Cambridge A Level Mathematics - Further 9231 — 2010 May/June Paper 2 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/21 Paper 2 May/June 2010 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. © UCLES 2010 [Turn over *9101962689*

2 1 A particle P, of mass 0.2 kg, moves in simple harmonic motion along a straight line under the action of a resultant force of magnitude F N. The distance between the end-points of the motion is 0.6 m, and the period of the motion is 0.5 s. Find the greatest value of F during the motion. [5] 2 A B P 60° 30° A uniform rod AB of weight W rests in equilibrium with A in contact with a rough vertical wall. The rod is in a vertical plane perpendicular to the wall, and is supported by a force of magnitude P acting at B in this vertical plane. The rod makes an angle of 60◦with the wall, and the force makes an angle of 30◦with the rod (see diagram). Find the value of P. [3] Find also the set of possible values of the coefficient of friction between the rod and the wall. [4] 3 A 3m m B a Two perfectly elastic small smooth spheres A and B have masses 3m and m respectively. They lie at rest on a smooth horizontal plane with B at a distance a from a smooth vertical barrier. The line of centres of the spheres is perpendicular to the barrier, and B is between A and the barrier (see diagram). Sphere A is projected towards sphere B with speed u and, after the collision between the spheres, B hits the barrier. The coefficient of restitution between B and the barrier is 1 2. Find the speeds of A and B immediately after they first collide, and the distance from the barrier of the point where they collide for the second time. [9] © UCLES 2010 9231/21/M/J/10

3 4 0.5 m 0.3 m A B P t rad s–1 1 2 Two coplanar discs, of radii 0.5 m and 0.3 m, rotate about their centres A and B respectively, where AB = 0.8 m. At time t seconds the angular speed of the larger disc is 1 2t rad s−1 (see diagram). There is no slipping at the point of contact. For the instant when t = 2, find (i) the angular speed of the smaller disc, [2] (ii) the magnitude of the acceleration of a point P on the circumference of the larger disc, and the angle between the direction of this acceleration and PA. [7] 5 b b a A light elastic band, of total natural length a and modulus of elasticity 1 2mg, is stretched over two small smooth pins fixed at the same horizontal level and at a distance a apart. A particle of mass m is attached to the lower part of the band and when the particle is in equilibrium the sloping parts of the band each make an angle β with the vertical (see diagram). Express the tension in the band in terms of m, g and β, and hence show that β = 1 4π. [4] The particle is given a velocity of magnitude √(ag) vertically downwards. At time t the displacement of the particle from its equilibrium position is x. Show that, neglecting air resistance, ¨x = −2g a x. [3] Show that the particle passes through the level of the pins in the subsequent motion, and find the time taken to reach this level for the first time. [6] © UCLES 2010 9231/21/M/J/10 [Turn over

