Cambridge A Level Mathematics - Further 9231 — 2011 May/June Paper 2 · Variant 1

*2465618767* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/21 Paper 2 May/June 2011 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. JC11 06_9231_21/2R © UCLES 2011 [Turn over

2 1 A particle oscillates in simple harmonic motion with centre O. When its distance from O is 3 m its speed is 16 m s−1, and when its distance from O is 4 m its speed is 12 m s−1. Find the period and amplitude of the motion. [5] 2 A particle of mass m is attached to the mid-point of a light elastic string. The string is stretched between two points A and B on a smooth horizontal surface, where AB = 2a. The string has modulus of elasticity λ and natural length 2l, where l < a. The particle is in motion on the surface along a line passing through the mid-point of AB and perpendicular to AB. When the displacement of the particle from AB is x, the tension in the string is T. Given that x is small enough for x2 to be neglected, show that T = λ l (a −l). [3] The particle is slightly disturbed from its equilibrium position. Show that it will perform approximate simple harmonic motion and find the period of the motion. [4] 3 P a a a A uniform solid hemisphere, of radius a and mass M, is placed with its curved surface in contact with a rough plane that is inclined at an angle α to the horizontal. A particle P of mass m is attached to the rim of the hemisphere. The system rests in equilibrium with the rim of the hemisphere horizontal and P at the point on the rim that is closest to the inclined plane (see diagram). Given that the coefficient of friction between the plane and the hemisphere is 1 2, show that (i) tan α ≤1 2, [4] (ii) m ≤M(1 + √5) 4 . [5] 4 Two uniform spheres A and B, of equal radius, are at rest on a smooth horizontal table. Sphere A has mass 3m and sphere B has mass m. Sphere A is projected directly towards B, with speed u. The coefficient of restitution between the spheres is 0.6. Find the speeds of A and B after they collide. [5] Sphere B now strikes a wall that is perpendicular to its path, rebounds and collides with A again. The coefficient of restitution between B and the wall is e. Given that the second collision between A and B brings A to rest, find e. [5] © UCLES 2011 9231/21/M/J/11

3 5 A particle P of mass m is placed at the point Q on the outer surface of a fixed smooth sphere with centre O and radius a. The acute angle between OQ and the upward vertical is α, where cos α = 9 10. The particle is released from rest and begins to move in a vertical circle on the surface of the sphere. Show that P loses contact with the sphere when OP makes an angle θ with the upward vertical, where cos θ = 3 5, and find the speed of P at this instant. [6] Show that, in the subsequent motion, when P is at a distance 7 5a from the vertical diameter through O, its distance below the horizontal through O is 31 30a. [6] 6 A random sample of residents in a town took part in a survey. They were asked whether they would prefer the local council to spend money on improving the local bus service or on improving the quality of road surfaces. The responses are shown in the following table, classified according to the area of the town in which the residents live. Area 1 Area 2 Area 3 Local bus service 73 36 30 Road surfaces 47 44 20 Using a 5% significance level, test whether there is an association between the area lived in and preference for improving the local bus service or improving the quality of road surfaces. [7] 7 A greengrocer claims that his cabbages have a mean mass of more than 1.2 kg. In order to check his claim, he weighs 10 cabbages, chosen at random from his stock. The masses, in kg, are as follows. 1.26 1.24 1.17 1.23 1.18 1.25 1.19 1.20 1.21 1.18 Stating any assumption that you make, test at the 10% significance level whether the greengrocer’s claim is supported by this evidence. [8] 8 In a crossword competition the times, x minutes, taken by a random sample of 6 entrants to complete a crossword are summarised as follows. Σx = 210.9 Σ(x −x)2 = 151.2 The time to complete a crossword has a normal distribution with mean µ minutes. Calculate a 95% confidence interval for µ. [5] Assume now that the standard deviation of the population is known to be 5.6 minutes. Find the smallest sample size that would lead to a 95% confidence interval for µ of width at most 5 minutes. [4] 9 Mr Lee asserts that boys are slower than girls at completing a particular mathematical puzzle. In order to test his assertion, a random sample of 40 boys and a random sample of 60 girls are selected from a large group of students who attempted the puzzle. The times taken by the boys, b minutes, and the times taken by the girls, g minutes, are summarised as follows. Σb = 92.0 Σb2 = 216.5 Σg = 129.8 Σg2 = 288.8 Test at the 2.5% significance level whether this evidence supports Mr Lee’s assertion. [9] © UCLES 2011 9231/21/M/J/11 [Turn over

