Cambridge A Level Mathematics - Further 9231 — 2010 Oct/Nov Paper 2 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/02 Paper 2 October/November 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 *7842478686*

2 1 A particle P is describing simple harmonic motion of amplitude 5 m. Its speed is 6 m s−1 when it is 3 m from the centre of the motion. Find, in terms of π, the period of the motion. [2] Find also (i) the maximum speed of P, [2] (ii) the magnitude of the maximum acceleration of P. [2] 2 A particle P of mass m is projected horizontally with speed u from the lowest point on the inside of a fixed hollow sphere with centre O. The sphere has a smooth internal surface of radius a. Assuming that the particle does not lose contact with the sphere, show that when the speed of the particle has been reduced to 1 2u the angle θ between OP and the downward vertical satisfies the equation 8ga(1 −cos θ) = 3u2. [2] Find, in terms of m, u, a and g, an expression for the magnitude of the contact force acting on the particle in this position. [4] 3 Two smooth spheres A and B, of equal radius, are moving in the same direction in the same straight line on a smooth horizontal table. Sphere A has mass m and speed u and sphere B has mass αm and speed 1 4u. The spheres collide and A is brought to rest by the collision. Find the coefficient of restitution in terms of α. [6] Deduce that α ≥2. [2] 4 A B C q q q 2 A hemispherical bowl of radius r is fixed with its rim horizontal. A thin uniform rod rests in equilibrium on the rim of the bowl with one end resting on the inner surface of the bowl at A, as shown in the diagram. The rod has length 2a and weight W. The point of contact between the rod and the rim is B, and the rim has centre C. The rod is in a vertical plane containing C. The rod is inclined at θ to the horizontal and the line AC is inclined at 2θ to the horizontal. The contacts at A and B are smooth. In any order, show that (i) the contact force acting on the rod at A has magnitude W tan θ, (ii) the contact force acting on the rod at B has magnitude W cos 2θ cos θ , (iii) 2r cos 2θ = a cos θ. [9] © UCLES 2010 9231/02/O/N/10

3 5 A uniform circular disc has diameter AB, mass 2m and radius a. A particle of mass m is attached to the disc at B. The disc is able to rotate about a smooth fixed horizontal axis through A. The axis is tangential to the disc. Show that the moment of inertia of the system about the axis is 13 2 ma2. [4] The disc is held with AB horizontal and released. Find the angular speed of the system when B is directly below A. [5] The disc is slightly displaced from the position of equilibrium in which B is below A. At time t the angle between AB and the vertical is θ. Write down the equation of motion, and find the approximate period of small oscillations about the equilibrium position. [5] 6 The mean Intelligence Quotient (IQ) of a random sample of 15 pupils at School A is 109. The mean IQ of a random sample of 20 pupils at School B is 112. You may assume that the IQs for the populations from which these samples are taken are normally distributed, and that both distributions have standard deviation 15. Find a 90% confidence interval for µB −µA, where µA and µB are the population mean IQs. [6] 7 The discrete random variable X has a geometric distribution with mean 4. Find (i) P(X = 5), [3] (ii) P(X ≥5), [2] (iii) the least integer N such that P(X ≤N) > 0.9995. [2] 8 The owner of three driving schools, A, B and C, wished to assess whether there was an association between passing the driving test and the school attended. He selected a random sample of learner drivers from each of his schools and recorded the numbers of passes and failures at each school. The results that he obtained are shown in the table below. Driving school attended A B C Passes 23 15 17 Failures 27 25 43 Using a χ2-test and a 5% level of significance, test whether there is an association between passing or failing the driving test and the driving school attended. [7] © UCLES 2010 9231/02/O/N/10 [Turn over

