Cambridge A Level Mathematics - Further 9231 — 2008 May/June Paper 2 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/02 Paper 2 May/June 2008 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 ms−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 2008 [Turn over *6894056850*

2 1 A bullet of mass 20 grams is fired horizontally into a fixed vertical barrier. It enters the barrier horizontally with speed 240 m s−1 and emerges 0.004 s later with speed 10 m s−1. Assuming that the motion occurs in a horizontal straight line, find (i) the magnitude of the impulse that acts on the bullet, [2] (ii) the magnitude of the resisting force, assuming that it is constant. [2] 2 A particle P moves on a straight line AOA′ in simple harmonic motion with period T. The centre of the motion is O and P is instantaneously at rest at A and A′. The mid-points of OA and OA′ are B and B′ respectively. (i) Find, in terms of T, the time for P to move directly from B to B′. [5] (ii) Find the ratio of the speed of P at B to its speed at O. [3] 3 A B 10 N x m 60° A light rod AB of length 0.8 m has a particle of weight 10 N attached to a point at a distance x m from A. The rod rests in equilibrium in a horizontal position with A against a rough vertical wall and B on a smooth plane inclined at 60◦to the horizontal (see diagram). Find, in terms of x, the magnitude of the force that the inclined plane exerts on the rod. [2] The rod can remain in equilibrium when the particle is attached to any point on the half of the rod nearer to B. Find the least possible value of the coefficient of friction between the rod and the wall. [6] © UCLES 2008 9231/02/M/J/08

3 4 P Q u ku tan–1 4 3( ( tan–1 3 4( ( Two identical smooth balls, P and Q, are projected simultaneously towards each other from two points on horizontal ground. P is projected with speed u at an angle tan−1
4 3 to the horizontal and Q is projected with speed ku at an angle tan−1
3 4 to the horizontal (see diagram). The balls collide when they are moving horizontally. It may be assumed that there is no air resistance. Find the value of k. [5] The coefficient of restitution between the balls is e. Find, in terms of e, u and g, the distance between the points where the balls first hit the ground. [5] 5 A B C D O A uniform circular disc has radius 0.5 m, centre O and mass per unit area 6.20 kg m−2. A rectangle ABCD, where AC and BD intersect at O, is removed (see diagram). The lengths of AB and BC are 0.4 m and 0.3 m respectively. Show that the moment of inertia of the resulting lamina, about an axis through A perpendicular to its plane, is 0.851 kg m2, correct to 3 significant figures. [8] The lamina is free to rotate in a vertical plane about a smooth horizontal axis through A. It is held with AB horizontal and with DC below AB. It is then given an initial angular speed ω rad s−1, so that B moves upwards. Find the least possible value of ω for the lamina to make complete revolutions. [5] 6 Cylindrical metal tubes have internal diameters which are normally distributed with mean 1.275 cm and standard deviation 0.015 cm. Rods, which are intended to be inserted in the tubes, are made from a different metal and have diameters which are normally distributed with mean 1.245 cm and standard deviation 0.028 cm. Both tubes and rods are kept in a temperature-controlled store. On a particular day there is a system failure and the diameters of all tubes increase by 2% and the diameters of all rods increase by 4%. Find the probability that a randomly chosen rod cannot be inserted into a randomly chosen tube. [6] © UCLES 2008 9231/02/M/J/08 [Turn over

