Cambridge A Level Mathematics - Further 9231 — 2008 Oct/Nov Paper 2 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/02 Paper 2 October/November 2008 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 2008 [Turn over *7716097040*

2 1 A B C 8a 6a A uniform wire, of length24a and mass m, is bent into the form of a triangle ABC with angleABC = 90◦, AB = 6a and BC = 8a (see diagram). Find the moment of inertia of the wire about an axis through A perpendicular to the plane of the wire. [5] 2 A small bead B of mass m is threaded on a smooth wire fixed in a vertical plane. The wire forms a circle of radius a and centre O. The highest point of the circle is A. The bead is slightly displaced from rest at A. When angle AOB = θ, where θ < cos−1
2 3, the force exerted on the bead by the wire has magnitude R1. When angle AOB = π + θ, the force exerted on the bead by the wire has magnitude R2. Show that R2 −R1 = 4mg. [8] 3 a 2m A uniform disc, of mass m and radius a, is free to rotate without resistance in a vertical plane about a horizontal axis through its centre. A light inextensible string has one end fixed to the rim of the disc, and is wrapped round the rim. A block of mass 2m is attached to the other end of the string (see diagram). The system is released from rest with the block hanging vertically. While the block moves it experiences a constant resistance to motion of magnitude 1 10mg. Find the angular acceleration of the disc, and find also the angular speed of the disc when it has turned through one complete revolution. [9] © UCLES 2008 9231/02/O/N/08

3 4 Two smooth spheres A and B, of equal radii, have masses 0.1 kg and m kg respectively. They are moving towards each other in a straight line on a smooth horizontal table and collide directly. Immediately before collision the speed of A is 5 m s−1 and the speed of B is 2 m s−1. (i) Assume that in the collision A does not change direction. The speeds of A and B after the collision are vA m s−1 and vB m s−1 respectively. Express m in terms of vA and vB, and hence show that m < 0.25. [4] (ii) Assume instead that m = 0.2 and that the coefficient of restitution between the spheres is 1 2. Find the magnitude of the impulse acting on A in the collision. [6] 5 A particle of mass m moves in a straight line AB of length 2a. When the particle is at a general point P there are two forces acting, one in the direction −−→ PA with magnitude mgPA a  −1 4 and the other in the direction −−→ PB with magnitude mgPB a  1 2. At time t = 0 the particle is released from rest at the point C, where AC = 1.04a. At time t the distance AP is a + x. Show that the particle moves in approximate simple harmonic motion. [4] Using the approximate simple harmonic motion, find the speed of P when it first reaches the mid-point of AB and the time taken for P to first reach half of this speed. [7] 6 The independent random variables X and Y have normal distributions with the same variance σ2. Samples of 5 observations of X and 10 observations of Y are made, and the results are summarised by Σ x = 15, Σ x2 = 128, Σ y = 36 and Σ y2 = 980. Find a pooled estimate of σ2. [3] 7 The pulse rate of each member of a random sample of 25 adult UK males who exercise for a given period each week is measured in beats per minute. A 98% confidence interval for the mean pulse rate, µ beats per minute, for all such UK males was calculated as 61.21 < µ < 64.39, based on a t-distribution. (i) Calculate the sample mean pulse rate and the standard deviation used in the calculation. [5] (ii) State an assumption necessary for the validity of the confidence interval. [1] (iii) The mean pulse rate for all UK males is 72 beats per minute. State, giving a reason, if it can be concluded that, on average, UK males who exercise have a reduced pulse rate. [2] 8 The equations of the regression lines for a random sample of 25 pairs of data (x, y) from a bivariate population are y on x : y = 1.28 −0.425x, x on y : x = 1.05 −0.516y. (i) Find the sample means, x and y. [3] (ii) Find the product moment correlation coefficient for the sample. [2] (iii) Test at the 5% significance level whether the population correlation coefficient differs from zero. [4] © UCLES 2008 9231/02/O/N/08 [Turn over