4 9 A national athletics coach suspects that, on average, 200-metre runners’ indoor times exceed their outdoor times by more than 0.1 seconds. In order to test this, the coach randomly selects eight 200-metre runners and records their indoor and outdoor times. The results, in seconds, are shown in the table. Runner A B C D E F G H Indoor time 21.5 21.8 20.9 21.2 21.4 21.4 21.2 21.0 Outdoor time 21.1 21.7 20.7 20.9 21.3 21.0 21.1 20.8 Stating suitable hypotheses and any necessary assumption that you make, test the coach’s suspicion at the 2.5% level of significance. [10] 10 For each month of a certain year, a weather station recorded the average rainfall per day, x mm, and the average amount of sunshine per day, y hours. The results are summarised below. n = 12, Σx = 24.29, Σx2 = 50.146, Σy = 45.8, Σy2 = 211.16, Σxy = 88.415. (i) Find the mean values, x and y. [1] (ii) Calculate the gradient of the line of regression of y on x. [2] (iii) Use the answers to parts (i) and (ii) to obtain the equation of the line of regression of y on x. [2] (iv) Find the product moment correlation coefficient and comment, in context, on its value. [4] (v) Stating your hypotheses, test at the 1% level of significance whether there is negative correlation between average rainfall per day and average amount of sunshine per day. [4] © UCLES 2010 9231/02/O/N/10

5 11 Answer only one of the following two alternatives. EITHER A particle of mass 0.1 kg lies on a smooth horizontal table on the line between two points A and B on the table, which are 6 m apart. The particle is joined to A by a light elastic string of natural length 2 m and modulus of elasticity 60 N, and to B by a light elastic string of natural length 1 m and modulus of elasticity 20 N. The mid-point of AB is M, and O is the point between M and B at which the particle can rest in equilibrium. Show that MO = 0.2 m. [4] The particle is held at M and then released. Show that the equation of motion is d2y dt2 = −500y, where y metres is the displacement from O in the direction OB at time t seconds, and state the period of the motion. [5] For the instant when the particle is 0.3 m from M for the first time, find (i) the speed of the particle, [2] (ii) the time taken, after release, to reach this position. [3] OR The continuous random variable T has a negative exponential distribution with probability density function given by f(t) =  λe−λt t ≥0, 0 otherwise. Show that for t ≥0 the distribution function is given by F(t) = 1 −e−λt. [2] The table below shows some values of F(t) for the case when the mean is 20. Find the missing value. [2] t 0 5 10 15 20 25 30 35 40 F(t) 0 0.2212 0.3935 0.6321 0.7135 0.7769 0.8262 0.8647 It is thought that the lifetime of a species of insect under laboratory conditions has a negative exponential distribution with mean 20 hours. When observation starts there are 100 insects, which have been randomly selected. The lifetimes of the insects, in hours, are summarised in the table below. Lifetime (hours) 0 −5 5 −10 10 −15 15 −20 20 −25 25 −30 30 −35 35 −40 ≥40 Frequency 20 20 11 9 9 8 5 1 17 Calculate the expected values for each interval, assuming a negative exponential model with a mean of 20 hours, giving your values correct to 2 decimal places. [3] Perform a χ2-test of goodness of fit, at the 5% level of significance, in order to test whether a negative exponential distribution, with a mean of 20 hours, is a suitable model for the lifetime of this species of insect under laboratory conditions. [7] © UCLES 2010 9231/02/O/N/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. 9231/02/O/N/10

