Cambridge A Level Mathematics - Further 9231 — 2010 May/June Paper 2 · Variant 3

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 May/June 2010 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. © UCLES 2010 [Turn over *0883584005*

2 1 A uniform disc with centre O has mass m and radius a. It is free to rotate in a vertical plane about a smooth fixed horizontal axis passing through O. One end of a light inextensible string is attached to a point on the circumference and is wrapped several times round the circumference. A particle P, of mass 2m, is attached to the free end of the string and the disc is held at rest with P hanging freely. The system is released from rest. Assuming that resistances may be neglected, find the acceleration of P. [5] 2 V A B a 30° 30° 60° A particle of mass m is attached to the end B of a light inextensible string. The other end of the string is attached to a fixed point A which is at a distance a above the vertex V of a circular cone of semi-vertical angle 60◦. The axis of the cone is vertical. The particle moves with constant speed u in a horizontal circle on the smooth surface of the cone. The string makes a constant angle of 30◦with the vertical (see diagram). The tension in the string and the magnitude of the normal force acting on the particle are denoted by T and R respectively. Show that T = m √3g + 2u2 a , and find a similar expression for R. [6] Deduce that u2 ≤1 2ga. [1] © UCLES 2010 9231/23/M/J/10

3 3 A B A spring balance is modelled by a vertical light elastic spring AB, of natural length 0.25 m and modulus of elasticity λ N. The bottom end B of the spring is fixed, and the top end A is attached to a small tray of mass 0.1 kg which is free to move vertically (see diagram). When in the equilibrium position, AB = 0.24 m. Show that λ = 25. [2] The tray is pushed down by 0.02 m to the point C and released from rest. At time t seconds after release the displacement of the tray from the equilibrium position is x m. Show that ¨x = −1000x. [2] Find the time taken for the tray to move a distance of 0.03 m from C. [5] 4 A small ball P, of mass 40 grams, is dropped from rest at a point A which is 10 m above a fixed horizontal plane. At the same instant an identical ball Q is dropped from rest at the point B, which is vertically below A and at a height of 5 m above the plane. The coefficient of restitution between Q and the plane is 1 2. Find the magnitude of the impulse exerted on Q by the plane. [4] The balls collide after Q rebounds from the plane and before Q hits the plane again. Find the height above the plane of the point at which the collision occurs. [6] 5 L X M N Y K a u q f 2a A rectangular pool table KLMN has KL = a and KN = 2a. A ball lies at rest on the table just outside the pocket at L and is projected along the table with speed u in a direction making an angle θ with the edge LM. The ball hits the edge KN at Y, rebounds to hit the edge LM at X and then rebounds into the pocket at N. Angle LXY is denoted by φ (see diagram). The coefficient of restitution between the ball and an edge is 3 4, and all resistances to motion may be neglected. Show that tan φ = 3 4 tan θ. [3] Show that XM = (2 −7 3 cot θ)a, and find the value of θ. [6] Find the speed with which the ball reaches N, giving the answer in the form ku, where k is correct to 3 significant figures. [3] © UCLES 2010 9231/23/M/J/10 [Turn over

