Cambridge A Level Mathematics - Further 9231 — 2012 Oct/Nov Paper 2 · Variant 3

*0251397462* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 October/November 2012 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. JC12 11_9231_23/RP © UCLES 2012 [Turn over

2 1 A particle P is moving in a circle of radius 1.5 m. At time t s its velocity is (k −t2) m s−1, where k is a positive constant. When t = 3, the magnitudes of the radial and transverse components of the acceleration of P are equal. Find the possible values of k. [4] 2 A small bead of mass m is threaded on a thin smooth wire which forms a circle of radius a. The wire is fixed in a vertical plane. A light inextensible string is attached to the bead and passes through a small smooth ring fixed at the centre of the circle. The other end of the string is attached to a particle of mass 4m which hangs freely under gravity. The bead is projected from the lowest point of the wire with speed √(kga). Show that, when the angle between the two parts of the string is θ, the normal force exerted on the bead by the wire is mg(3 cos θ + k −6), towards the centre. [5] Given that the bead reaches the highest point of the wire, find an inequality which must be satisfied by k. [2] 3 A B C a b Two uniform rods AB and BC, each of length 2a and mass m, are smoothly hinged at B. They rest in equilibrium with C in contact with a smooth vertical wall and A in contact with a rough horizontal floor. The rods are in a vertical plane perpendicular to the wall. The rods AB and BC make angles α and β respectively with the horizontal (see diagram). Show that (i) the reaction at C has magnitude 1 2mg cot β, [2] (ii) tan α = 3 tan β. [5] The coefficient of friction at A is µ. Given that α = 60◦, find the least possible value of µ. [2] 4 Three particles A, B and C have masses m, 2m and m respectively. The particles are able to move on a smooth horizontal surface in a straight line, and B is between A and C. Initially A is moving towards B with speed 2u and C is moving towards B with speed u. The particle B is at rest. The coefficient of restitution between any pair of particles is e. The first collision is between A and B. (i) Show that the speed of B immediately before its collision with C is 2 3u(1 + e). [4] (ii) Find the velocity of B immediately after its collision with C. [3] (iii) Given that e > 1 2, show that there are no further collisions between the particles. [4] © UCLES 2012 9231/23/O/N/12

3 5 Four identical uniform rods, each of mass m and length 2a, are rigidly joined to form a square frame ABCD. Show that the moment of inertia of the frame about an axis through A perpendicular to the plane of the frame is 40 3 ma2. [4] The frame is suspended from A and is able to rotate freely under gravity in a vertical plane, about a horizontal axis through A. When the frame is at rest with C vertically below A, it is given an angular velocity q 6g 5a. Find the angular velocity of the frame when AC makes an angle θ with the downward vertical through A. [5] When AC is horizontal, the speed of C is k√(ga). Find the value of k correct to 3 significant figures. [3] 6 The random variable X has probability density function f given by f(x) = ( 1 6e −1 6x x ≥0, 0 otherwise. Find (i) the distribution function of X, [2] (ii) the probability that X lies between the median and the mean. [4] 7 The speed v at which a javelin is thrown by an athlete is measured in km h−1. The results for 10 randomly chosen throws are summarised by Σv = 1110.8, Σ(v −v)2 = 333.9, where v is the sample mean. (i) Stating any necessary assumption, calculate a 99% confidence interval for the mean speed of a throw. [6] The results for a further 5 randomly chosen throws are now combined with the above results. It is found that the sample variance is smaller than that used in part (i). (ii) State, with reasons, whether a 95% confidence interval calculated from the combined 15 results will be wider or less wide than that found in part (i). [2] 8 Drinking glasses are sold in packs of 4. The manufacturer conducts a survey to assess the quality of the glasses. The results from a sample of 50 randomly chosen packs are summarised in the following table. Number of perfect glasses 0 1 2 3 4 Number of packs 1 3 10 17 19 Fit a binomial distribution to the data and carry out a goodness of fit test at the 10% significance level. [9] © UCLES 2012 9231/23/O/N/12 [Turn over

4 9 Experiments are conducted to test the breaking strength of each of two types of rope, P and Q. A random sample of 50 ropes of type P and a random sample of 70 ropes of type Q are selected. The breaking strengths, p and q, measured in appropriate units, are summarised as follows. Σp = 321.2 Σp2 = 2120.0 Σq = 475.3 Σq2 = 3310.0 Test, at the 10% significance level, whether the mean breaking strengths of type P and type Q ropes are the same. [10] 10 Delegates who travelled to a conference were asked to report the distance, y km, that they had travelled and the time taken, x minutes. The values reported by a random sample of 8 delegates are given in the following table. Delegate A B C D E F G H x 90 46 72 98 52 65 105 82 y 90 55 69 85 45 50 110 74 [Σx = 610, Σx2 = 49 682, Σy = 578, Σy2 = 45 212, Σxy = 47 136.] Find the equations of the regression lines of y on x and of x on y. [6] Estimate the time taken by a delegate who travelled 100 km to the conference. [2] Calculate the product moment correlation coefficient for this sample. [2] © UCLES 2012 9231/23/O/N/12

