Cambridge A Level Mathematics - Further 9231 — 2013 May/June Paper 2 · Variant 2

*3195746375* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/22 Paper 2 May/June 2013 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. JC13 06_9231_22/FP © UCLES 2013 [Turn over

2 1 P B C A 5a 3a a A uniform rod AB, of mass m and length 4a, rests with the end A on rough horizontal ground. The point C on AB is such that AC = 3a. A light inextensible string has one end attached to the point P which is at a distance 5a vertically above A, and the other end attached to C. The rod and the string are in the same vertical plane and the system is in equilibrium with angle ACP equal to 90Å (see diagram). The coefficient of friction between the rod and the ground is -. Show that the least possible value of - is 24 43. [8] 2 Three uniform small smooth spheres, A, B and C, have equal radii. Their masses are 4m, 2m and m respectively. They lie in a straight line on a smooth horizontal surface with B between A and C. Initially A is moving towards B with speed u, B is at rest and C is moving in the same direction as A with speed 1 2u. The coefficient of restitution between any two of the spheres is e. The first collision is between A and B. In this collision sphere A loses three-quarters of its kinetic energy. Show that e = 1 2. [6] Find the speed of B after its collision with C and deduce that there are no further collisions between the spheres. [5] 3 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. When P is hanging vertically below O, it is given a horizontal speed u. In the subsequent motion, P moves in a complete circle. When OP makes an angle 1 with the downward vertical, the tension in the string is T. Show that T = mu2 a + mg3 cos 1 −2. 5 Given that the ratio of the maximum value of T to the minimum value of T is 3 : 1, find u in terms of a and g. [4] Assuming this value of u, find the value of cos 1 when the tension is half of its maximum value. [3] © UCLES 2013 9231/22/M/J/13

3 4 a 1 2 A B O C 3a 3a The end A of a uniform rod AB, of mass 4m and length 3a, is rigidly attached to a point on a uniform spherical shell, of mass ,m and radius 3a. The end B of the rod is rigidly attached to a point on a uniform ring. The ring has centre O, mass 4m and radius 1 2a. The ring and the rod are in the same vertical plane. The line OBA, extended, passes through the centre of the spherical shell. BC is a diameter of the ring (see diagram). Show that the moment of inertia of this system, about a fixed horizontal axis through C perpendicular to the plane of the ring, is 30 + 55,ma2. [6] Given that the system performs small oscillations of period 20 O@5a g A about this axis, find the value of ,. [6] 5 For a random sample of 12 observations of pairs of values x, y, the product moment correlation coefficient is −0.456. Test, at the 5% significance level, whether there is evidence of negative correlation between the variables. [4] 6 The random variable X has distribution function F given by Fx = D 1 −e−0.6x x ≥0, 0 otherwise. Identify the distribution of X and state its mean. [2] Find (i) PX > 4, [2] (ii) the median of X. [3] © UCLES 2013 9231/22/M/J/13 [Turn over

4 7 A random sample of 80 observations of the continuous random variable X was taken and the values are summarised in the following table. Interval 2 ≤x < 3 3 ≤x < 4 4 ≤x < 5 5 ≤x < 6 Observed frequency 36 29 9 6 It is required to test the goodness of fit of the distribution having probability density function f given by fx = t 3 x2 2 ≤x < 6, 0 otherwise. Show that the expected frequency for the interval 2 ≤x < 3 is 40 and calculate the remaining expected frequencies. [4] Carry out a goodness of fit test, at the 10% significance level. [5] 8 The continuous random variable X has probability density function f given by fx = d 1 6x 2 ≤x ≤4, 0 otherwise. The random variable Y is defined by Y = X3. Show that Y has probability density function g given by gy = d 1 18y −1 3 8 ≤y ≤64, 0 otherwise. [6] Find EY. [3] 9 A gardener P claims that a new type of fruit tree produces a higher annual mass of fruit than the type that he has previously grown. The old type of tree produced 5.2 kg of fruit per tree, on average. A random sample of 10 trees of the new type is chosen. The masses, x kg, of fruit produced are summarised as follows. Σ x = 61.0 Σ x2 = 384.0 Test, at the 5% significance level, whether gardener P’s claim is justified, assuming a normal distribution. [6] Another gardener Q has his own type of fruit tree. The masses, y kg, of fruit produced by a random sample of 10 trees grown by gardener Q are summarised as follows. Σ y = 70.0 Σ y2 = 500.6 Test, at the 5% significance level, whether the mean mass of fruit produced by gardener Q’s trees is greater than the mean mass of fruit produced by gardener P’s trees. You may assume that both distributions are normal and you should state any additional assumption. [8] © UCLES 2013 9231/22/M/J/13

