of the programme and after following the programme for 6 weeks are shown in the table. Employee A B C D E F G H Weight before (kg) 98.6 87.3 90.4 85.2 100.5 92.4 89.9 91.3 Weight after (kg) 93.5 85.2 88.2 84.6 95.4 89.3 86.0 87.6 Assuming that loss in weight is normally distributed, find a 95% confidence interval for the mean loss in weight of the company’s employees. [6] Test at the 5% significance level whether, after the exercise programme, there is a reduction of more than 2.5 kg in the population mean weight. [5] © UCLES 2011 9231/23/M/J/11 [Turn over

4 9 The marks achieved by a random sample of 15 college students in a Physics examination (x) and in a General Studies examination (y) are summarised as follows. Σx = 752 Σx2 = 38 814 Σy = 773 Σy2 = 45 351 Σ xy = 40 236 (i) Find the mean values, x and y. [1] (ii) Another college student achieved a mark of 56 in the General Studies examination, but was unable to take the Physics examination. Use the equation of a suitable regression line to estimate the mark that the student would have obtained in the Physics examination. [4] (iii) Find the product moment correlation coefficient for the given data. [2] (iv) Stating your hypotheses, test at the 5% level of significance whether there is a non-zero product moment correlation coefficient between examination marks in Physics and in General Studies achieved by college students. [4] © UCLES 2011 9231/23/M/J/11

5 10 Answer only one of the following two alternatives. EITHER One end of a light elastic string is attached to a fixed point O. A particle of mass m is attached to the other end of the string and hangs freely under gravity. In the equilibrium position, the extension of the string is d. Show that the period of small vertical oscillations about the equilibrium position is 2π r d g . [5] The particle is now pulled down and released from rest at a distance 2d below the equilibrium position. Given that the particle does not reach O in the subsequent motion, show that the time taken until the particle first comes to instantaneous rest is √3 + 2 3π r d g . [9] OR A family was asked to record the number of letters delivered to their house on each of 200 randomly chosen weekdays. The results are summarised in the following table. Number of letters 0 1 2 3 4 5 ≥6 Number of days 57 60 53 25 4 1 0 It is suggested that the number of letters delivered each weekday has a Poisson distribution. By finding the mean and variance for this sample, comment on the appropriateness of this suggestion. [3] The following table includes some of the expected values, correct to 3 decimal places, using a Poisson distribution with mean equal to the sample mean for the above data. Number of letters 0 1 2 3 4 5 ≥6 Expected number of days 53.964 70.693 p q 6.622 1.735 0.463 (i) Show that p = 46.304, correct to 3 decimal places, and find q. [2] (ii) Carry out a goodness of fit test at the 10% significance level. [9] © UCLES 2011 9231/23/M/J/11

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/11

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2011 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. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2011 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 2011 9231 23 © University of Cambridge International Examinations 2011 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 2011 9231 23 © University of Cambridge International Examinations 2011 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 2011 9231 23 © University of Cambridge International Examinations 2011 1 Use conservation of momentum for 1st collision: kmuB = mu B1 Use Newton’s law of restitution: uB = ½u B1 Eliminate uB to find k: k = 2 A.G. B1 Use conservation of momentum for 2nd collision: kmvB + 6mvC = kmuB M1 Use Newton’s law of restitution: vB – vC = – euB M1 Substitute and solve for vB : 2vB + 6vC = u, vB – vC = – ½ eu vB = (1 – 3e)u/8 [vC = (1+e)u/8] M1 A1 Use vB ≥ 0 if no further collisions: 1 – 3e ≥ 0, e ≤ ⅓ A.G. B1 S.R. Taking vB = 0 throughout: e = ⅓ (M1 A1) 3 5 (2) 8 2 Find MI of large disc about O: ½M (3a)2 + M (5a)2 [= 59Ma2/2] M1 A1 Find MI of small disc about O: ½(M/9)a2 + (M/9)a2 [= Ma2/6] M1 A1 Find MI of particle about O: (M/3) (8a)2 [= 64Ma2/3] B1 Sum to find MI of system about O: A.G. I = (177+1+128) Ma2/6 = 51 Ma2 A1 State or imply that speed is max when OP vertical M1 Use energy when OP vertical (or at general point): ½Iω2 = (5+1/9+⅓8)Mga = 70Mga/9 M1 A1 Substitute for a, I and find max speed 8aω: ω = √6⋅10 = 2⋅47, 8aω = 9⋅9 [ms–1] M1 A1 6 5 11 3 Moments for system about C, denoting ACB by θ: NB × BC = 2W × 3a cos θ + W (BC – 4a sin θ) (A.E.F.) M1 A1 Substitute for BC, θ: NB × 10a = 2W × 9a/5 + W × 34a/5 A1 Simplify to give NB: NB = (26/25) W A.G. A1 Find NC by vertical resolution or moments: NC = 3W – NB = (49/25) W M1 A1 Find FB (or FC ) by moments about A: FB × 24a/5 = NB × 32a/5 – W × 16a/5 M1 FB = (18/25) W or 0⋅72W A1 Find limiting value for µ at B [or C] (A.E.F.): 18/26 [= 0⋅692 or 18/49 = 0⋅367] M1 A1 Relate FB , FC by e.g. horizontal resolution: FC = FB [= (18/25) W] B1 Deduce least possible value of µ for system: µ min = 9/13 or 0⋅692 B1 4 8 12