4 7 The time for a certain kind of plaster to dry has a normal distribution with mean µ minutes. In order to find a confidence interval for µ, the times for a random sample of 10 specimens to dry were measured. The results, x minutes, are summarised by Σ x = 285.4, Σ(x −x)2 = 448.65. Calculate a 95% confidence interval for µ. [5] Estimate the width of a 95% confidence interval for µ based on a random sample of 20 specimens. [3] 8 The number of serious faults on a randomly chosen 1 km stretch of motorway has a Poisson distribution with mean 2.1. The random variable X km is the distance between two successive serious faults on the motorway. (i) Explain, using a Poisson distribution, why, for x ≥0, P(X > x) = e−2.1x. [3] (ii) Obtain the distribution function and the probability density function of X. [3] (iii) Obtain the mean distance between successive serious faults. [1] (iv) Find the median distance between successive serious faults. [2] 9 A study was carried out to investigate the effect of moderate exercise on blood cholesterol levels. The levels were measured before and after a training programme administered to a random sample of 8 people. The results, in suitable units, are given in the following table. Person 1 2 3 4 5 6 7 8 Before 195 209 184 215 295 198 310 191 After 197 201 192 204 266 183 272 168 Stating any assumption you need to make, test at the 10% significance level whether, after the training programme, (i) there is a reduction in the population mean cholesterol level, [7] (ii) there is a reduction of more than 5 units in the population mean cholesterol level. [3] 10 The time in the air, x seconds, and the score, y, for 12 randomly chosen divers competing at a diving competition for a reverse half-somersault event are summarised as follows. Σ x = 5.22 Σ x2 = 2.2774 Σ y = 85.9 Σ y2 = 618.11 Σ xy = 37.338 Obtain an estimate for the score of a diver who takes 0.40 seconds. [4] Find the product moment correlation coefficient for the sample. [2] Stating your hypotheses, test at the 5% significance level whether there is a non-zero correlation between the two variables. [4] © UCLES 2008 9231/02/M/J/08

5 11 Answer only one of the following two alternatives. EITHER A particle P of mass m moves on an arc of a circle with centre O and radius a. At time t = 0 the particle is at the point A and at time t the angle POA = θ, where θ = sin2kt and k is a positive constant. Find constants b and c such that ¨θ = b + cθ, and deduce that P moves in simple harmonic motion along the arc. State the centre and amplitude of the motion. [9] Show that the magnitude of the resultant force acting on P at time t is mak2(1 + cos2 2kt). [5] OR Each of 100 coins is tossed repeatedly until a head is obtained, and the number, x, of tosses needed is recorded. The results are summarised in the following table. x 1 2 3 4 5 ≥6 Frequency 56 20 11 10 3 0 Assume that, for each coin, the probability of obtaining a head has the same value p. State a distribution which could be used to fit the above data, and use the sample mean to obtain 0.5435, correct to 4 decimal places, as an estimate of p. [4] Carry out a goodness of fit test at the 21 2% level. [8] Without further calculation, state two changes which would be necessary if a similar method is to be used on the data above to test whether all the coins are fair. [2] © UCLES 2008 9231/02/M/J/08

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. 9231/02/M/J/08

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2008 question paper 9231 FURTHER MATHEMATICS 9231/02 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. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. 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 2008 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2008 9231 02 © UCLES 2008 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 – May/June 2008 9231 02 © UCLES 2008 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 – May/June 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total 1 (i) (ii) Find impulse I from change in momentum: ± I = 0.02 (240 – 10) = 4.6 [N s] M1 A1 Find force F from I/t: ± F = I / 0.004 = 1150 [N] M1 A1 2 2 [4] 2 (i) (ii) State or use relation between T and ω: T = 2π /ω B1 Use trigonometric form for SHM, e.g.: x = a sin ωt or a cos ωt M1 Find time to move BO or OB′, e.g.: (1/ω) sin–1 ½ or (1/ω) (cos–1 0 – cos–1 ½) A1 Replace sin–1 or cos–1 and relate to T: (T/2π) (π/6) or (T/2π) (π/2 – π/3) A1 Find time to move BB′: 2 × T/12 = T/6 A1 Use SHM formula for v or v2, e.g.: v = aω cos ωt or v2 = ω2 (a2 – x2) M1 Form ratio vB / vO or vB 2 / vO 2: aω cos (sin–1 ½) / aω cos (sin–1 0) or (a2 – a2/4) / (a2 – 0) A1 Evaluate ratio: cos (π/6) / cos 0 = √3/2 or 0.866 A1 5 3 [8] 3 Take moments for rod about A: RB sin 30° × 0.8 = 10x M1 Evaluate force RB at B: RB = 25x A1 Resolve horizontally and vertically for rod: FA = 10 – RB sin 30° and RA = RB cos 30° M1 A1 Use FA Ğ µ RA (or <; A.E.F.): µ [ (10 – 12.5x) / 25x √3/2 M1 A1 Find µmin by putting x = 0.4: µmin = 5/10√3/2 = 1/√3 or 0.577 M1 A1 S.R. Using FA = µ RA without justification: µ = (10 – 12.5x) / 25x √3/2 (B1) Using x [ 0.4 to give: µ Ğ 1/√3 (B1) 2 6 [8] 4 Equate vertical speeds to zero (here tan α = 4/3): u sin α – gt = 0 = ku cos α – gt or: (u sin α)2 – 2gs = (ku cos α)2 – 2gs or equate vertical distances at collision: ut sin α – ½gt2 = kut cos α – ½gt2 M1 A1 Simplify: u sin α = ku cos α A1 Evaluate k: k = 4/3 M1 A1 Find time t to reach ground: t = (u sin α) / g = 4u/5g B1 Find speed of separation (ignore sign): vP – vQ = – e (uP – uQ) M1 Substitute for uP , uQ: vP – vQ = – e (u cos α + ku sin α) (ignore sign) = – eu (3 + 4k)/5 [= – 5eu/3] A1 Find distance apart: |vP – vQ | t = 4eu2/3g M1 A1 5 5 [10]