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 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, 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 October/November 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 – October/November 2010 9231 02 © 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 – October/November 2010 9231 02 © 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 – October/November 2010 9231 02 © UCLES 2010 Question Number Mark Scheme Details Part Mark Total 1 Find period T using v2 = ω2 (A2 – x2) and T = 2π/ω: ω = 6/4, T = 4π/3 or 4⋅19 [s] M1 A1 (i) Find max speed using vmax = ωA: vmax = 15/2 or 7⋅5 [ms-1] M1 A1 (ii) Find mag. of max accel. using amax = ω2A: amax = 45/4 or 11⋅2[5] [ms-2] M1 A1 2 2 2 [6] 2 Apply conservation of energy: ½mv2 = ½mu2 – mga(1 – cos θ) M1 Put v = ½u and simplify: 8ga(1 – cos θ) = 3u2 A.G. A1 Equate radial forces to find contact force N: N = mg cos θ + m(½u)2/a M1 A1 Replace cos θ by 1 – 3u2/8ga (A.E.F.): N = mg – mu2/8a M1 A1 2 4 [6] 3 Use conservation of momentum: mu + ¼α mu = α mvB M1 A1 Use Newton’s law of restitution: – vB = – e(u – ¼u) [vB = ¾eu] M1 A1 Eliminate vB to find e (A.E.F.): e = (1 + ¼α)/¾α or (4 + α)/3α M1 A1 Use e ≤ 1 to find inequality for α: 4 + α ≤ 3α so α ≥ 2 A.G. M1 A1 6 2 [8] 4 Resolve in any two dirns. for rod, e.g. vertically: RA sin 2θ + RB cos θ = W or horizontally: RA cos 2θ – RB sin θ = 0 or parallel to rod: RA cos θ = W sin θ or normal to rod: RA sin θ + RB = W cos θ B1 B1 (i) Solve to find RA, e.g.: RA = W sin θ / (cos 2θ cos θ – sin 2θ sin θ) = W tan θ A.G. M1 A1 (ii) Solve to find RB, e.g.: RB = W tan θ cos 2θ / sin θ = W cos 2θ / cos θ A.G. M1 A1 (iii) Take moments for rod, e.g. about A: RB 2r cos θ = W a cos θ or about B: RA 2r cos θ sin θ = W (2r cos θ – a) cos θ M1 A1 Substitute and simplify: 2r cos 2θ = a cos θ A.G. A1 2 2 2 3 [9] 5 Find MI of disc about axis at A by par. axes thm: Idisc = ¼ 2ma2 + 2ma2 [= 5ma2 /2] M1 A1 Find MI of particle about axis at A: Im = m(2a)2 [= 4ma2 ] B1 Combine to find MI of system : I = 13ma2 /2 A.G. A1 Use conservation of energy (lose A1 for one error): ½IΩ2 = 2mg × a + mg × 2a M1 A2 Substitute for I to find angular speed Ω: Ω = √(16g/13a) A.E.F. M1 A1 State eqn of motion (A.E.F.): I d2θ /dt2 = – 4mga sin θ M1 A1 Approximate sin θ by θ (implied by use of SHM): I d2θ /dt2 = – 4mga θ M1 Find approx. period T from SHM formula: T = 2π /√(8mga/13ma2) = 2π √(13a/8g) A.E.F. M1 A1 4 5 5 [14] 6 Use valid formula for C.I.: x B – x A ± zσ √(1/nA + 1/nB) M2 = 112 – 109 ± z 15 √(1/15 + 1/20) A1 = 3 ± 5⋅123 z A1 Use of correct tabular value: z 0.995 = 1⋅64[5] *A1 C.I. correct to 3 s.f. (dep *A1): 3 ± 8⋅43 or [-5⋅43, 11⋅4] A1 6 [6] 7 (i) Find or imply value of p: p = ¼ or 0⋅25 B1 Find P(X = 5): (1 – p)4 p or q4 p = 0⋅0791 M1 A1 (ii) Find P(X ≥ 5): 1 – (1 + q + q2 + q3) p or q4 or q4p + q5 = 0⋅316 M1 A1 (iii) Find least N with P(X ≤ N) > 0⋅9995: 1 – qN > 0⋅9995, qN < 0⋅0005 N > 26⋅4, Nmin = 27 M1 A1 3 2 2 [7]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2010 9231 02 © UCLES 2010 8 Find expected values to (at least) 1 dp: A B C (lose A1 if one or more errors Passes 18⋅33 14⋅67 22⋅00 or if rounded to integers) Failures 31⋅67 25⋅33 38⋅00 M1 A1 State (at least) null hypothesis (A.E.F.): H0: Test result indep of school B1 Calculate value of χ2 : χ2 = 3⋅7 ± 0⋅02 B1 S.R. If rounded to integers above allow: χ2 = 3⋅96 or 4⋅0 (earns max 6/7) (B1) Compare with tabular value (to 2 dp): χ2, 0.95 2 = 5⋅99 B1 Valid method for reaching conclusion: Reject H0 if χ2 > tabular value M1 Correct conclusion (A.E.F., requires correct values): No association A1 7 [7] 9 State both hypotheses (A.E.F.): H0: µI – µO = 0⋅1, H1: µI – µO > 0⋅1 B1 State valid assumption for