Cambridge A Level Mathematics - Further 9231 — 2009 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 2009 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 2009 [Turn over *9759995938*

2 1 A line OP of fixed length l rotates in a plane about the fixed point O. At time t = 0, the line is at the position OA. At time t, angle AOP = θ radians and dθ dt = sin θ. Show that, for all t, the magnitude of the acceleration of P is equal to the magnitude of its velocity. [5] 2 The tip of a sewing-machine needle oscillates vertically in simple harmonic motion through a distance of 2.10 cm. It takes 2.25 s to perform 100 complete oscillations. Find, in m s−1, the maximum speed of the tip of the needle. [5] Show that the speed of the tip when it is at a distance of 0.5 cm from a position of instantaneous rest is 2.50 m s−1, correct to 3 significant figures. [2] 3 T O 2a a A uniform lamina of mass m is bounded by concentric circles with centre O and radii a and 2a. The lamina is free to rotate about a fixed smooth horizontal axis T which is tangential to the outer rim (see diagram). Show that the moment of inertia of the lamina about T is 21 4 ma2. [5] When hanging at rest, with O vertically below T, the lamina is given an angular speed ω about T. The lamina comes to instantaneous rest in the subsequent motion. Neglecting air resistance, find the set of possible values of ω. [3] © UCLES 2009 9231/02/M/J/09

3 4 O B C A 0.6 m 0.6 m 1 m A uniform sphere rests on a horizontal plane. The sphere has centre O, radius 0.6 m and weight 36 N. A uniform rod AB, of weight 14 N and length 1 m, rests with A in contact with the plane and B in contact with the sphere at the end of a horizontal diameter. The point of contact of the sphere with the plane is C, and A, B, C and O lie in the same vertical plane (see diagram). The contacts at A, B and C are rough and the system is in equilibrium. By taking moments about C for the system, show that the magnitude of the normal contact force at A is 10 N. [2] Show that the magnitudes of the frictional forces at A, B and C are equal. [2] The coefficients of friction at A, B and C are all equal to µ. Find the smallest possible value of µ. [7] 5 Two spheres A and B, of equal radius, have masses m1 and m2 respectively. They lie at rest on a smooth horizontal plane. Sphere A is projected directly towards sphere B with speed u and, as a result of the collision, A is brought to rest. Show that (i) the speed of B immediately after the collision cannot exceed u, [2] (ii) m1 ≤m2. [2] B 60° After the collision, B hits a smooth vertical wall which is at an angle of 60◦to the direction of motion of B (see diagram). In the impact with the wall B loses 2 3 of its kinetic energy. Find the coefficient of restitution between B and the wall and show that the direction of motion of B turns through 90◦. [8] 6 The times taken by employees in a factory to complete a certain task have a normal distribution with mean µ seconds and standard deviation σ seconds, both of which are unknown. Based on a random sample of 20 employees, the symmetric 95% confidence interval for µ is (481, 509). Calculate a symmetric 90% confidence interval for µ. [6] © UCLES 2009 9231/02/M/J/09 [Turn over

4 7 An experiment was carried out to determine how much weedkiller to apply per 100 m2 in a large field. Ten 100 m2 areas of the field were randomly chosen and sprayed with predetermined volumes of the weedkiller. The volume of the weedkiller is denoted by x litres and the number of weeds that survived is denoted by y. The results are given in the table. x 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 y 48 40 44 35 39 24 10 13 9 6 [Σ x = 3.25, Σ x2 = 1.2625, Σy = 268, Σy2 = 9548, Σ xy = 66.10.] It is given that the product moment correlation coefficient for the data is −0.951, correct to 3 decimal places. (i) Calculate the equation of a suitable