5 10 Answer only one of the following two alternatives. EITHER A light elastic string has modulus of elasticity 3 2mg and natural length a. A particle of mass m is attached to one end of the string. The other end of the string is attached to a fixed point A. The particle is released from rest at A. Show that when the particle has fallen a distance ka from A, where k > 1, its kinetic energy is 1 4mga10k −3 −3k2. 3 Show that the particle first comes to instantaneous rest at the point B which is at a distance 3a vertically below A. [3] Show that the time taken by the particle to travel from A to B is O@2a g A + 20 3 O@2a 3g A . 8 OR The regression line of y on x, obtained from a random sample of five pairs of values of x and y, has equation y = x + k, where k is a constant. The following table shows the data. x 2 3 3 4 p y 4 5 8 4 2 Find the two possible values of p. [8] For the smaller of these two values of p, find (i) the product moment correlation coefficient, [3] (ii) the equation of the regression line of x on y. [3] © UCLES 2013 9231/22/M/J/13

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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. © UCLES 2013 9231/22/M/J/13

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2013 series 9231 FURTHER MATHEMATICS 9231/22 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9231 22 © Cambridge International Examinations 2013 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 2013 9231 22 © Cambridge International Examinations 2013 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 2013 9231 22 © Cambridge International Examinations 2013 Question Number Mark Scheme Details Part Mark Total 1 State or imply length of CP or equivalent: e.g. for angle CPA = θ EITHER: 2 moment. eqns for R and F, e.g. about P: about C: Solve for R: OR: Take moments about A to give tension T: Resolve horizontally to find friction at A: Resolve horizontally to find reaction at A: Use F = µmin R to find µmin: CP = 4a or sin θ = 3/5 or cos θ = 4/5 B1 5aF = 2a mg cos θ ; F = 8mg/25 M1;A1 3aR cos θ – 3aF sin θ = a mg cos θ M1 R = (4mg + 9F)/12 = 43mg/75 M1 A1 3aT = 2a mg cos θ [T = 8mg/15] (M1) F = T sin θ ; = 8mg/25 (M1;A1) R = mg – T cos θ ;= = 43mg/75 (M1;A1) µmin = 24/43 A.G. M1 A1 8 8 2 Use conservation of momentum, e.g.: Use Newton’s law of restitution (consistent signs) Relate vA to u using K.E.: EITHER : Consider one possible value of vA: Consider other value of vA: OR: Combine first 2 eqns to find vA: Find 2 possible values of e from K.E: Select one value (stating reason): 4mvA + 2mvB = 4mu M1 vA – vB = – eu M1 ½ 4mvA 2 = ¼ ½ 4mu2 [vA 2 = ¼ u2] M1 vA = ½ u, vB = u, e = ½ B1 vA = –½ u, vB = 3u, e = 3½ (B1) vA = ⅓(2 – e)u, [vB = ⅔(1 + e)u] (2 – e)2 = 9/4, e2 – 4e + 7/4 = 0 e = ½ or 3½ (B1) e ≤ 1 (or < 1) so e = ½ A.G. B1 6 Use conservation of momentum, e.g. Use Newton’s law of restitution (consistent signs): Substitute vB = u and solve for vB′ : State why no further collisions ( on vB′, vA provided vB′ > vA): 2mvB′ + mvC = 2mvB + ½ mu M1 vB′ – vC = – e(vB – ½ u) M1 2vB′ + vC = 5u/2 and vB′ – vC = – ¼ u so vB′ = ¾ u M1 A1 ¾ u > ½ u [vC = u not reqd.] B1 5 11