5 11 Answer only one of the following two alternatives. EITHER A particle P of mass m is attached to one end of a light elastic string of modulus of elasticity 8mg and natural length a. The other end of the string is attached to a fixed point O. The particle is pulled vertically downwards a distance 1 4a from its equilibrium position and released from rest. Show that the string first becomes slack after a time 2π 3 q  a 8g. [8] Find, in terms of a, the total distance travelled by P from its release until it subsequently comes to instantaneous rest for the first time. [6] OR x f( )x 0 2 5 k The continuous random variable X takes values in the interval 0 ≤x ≤5 only. For 0 ≤x ≤5 the graph of its probability density function f consists of two straight line segments, as shown in the diagram. Find k and show that f is given by f(x) =   1 8x 0 ≤x ≤2, 1 4 2 < x ≤5, 0 otherwise. [3] The random variable Y is given by Y = X2. (i) Find the probability density function of Y. [6] (ii) Show that E(Y) = 10.25. [3] (iii) Show that the median of Y is the square of the median of X. [2] © UCLES 2012 9231/23/O/N/12

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. 9231/23/O/N/12

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2012 series 9231 FURTHER MATHEMATICS 9231/23 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 October/November 2012 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 – October/November 2012 9231 23 © Cambridge International Examinations 2012 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 – October/November 2012 9231 23 © Cambridge International Examinations 2012 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 – October/November 2012 9231 23 © Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 1 Find radial acceleration when t = 3: (k – 32)2 / 1⋅5 [m s–2] B1 Find transverse accel. (ignoring sign) when t = 3: 2t = 6 [m s–2] B1 Equate magnitudes to find k: (k – 9)2 = 9, k = 6 or 12 M1 A1 4 [4] 2 Use conservation of energy: ½mv2 = ½mkga – mga(1 – cos θ) B1 Use F = ma radially: R + 4mg – mg cos θ = mv2/a M1 A1 Eliminate v to find R: R = mg(3 cos θ + k – 6) A.G. M1 A1 Find k from v ≥ 0 (or > 0) when θ = π: k ≥ 4 (or k > 4) M1 A1 5 2 [7] 3 (i) (ii) (iii) Find RC by moments for BC about B: RC 2a sin β = mg a cos β RC = ½ mg cot β A.G. M1 A1 EITHER: Moments for system about A: RC (2a sin α + 2a sin β) = mg (3a cos α + a cos β) M1 A1 Substitute for RC from (i): ½ cos β (2 sin α + 2 sin β) = sin β (3 cos α + cos β) M1 A1 tan α = 3 tan β A.G. A1 OR: Moments for AB about B: RA 2a cos α = FA 2a sin α + mg a cos α (M1 A1) Substitute RA = 2mg, FA = RC: 4 cos α = (½ cot β) sin α + cos α (M1 A1) tan α = 3 tan β A.G. (A1) Find µmin using FA ≤ µRA: µmin = ¼ cot β = ¾ cot α = ¼√3 M1 A1 2 5 2 [9]

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 23 © Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 4 (i) (ii) (iii) Use cons. of momentum for 1st collision: muA + 2muB = 2mu B1 Use Newton’s law of restitution: uA – uB = – e 2u B1 Eliminate uA to find uB: uB = 2u(1 + e)/3 A.G. M1 A1 Use cons. of momentum for 2nd collision: 2mvB + mvC = 2muB – mu M1 Use Newton’s law of restitution: vB – vC = – e (uB + u) M1 Substitute and solve for vB : vB = u(1 + e)(1 – 2e)/9 (A.E.F.) A1 Find uA: uA = ⅔u(1 – 2e) B1 State or imply dirns. in which A, B move: e > ½ so A/B change direction (needs uA, vB correct) in 1st/2nd collision (A.E.F.) B1 Show |uA| > |vB |: (needs uA, vB correct): |uA| / |vB | = ⅔/(1 + e)/9 = 6/(1 + e) > 1 (A.E.F.) M1 A1 4 3 4 [11] 5 State or find MI of rod AB (or AD) about A: IAB = ⅓ma2 + ma2 = (4/3)ma2 B1 State or find MI of rod BC (or CD) about A: IBC = ⅓ma2 + m5a2 [=(16/3)ma2] M1 Find MI of frame about A: I = 2(IAB + IBC) = 40ma2/3 A.G. M1 A1 Use energy to find ang. vel. ω at angle θ: ½Iω2 = ½I (6g/5a) (lose A1 for one incorrect term) – 4mg a√2 (1 – cos θ) M1 A2 Substitute for I and simplify (A.E.F.): ω = √{(3g/5a)(2 – √2(1 – cos θ))} M1 A1 Equate AC ω to k√(ga) to find k when θ = 90°: k√(ga) = 2√2a √{(3g/5a)(2 – √2)} M1 A1 k = 2√{6(2 – √2)/5} = 1⋅68 A1 4 5 3 [12] 6 (i) (ii) State or find by integration F(x): F(x) = 1 – e-x/6 (x ≥ 0), 0 otherwise M1 A1 State or find mean µ: µ = 1/(1/6) = 6 B1 Find ±P(m ≤ X ≤ µ) [m = 4⋅16 not reqd]: F(µ) – ½ = 1 – e-1 – ½ M1 A1 Reqd. prob. = 0⋅132 A1 2 4 [6]