regression line, giving a reason for your choice of line. [5] (ii) Estimate the best volume of weedkiller to apply, and comment on the reliability of your estimate. [3] 8 Part of a research study of identical twins who had been separated at birth involved a random sample of 9 pairs, in which one twin had been raised by the natural parents and the other by adoptive parents. The IQ scores of these twins were measured, with the following results. Twin pair 1 2 3 4 5 6 7 8 9 IQ of twin raised by natural parents 82 92 115 132 88 95 112 83 123 IQ of twin raised by adoptive parents 92 88 115 134 97 94 107 88 130 It may be assumed that the difference in IQ scores has a normal distribution. The mean IQ scores of separated twins raised by natural parents and by adoptive parents are denoted by µN and µA respectively. Obtain a 90% confidence interval for µN −µA. [6] One of the researchers claimed that there was no evidence of a difference between the two population means. State, giving a reason, whether the confidence interval supports this claim. [2] 9 The proportions of blood types A, B, AB and O in the Australian population are 38%, 10%, 3% and 49% respectively. In order to test whether the population in Sydney conforms to these figures, a random sample of 200 residents is selected. The table shows the observed frequencies of these types in the sample. Blood Type A B AB O Frequency 57 24 9 110 Carry out a suitable test at the 5% significance level. [7] Find the smallest sample size that could be used for the test. [2] © UCLES 2009 9231/02/M/J/09

5 10 The number of hits per minute on a particular website has a Poisson distribution with mean 0.8. The time between successive hits is denoted by T minutes. Show that P(T > t) = e−0.8t and hence show that T has a negative exponential distribution. [5] Using a suitable approximation, which should be justified, find the probability that the time interval between the 1st hit and the 51st hit exceeds one hour. [7] 11 Answer only one of the following two alternatives. EITHER a x P A B Two particles A and B, of equal mass m, are connected by a light elastic string of natural length a and modulus of elasticity 4mg. Particle A rests on a rough horizontal table at a distance a from the edge of the table. The string passes over a small smooth pulley P fixed at the edge of the table. At time t = 0, B is released from rest at P and falls vertically. At time t, B has fallen a distance x, without A slipping (see diagram). Show that ¨x = −g a(4x −a). [3] Deduce that, while A does not slip, B moves in simple harmonic motion and identify the centre of the motion. [3] Given that the coefficient of friction between A and the table is 1 3, find the value of x when A starts to slip, and the corresponding value of t, expressing this answer in the form k q a g. Give the value of k correct to 3 decimal places. [8] OR A study was made of the acidity levels in farmland on opposite sides of an island. The levels were measured at six randomly chosen points on the eastern side and at five randomly chosen points on the western side. The values obtained, in suitable units, are denoted by xE and xW respectively. The sample means xE and xW, and unbiased estimates of the two population variances, s2 E and s2 W, are as follows. xE = 5.035, s2 E = 0.0231, xW = 4.782, s2 W = 0.0195. The population means on the eastern and western sides are denoted by µE and µW respectively. State suitable hypotheses for a test for a difference between the mean acidity levels on the two sides of the island. [1] Stating any required assumptions, obtain the rejection region for a test at the 5% significance level of whether the mean acidity levels differ on the two sides of the island. Give the conclusion of the test. [9] Find the largest value of a for which the samples above provide evidence at the 5% significance level that µE −µW > a. [4] © UCLES 2009 9231/02/M/J/09