4 10 The mid-day temperature, x ◦C, and the amount of sunshine, y hours, were recorded at a winter holiday resort on each of 12 days, chosen at random during the winter season. The results are summarised as follows. Σ x = 18.7 Σx2 = 106.43 Σy = 34.7 Σy2 = 133.43 Σxy = 92.01 (i) Find the product moment correlation coefficient for the data. [3] (ii) Stating your hypotheses, test at the 1% significance level whether there is a non-zero correlation between mid-day temperature and amount of sunshine. [4] (iii) Use the equation of a suitable regression line to estimate the number of hours of sunshine on a day when the mid-day temperature is 2 ◦C. [3] © UCLES 2011 9231/21/M/J/11

5 11 Answer only one of the following two alternatives. EITHER A B C D O a a A rigid body is made from uniform wire of negligible thickness and is in the form of a square ABCD of mass M enclosed within a circular ring of radius a and mass 2M. The centres of the square and the circle coincide at O and the corners of the square are joined to the circle (see diagram). Show that the moment of inertia of the body about an axis through O, perpendicular to the plane of the body, is 8 3Ma2. [6] Hence find the moment of inertia of the body about an axis l, through A, in the plane of the body and tangential to the circle. [2] A particle P of mass M is now attached to the body at C. The system is able to rotate freely about the fixed axis l, which is horizontal. The system is released from rest with AC making an angle of 60◦ with the upward vertical. Find, in terms of a and g, the greatest speed of P in the subsequent motion. [6] OR 0 1 3 0 k x f( )x The continuous random variable X takes values in the interval 0 ≤x ≤3 only. For 0 ≤x ≤3 the graph of its probability density function f consists of two straight line segments meeting at the point (1, k), as shown in the diagram. Find k and hence show that the distribution function F is given by F(x) =   0 x ≤0, 1 3x2 0 < x ≤1, x −1 2 −1 6x2 1 < x ≤3, 1 x > 3. [6] The random variable Y is given by Y = X2. Find (i) the probability density function of Y, [4] (ii) the median value of Y. [4] © UCLES 2011 9231/21/M/J/11

