Cambridge A Level Mathematics - Further 9231 — 2013 Oct/Nov Paper 2 · Variant 2

*9729440533* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/22 Paper 2 October/November 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 11_9231_22/FP © UCLES 2013 [Turn over

2 1 A B C D 2a 2a a Three identical uniform rods, AB, BC and CD, each of mass M and length 2a, are rigidly joined to form three sides of a square. A uniform circular disc, of mass 2 3M and radius a, has the opposite ends of one of its diameters attached to A and D respectively. The disc and the rods all lie in the same plane (see diagram). Find the moment of inertia of the system about the axis AD. [6] 2 The point O is on the fixed line l. The point A on l is such that OA = 3 m. A particle P oscillates on l in simple harmonic motion with centre O and period
seconds. When P is at A its speed is 12 m s−1. Find the speed of P when it is at the point B on l, where OB = 6 m and B is on the same side of O as A. [4] Find, correct to 2 decimal places, the time, in seconds, taken for P to travel directly from A to B. [4] 3 0.2 m 4 kg A uniform disc, of mass 2 kg and radius 0.2 m, is free to rotate in a vertical plane about a smooth horizontal axis through its centre. One end of a light inextensible string is attached to a point on the rim of the disc and the string is wound round the rim. The other end of the string is attached to a small block of mass 4 kg, which hangs freely (see diagram). The system is released from rest. During the subsequent motion, the block experiences a constant resistance to its motion, of magnitude R N. Given that the angular speed of the disc after it has turned through 2 radians is 5 rad s−1, find R and the tension in the string. [9] © UCLES 2013 9231/22/O/N/13

3 4 B A C O P
A uniform circular disc, with centre O and weight W, rests in equilibrium on a horizontal floor and against a vertical wall. The plane of the disc is vertical and perpendicular to the wall. The disc is in contact with the floor at A and with the wall at B. A force of magnitude P acts tangentially on the disc at the point C on the edge of the disc, where the radius OC makes an angle
with the upward vertical, and tan
= 4 3 (see diagram). The coefficient of friction between the disc and the floor and between the disc and the wall is 1 2. Show that the sum of the magnitudes of the frictional forces at A and B is equal to P. [2] Given that the equilibrium is limiting at both A and B, (i) show that P = 15 34W, [6] (ii) find the ratio of the magnitude of the normal reaction at A to the magnitude of the normal reaction at B. [2] 5 Two uniform small smooth spheres A and B, of equal radii, have masses 2m and m respectively. They lie at rest on a smooth horizontal plane. Sphere A is projected directly towards B with speed u. After the collision B goes on to collide directly with a fixed smooth vertical barrier, before colliding with A again. The coefficient of restitution between A and B is 2 3 and the coefficient of restitution between B and the barrier is e. After the second collision between A and B, the speed of B is five times the speed of A. Find the two possible values of e. [11] 6 A fair die is thrown until a 5 or a 6 is obtained. The number of throws taken is denoted by the random variable X. State the mean value of X. [1] Find the probability that obtaining a 5 or a 6 takes more than 8 throws. [2] Find the least integer n such that the probability of obtaining a 5 or a 6 in fewer than n throws is more than 0.99. [3] 7 A random sample of 10 observations of a normally distributed random variable X gave the following summarised data, where x denotes the sample mean. Σx = 70.4 Σ
x −x2 = 8.48 Test, at the 10% significance level, whether the population mean of X is less than 7.5. [7] © UCLES 2013 9231/22/O/N/13 [Turn over

4 8 The lifetime, in years, of an electrical component is the random variable T, with probability density function f given by f
t =  Ae−
t t ≥0, 0 otherwise, where A and
are positive constants. (i) Show that A =
. [2] It is known that out of 100 randomly chosen components, 16 failed within the first year. (ii) Find an estimate for the value of
, and hence find an estimate for the median value of T. [6] 9 For a random sample of 10 observations of pairs of values
x, y, the equations of the regression lines of y on x and of x on y are y = 4.21x −0.862 and x = 0.043y + 6.36, respectively. (i) Find the value of the product moment correlation coefficient for the sample. [3] (ii) Test, at the 10% significance level, whether there is evidence of non-zero correlation between the variables. [4] (iii) Find the mean values of x and y for this sample. [2] (iv) Estimate the value of x when y = 2.3 and comment on the reliability of your answer. [2] 10 Customers were asked which of three brands of coffee, A, B and C, they prefer. For a random sample of 80 male customers and 60 female customers, the numbers preferring each brand are shown in the following table. A B C Male 32 36 12 Female 18 30 12 Test, at the 5% significance level, whether there is a difference between coffee preferences of male and female customers. [8] A larger random sample is now taken. It consists of 80n male customers and 60n female customers, where n is a positive integer. It is found that the proportions choosing each brand are identical to those in the smaller sample. Find the least value of n that would lead to a different conclusion for the 5% significance level hypothesis test. [4] © UCLES 2013 9231/22/O/N/13