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9231 22 © Cambridge International Examinations 2013 3 Use conservation of energy: (allow B1 here if found by v2 = u2 – 2gh) Use F = ma radially: Eliminate v2 to find T: ½mv2 = ½mu2 – mga(1 – cos θ) M1 A1 T – mg cos θ = mv2/a B1 T = mu2/a + mg(3 cos θ – 2) A.G. M1 A1 5 Find or use max. and min. values of T: Find u from Tmax / Tmin = 3: Tmax = mu2/a + mg B1 Tmin = mu2/a – 5mg B1 u2/a + g = 3u2/a – 15g, 2u2/a = 16g, u = √(8ag) A.E.F. M1 A1 4 Find cos θ from T = ½Tmax [= ½ 9mg]: 3 cos θ – 2 = – ½ u2/ag + ½ M1 A1 cos θ = ⅓ (2 – 4 + ½) = – ½ A1 3 12 4 Find MI of sphere about C: Find MI of rod about C: Find MI of ring about C: Combine to give MI of system about C: ISphere = ⅔ λm (3a)2 + λm (7a)2 M1 A1 IRod = ⅓ 4m (3a/2)2 + 4m (5a/2)2 M1 A1 IRing = 4m (½ a)2 + 4m (½ a)2 M1 I = (55λ + 28 + 2) ma2 = (30 + 55λ) ma2 A.G. A1 6 Find eqn of motion for system: Approximate sin θ by θ and find ω2 in SHM eqn: Equate ω2 to 4π2/T2 to find λ: I d2θ/dt2 = – (λ × 7 + 4 × 5/2 + 4 × ½) mga sin θ M1 A1 ω2 = (12 + 7λ)g / (30 + 55λ)a M1 A1 5(12 + 7λ) = 30 + 55λ; λ = 3/2 M1 A1 6 12 5 State both hypotheses (B0 if r not ρ used): State or use correct tabular one-tail r value: Valid method for reaching conclusion, e.g.: Correct conclusion (AEF, dep *B1): H0: ρ = 0, H1: ρ < 0 B1 r12, 5% = 0⋅497 *B1 Accept H0 if |r| < tabular value M1 No evidence of neg. correlation A1 4 4 6 Identify distribution of X: State mean of X: Negative exponential B1 5/3 or 1⋅67 B1 2 7 (i) Find P(X > 4): P(X > 4) = 1 – F(4) = e-2⋅4 = 0⋅0907 M1 A1 2 (ii) State or use eqn. for median m of X: Find value of m: F(m) [or 1 – F(m)] = ½ M1 e–0⋅6m = ½, m = (5/3) ln 2 or 1⋅16 M1 A1 3 7 Find 1st expected frequency E2: Find other expected frequencies: ∫ − = 3 2 2 2 d 3 80 x x E [ ] 1 3 2 3 80 − − = x = 40 A.G. M1 A1 E3 = 20, E4 = 12, E5 = 8 M1 A1 4 State (at least) null hypothesis: Calculate χ2 (to 2 d.p.): H0: f(x) fits data (A.E.F.) B1 χ2 = 0⋅4 + 4⋅05 + 0⋅75 + 0⋅5 = 5⋅7 M1 A1

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9231 22 © Cambridge International Examinations 2013 State or use consistent tabular value: Conclusion (A.E.F., on χ2, dep *B1): χ3, 0.9 2 = 6⋅25[1] *B1 χ2 < 6⋅25 so f(x) does fit B1 5 9 8 EITHER Integrate f(x) to find F(x) for 2 ≤ x ≤ 4: Find G(y) for 8 ≤ y ≤ 64: OR Find G(y) for 8 ≤ y ≤ 64: State G(y) for other values of y: Differentiate to find g(y) for 8 ≤ y ≤ 64: ( ) ( ) x x x x d f F 2∫ = M1 A1 G(y) = P(Y < y) = P(X3 < y) = P(X < y1/3) = F(y1/3) = (y2/3 – 4)/12 M1 A1 G(y) = P(Y < y) = P(X3 < y) = P(X < y1/3) = ∫ 2 y1/3 f(x) dx (M1 A1) [ ] ( ) 12 4 12 3 2 3 1 2 2 / y / x / / y − = = (M1 A1) G(y) = 0 (y < 8), 1 (y > 64) B1 g(y) = y–1/3 /18 A.G. B1 6 EITHER: Find E(Y) from ∫yg(y): OR: Find E(Y) from E(X3): Evaluate: ( ) ( )[ ] 64 8 3 5 64 8 3 2 30 1 d 18 1 / / y / y y / = ∫ M1 A1 ( ) ( )[ ] 64 8 5 4 2 4 30 1 d 6 1 x / x x / = ∫ (M1 A1) = (45 – 25)/30 = 992/30 = 496/15 or 33⋅1 A1 3 9 9 Find sample mean and estimate population variance: (allow biased here: 1⋅19 or 1⋅0912) State hypotheses (B0 forx …): Calculate value of t (to 3 s.f.): State or use correct tabular t value: (or can compare x = 6⋅1 with 5⋅2 + 0⋅667 = 5⋅87) Correct conclusion (A.E.F., on t, dep *B1): x = 6⋅1, sx 2 = (384 – 612/10) / 9 = 1⋅322 or 119/90 or 1⋅152 M1 H0: µ = 5⋅2, H1: µ > 5⋅2 B1 t = ( x – 5⋅2)/(sx /√10) = 2⋅47[5] M1 A1 t9, 0.95 = 1⋅83[3] *B1 t > 1⋅83 so mean is greater B1 6 State hypotheses (B0 forx …), e.g.: State assumption (A.E.F.): Find sample mean [and estimate variance] for Y : EITHER Estimate (pooled) common varian OR (Equivalently) estimate common variance: Calculate value of t (to 3 s.f.): State or use correct tabular t value: (or can comparey –x = 0⋅9 with 0⋅867) Correct conclusion (A.E.F., on t, dep *B1): H0: µP = µQ , H1: µP < µQ B1 Distributions have equal variances B1 y = 7⋅0, [sy 2 = (500⋅6 – 702/10) / 9 = 1⋅178 or 53/45 or 1⋅0852 ] s2 = (9 × 1⋅322 + 9 × 1⋅178) / 18 or (384 – 612/10 + 500⋅6 – 702/10) / 18 = 1.25 or 1.182 M1 A1 s2 = (1⋅322 + 1⋅178) / 10 = 0⋅25 (M1 A1) t = ( y–x) / s = 1⋅8 (M1 A1) t18, 0.95 = 1⋅73[4] *B1 t > 1⋅73 so Q’s mean is greater B1 8 14