4 9 A sample of 100 observations of the continuous random variable T was obtained and the values are summarised in the following table. Interval 1 ≤t < 1.5 1.5 ≤t < 2 2 ≤t < 2.5 2.5 ≤t < 3 Frequency 64 17 16 3 It is required to test the goodness of fit of the distribution with probability density function given by f(t) = ⎧⎪⎨ ⎪⎩ 9 4t3 1 ≤t < 3, 0 otherwise. The relevant expected values are as follows. Interval 1 ≤t < 1.5 1.5 ≤t < 2 2 ≤t < 2.5 2.5 ≤t < 3 Expected frequency 62.5 21.875 10.125 5.5 Show how the expected value 10.125 is obtained. [3] Carry out the test, at the 10% significance level. [7] 10 The continuous random variable X has probability density function given by f(x) = 0 x < 0, a 2x x ≥0, where a is a positive constant. By expressing 2x in the form ekx, where k is a constant, show that X has a negative exponential distribution, and find the value of a. [4] State the value of E(X). [1] The variable Y is related to X by Y = 2X. Find the distribution function of Y and hence find its probability density function. [8] © UCLES 2008 9231/02/O/N/08

5 11 Answer only one of the following two alternatives. EITHER W kW 5a 5a 4a 4a a 3a 5a C F B D E A The diagram shows a central cross-section CDEF of a uniform solid cube of weight kW with edges of length 4a. The cube rests on a rough horizontal floor. One of the vertical faces of the cube is parallel to a smooth vertical wall and at a distance 5a from it. A uniform ladder, of length 10a and weight W, is represented by AB. The ladder rests in equilibrium with A in contact with the rough top surface of the cube and B in contact with the wall. The distance AC is a and the vertical plane containing AB is perpendicular to the wall. The coefficients of friction between the ladder and the cube, and between the cube and the floor, are both equal to µ. A small dog of weight 1 4W climbs the ladder and reaches the top without the ladder sliding or the cube turning about the edge through D. Show that µ ≥4 5 . [7] Show that the cube does not slide whatever the value of k. [4] Find the smallest possible value of k for which equilibrium is not broken. [3] OR A perfume manufacturer had one bottle-filling machine, but because of increased sales a second machine was obtained. In order to compare the performance of the two machines, a random sample of 50 bottles filled by the first machine and a random sample of 60 bottles filled by the second machine were checked. The volumes of the contents from the first machine, x1 ml, and from the second machine, x2 ml, are summarised by Σ x1 = 1492.0, Σ x2 1 = 44 529.52, Σ x2 = 1803.6, Σ x2 2 = 54 220.84. The volumes have distributions with means µ1 ml and µ2 ml for the first and second machines respectively. Test, at the 2% significance level, whether µ2 is greater than µ1. [10] Find the set of values of α for which there would be evidence at the α% significance level that µ2 −µ1 > 0.1. [4] © UCLES 2008 9231/02/O/N/08

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/O/N/08

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2008 question paper 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. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. 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 2008 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2008 9231 02 © UCLES 2008 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 2008 9231 02 © UCLES 2008 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 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total 1 Find MI about A of AB, AC (M1 for either): IAB = (6m/24)(⅓32 + 32)a2 = 3ma2 M1 IAC = (10m/24)(⅓52 + 52)a2 = (125/9) ma2 A1 Find MI about A of BC: IBC = (8m/24)(⅓42 + 62 + 42)a2 = (172/9)ma2 M1 A1 Sum to find MI about A of wire: I = (324/9)ma2 = 36ma2 A1 5 [5] 2 Find speeds from cons. of energy (M1 for either): ½mv1 2 = mga (1 – cos θ) M1 A1 ½mv2 2 = mga (1 + cos θ) A1 Find R1, R2 by radial resolution (M1 for either): R1 = mg cos θ – mv1 2 /a M1 A1 R2 = mg cos θ + mv2 2 /a A1 EITHER: Substitute in R1, R2 and combine: R1 = 3mg cos θ – 2mg R2 = 3mg cos θ + 2mg R2 – R1 = 4mg A.G. M1 A1 OR: Combine R1, R2 and substitute: R2 – R1 = m(v2 2 + v1 2)/a = 4mg A.G. (M1 A1) 8 [8] 3 EITHER: Relate angular acceln. to tension for block: 2ma d2θ/dt2 = 2mg – T – mg/10 M1 A1 Relate angular acceln. to tension for disc: I d2θ/dt2 = aT, I = ½ ma2 M1 A1 Eliminate tension T: (½ + 2)ma2 d2θ/dt2 = (2 – 0⋅1)mga M1 Find d2θ/dt2 : 19g/25a or 0⋅76g/a or 7⋅6/a A1 Use (dθ/dt)2 = 2 d2θ/dt2 2π (√ on d2θ/dt2): (dθ/dt)2 = 76πg/25a A.E.F. M1 A1√ Find dθ/dt (A.E.F.): dθ/dt = 3⋅09√(g/a) or 9⋅77/√a A1 OR: Use conservation of energy for system: ½I (dθ/dt)2 + ½ 2m (a dθ/dt)2 (M1 A1) = 2mgaθ – 0⋅1 mgaθ (M1 A1) Put θ = 2π and find dθ/dt (A.E.F.): dθ/dt = 3⋅09√(g/a) or 9⋅77/√a (M1 A1) Differentiate energy eqn w.r.t. t: (5ma2/4) 2 d2θ/dt2 = 1⋅9 mga (M1 A1) Find d2θ/dt2 : 19g/25a or 0⋅76g/a or 7⋅6/a (A1) 9 (9) [9]