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total 5 Find moment of inertia of disc: Io = ½ (π × 0.52 × 6.2) 0.52 M1 A1 Find moment of inertia of rectangle: I□ = ⅓(0.4 × 0.3 × 6.2)(0.22 + 0.152) M1 A1 Combine to give M.I. of lamina about O: I = Io – I□ [= 0.6087 – 0.0155] M1 Apply parallel axes theorem for M.I. about A: I′ = I + M′ AO2 [= I + M′ 0.252], M1 and M′ = (π 0.5² – 0.4 × 0.3) 6.2 A1 [= 4.869 – 0.744 = 4.1255] Evaluate I′: 0.5932 + 0.2578 or 0.913 – 0.062 = 0.851 A.G. A1 Consider energy E when O rises by some h: E = ½ I′ω2 – M′gh M1 Find ωmin from E = 0 or E [ 0 when AO vertical M1 Find h when AO vertical: h = ½AD + AO = 0.4 A1 Evaluate ωmin: ωmin = √ (4.126gh / ½ 0.851) A1 = 6.23 A1 8 5 [13] 6 Consider distn. Y of differences in diameters (or –Y): Y = 1.02 T – 1.04 R M1 Find E(Y): E(Y) = 1.3005 – 1.2948 [= 0.0057] A1 Find Var(Y): Var(Y) = 1.022 × 0.0152 + 1.042 × 0.0282 [=0.001082] A1 Find P(Y < 0): P(Y < 0) = Φ(–E(Y) / √Var(Y)) M1 (Condone incorrect sign here) = Φ(–0.0057 / √0.001082) or Φ(–0.0057 / 0.03289) A1 (0.569 earns 5/6) = Φ(–0.1733) = 0.431 A1 6 [6] 7 Use valid formula for C.I. with any t (or z) value: x ± t√(s2/n) with n = 10 (or 9) M1 Find mean of sample: x = 28.54 A1 Find population variance consistent with n (to 2 dp): s2 = 49.85 or 7.062 (n = 10) or 44.865 or 6.6982 (n = 9) A1 Use correct tabular t value: t9, 0.975 = 2.262 (to 3 sf) *A1 Evaluate C.I. (dep *A1): 28.54 ± 5.05 or [23.5, 33.6] A1 Use valid formula for width (or CI) with any t value: 2 t√((448.65 / 9) / 20) M1 Use correct tabular t value: t19, 0.975 = 2.093 (to 3 sf) *A1 Evaluate width of C.I. (dep *A1): 6.61 A1 5 3 [8]