5 11 Answer only one of the following two alternatives. EITHER A smooth sphere, with centre O and radius a, is fixed on a smooth horizontal plane
. A particle P of mass m is projected horizontally from the highest point of the sphere with speed
2 5ga. While P remains in contact with the sphere, the angle between OP and the upward vertical is denoted by . Show that P loses contact with the sphere when cos  = 4 5. [6] Subsequently the particle collides with the plane
. The coefficient of restitution between P and
is 5 9. Find the vertical height of P above
when the vertical component of the velocity of P first becomes zero. [6] OR A factory produces bottles of spring water. The manager decides to assess the performance of the two machines that are used to fill the bottles with water. He selects a random sample of 60 bottles filled by the first machine X and a random sample of 80 bottles filled by the second machine Y. The volumes of water, x and y, measured in appropriate units, are summarised as follows. Σx = 58.2 Σx2 = 85.8 Σy = 97.6 Σy2 = 188.6 A test at the % significance level shows that the mean volume of water in bottles filled by machine X is less than the mean volume of water in bottles filled by machine Y. Find the set of possible values of . [12] © UCLES 2013 9231/22/O/N/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/O/N/13

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 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 October/November 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 AS/A LEVEL – October/November 2013 9231 22 © 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 AS/A LEVEL – October/November 2013 9231 22 © 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 2013 9231 21 © Cambridge International Examinations 2013 Question Number Mark Scheme Details Part Mark Total 1 Find MI of rod AB or CD about AD: IAB = ⅓ Ma2 + Ma2 [= 4Ma2/3] B1 Find MI of rod BC about AD: IBC = M(2a)2 [= 4Ma2] B1 Find MI of disc about AD by perpendicular axes: IDisc = ½ ½ ⅔ Ma2 = Ma2/6 M1 A1 Find MI of system about A: I = 2 IAB + IBC + IDisc I = (41/6) Ma2 M1 A1 6 [6] 2 Use T = 2π/ω to find ω: ω = 2 B1 Use v2 = ω2 (A2 – x2) to find A2: 122 = ω2 (A2 – 32), A2 = 45 M1 A1 Use v2 = ω2 (A2 – x2) to find v at x = 6: [√ on ω] vB 2 = 22 (45 – 62), vB = 6 [m s-1] B1√ Find time at A ½ sin-1 (3/√45) or ½ cos-1 (3/√45) or at B, e.g.: ½ sin-1 (6/√45) or ½ cos-1 (6/√45) M1 A1 Combine correctly to find time from A to B: ½ sin-1 (6/√45) – ½ sin-1 (3/√45) or ½ cos-1 (3/√45) – ½ cos-1 (6/√45) M1 Evaluate to 2 d.p.: [√ on ω] = 0⋅554 – 0⋅232 = 0⋅32 [s] A1√ 4 4 [8] 3 EITHER: Use ω2 = 2 (dω /dt) θ to find dω /dt [or a]: 52 = 2 × dω /dt × 2 M1 A1 dω /dt = 25/4 or 6⋅25 [a = 1⋅25] A1 Find eqn of motion for disc: T × 0⋅2 = (½ × 2 × 0⋅22) dω /dt M1 A1 Substitute to find T: T = 0⋅2 dω /dt = 5/4 or 1⋅25 A1 Find eqn of motion for block: 4g – R – T = 4 × 0⋅2 dω /dt or 4a M1 A1 Substitute to find R: R = 4g – T – 0⋅8 dω /dt (g = 9⋅81 gives R = 32⋅99) = 135/4 or 33⋅7[5] A1 S.R.: B1 only for (4g – R) × 0⋅2 = 0⋅22 dω /dt OR: Use energy to find R (without dω /dt): ½ 0⋅22 ω2 + ½ 4 (0⋅2 ω)2 (or in terms of v = 0⋅2 ω = 1) = (4g – R) × 0⋅2 θ (M1 A1) Simplify: ½ + 2 = (4g – R) × 0⋅4 R = 4g – 5/4 – 5 = 135/4 or 33⋅7[5] (A1) Find 2 eqns of motion (or energy for block) (2 × M1 A1) Eliminate dω /dt to find T: 4g – R – T = 4T, T = 1⋅25 (M1 A1) 9 [9] 4 (i) (ii) Take moments about O (a may be cancelled out): FA + FB = P A.G. M1 A1 Find 2 indep. moment or resln. eqns, e.g.: RA + FB = W – P sin θ (M1 needs 2 eqns) RB – FA = P cos θ M1 Relate RA , FA , RB and FB : RA = 2FA and RB = 2FB B1 Substitute in above eqns to give e.g.: 2FA + FB = W – 4P/5 A1 2FB – FA = 3P/5 A1 Solve for P: [FA = 7P/15 = 7W/34 FB = 8P/15 = 8W/34] P = 15W/34 A.G. M1 A1 Find RA / RB by any valid method, e.g.: RA / RB = FA / FB = 7/8 M1 A1 2 6 2 [10]