4 6 The lifetime, X days, of a particular insect is such that log10 X has a normal distribution with mean 1.5 and standard deviation 0.2. Find the median lifetime. [3] Find also P(X ≥50). [2] 7 The continuous random variable X has distribution function given by F(x) = ( 0 x < 0, 1 −e −1 2x x ≥0. For a random value of X, find the probability that 2 lies between X and 4X. [3] Find also the expected value of the width of the interval (X, 4X). [4] 8 An examination involved writing an essay. In order to compare the time taken to write the essay by students in two large colleges, a sample of 12 students from college A and a sample of 8 students from college B were randomly selected. The times, tA and tB, taken for these students to write the essay were measured, correct to the nearest minute, and are summarised by nA = 12, Σ tA = 257, Σt2 A = 5629, nB = 8, ΣtB = 206, Σ t2 B = 5359. Stating any required assumptions, calculate a 95% confidence interval for the difference in the population means. [8] State, giving a reason, whether your confidence interval supports the statement that the population means, for the two colleges, are equal. [1] 9 A set of 20 pairs of bivariate data (x, y) is summarised by Σx = 200, Σx2 = 2125, Σy = 240, Σy2 = 8245. The product moment correlation coefficient is −0.992. (i) What does the value of the product moment correlation coefficient indicate about a scatter diagram of the data points? [1] (ii) Find the equation of the regression line of y on x. [6] (iii) The equation of the regression line of x on y is x = a′ + b′y. Find the value of b′. [2] 10 Three new flu vaccines, A, B and C, were tested on 500 volunteers. The vaccines were assigned randomly to the volunteers and 178 received A, 149 received B and 173 received C. During the following year, 30 of the volunteers given A caught flu, 29 of the volunteers given B caught flu, and 16 of the volunteers given C caught flu. Carry out a suitable test for independence at the 5% significance level. [10] Without using a statistical test, decide which of the vaccines appears to be most effective. [3] © UCLES 2010 9231/21/M/J/10

5 11 Answer only one of the following two alternatives. EITHER A B C O a a A uniform disc, of mass 4m and radius a, and a uniform ring, of mass m and radius 2a, each have centre O. A wheel is made by fixing three uniform rods, OA, OB and OC, each of mass m and length 2a, to the disc and the ring, as shown in the diagram. Show that the moment of inertia of the wheel about an axis through A, perpendicular to the plane of the wheel, is 42ma2. [5] The axis through A is horizontal, and the wheel can rotate freely about this axis. The wheel is released from rest with O above the level of A and AO making an angle of 30◦with the horizontal. Find the angular speed of the wheel when AO is horizontal. [3] When AO is horizontal the disc becomes detached from the wheel. Find the angle that AO makes with the horizontal when the wheel first comes to instantaneous rest. [6] OR The continuous random variable T has probability density function given by f(t) =   0 t < 2, 2 (t −1)3 t ≥2. (i) Find the distribution function of T, and find also P(T > 5). [3] (ii) Consecutive independent observations of T are made until the first observation that exceeds 5 is obtained. The random variable N is the total number of observations that have been made up to and including the observation exceeding 5. Find PN > E(N). [3] (iii) Find the probability density function of Y, where Y = 1 T −1. [8] © UCLES 2010 9231/21/M/J/10

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. 9231/21/M/J/10