paired-sample test: Popln. of diffs. has Normal distn. B1 Consider differences eg: 0⋅4 0⋅1 0⋅2 0⋅3 0⋅1 0⋅4 0⋅1 0⋅2 M1 Calculate sample mean: d = 1⋅8 / 8 [= 0⋅225] M1 Estimate population variance: s2 = (0⋅52 – 1⋅82/8) / 7 (allow biased: 0⋅0144 or 0⋅1202) [= 0⋅0164 or 0⋅1282 ] M1 Calculate value of t (to 2 dp): t = ( d – 0⋅1) / s√(1/8) = 2⋅76 M1 *A1 Compare with correct tabular t value: t7, 0.975 = 2⋅36[5] *B1 Valid method for reaching conclusion: Reject H0 if χ2 > tabular value M1 Correct conclusion (AEF, dep *A1, *B1): Coach’s suspicion is correct A1 S.R.: State both hypotheses: H0: µI – µO = 0⋅1, H1: µI – µO > 0⋅1 (B1) State valid assumption for 2-sample test: Both poplns. have Normal distns. and a common variance (B1) Calculate sample means: 170⋅4/8, 168⋅6/8 [= 21⋅3, 21⋅075] and estimate population variance: s2 = (3630⋅1 – 170⋅42/8 + 3553⋅94 – 168⋅62/8)/14 [= 0⋅09107] (M1) Calculate value of t (to 2 dp): (0⋅225 – 0⋅1)/s√(1/8 + 1/8) = 0⋅828 (M1*A1) Compare with correct tabular t value: t14, 0.975 = 2⋅14[5] (*B1) Correct conclusion (AEF, dep *A1, *B1): Coach’s suspicion is not correct (B1 max 7) 10 [10] 10 (i) Find mean values to 3 s.f.: x = 2⋅024, y = 3⋅817 B1 (ii) Calculate gradient b in y –y = b(x –x): b = (88⋅415 – 24⋅29 × 45⋅8/12) / (50⋅146 – 24⋅292/12) M1 = – 4⋅292 / 0⋅979 or – 0⋅358 / 0⋅0816 = – 4⋅38[4] A1 (iii) Find regression line: y – 3⋅817 = – 4⋅384 (x – 2⋅024) M1 y = 12⋅7 – 4⋅38x A1 (iv) Find correlation coefficient r: r = (88⋅415 – 24⋅29 × 45⋅8/12) / √{(50⋅146 – 24⋅292/12) (211⋅16 – 45⋅82/12)} M1 = – 4⋅292 / √(0⋅979 × 36⋅36) A1 or – 0⋅358 / √(0⋅0816 × 3⋅03) = – 0⋅719 A1 State valid comment in context (A.E.F.): [Moderate] negative correlation between rainfall and sunshine A1 (v) State both hypotheses: H0: ρ = 0, H1: ρ < 0 B1 Use correct tabular r value: r12, 1% = 0⋅658 B1 Valid method for reaching conclusion: Reject H0 if |r| > tabular value M1 S.R. Calculate t-value: t = r√10 / √(1 – r2) = – 3⋅27 (B1) Use correct tabular t value: t10, 0.99 = 2⋅76[4] (B1) Correct conclusion (needs values correct): There is negative corrln. (A.E.F.) A1 1 2 2 4 4 [13]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2010 9231 02 © UCLES 2010 11 EITHER Equate tensions at O (A1 for each): 60 (1 + MO)/2 = 20 (2 – MO)/1 M1 A1 A1 Simplify to evaluate MO: MO = 10/50 = 0⋅2 A.G. A1 Apply Newton’s law at general point: 0⋅1 d2y/dt2 = (lose A1 for each incorrect term) 20 (1⋅8 – y) – 60 (1⋅2 + y)/2 M1 A2 Simplify: d2y/dt2 = – 500y A.G. A1 State period (A.E.F.): T = 2π/√500 or π /5√5 or 0⋅281 [s] B1 (i) Find speed v when first at 0⋅3 from M: v2 = 500 (0⋅22 – 0⋅12) M1 v = √15 or 3⋅87 [m s-1] A1 (ii) Find time t to reach this point: t = (1/ω) cos-1 (-0⋅1/0⋅2) (A.E.F.) or ¼ T + (1/ω) sin-1 (0⋅1/0⋅2) M1 = (2π/3) /ω or (π/2 + π/6) /ω A1 = 2⋅094/√500 [or = 0⋅07025 + 0⋅02342] = 0⋅0937 [s] A1 4 5 2 3 [14] 11 OR Integrate f(t) to find F(t): F(t) = ∫0 t λe-λx dx = [– e-λx] 0 t = 1 – e-λt A.G. M1 A1 EITHER: Deduce λ directly from mean: λ = 1/20 or 0⋅05 OR: Deduce λ from a tabular value, e.g.: 1 – e-40λ = 0⋅8647, λ = 0⋅05 M1 Substitute for λ and put t = 15 to give F(15) to 4 dp: 1 – e-15/20 = 0⋅5276 [or 0⋅5277] A1 Calculate expected values to 2 dp (5 values earn A1): 22⋅12 17⋅23 13⋅41 10.45 8.14 6.34 4.93 3⋅85 13⋅53 M1 A2 State (at least) null hypothesis: H0: 1 – e-t/20 fits data (A.E.F.) B1 Combine two adjacent cells with exp. value < 5: O: . . . 8 6 17 E: . . . 6⋅34 8⋅78 13⋅53 M1 Calculate value of χ2 (to 2 dp): χ2 = 3⋅58 M1 A1 (Cells not combined gives 4⋅81 earning M1 A0, max 4/7) Compare with consistent tabular value (to 2 dp): χ7, 0.95 2 = 14⋅07 (cells combined) χ8, 0.95 2 = 15⋅51 (not combined) B1 Valid method for reaching conclusion: Reject H0 if χ2 > tabular value M1 Correct conclusion (A.E.F., requires correct values): 3⋅58 < 14⋅07 so suitable model A1 2 2 3 7 [14]