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total 4 (i) Use conservation of momentum: 0⋅1vA + mvB = 0⋅1 × 5 – m × 2 M1 Find m: m = (0⋅5 – 0⋅1 vA) / (2 + vB) A1 Use vA > 0 to find lower bound on m: vB > 0, m < 0⋅5/2 = 0⋅25 A.G. M1 A1 (ii) Use Newton’s law of restitution: vB – vA = ½( 2 + 5) = 7/2 M1 A1 Put m = 0⋅2 and find one of vA, vB: 2 vB + vA = 1, vA = –2 or vB = 1⋅5 M1 A1 Find magnitude of impulse [N s]: 0⋅1 (5 – vA) or 0⋅2 (1⋅5 + 2) = 0⋅7 M1 A1 4 6 [10] 5 Find equation of motion at general point: m d²x/dt² = mg ((a – x)/a)½ – mg ((a + x)/a)–¼ M1 Expand terms and approximate: ≈ mg ((1 – x/2a) – (1 – x/4a)) M1 A1 Simplify to give SHM eqn: d²x/dt² = – gx/4a A1 Use SHM eqn to find speed when x = 0: vmax² = (g/4a) (0⋅04a)² M1 A1 Simplify (A.E.F.): vmax = 0⋅02 √(ag) or 0⋅0632 √a A1 Use SHM eqn to find time when v = ½vmax: ½aω = aω sin ωt (A.E.F.) M1 A1 Substitute ω = √(g/4a) and simplify: t = √(4a/g) sin–1 ½ M1 = (π/3) √(a/g) (A.E.F.) A1 4 7 [11] 6 Use standard formula for pooled estimate, e.g.: ((128 – 15²/5) + (980 – 36²/10))/13 M1 Award A1 for one term in numerator, e.g.: 5 × 16⋅6 or 10 × 85⋅04 or 83 or 850⋅4 or 4 × 20⋅75 or 9 × 94⋅5 A1 Calculate value of pooled estimate: 71⋅8 A1 3 [3] 7 (i) Find sample mean: x = ½(61⋅21 + 64⋅39) = 62⋅8 M1 A1 Use confidence interval formula: x ± ts/√n for any t M1 Use correct tabular t: t24,099 = 2⋅492 A1 Calculate standard deviation: s = 1⋅59 × 5 / 2⋅492 = 3⋅19 A1 (ii) State assumption (A.E.F.): Population has normal distribution B1 (iii) State valid reason (A.E.F.): 72 exceeds upper limit of interval *B1 State conclusion (A.E.F., dep *B1): Yes, it does reduce pulse rate B1 5 1 2 [8]