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9231 22 © Cambridge International Examinations 2013 10 (a) Find K.E. after falling ka from P.E. and E.P.E.: Find eqn. for k (or ka) at B using K.E. = 0: mg × ka – ½(3mg/2)(ka – a)2/a M1 =[A1 = mga(k – 3k2/4 + 3k/2 – ¾) = ¼mga(10k – 3 – 3k2) A.G. A1 3k2 – 10k + 3 = 0 (A.E.F.) M1 3 Find both roots and select root > 1: Find time t1 to fall a from A (under no tension): Roots 3 and ⅓, so k = 3 A.G. M1 A1 a = ½ gt1 2, t1 = √(2a/g) B1 3 Find ext. e or distance fallen at equilibrium pt O: State or find SHM eqn. at x below (or above) O: State or use correct amplitude x0 and ω2: Find time t2 to fall from a to 3a below A, e.g.: Find total time t1 + t2: mg = 3mge/2a, e = 2a/3 or AO = 5a/3 B1 md2x/dt2 = mg – 3mg(x + e)/2a M1 d2x/dt2 = – 3gx/2a A1 x0 = 4a/3 and ω2 = 3g/2a B1 [–] e = x0 cos ωt2 M1 t2 = √(2a/3g) cos-1 (-½) (A.E.F.) A1 √(2a/g) + (2π/3) √(2a/3g) A.G. A1 8 14 (b) Find 4 summations reqd. in this part: EITHER: Substitute b1 = 1 in formula for gradient: (A.E.F., A1 for each side) OR Substitute in normal eqns. or equivalent: (A.E.F., A1 for each eqn.) Obtain and solve quadratic for p (A.E.F): Σx = 12 + p, Σx2 = 38 + p2, Σy = 23, [Σy2 = 125] Σxy = 63 + 2p M1 63 + 2p – (12 + p) × 23/5 = 38 + p2 – (12 + p)2/5 or 39 – 13p = 4p2 – 24p + 46 M1 A1 A1 5k + (12 + p) = 23 and (12 + p)k + (38 + p2) = 63 + 2p (M1 A1 A1) 0⋅8p2 – 2⋅2p + 1⋅4 = 0 or 4p2 – 11p + 7 = 0 M1 A1 p = 1, 1⋅75 (or 7/4) A1, A1 8

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9231 22 © Cambridge International Examinations 2013 (i) Find eqn, for correlation coefficient with p = 1: 3 EITHER: OR Evaluate r r = (65 – 13 × 23/5) / √{(39 – 132/5) (125 – 232/5)} M1 A1 = 5⋅2 / √(5⋅2 × 19⋅2) or 1⋅04 / √(1⋅04 × 3⋅84) r2 = 1 × (65 – 13 × 23/5) / (125 – 232/5) or 1 × (39 – 132/5) / (125 – 232/5) = 5⋅2 / 19⋅2 or 1⋅04 / 3⋅84 M1 A1 r = √39/12 or 0⋅52 A1 (ii) EITHER: Recall or find gradient b2 of line: Find regression line of x on y: OR Use normal eqns for x = a2 + b2y: Solve for a2, b2: b2 = 5⋅2/19⋅2 or r2 [= 0⋅2708] M1 x – 13/5 = b2 (y – 23/5) M1 x = (13/48)y + 65/48 or 0⋅271y + 1⋅35 A1 5a2 + 23b2 = 13 and (M1) 23a2 + 125b2 = 65 (M1) b2 = 13/48 or 0⋅271 And a2 = 65/48 or 1⋅35 (A1) 3 14