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 23 © University of Cambridge International Examinations 2011 4 Use conservation of energy at general point: ½mv2 = ½mu2 – mga (1 – cos θ) B1 Equate radial forces to find tension T: T = mg cos θ + mv2/a B1 Eliminate v2 , replace u2 by 3ag and simplify: T = mg (1 + 3 cos θ) A.G. M1 A1 Use energy to find speed v when PQ horizontal: ½mv2 = ½mu2 – mga, v2 = ga M1 A1 Use energy to find speed w when P above Q: ½mw2 = ½mv2 – mg(a – x) M1 A1 (note that v need not be found) [mw2 = mg(2x – a) ] EITHER: Consider tension to find reqd. condition: T = mw2/(a – x) – mg ≥ (or >) 0 M1 A1 Combine to find least value of x: mg(3x – 2a)/(a – x) ≥ 0 x ≥ 2a/3, xmin = 2a/3 M1 A1 OR: Find x for which T becomes zero: mw2/(a – x) = mg, x = 2a/3 (M1 A1) Show this is least possible value of x, e.g.: T = mg(3x–2a)/(a–x) ≥ 0 if x ≥ 2a/3 (M1 A1) 4 8 12 5 (i) (ii) (iii) State or find E(X): E(X) = 1/ 0⋅01 or 100 B1 Integrate f(x) to find median m: ∫ m 0 f(x) dx = 1 – e–0⋅01m = ½ M1 A1 Solve for m: m = 100 ln 2 or 69⋅3 A1 Integrate f(x) to find probability: ∫ 100 m f(x) dx = ½ – e–1 = 0⋅132 M1 A1 1 3 2 6 6 Find pooled estimate: (15⋅05 – 5⋅52/5 + 36⋅4 – 82/n)/(3+n) M1 A1 Equate to 3 and rearrange: 45⋅4 – 64/n = 9 + 3n M1 A1 3n2 – 36⋅4n + 64 = 0 A1 Solve for n: n = (36⋅4 ± 23⋅6) / 6 = 10 M1 A1 7 7 7 (i) (ii) (iii) Find probability for needing 5 throws: p(1 – p)4 with p = 1/6; = 0⋅0804 M1 A1; A1 Find probability for needing < 8 throws: 1 – (1 – p)7 = 0⋅721 M1 A1 Relate prob. to 0⋅99 (allow > but not =): 1 – (1 – p)n–1 ≥ 0⋅99 B1 Find least integer n: (n – 1) log 5/6 ≤ log 0⋅01 M1 (Allow M1 A1 even if equality used) n – 1 ≥ 25⋅3, nmin = 27 A1 3 2 3 8

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 23 © University of Cambridge International Examinations 2011 8 Consider differences e.g.: 5⋅1 2⋅1 2⋅2 0⋅6 5⋅1 3⋅1 3⋅9 3⋅7 M1 Calculate sample mean: d = 25⋅8 / 8 = 3⋅225 M1 Estimate population variance: s2 = (100⋅14 – 25⋅82/8) / 7 (allow biased here: 2⋅117 or 1⋅4552) [= 2⋅419 or 1⋅5552 ] M1 Find confidence interval (allow z in place of t) e.g.: 3⋅225 ± t √(2⋅419/8) M1 (inconsistent use of 7 or 8 loses M1) Use of correct tabular value: t7, 0.975 = 2⋅36[5] A1 Evaluate C.I. correct to 3 s.f. (in kg): 3⋅225 ± 1⋅301 or [1⋅92, 4⋅53] A1 State hypotheses: H0: µb – µa = 2⋅5, H1: µb – µa > 2⋅5 B1 Calculate value of t (to 2 dp): t = (d – 2⋅5)/(s/√8) = 1⋅32 M1 *A1 Compare with correct tabular t value: t7, 0.95 = 1⋅89[5] *B1 Correct conclusion (AEF, dep *A1, *B1): Reduction not more than 2⋅5 B1 6 5 11