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 21 © Cambridge International Examinations 2013 Question Number Mark Scheme Details Part Mark Total 5 Use conservation of momentum, e.g.: 2mvA + mvB = 2mu B1 Use restitution (must be consistent with eqn.): vA – vB = – ⅔ u B1 Solve for vA and vB (A.E.F.): vA = 4u/9, vB = 10u/9 M1 A1 Find speed of B after striking barrier (ignore sign): vB′ = – evB [= – 10eu/9] M1 Use conservation of momentum, e.g.: 2mwA + mwB = 2mvA + m vB′ B1 Use restitution (must be consistent with prev. eqn.): wA – wB = – ⅔(vA – vB′) B1 Substitute and take wB = 5wA or – 5wA: 2wA ± 5wA = 8u/9 – 10eu/9 (M1 needs 2 eqns with +5wA or –5wA): wA – (±5wA) = – ⅔(4u/9 + 10eu/9) M1 Find values of e (M1 for either): +5wA ⇒ 2/13, –5wA ⇒ ½ M1 A1, A1 11 [11] 6 State or find the mean of X: E(X) = 1/⅓ = 3 B1 Find P(X > 8): P(X > 8) = (⅔)8 M1 = 256/6561 or 0⋅0390 A1 Formulate condition for n (M0 here if equality used): 1 – (⅔)n-1 > 0⋅99, (⅔)n-1 < 0⋅01 M1 Take logs (any base) to give inequality for n: n – 1 > log 0⋅01 / log ⅔ M1 Find nmin: n – 1 > 11⋅4, nmin = 13 A1 1 2 3 [6] 7 State hypotheses (A.E.F.): H0: µ = 7⋅5, H1: µ < 7⋅5 B1 Calculate sample mean: x = 70⋅4 / 10 = 7⋅04 M1 Estimate population variance: s2 = 8⋅48 / 9 = 211/225 or (allow biased here: 0⋅848 or 0⋅92092) 0⋅9422 or 0⋅97072 M1 Calculate value of t (to 3 sf): t = ( x – 7⋅5)/(s/√10) = ±1⋅49[9] M1 *A1 State or use correct tabular t value: t9, 0.9 = 1⋅38[3] *B1 (or can compare x with 7⋅5 – 0⋅425 = 7⋅07[5]) correct conclusion (AEF, dep *A1, *B1): Mean is less than 7⋅5 B1 7 [7] 8 (i) (ii) Show that A = λ: [ ] ∞ ∞ − − ∫ − = 0 0 e ) / ( d e t t A t A λ λ λ = A/λ = 1 if A = λ A.G. M1 A1 Find estimate for λ: [ ] 1 0 1 0 e d e ∫ − − − t t or t λ λ λ = 16/100 M1 1 – e-λ = 0⋅16 λ = – ln 0⋅84 = 0⋅174 M1 A1 State or use eqn. for median m of T: [ ] m t 0 e λ − − = ½ (A.E.F.) M1 Find value of m: e-λm = ½, m = λ-1 ln 2 = 3⋅98 M1 A1 2 6 [8]