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total 8 (i) Formulate two eqns for means: y + 0⋅425 x = 1⋅28 and x + 0⋅516y = 1⋅05 M1 Solve for means: x = 0⋅499, y = 1⋅068 or 1⋅07 M1 A1 (ii) Find correlation coefficient for sample: r² = 0⋅425 × 0⋅516; r = – 0⋅468 M1; *A1 (iii) State hypotheses: H0: ρ = 0, H1: ρ ≠ 0 B1 Valid method for reaching conclusion: ρ ≠ 0 if |r| > tabular value M1 Use of correct tabular value: ρ25,2⋅5% = 0⋅396 *B1 Correct conclusion (A.E.F., dep *A1, *B1): Coefficient does differ from zero A1 3 2 4 [9] 9 Integrate f(t) to give F(t): F(t) = – 9/8t2 M1 Apply limits: F(2⋅5) – F(2) = – (9/8) (2⋅5–2 – 2–2) A1 Evaluate and multiply by 100: 10⋅125 A.G. A1 State hypotheses (A.E.F.): H0: f(t) fits data, H1: doesn’t fit B1 Find χ2 (A1 if at least 3 terms correct): χ2 = 1⋅52/62⋅5 + 4⋅8752/21⋅875 + 5⋅8752/10⋅125 + 2⋅52/5⋅5 M1 A1 Evaluate χ2: = 0⋅036 + 1⋅086 + 3⋅409 + 1⋅136 = 5⋅67 [± 0⋅01] *A1 Compare with consistent tab. value (to 2 dp): χ3, 0.9 2 = 6⋅251 *B1√ (χ2, 0.9 2 = 4⋅605, χ1, 0.9 2 = 2⋅706) Consistent conclusion (A1 dep *A1, *B1): Distribution fits data (A.E.F.) M1√ A1 3 7 [10] 10 Replace 2x by ekx to find k: f(x) = ae–kx; k = ln 2 M1; A1 Show a = k by e.g. ∫ ∞ = 0 1 ) f(x : [ – (a/k) e–kx] 0 ∞ = 1, a = k or ln 2 M1 A1 State value of E(X): 1 / ln 2 or 1⋅44 B1 Find distribution fn G of Y: G(y) = P(Y ≤ y) = P(X ≤ k–1 ln y) M1 A1 = F(k–1ln y) = (a/k)(1 – e–ln y ) M1 A1 = 1 – 1/y (CAO) A1 Find probability density function g of Y: g(y) = 1/y2 (CAO) M1 A1 State interval for either G or g: y ≥ 1 B1 4 1 8 [13]

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2008 9231 02 © UCLES 2008 Qu No Mark Scheme Details Part Mark Total 11 a EITHER: Observe or deduce when RB is maximised: RB, max occurs when dog at B M1 Moments for ladder about A: 6a RB, max = 4aW + 8a × ¼W M1 OR: Moments when dog is x hor. from wall: 6a RB = 4aW + (8a – x) ¼W (M1) Deduce limit on RB : RB, max occurs when dog at B (M1) Find max. value of RB: RB, max = W A1 Resolve horizontally for ladder AB: FA = RB B1 Resolve vertically for ladder AB: RA = W + ¼W [= 5W/4] B1 Find bound on µ from FA ≤ µRA : µ ≥ FA / RA ≥ RB, max / (5W/4) M1 µ ≥ 4/5 A.G. A1 Find friction Fcube along DE by hor. resolution: Fcube = FA or RB B1 Find reaction Rcube from floor by vert. resolution: Rcube = W + ¼W + kW B1 Show that Fcube ≤ µ Rcube : µ Rcube ≥ W + 4kW/5 ≥ W = RB, max ≥ Fcube M1 A1 Find moments about D opposing effect of Rcube : 2akW + a5W/4 – 4a FA M1 Find smallest value of k for which moments ≥ 0: (4W – 5W/4) / 2W = 11/8 M1 A1 7 4 3 [14] 11 b State hypotheses (A.E.F.): H0: µ 2 = µ 1, H1: µ 2 > µ 1 B1 Calculate Σ (xi – x )2 (M1 for either) 8⋅24, 4⋅62[4] or estimate variances: 0⋅168 or 0⋅165, 0⋅0784 or 0⋅0771 M1 A1 A1 Find s2 (M0 if inconsistent denominators used): s2 = 0⋅168/50 + 0⋅0784/60 or 0⋅165/49 + 0⋅0771/59 [= 0⋅00467] *M1 Find value of z (dep *M1): z = (1803⋅6/60 – 1492⋅0/50) / s M1 = (30⋅06 – 29⋅84)/0⋅0683 = 3⋅22 *A1 S.R. Using pooled estimate of variance: s2 = (8⋅24 + 4⋅62)/108 = 0⋅119 (M0) z = 0⋅22 / s√ (1/50 + 1/60) = 3⋅33 (B1) Find tabular. value (to 2 dp): Φ–1(0⋅98) = 2⋅05[4] *B1 Compare values for conclusion (A1 dep *A1, *B1): µ 2 > µ 1 (A.E.F., M1√ on values) M1√ A1 Find limiting value of z (to 2 dp): z = (0⋅22 – 0⋅1)/s = 1⋅756 M1 A1 Find Φ(z) and hence values of α (to 1 dp): Φ(z) = 0⋅9604, α ≥ 3⋅9 or 4⋅0 M1 A1 s2 = 0⋅119 gives: z = 1⋅816, α ≥ 3⋅47 (M1 M1) 10 4 [14]