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 23 © University of Cambridge International Examinations 2011 9 (i) (ii) (iii) (iv) Find mean values: x = 50⋅1[3], y = 51⋅5[3] B1 Calculate gradient b′ in x –x = b′ (y –y): b′ = (40236 – 752 × 773/15) / (45351 – 7732/15) M1 = 1482⋅9 / 5515⋅7 = 0⋅268[9] A1 Use regression line to estimate x at y = 56: x = 50⋅13 + 0⋅2689 (56 – 51⋅53) M1 [x = 36⋅28 + 0⋅2689y] = 51[⋅3] A1 OR Calculate gradient b in y –y = b(x –x): b = (40236 – 752 × 773/15) / (38814 – 7522/15) (M1) = 1482⋅9 / 1113⋅7 = 1⋅33[15] (A1) Use regression line to estimate x at y = 56: x = 50⋅13 + (56 – 51⋅53)/ 1⋅332 (M1) [y = – 15⋅22 + 1⋅332x] = 53[⋅49] (A1) Find correlation coefficient r: r = (40236 – 752 × 773/15) / √{(38814 – 7522/15) (45351 – 7732/15)} M1 = 1482⋅9 / √(1113⋅7 × 5515⋅7) = 1483 / (33⋅37 × 74⋅27) or 98⋅86 / √(74⋅25 × 367⋅7) = 98⋅86 / (8⋅617 × 19⋅18) = 0⋅598 *A1 State both hypotheses: H0: ρ = 0, H1: ρ ≠ 0 B1 Use correct tabular 2-tail r value: r15, 5% = 0⋅514 (to 2 dp) *B1 Valid method for reaching conclusion: Reject H0 if |r| > tabular value M1 Correct conclusion (AEF, dep *A1, *B1): There is a non-zero coefficient A1 1 4 2 4 11

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 23 © University of Cambridge International Examinations 2011 10 (a) Resolve vertically at equilibrium: λd/a = mg [λ/a = mg/d] B1 Use Newton’s Law at general point: m d2x/dt2 = mg – λ(d+x)/a [or – mg + λ(d–x)/a] M1 A1 Simplify: d2x/dt2 = – (λ/ma) x or – (g/d) x A1 S.R.: Stating this without derivation (max 3/5): (B1) Find period T using SHM with ω = √(g/d): T [= 2π√(ma/λ)] = 2π√(d/g) A.G. B1 Use SHM formula for x with amplitude 2d: x = 2d cos (ωt) [or sin] M1 Find time t1 to string becoming slack: t1 = (1/ω) cos–1(–1/2) or T/4 + (1/ω) sin–1(1/2) M1 A1 Evaluate: A.G. t1 = (1/ω) 2π/3 = (2π/3)√(d/g) A1 Find speed v when string becomes slack: v =ω√(4d2 – d2) = ωd√3 or √ (3dg) M1 A1 Find further time t2 to instantaneous rest: t2 = v/g B1 Substitute and simplify: A.G. t2 = √(3dg) / g = √3√(d/g) M1 A1 5 9 14 (b) (i) (ii) Find mean and variance of sample: 262/200 = 1⋅31 and (586 – 2622/200) / 200 = 1⋅21[39] M1 A1 Valid comment (AEF, needs values approx correct): Values close, so distn. appropriate B1 State and evaluate expression for p A.G.: p = 200 (1⋅312/2)e–1⋅31 = 46⋅304 B1 Find q (can use Σ Ei = 200): q = 200 (1⋅313/6)e–1⋅31 = 20⋅2[19] B1 State (at least) null hypothesis: H0: Poisson fits data (A.E.F.) B1 Combine last 3 cells since exp. value < 5: O: . . . 5 E: . . . 8⋅82 *M1 A1 Calculate χ2 (to 2 dp ; A1 dep *M1): χ2 = 5⋅54 M1 A1 Compare consistent tabular value (to 2 dp): χ3, 0.9 2 = 6⋅251 M1 A1 (A1 dep *M1) [χ4, 0.9 2 = 7⋅779, χ5, 0.9 2 = 9⋅236] Valid method for reaching conclusion: Accept H0 if χ2 < tabular value M1 Conclusion (A.E.F., needs correct values): 5⋅54 < 6⋅25 so Poisson does fit A1 3 2 9 14