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 21 © Cambridge International Examinations 2013 Question Number Mark Scheme Details Part Mark Total 9 (i) (ii) (iii) (iv) Find sample coefficient using r2 = b1 b2: r2 = 4⋅21 × 0⋅043 = 0⋅181 or 0⋅4252M1 A1 r = 0⋅425 *A1 State both hypotheses: H0: ρ = 0, H1: ρ ≠ 0 B1 State or use correct tabular one-tail r value: r10, 5% = 0⋅549 *B1 Valid method for reaching conclusion: Accept H0 if |r| < tabular value M1 Correct conclusion (AEF, dep *A1, *B1): There is no non-zero correlation A1 Solve regression eqns for mean values: x = 7⋅72 and y = 31⋅6 M1 A1 Estimate x from either eqn: x = 6⋅46 or 0⋅751 B1 State valid comment on reliability: Not reliable because e.g. value of r is small or range of data is unknown or two estimates of x very different B1 3 4 2 2 [11] 10 State (at least) null hypothesis (A.E.F.): H0: No difference in preferences [or independent] B1 Find expected values (to 1 d.p.): 28⋅57 37⋅71 13⋅71 (lose A1 if rounded to integers) 21⋅43 28⋅29 10⋅29 M1 A1 Calculate value of χ2: χ2 = 0⋅411 + 0⋅078 + 0⋅214 + 0⋅549 + 0⋅104 + 0⋅286 = 1⋅64 (or 1⋅61 if 1 d.p. used) M1 A1 State or use correct tabular χ2 value (to 3 s.f.): χ2, 0.95 2 = 5⋅99[1] B1 Valid method for reaching conclusion: Accept H0 if χ2 ≤ tabular value M1 Conclusion consistent with correct values (A.E.F): No difference in preferences A1 Calculate new value χnew 2 of χ2: χnew 2 = n ×χ2 M1 State or use correct tabular χ2 value: χ2, 0.95 2 = 5⋅99[1] B1 Find nmin: n > 5⋅99/1⋅64, nmin = 4 M1 A1 8 4 [12] 11a Use conservation of energy (B0 for v2 = …): ½mv2 = ½mu2 + mga(1 – cos θ) M1 A1 [v2 = u2 + 2ga(1 – cos θ)] Equate radial forces (R may be taken as zero): R = mg cos θ – mv2/a M1 A1 Eliminate v2 to find cos θ when R = 0: 0 = mg(3 cos θ – 2) – mu2/a (denoting this θ by θ1) cos θ1 = ⅓(2 + 2/5) = 4/5 A.G. M1 A1 Find v2 at this point: v1 2 = 2ga/5 + 2ga/5 = 4ga/5 B1 Find vertical comp. v2 of v at plane: v2 2 = v1 2 sin2 θ1 + 2ga(1 + cos θ1) M1 (using u in place of v1 can earn M1 A0) = (4/5 × 32/52 + 2 × 9/5)ga (A.E.F.) = 486ga/125 or 3⋅89ga A1 Find vertical comp. v3 of rebound speed: v3 = (5/9) v2 [v3 2 = 6ga/5] M1 Find vertical height reached: v3 2/2g = (5/9)2 (486ga/125) / 2g M1 = 3a/5 A1 6 6 [12]

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 21 © Cambridge International Examinations 2013 Question Number Mark Scheme Details Part Mark Total 11b State hypotheses: H0: µX = µY , H1: µX < µY B1 Estimate population variance using X’s sample: sX 2 = (85⋅8 – 58⋅22/60) / 59 allow use of biased: σ X,60 2 = 0⋅4891 or 0⋅69942) [= 0⋅4974 or 0⋅70532] M1 A1 Estimate population variance using Y’s sample: sY 2 = (188⋅6 – 97⋅62/80) / 79 (allow use of biased: σ Y,80 2 = 0⋅8691 or 0⋅93232) [= 0⋅8801 or 0⋅93812] M1 A1 Estimate population variance for combined sample: s2 = sX 2 /60 + sY 2 /80 = 0⋅01929 or 0⋅13892 (allow use of σ X,60 2, σ Y,80 2) (or 0⋅01901 or 0⋅13792) M1 A1 Calculate value of z (to 2 d.p., either sign): z = (0⋅97 – 1⋅22) / s M1 = –0⋅25/0⋅1389 = –1⋅80 (or –1⋅813) A1 Find Φ(z): Φ(z) = 0⋅9641 (or 0⋅9651) M1 Find corresponding significance level: 3⋅59 (or 3⋅49) M1 Find set of possible values of α (to 1 d.p.): α > 3⋅6 A1 S.R. Allow (implicit) assumption of equal variances as follows, but deduct A1 if not explicit: Find pooled estimate of common variance s2 : (60σ X,60 2 + 80σ Y,80 2 )/138 = 0⋅7165 or 0⋅84652 Calculate value of z (to 2 d.p.): z = (0⋅97 – 1⋅22)/s√(1/60+1/80) = –1⋅73 Find set of possible values of α (to 1 d.p.): Φ(z) = 0⋅9582, α > 4⋅2 12 [12]