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. 9231/21/M/J/11

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2011 question paper for the guidance of teachers 9231 FURTHER MATHEMATICS 9231/21 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. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2011 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 2011 9231 21 © University of Cambridge International Examinations 2011 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 2011 9231 21 © University of Cambridge International Examinations 2011 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 2011 9231 21 © University of Cambridge International Examinations 2011 Question Number Mark Scheme Details Part Mark Total 1 Apply v2 = ω2 (A2 – x2) at both points: 162 = ω2(A2 – 32), 122 = ω2(A2 – 42) M1 A1 Combine to find A and ω (in either order): A = 5 [m], ω = 4 B1 B1 Find period T using ω (√ on ω): T = 2π/ω = π /2 or 1⋅57 [s] B1√ 5 [5] 2 Find T by Hooke’s Law for whole [or half] string: T = λ{2√(a2 + x2) – 2l} / 2l M1 A1 Neglect x2: [M0 here for motion along string] = (λ/l) (a – l) A.G. A1 Find eqn of motion at general point: md2x/dt2 = – 2T x/√(a2 + x2) M1 Neglect x2 to give SHM eqn for d2x/dt2 : – (2T/ma) x or – {2λ (a–l)/alm}x M1 A1 State period of motion using 2π/ω (A.E.F.): 2π √{alm / 2λ (a – l)} B1 M.R. Vertical motion loses one A/B mark S.R. Motion along string earns max 1 mark: d2x/dt2 = – 2λx/lm (B1) 3 4 [7] 3 (i) (ii) EITHER: Resolve horizontally: F cos α = R sin α B2 OR: Resolve along plane to find friction F: F = (M + m)g sin α (B1) Resolve normal to plane for reaction R: R = (M + m)g cos α (B1) Use F/R ≤ ½ : tan α ≤ ½ A.G. M1 A1 EITHER: Take moments about pt of contact: mg (a – a sin α) = Mga sin α OR: Take moments about centre: mg = F = (M + m)g sin α M1 A1 Find inequality for sin α: sin α ≤ 1/√5 B1 Combine: m = M/(1/sin α – 1) ≤ M/(√5 – 1) m ≤ M/(1 + √5) / 4 A.G. M1 A1 4 5 [9] 4 Use conservation of momentum for 1st collision: 3muA + muB = 3mu M1 Use Newton’s law of restitution (A1 if both correct): uA – uB = – 0⋅6 u M1 A1 Solve for uA and uB : uA = 0⋅6 u and uB = 1⋅2u M1 A1 Find speed u′B of B after striking wall: u′B = euB [= 1⋅2eu] M1 Use conservation of momentum for 2nd collision: mvB = 3muA – mu′B B1 Use Newton’s law of restitution: vB = 0⋅6 (uA + u′B) B1 Combine to find e: 1⋅6e × 1⋅2u = 2⋅4 × 0⋅6u, e = ¾ M1 A1 5 5 [10]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 21 © University of Cambridge International Examinations 2011 Question Number Mark Scheme Details Part Mark Total 5 Use conservation of energy: ½mv2 = mga (cos α – cos θ) M1 A1 Equate radial forces [may imply R = 0]: mv2/a + R = mg cos θ M1 Take R = 0 when contact lost: mv2/a = mg cos θ A1 Eliminate v2 to find show cos θ = 3/5 when R = 0: 2 (cos α – cos θ) = cos θ [or can show R = 0 when cos θ = 3/5] cos θ = (2/3) cos α = 3/5 A.G. B1 Find v: v = √(3ag/5) or √(6a) B1 Find horizontal distance moved: x = 7a/5 – 4a/5 [= 3a/5] B1 EITHER: Find time of subsequent motion: t = x / v cos θ [= a/v = √(5a/3g) = √(a/6)] M1 Find distance h fallen in time t: h = (v sin θ) t + ½gt2 M1 OR: Use trajectory eqn to find h: h = x tan θ + ½g(x / v cos θ)2 (M2) Substitute and simplify h: h = 4a/5 + 5a/6 = 49a/30 M1 A1 Find distance below horizontal through O: h – 3a/5 = 31a/30 A.G. B1 6 6 [12] 6 Find expected values (to 1 dp): 66⋅72 44⋅48 27⋅80 (lose A1 if rounded to integers) 53⋅28 35⋅52 22⋅20 M1 A1 State (at least) null hypothesis (A.E.F.): H0: No association between them B1 Calculate value of χ2: χ2 = 5⋅36 ± 0⋅03 M1 A1 Compare with consistent tabular value (to 2 dp): χ2, 0.95 2 = 5⋅991 B1 Conclusion consistent with values (A.E.F): No association A1 √ 7 [7] 7 State valid assumption (A.E.F.): Popln. of masses has normal distn. B1 State both hypotheses: H0: µ = 1⋅2, H1: µ > 1⋅2 B1 Calculate sample mean: x = 12⋅11 / 10 [= 1⋅211] B1 Estimate population variance: s2 = (14⋅6745 – 12⋅112/10) / 9 (allow biased: 0⋅000929 or 0⋅03052) [= 0⋅00103 or 0⋅03212 ] M1 Calculate value of t (to 2 dp): t = (x – 1⋅2) / (s/√10) = 1⋅08 M1 A1 Compare with correct tabular t value: t9, 0.90 = 1⋅38[3] B1 Conclusion consistent with values (A.E.F): Claim is not correct A1 √ 8 [8] 8 Calculate sample