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2010 question paper for the guidance of teachers 9231 FURTHER MATHEMATICS 9231/21 Paper 21, 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 2010 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 2010 9231 21 © UCLES 2010 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 2010 9231 21 © UCLES 2010 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 2010 9231 21 © UCLES 2010 Question Number Mark Scheme Details Part Mark Total 1 Find ω or ω2 from 2π/T: ω = 2π / 0⋅5 [= 4π = 12⋅57] B1 Relate F to acceleration: F = 0⋅2 d2x/dt2 M1 Relate acceleration to ω and x: d2x/dt2 = [–]ω 2x M1 State or use value of x giving max of F [or d2x/dt2]: Maximum when x = [±] 0⋅3 M1 Evaluate maximum Fmax of F: 0⋅2 (4π)2 0⋅3 = 0⋅96π2 or 9⋅47 A1 5 [5] 2 Take moments for rod about A (A.E.F.): 2P cos 60° = W sin 60° M1 A1 Find P: P = W √3/2 or 0⋅866W A1 Resolve vertically for friction F at A (√ on P): F = W – P cos 30° = W/4 M1 A1√ Resolve horizontally for reaction R at A (√ on P): R = P sin 30° = W√3/4 or 0433W B1√ Use F ≤ µR to find values of µ: µ ≥ 1/ √3 or µ ≥ 0⋅577 B1 3 4 [7] 3 Use conservation of momentum: 3mvA + mvB = 3mu B1 Use Newton’s law of restitution: vA – vB = – u B1 Solve for vA and vB: vA = ½ u and vB = 3u/2 M1 A1 Find rebound speed of B after collision with barrier: wB = ½ vB [= 3u/4] M1 EITHER: Find time for B to colln. at d from barrier: t1 = a / vB [= 2a/3u] and t2 = d / wB [= 8a/15u] B1 Find time for A to same collision: t1 + t2 = (a – d) / vA B1 Equate times and solve for d: 2(a – d) = 2a/3 + 4d/3, d = 2a/5 M1 A1 OR: Find dist. A moves in time t1: sA = vA × (a/vB) [= a/3] (B1) Find t2 from both A and B: t2 = (a – sA – d) / vA , t2 = d / wB (B1) Equate times and solve for d: 2(2a/3 – d)/u = 4d/3u, d = 2a/5 (M1 A1) MR: Taking vA – vB = – ½ u gives: vA = 5u/8, vB = 9u/8, wB = 9u/16 t1 = 8a/9u , t2 = 64a/171u, d = 4a/19 or taking vA – vB = – e u gives: vA = (3 – e)u/4, vB = 3(e + 1)u/4 wB = 3(e + 1)u/8, t1 = 4a/3(e + 1)u t2 = 32ea/3(e + 1)(e + 9)u d = 4ea/(e + 9) (max 8) 5 4 [9] 4 (i) (ii) Equate tangential speeds to find ωB: 0⋅5ωA = 0⋅3ωB , ωB = 5/3 [rad s-1] M1 A1 Find tangential acceleration, r d2θ/dt2: 0⋅5 × ½ = 0⋅25 M1 A1 Find radial acceleration, r (dθ/dt)2: 0⋅5 × 12 = 0⋅5 B1 Combine to give mag. of acceln: √(0⋅252 + 0⋅52) M1 = √5/4 or 0⋅559 [m s-2] A1 Find angle made with PA (A.E.F.): tan-1 (0⋅25/0⋅5) M1 = 0⋅464 rad or 26⋅6° A1 2 7 [9]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 21 © UCLES 2010 Question Number Mark Scheme Details Part Mark Total 5 Find extension and apply Hooke’s Law: T = ½mg (a / sin β) / a M1 = mg / 2 sin β A1 Resolve vertically for particle: mg = 2T cos β [T = mg / 2 cos β] B1 Equate two expressions for T: sin β = cos β, β = ¼π A.G. B1 Use Newton’s Law for vertical motion (A.E.F.): m d2x/dt2 = mg – 2T′ cos β′ = mg – (mg / sin β′) cos β′ M1 A1 Simplify: d2x/dt2 = g – g (½a + x) / ½a = – 2gx/a A.G. A1 Use v = Aω to find amplitude A of motion: A = √(ag) √(2g/a) = a/√2 M1 A1 Hence show particle reaches pins: a/√2 > a/2 A1 Use x = A sin ωt to find time t: t = (sin-1 (-½a/A))/ω or ½T + (sin-1 (½a/A))/ω M1 = (sin-1 (–1/√2)) / √(2g/a) or (π + sin-1 (1/√2)) / √(2g/a) A1 Simplify (A.E.F.): = (5π/4)√(a/2g) or 0⋅878√a A1 4 3 6 [13] 6 Find relation for median M: log10 M = 1⋅5 M1 A1 Evaluate M: M = 101⋅5 = 31⋅6 A1 Relate P(X ≥ 50) to Normal distribution: P(X ≥ 50) = P(log X ≥ log 50) = 1 – Φ((log 50 – 1⋅5) / 0⋅2) M1 [log 50 = 1⋅699] = 1 – Φ(0⋅995) = 0⋅160 A1 3 2 [5] 7 Relate P(X ≤ 2 ≤ 4X) to F(x): = P(½ ≤ X ≤ 2) = F(2) – F(½) M1 A1 Evaluate: = (1 – e-1) – (1 – e-¼) = 0⋅632 – 0⋅221 = 0⋅411 A1 EITHER: State E(X) or find using f(x) for x > 0: f(x) = ½e-½x , E(X) = 2 M1 A1 Find width of interval (X, 4X): E(3X) = 3 × E(X) = 6 M1 A1 OR: Find f(y) for Y = 4X – X: F(y) = P(X < y/3) = 1 – e-y/6 f(y) = e-y/6 /6 (M1 A1) Find width of interval (X, 4X): E(Y) = ∫ y(e-y/6 /6) dy = 6 (M1 A1) 3 4 [7] 8 State assumptions (A.E.F.): Equal variances B1 Normal populations B1 Find