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 23 © Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 7 (i) (ii) State suitable assumption (A.E.F.): Population is Normal B1 Find confidence interval: 1110⋅8/10 ± t √(333⋅9 /90) M1 A1 = 111⋅1 ± t √3⋅71 A1 State or use correct tabular value of t: t9,0.995 = 3⋅25 A1 Evaluate C.I.: 111 ± 6 or [105, 117] A1 Compare t , est. variance s and n: t and s smaller, n larger M1 Deduce effect on width of C.I. (A.E.F.): Width is less than in (i) A1 S.R. B1 if valid apart from considering n 6 2 [8] 8 Find value of p for binomial dist.: mean = 150/50 = 3, p = ¾ M1 A1 Find expected binomial values (to 2 d.p.): 0⋅20 2⋅34 10⋅55 21⋅09 15⋅82 M1 A1 Combine adjacent cells since exp. value < 5: O: 14 17 19 E: 13⋅09 21⋅09 15⋅82 *M1 Calculate value of χ2 (to 2 d.p. ; A1 dep *M1): χ2 = 1⋅50 M1 *A1 State or use consistent tabular value (to 2 d.p.): χ1, 0.9 2 = 2⋅706 (cells combined) *B1 [χ2, 0.9 2 = 4⋅605, χ3, 0.9 2 = 6⋅251] Correct conclusion (A.E.F., dep *A1, *B1): 1⋅50 < 2⋅71 so distn. does fit A1 9 [9]

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 23 © Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 9 State hypotheses: H0: µP = µQ , H1: µP ≠ µQ B1 Estimate population variance using P’s sample: sP 2 = (2120 – 321⋅22/50) / 49 (allow use of biased: σ P,50 2 = 1⋅132 or 1⋅0642) [= 1⋅155 or 1⋅0752] M1 Estimate population variance using Q’s sample: sQ 2 = (3310 – 475⋅32/70) / 69 (allow use of biased: σ Q,70 2 = 1⋅182 or 1⋅0872) [= 1⋅199 or 1⋅0952] M1 Estimate population variance for combined sample: s2 = sP 2 /50 + sQ 2 /70 = 0⋅04023 or 0⋅20062 (allow use of σ P,50 2, σ Q,70 2) (or 0⋅03949 or 0⋅19872) M1 A1 Calculate value of z (to 2 d.p., either sign): z = (6⋅424 – 6⋅79) / s M1 A1 = –0⋅366/0⋅2006 = –1⋅82[5] (or –1⋅84) A1 S.R. Allow (implicit) assumption of equal variances, but deduct A1 if not explicit: Find pooled estimate of common variance s2 : (50σ P,50 2 + 70σ Q,70 2 )/118 = 1⋅180 or 1⋅0862 (M1A1) Calculate value of z (to 2 d.p.): z = (6⋅424 – 6⋅79)/s√(1/50+1/70) (M1 A1) = –1⋅82 (A1) State or use correct tabular z value: z 0.95 = 1⋅645 (to 2 d.p.) B1 Conclusion consistent with values (A.E.F): Breaking strengths not the same A1 10 [10]