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/09

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2009 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 May/June 2009 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 2009 9231 02 © UCLES 2009 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 2009 9231 02 © UCLES 2009 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 2009 9231 02 © UCLES 2009 Qu No Mark Scheme Details Part Mark Total 1 Find tangential acceleration, l d2θ /dt2: l cos θ dθ /dt = l cos θ sin θ M1 A1 Find radial acceleration, l (dθ /dt)2: l sin2 θ M1 A1 Combine to give l dθ /dt (ignore magnitudes): A.G. l √(cos2 θ + sin2 θ) sin θ = l sin θ B1 5 [5] 2 Find frequency ω using T = 2π /ω : ω = 2π / 0⋅0225 [= 800π / 9 = 279⋅25] M1 A1 Find vmax using vmax = aω : vmax = 0⋅0105 ω ; = 2⋅93[2] M1 A1; A1 Find v using v2 = ω 2 (a2 – x2) v = ω √(0⋅01052 – 0⋅00552) or ω t = sin–1 (x/a) = [0⋅5513], v = aω cos ω t: [t = 0⋅00197], v = 2⋅50 A.G. M1 A1 5 2 [7] 3 Use perp. axes theorem for both discs (or lamina): I2a = m2aa2 or Ia = ¼maa2 M1 Combine to find MI of lamina about diameter (or T): I = I2a – Ia [= a2 (m2a – ¼ma)] M1 Use par. axes theorem for lamina (or both discs): IT = I + 4ma2 M1 Find masses of both discs in terms of m: m2a = 4m/3 and ma = m/3 B1 Combine to find MI of lamina about T: IT = a2 (4 – ¼) m/3 + 4ma2 = 5ma2/4 + 4ma2 = 21ma2/4 A.G. A1 Relate initial KE to change in PE at highest pt: ½ IT ω 2 and 4mga M1 A1 Find set of values [or max. value] of ω (A.E.F.): ω < √(32g/21a) or 1⋅23√(g/a) or 3⋅90/√a A1 5 3 [8] 4 Find RA by taking moments about C for system: 1⋅4 RA = 1⋅0 × 14, RA = 10 A.G. M1 A1 Deduce by taking moments about O for sphere: FB = FC A.G. B1 Resolve horizontally for system: FA = FC A.G. B1 Find any F by e.g. vertical resolution for AB FB = 14 – RA = 4 M1 A1 or taking moments about B for AB: FA = (0⋅8 RA – 0⋅4 × 14)/0⋅6 = 4 (M1 A1) Find RB by e.g. hor. resolution for rod or sphere: RB = FA or FC [= 4] M1 Find RC by e.g. vert. resolution for sphere or system: RC = 36 + FB or 50 – RA = 40 M1 A1 Find µmin: µmin = max{FA /RA, FB /RB, FC /RC} M1 = max{4/10, 4/4, 4/40} = 1 A1 2 2 7 [11]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 02 © UCLES 2009 Qu No Mark Scheme Details Part Mark Total 5 (i) Find max. speed of B using elasticity: v = e1u ≤ u [or < u] A.G. M1 A1 (ii) Use conservation of momentum: m1 u = m2 v ≤ m2 u A.G. M1 A1 Equate speeds normal to wall, e.g.: V sin α = ev sin 60° or ev √3/2 M1 Equate speeds parallel to wall, e.g.: V cos α = v cos 60° or v/2 M1 Eliminate α by squaring and adding: V ² = v² (e² sin² 60° + cos² 60°) M1 Relate KEs: ½mV ² = ⅓ (½mv²) B1 Hence eliminate speeds to find e: e² = (⅓ – ¼) / ¾ = 1/9, e = ⅓ M1 A1 Show that B leaves wall at 30°: A.G. tan α = e tan 60° = 1/√3, α = 30° M1 A1 2 2 8 [12] 6 Find sample mean: x = ½(481 + 509) = 495 M1 A1 Use or imply confidence interval formula: x ± ts/√n , any t or z [s = 29⋅9] M1 Find 90% interval semi-width: (t19, 0.95 / t19, 0.975) 14 M1 (1⋅725/2⋅086 or 1⋅645/1⋅96 lose A1 only) = (1⋅729 / 2⋅093) 14 = 11⋅6 A1 Hence 90% confidence interval: [483⋅4, 506⋅6] or [483, 507] A1 6 [6] 7 (i) State choice of line with reason (A.E.F.): y depends on x so choose y on x B1 Find coefficient b in regression line for y: b = (66⋅1 – 3⋅25 × 268/10) / (1⋅2625 – 3⋅252/10) = –21/ 0⋅20625 = –101⋅8 or –102 M1 A1 Find equation of regression line: y = b(x – 0⋅325) + 26⋅8 = 59⋅9 – 102 x M1 A1 SR: M1 A1 for finding x on y: x = 0⋅563 – 0⋅00888 y (M1 A1) (ii) Find x when y = 0: 59⋅9/102 = 0⋅587 or 0⋅588 or 0⋅59 M1 A1 SR: If using eqn of x on y: 0⋅563 (B1) Valid comment on reliability: OK since point just outside range or OK as r ≈ –1 or |r| ≈ 1 (A.E.F.) B1 5 3 [8]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 02 © UCLES 2009 Qu No Mark Scheme Details Part Mark Total 8 Consider differences N – A (or A – N): –10 4 0 –2 –9 1 5 –5 –7 M1 Estimate mean and population variance (to 3 sf): x = – 23/9 [= – 2⋅556; allow +] and s2 = (301 – 232/9) / 8 = 30⋅28 or 5⋅5032 (allow biased) (26⋅91 or 5⋅1882) M1 A1 Use valid formula (M1 needs n consistent with s): x ± ts/√n for any t M1 Use correct tabular t value: t8, 0.95 = 1⋅86[0] *B1 Evaluate confidence interval (dep *B1): – 2⋅56 ± 3⋅41 or [– 5⋅97, 0⋅86] *A1 Use valid argument consistent with interval: 0 lies in interval (A.E.F.) M1 State conclusion (dep *A1 apart from rounding): Yes, supports claim (A.E.F.) A1 6 2 [8] 9 State (at least) null hypothesis (A.E.F.): H0: Sydney popn. has same freq. [0⋅38, 0⋅10, 0⋅03, 0⋅49] B1 Calculate expected values: 76 20 6 98 M1 A1 Calculate value of χ 2 (M1 even if cells combined): χ 2 ≈ 8⋅5[2] M1 *A1 Compare with consistent tab. value (to 2 dp): χ3, 0.95 2 = 7⋅815 [χ2 2 = 5⋅991] *B1√ Correct conclusion (dep *A1, *B1): Sydney does not conform (AEF) B1 Find smallest size n with expected values ≥ 5 [or >]: n ≥ 5/0⋅03 = 166⋅7 so nmin is 167 M1 A1 7 2 [9]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 02 © UCLES 2009 Qu No Mark Scheme Details Part Mark Total 10 Relate P(T > t) to number of hits: P(T > t) = P(hits in t mins = 0) M1 Relate to Poisson distribution: = P0(0⋅8t) = e–0⋅8 t A.G. M1 A1 Find distribution F(t) of T: F(t) = P(T < t) = 1 – e–0⋅8 t B1 Differentiate to find f(t) in required form: f (t) = 0⋅8 e–0⋅8 t B1 EITHER: State or imply required probability: S = Σ 50 1 = i Ti M1 State or use mean of S: µ S = 50 (1/0⋅8) = 62⋅5 A1 State or use variance of S: σ S 2 = 50 (1/0⋅8) 2 = 78⋅1[25] A1 Justify use of Normal approxn. (A.E.F.): By Central Limit Theorem or 50 is large or 50 > e.g. 30 B1 Evaluate approximate probability (A.E.F): 1 – Φ((60 – µ S)/σ S) M1 = Φ(0⋅283); = 0⋅611 A1; A1 OR: State or imply required probability: P(hits in 60 mins < 50) [allow ≤ 50] (M1) State or use mean: µ = 60 × 0⋅8 = 48 (A1) State or use variance: σ 2 = 60 × 0⋅8 = 48 (A1) Justify use of Normal approxn. (A.E.F.): 48 is large or 48 > e.g. 15 (B1) Evaluate approximate probability (A.E.F): Φ((49⋅5 – µ )/σ) (M1) = Φ(0⋅216[5]); = 0⋅586 (A1; A1) S.R. Omission of continuity correction: Φ((50 – µ )/σ) = 0⋅614 earns B1 5 7 [12]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 02 © UCLES 2009 Qu No Mark Scheme Details Part Mark Total 11 EITHER Find tension T: T = 4mgx/a B1 Apply Newton’s law of motion to B: m d2x/dt2 = mg – T M1 Combine: d2x/dt2 = – (g/a)(4x – a) A.G. A1 Substitute e.g. y = x – ¼ a and rearrange: d2y/dt2 = – (4g/a) y M1 A1 State centre of motion, or derive from y = 0: xc = ¼ a A1 Find x when A starts to slip using F = µR: T = ⅓ mg, xs = a/12 M1 A1 Valid use of SHM eqn to find time ts to slipping: y = ymax cos ω t or ymax sin ω t M1 EITHER: Valid use of cosine form: ys = ymax cos ω ts M1 OR: Valid use of sine form: ts = t1 – t2 , ymax = ymax sin ω t1 ys = ymax sin ω t2 (M1) Substitute for ys, ymax : ys / ymax = (xc – xs) / xc M1 = (a/6)/(a/4) = ⅔ A1 Find ts : ts = (cos–1 ⅔) /ω or (½π – sin–1 ⅔) /ω M1 Substitute ω = 2√(g/a) and evaluate: ts = 0⋅421 √(a/g) A1 3 3 8 [14] 11 OR State hypotheses: H0: µE = µW , H1: µE ≠ µW B1 State assumption [A.E.F.]: Two populations have Normal distns. and common variance B1 Estimate common variance: σ2 = (5 × 0⋅0231 + 4 × 0⋅0195) / 9 M1 = 0⋅0215 or 0⋅14732 A1 Use correct tabular value of t: t9, 0.975 = 2⋅26[2] B1 Formulate rejection region (with any t; allow >): | x E – x W | ≥ tσ √(1/6 + 1/5) M1 = 0⋅201 A1 Compare actual sample means with region: 0⋅253 > 0⋅201√ or compare calculated t with tabular t: 2⋅85 > 2⋅26[2]√ M1 A1 Consistent conclusion (A.E.F.; dep values above): Mean acidity levels do differ A1 State condition on a (with any t; allow > or =): x E – x W – a ≥ tσ √(1/6 + 1/5) M1 Use correct tabular value of t: t9, 0.95 = 1⋅83[3] A1 Substitute to find largest value of a: amax = 0⋅253 – 0⋅163 = 0⋅09 (2 dp) M1 A1 1 9 4 [14]