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total 8 (i) (ii) (iii) (iv) Relate P(X > x) to number of faults: Equivalent to 0 faults in x km M1 State or use distribution N of faults in x km: Poisson with mean 2.1x A1 Find P(X > x): P(X > x) = P(N = 0) = e–2.1x A.G. A1 Find distribution function of X: F(x) = 1 – P(X > x) = 1 – e–2.1x B1 Find probability density function of X: f(x) = F′(x) = 2.1 e–2.1x M1 A1 Find mean distance: 1/2.1 = 10/21 or 0.476 [km] B1 Find median m from F(m) or ∫0 m f(x) = ½: 1 – e–2.1m = ½, m = 0.330 [km] M1 A1 3 3 1 2 [9] 9 (i) (ii) State hypotheses: H0: µB – µA = 0, H1: µB – µA > 0 B1 State assumption (AEF): Population of differences is normal B1 Consider differences B – A (or A – B): –2 8 –8 11 29 15 38 23 M1 Estimate mean and population variance: x = 114/8 [= 14.25] and (allow biased: 208.44 or 14.4372) s2 = (3292 – 8 × 14.252) / 7 M1 [= 238.21 or 15.4342 ] Calculate value of t: t = x /(s/√8) = 2.61[1] *B1 Use correct tabular t value: t7, 0.9 = 1.41[5] *B1 Correct conclusion (AEF, dep both *B1): There is a reduction B1 Calculate t (M0 if inconsistent with s): t = ( x – 5) /(s/√8) = 1.69[5] M1*A1 Correct conclusion by comparing with t7, 0.9 = 1.41[5] (AEF, dep *A1): There is a reduction > 5 units B1 7 3 [10]

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total 10 Find coefficient b in regression line for y: b = (37.338 – 5.22 × 85.9/12) / (2.2774 – 5.222/12) SR: Penalise rounding error in b here only = –0.002375 /0.00055833 or –0.0285 /0.0067 = –4.25[4] B1 Find equation of regression line: y = b(x – 0.435) + 7.158 (√ on rounding error in b) = 9.01 – 4.25 x M1 A1√ SR: B1 for x on y: x = 0.499 – 0.008881 y Estimate y when x = 0.4 (√ on rounding error in b): y = 7.31 B1√ Find r: r = (37.338 – 5.22 × 85.9/12) / √{(2.2774 – 5.222/12)(618.11 – 85.92/12)} M1 = –0.194 ± 0.001 *A1 State hypotheses: H0: ρ = 0, H1: ρ ≠ 0 B1 Reject H0 if |r| exceeds critical value: r12, 2.5% = 0.576 M1 *A1 Correct conclusion (AEF, dep both *A1): Variables are not correlated A1 4 2 4 [10] 11 EITHER Reasonable attempt to differentiate θ twice: M1 dθ/dt = 2k sin kt cos kt or k sin 2kt A1 d2θ/dt2 = 2k2 (cos2 kt – sin2 kt) or 2k2 cos 2kt A1 Rearrange in the form b + cθ: d2θ/dt2 = 2k2 (1 – 2 sin2 kt) = 2k2 – 4k2 θ M1 A1 Show SHM by e.g. change of variable: φ = θ – ½, d2φ /dt2 = – 4k2 φ M1 A1 State centre and amplitude: Centre θ = ½ ; amplitude ½ B1; B1 Find tangential force (AEF in terms of sin or cos): ma d2θ/dt2 = 2mak2 cos 2kt B1 Find radial force (AEF in terms of sin or cos): ma (dθ/dt)2 = mak2 sin2 2kt B1 Find magnitude of resultant force: mak2 √(4 cos2 2kt + sin4 2kt) M1 Simplify: mak2 √(4 cos2 2kt + (1 – cos2 2kt)2) or mak2 √(4 – 4 sin2 2kt + sin4 2kt) M1 Hence result: mak2 (1 + cos2 2kt) A.G. A1 9 5 [14]

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total OR State distribution: Geometric B1 Find sample mean: x = 1.84 B1 Estimate p from sample mean: 1 / x = 0.5435 A.G. M1 A1 Calculate expected values (correct to ± 0.01): 54.35 24.81 11.33 5.17 2.36 1.98 *M1 A1 Combine values as appropriate: Last 3 to give 9.51 (dep *M1) M1√ Calculate value of χ2 : χ2 ≈ 2.27 [± 0.01] M1 *A1 Compare with consistent tab. value (to 2 dp): χ2, 0.975 2 = 7.378 [χ4 2 = 11.14] *B1√ Consistent conclusion (A1 dep *A1, *B1): Distribution fits data (A.E.F.) M1√ A1 State 2 changes (A.E.F.): Change estimated p to 0.5 B1 Increase degrees of freedom by 1 B1 4 8 2 [14]