difference in sample means to 2 dp, e.g.: xA –xB = 21⋅417 – 25⋅75 = –4⋅33 B1 Estimate common population variance: s2 = (5629 – 2572/12 + 5359 – 2062/8) / 18 = (124⋅9 + 54⋅5)/18 = 9⋅968 or 3⋅1572 (3 sf) M1 A1 Use of correct tabular t value: t18, 0.975 = 2⋅101 (to 3 sf) B1 Find confidence interval for e.g. µA –µB : xA –xB ± ts √(1/12 + 1/8) M1 Evaluate: –4⋅33 ± 3⋅03 or [–7⋅36, –1⋅31] A1* State reason and conclusion (A.E.F.): Interval does not include zero (dep *A1 apart from rounding) so statement not supported B1 8 1 [9]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 21 © UCLES 2010 Question Number Mark Scheme Details Part Mark Total 9 (i) (ii) (iii) Valid comment on scatter diagram (A.E.F.): Approx str. line with neg. gradient B1 EITHER: Find gradient b directly using r: b = r √(Syy / Sxx ) M2 = r √{(8245 – 2402/20) / (2125 – 2002/20)} A1 = – 0⋅992 √ (5365 / 125) OR: Find Sxy (to 3 sf): Sxy = r √(Sxx Syy ) = – 812⋅37 (M1 A1) Find b: b = Sxy / Sxx [= – 812⋅37/125] (M1) Evaluate b: [= – 6⋅499] = – 6⋅50 A1 Find equation of regression line: y = b(x – 10) + 12 M1 = 77⋅0 – 6⋅50 x A1 Find b′ using r2 = bb′: [or Sxy / Syy ] b′ = –0⋅9922/6⋅499 = –0⋅151 M1 A1 1 6 2 [9] 10 Tabulate observed data with totals: A B C Flu 30 29 16 75 No flu 148 120 157 425 178 149 173 M1 Find expected values: A B C (lose A1 if 1 or 2 errors; Flu 26⋅7 22⋅35 25⋅95 lose A1 if rounded to integers) No flu 151⋅3 126⋅65 147⋅05 M1 A2 State (at least) null hypothesis (A.E.F.): H0: Catching flu indep of vaccine B1 Calculate value of χ2 (to 2 dp): χ2 = 7⋅30 M1 A1 S.R. If rounded to integers above allow (to 2 dp): χ2 = 7⋅53 (earns max 8/10) (B1) Compare with consistent tabular value (to 2 dp): χ2, 0.95 2 = 5⋅991 B1 Valid method for reaching conclusion: Reject H0 if χ2 > tabular value M1 Correct conclusion (A.E.F., requires correct values): Catching flu depends on vaccine A1 Find proportions (or complements) for A,B,C: 0⋅169, 0⋅195, 0⋅092 (to 2 dp) M1 A1 Correct conclusion (A.E.F., requires correct values): C appears most effective A1 10 3 [13] 11 EITHER Find MI of disc about O [or A]: Idisc = ½ 4ma2 [= 2ma2 or 18ma2] B1 Find MI of ring about O [or A]: Iring = m(2a)2 [= 4ma2 or 8ma2] B1 Find MI of AO about O [or BO about A]: Irod = (4/3)ma2 [or 22ma2 /3] B1 Find MI of wheel about A: Iwheel = 10ma2 + 8m(2a)2 M1 = 42ma2 A.G. A1 Find angular speed ω using energy: ½ Iwheel ω2 = 8mg × 2a sin 30° M1 A1 ω2 = 8mga2 /21ma2 ω = √(8g/21a) or 1⋅95/√a (A.E.F.) A1 Find new MI about A: Inew = 8ma2 + 4m(2a)2 = 24ma2 M1 A1 Find reqd. angle θ using energy: ½ Inew ω2 = Mnewg × 2a sin θ M1 Find and use new mass: Mnew = m + 3m = 4m A1 Substitute for Inew, Mnew, ω2: (32/7)mga = 8mga sin θ (A.E.F.) A1 Solve for θ: θ = sin-1(4/7) = 0⋅608 rad or 34⋅8° A1 5 3 6 [14]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 21 © UCLES 2010 Question Number Mark Scheme Details Part Mark Total 11 OR (i) (ii) (iii) Integrate to find F(t) for t ≥ 2 [c needed]: F(t) = c – (t – 1)-2 M1 Use F(2) = 0 to find c: F(t) = 1 – (t – 1)-2 A1 Find p = P(T > 5): p = 1 – F(5) = 1 – (1 – 4-2) = 1/16 B1 State or imply distribution: P(N > n) = p (1 – p)n-1 or geometric distn. with par. p M1 Find P(N > E(N)): (1 – p)1/p = (15/16)16 = 0⋅356 M1 A1 Relate dist. fn. G(y) of Y to T: G(y) = P(Y < y) = P(1/(T – 1) < y) M1 Rearrange : = P(T > 1 + 1/y) A1 Relate to dist. fn. F: = 1 – F(1 + 1/y) M1 Substitute expression for F: = 1 – {1 – (1 + 1/y – 1) -2} A1 Simplify: = y2 A1 Differentiate to find prob. density fn: g(y) = 2y M1 A1 Give complete statement of g(y): g(y) = 2y (0 ≤ y ≤ 1), 0 otherwise A1 3 3 8 [14]