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 23 © Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 10 Calculate gradient b in y –y = b(x –x) : b = (47136 – 610 × 578/8) / (49682 – 6102/8) = 3063⋅5 / 3169⋅5 = 0⋅966[6] B1 Find regression line of y on x (A.E.F.): y = 578/8 + 0⋅967 (x – 610/8) M1 = 72⋅2[5] + 0⋅967 (x – 76⋅2[5]) or – 1⋅45 + 0⋅967x A1 Calculate gradient b′ in x –x = b′ (y –y): b′ = (47136 – 610 × 578/8) / (45212 – 5782/8) = 3063⋅5 / 3451⋅5 = 0⋅887[6] B1 Find regression line of x on y (A.E.F.): x = 610/8 + 0⋅888 (y – 578/8) M1 = 76⋅2[5] + 0⋅888 (y – 72⋅2[5]) or 12⋅1 + 0⋅888y A1 Use regression line for x on y at y = 100: x = 101 [mins] M1 A1 S.R. Using regression line for y on x at y = 100: x = 105 [mins] (B1) Find correlation coefficient r: EITHER: r2 = bb′ = 0⋅8580, r = 0⋅926 M1 A1 OR: r = (47136 – 610 × 578/8) / √{(49682 – 6102/8)(45212 – 5782/8)} = 3063⋅5 / √(3169⋅5 × 3451⋅5) = 0⋅926 (M1 A1) 6 2 2 [10]

Page 9 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 23 © Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total 11 (a) Resolve vertically at equilibrium with extn. e: 8mge / a = mg [e = a/8] B1 EITHER: Use Newton’s Law at general point: m d2x/dt2 = mg – 8mg(e+x)/a M1 A1 [ or – mg + 8mg(e–x)/a ] Simplify to give ω2 in d2x/dt2 = – ω2x : d2x/dt2 = – (8g/a) x or ω2 = 8g/a A1 (allow stating result without derivation) OR: Assume SHM and find ω2 from speed v when first slack, found from energy as below: v2 = ω2 {(¼a)2 – e2} (M1) 3ga/8 = ω2 (a2/16 – a2/64) (A1) ω2 = 8g/a (A1) Use x = ¼ a cos ωt or ¼ a sin ωt to find ωt: ωt = cos–1 (-½) or ½π + sin–1 (½) M1 A1 = 2π/3 A1 Substitute ω = √(8g/a): t = (2π/3)√(a/8g) A.G. A1 EITHER: Find v2 when first slack from an SHM eqn: v2 = ω2 (a2/16 – e2) = 3ga/8 or ¼aω sin 2π/3 = 3ga/8 M1 A1 OR: Find v2 when first slack using energy: ½mv2 = ½ 8mg(e + ¼a)2 / a – mg(e + ¼a) (this result may be used above) v2 = 9ga/8 – 3ga/4 = 3ga/8 (M1 A1) Find further distance s2 to rest: 2gs2 = v2, s2 = 3a/16 M1 A1 Find total distance: ¼a + e + s2 = 9a/16 or 0⋅562[5]a M1 A1 8 6 [14]

Page 10 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 23 © Cambridge International Examinations 2012 Question Number Mark Scheme Details Part Mark Total (b) (i) (ii) (iii) Find k by equating area under graph to 1: k + 3k = 1, k = ¼ M1 A1 Find f(x) for 0 < x ≤ 2 and 2 < x ≤ 5: ½kx = x/8 and k = ¼ A.G. B1 Integrate to find F(x): F(x) = x2/16 (0 ≤ x ≤ 2) ¼x – ¼ (2 < x ≤ 5) M1 A1 Relate dist. fn. G(y) of Y to X: G(y) = P(Y < y) = P(X 2 < y) (working may be omitted) = P(X < y1/2) = F(y1/2) = y/16 and ¼y1/2 – ¼ M1 A1 Differentiate to find g(y): g(y) = 1/16 or 0⋅0625 (0 ≤ y ≤ 4) (both results reqd. for M1) 1/8√y (4 < y ≤ 25) M1 A1 [0 otherwise] EITHER: Find E(Y) using ∫ y g(y) dy: E(Y) = (1/16)∫ y dy + (1/8)∫ y1/2 dy M1 Integrate and insert limits: = [y2/32] 4 0 + [y3/2/12] 25 4 A1 = ½ + 117/12 = 10⋅25 A.G. A1 OR: Find E(Y) using ∫ x2 f(x) dx: E(Y) = (1/8)∫ x3 dx + ¼ ∫ x2 dx (M1) Integrate and insert limits: = [x4/32] 2 0 + [x3/12] 5 2 (A1) = ½ + 117/12 = 10⋅25 A.G. (A1) EITHER: Find median mx of X and F(mx) = ¼ mx – ¼ = ½ , mx = 3 median my of Y (or √my): F(my) = ¼ my 1/2 – ¼ = ½ , my = 9 M1 A1 OR: Show my = mx 2 : P(Y < mx 2) = P(X2 < mx 2) = P(X < mx) (M1 A1) 3 6 3 2 [14]