mean: d = 35⋅15 B1 Estimate population variance (allow biased): s2 = 30⋅24 [5⋅52] (or 25⋅2 [5⋅022]) B1 Find confidence interval (allow z in place of t) e.g.: 35⋅15 ± t √(30⋅24/6) M1 (inconsistent use of 5 or 6 loses M1) Use of correct tabular value: t5, 0.975 = 2⋅571 (2 d.p.) A1 Evaluate C.I. correct to 3 s.f. (needs correct s, t): 35⋅15 ± 5⋅77 or [29⋅4, 40⋅9] A1 State inequality involving sample size n: 1⋅96 × 5⋅6 /√n ≤ (or <) 2⋅5 M1 A1 (Equality or wrong critical value loses A1) Solve for limiting value of n: 4⋅392 = 19⋅3 A1 State smallest sample size: nmin = 20 A1 5 4 [9]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 21 © University of Cambridge International Examinations 2011 Question Number Mark Scheme Details Part Mark Total 9 State hypotheses: H0: µb = µg , H1: µb > µg B1 Estimate population variance using boys’ sample: sb 2 = (216⋅5 – 92⋅02/40) / 39 (allow use of biased: σ b,40 2 = 0⋅1225 or 0⋅352) [= 0⋅126 or 0⋅3542] M1 Estimate population variance using girls’ sample: sg 2 = (288⋅8 – 129⋅82/60) / 59 (allow use of biased: σ g,60 2 = 0⋅133 or 0⋅3652) [= 0⋅136 or 0⋅3682] M1 Estimate population variance for combined sample: s2 = sb 2 /40 + sg 2 /60 = 0⋅00541 or 0⋅07362 (allow use of σ b,40 2, σ g,60 2) (or 0⋅00528 or 0⋅07272) M1 A1 Calculate value of z (to 2 dp): z = (2⋅3 – 2⋅163)/ s = 0⋅1367/0⋅0736 = 1⋅86 (or 1⋅88) M1 A1 S.R. Allow (implicit) assumption of equal variances: Find pooled estimate of common variance s2 : (40σ b,40 2 + 60σ g,60 2 )/98 = 0⋅132 (M1A1) Calculate value of z (to 2 dp): z = (2⋅3 – 2⋅163)/s√(1/40+1/60) = 1⋅85 (M1 A1) Compare with correct tabular t value: z 0.975 = 1⋅96 B1 Conclusion consistent with values (A.E.F): Claim is not correct A1 √ 9 [9] 10 (i) (ii) (iii) Find correlation coefficient r: r = (92⋅01 – 18⋅7 × 34⋅7/12) / √{(106⋅43 – 18⋅72/12) (133⋅43 – 34⋅72/12)} M1 = 37⋅94 / √(77⋅29 × 33⋅09) = 37⋅94 / (8⋅791 × 5⋅752) or 3⋅161 / √(6⋅441 × 2⋅757) = 3⋅161 / (2⋅538 × 1⋅661) A1 = 0⋅750 [allow 0⋅75] *A1 State both hypotheses: H0: ρ = 0, H1: ρ ≠ 0 B1 Use correct tabular 2-tail r value: r12, 1% = 0⋅708 (to 2 dp) *B1 Valid method for reaching conclusion: Reject H0 if |r| > tabular value M1 Correct conclusion (AEF, dep *A1, *B1): There is non-zero correlation A1 EITHER: Calculate gradient b in y –y = b(x –x): b = (92⋅01 – 18⋅7 × 34⋅7/12) / (106⋅43 – 18⋅72/12) = 37⋅94 / 77⋅29 = 0⋅491 B1 Use regression line for y at x = 2: y = 34⋅7/12 + 0⋅491(x – 18⋅7/12) M1 [y = 0⋅491x + 2⋅13] = 2⋅13 + 0⋅491x = 3⋅11 ± 0⋅01 A1 OR: Calculate gradient b′ in x –x = b′ (y –y): b′ = (92⋅01 – 18⋅7 × 34⋅7/12) / (133⋅43 – 34⋅72/12) = 37⋅94 / 33⋅09 = 1⋅15 (B1) Use regression line for y at x = 2: y = 34⋅7/12 + (x – 18⋅7/12)/1⋅15 (M1) [x = 1⋅15y – 1⋅76] = 1⋅53 + x/1⋅15 = 3⋅28 ± 0⋅01 (A1) 3 4 3 [10]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 21 © University of Cambridge International Examinations 2011 Question Number Mark Scheme Details Part Mark Total 11 (a) Find MI of one side of square about midpoint: ⅓(M/4) a2/2 = Ma2/24 M1 A1 Find MI of side of square (or of square) about O: Ma2/24 + (M/4) a2/2 = Ma2/6 M1 A1 Find MI of ring about O: 2Ma2 B1 Sum to find MI of system about O: A.G. IO = 4 × Ma2/6 + 2Ma2 = 8 Ma2/3 A1 Find MI of system about axis through A: IA = ½ IO + 3Ma2 = 13 Ma2/3 M1 A1 State or imply that speed is max when AC vertical M1 Use energy when AC vertical (or at general point): ½(IA + 4Ma2) ω2 [=25 Ma2/3] or ½IA ω2 + ½Mv2 (A1 for each side of eqn, A.E.F.) = 5Mga (1 + cos 60°) or 15Mga/2 M1 A1 A1 Substitute for IA and find max speed v = 2aω: ω2 = (15Mga/2) / (25Ma2/6) = 9g/5a v = 6√(ga/5) or 6√(2a) or 8⋅49√a M1 A1 6 2 6 [14] (b) Find k by equating area under graph to 1: ½k + k = 1, k = ⅔ M1 A1 Find f(x) for 0 < x ≤ 1 and 1 < x ≤ 3: kx [= ⅔x], ½k(3 – x) [= 1 – ⅓x] M1 A1 Integrate to find F(x) for 0 < x ≤ 1 and 1 < x ≤ 3: ⅓x2, x – ½ – x2/6 A.G. M1 A1 (i) 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) = ⅓y, y1/2 – ½ – y/6 M1 A1 Differentiate to find g(y): g(y) = ⅓ (0 < y ≤ 1), ½ y-1/2 – 1/6 (1 < y ≤ 9), [0 otherwise] M1 A1 (ii) Use G(m) = ½ to find eqn for median m: m1/2 – ½ – m/6 = ½ M1 Solve quadratic for √m: (√m) 2 – 6√m +6 = 0,√m = 3 ± √3 M1 A1 Select value of m in interval: m = (3 – √3) 2 = 12 – 6√3 or 1⋅61 B1 6 4 4 [14]