4 6 The amount of caffeine in a randomly selected cup of coffee dispensed by a machine has a normal distribution. The amount of caffeine in each of a random sample of 25 cups was measured. The sample mean was 110.4 mg and the unbiased estimate of the population variance was 50.42 mg2. Calculate a 90% confidence interval for the mean amount of caffeine dispensed. [4] 7 Benford’s Law states that, in many tables containing large numbers of numerical values, the probability distribution of the leading non-zero digit D is given by P(D = d) = log10d + 1 d , d = 1, 2, . . . , 9. The following table shows a summary of a random sample of 100 non-zero leading digits taken from a table of cumulative probabilities for the Poisson distribution. Leading digit 1 2 3 4 5 ≥6 Frequency 22 21 13 11 11 22 Carry out a suitable goodness of fit test at the 10% significance level. [8] 8 A certain mechanical component has a lifetime, T months, which has a negative exponential distribution with mean 2.5. (a) A machine is fitted with 5 of these components which function independently. (i) Find the probability that all 5 components are operating 3 months after being fitted. [2] (ii) Find also the probability that exactly two components fail within one month of being fitted. [3] (b) Show that the probability that n independent components are all operating c months after being fitted is equal to the probability that a single component is operating nc months after being fitted. [4] 9 (a) The following are values of the product moment correlation coefficient between the x and y values of three different large samples of bivariate data. State what each indicates about the appearance of a scatter diagram illustrating the data. (i) −1, (ii) 0.02, (iii) 0.92. [3] (b) In 1852 Dr William Farr published data on deaths due to cholera during an outbreak of the disease in London. The table shows the altitude (in feet, above the level of the river Thames) at which people lived and the corresponding number of deaths from cholera per 10 000 people. Altitude, x 10 30 50 70 90 100 350 Number of deaths, y 102 65 34 27 22 17 8 [Σx = 700, Σx2 = 149 000, Σ y = 275, Σ y2 = 17 351, Σxy = 13 040.] (i) Calculate the product moment correlation coefficient. [3] (ii) Test, at the 5% significance level, whether there is evidence of negative correlation. [4] © UCLES 2010 9231/23/M/J/10

5 10 Carpal Tunnel syndrome is a condition which affects a person’s ability to grip with their hands. Researchers tested a treatment for this syndrome which was applied to 8 randomly chosen patients. A pre-treatment and a post-treatment test of grip was given to each patient, with the following results, measured in kg. Patient 1 2 3 4 5 6 7 8 Pre-treatment grip 24.3 29.5 28.0 28.5 21.5 28.7 25.1 26.3 Post-treatment grip 28.3 34.6 30.3 31.6 21.5 29.8 26.0 27.5 Stating any required assumption, test, at the 1% significance level, whether the mean grip of people with the syndrome increases after undergoing the treatment. [9] It is given that there is evidence at the 10% significance level that the mean grip increases by more than w kg. Find an inequality for w. [3] 11 Answer only one of the following two alternatives. EITHER A B C a a 2a 2a W 2W b b Two uniform rods AB and AC have lengths 2a and 4a and weights W and 2W respectively. They are freely hinged together at A and rest in equilibrium in a vertical plane with B and C in contact with two rough parallel vertical walls. The plane containing the rods is perpendicular to the walls. The rods AB and AC each make an angle β with the vertical (see diagram). Show that the magnitude of the frictional force acting on AB at B is 5 4W. [7] Given that the coefficient of friction at B and at C is µ, find the set of possible values of µ in terms of β. [7] OR Aram is a packer at a supermarket checkout and the time he takes to pack a randomly chosen item has mean 1.5 s and standard deviation 0.4 s. Justifying any approximation that you make, find the probability that Aram will pack 50 randomly chosen items in less than 70 s. [4] Find the greatest number of items that Aram could pack within 70 s with probability at least 90%. [6] Huldu is also a packer at the supermarket. The time that she takes to pack a randomly chosen item has mean 1.3 s and standard deviation 0.5 s. Aram and Huldu each have 50 items to pack. Find the probability that Huldu takes a shorter time than Aram. [4] © UCLES 2010 9231/23/M/J/10

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/M/J/10

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2010 question paper for the guidance of teachers 9231 FURTHER MATHEMATICS 9231/23 Paper 23, 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. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2010 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 2010 9231 23 © UCLES 2010 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 2010 9231 23 © UCLES 2010 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 2010 9231 23 © UCLES 2010 Question Number Mark Scheme Details Part Mark Total 1 Find eqn of motion for P, denoting P’s accel. by A: 2mg – T = 2mA M1 Find eqn of motion for disc: Ta = IdiscA/a M1 (2mga = IdiscA/a can earn this M1; usually 1/5) Substitute ½ma2 for MI Idisc of disc: T = ½mA A1 Eliminate T to find accel. A: 2mg = (5/2)mA M1 A = 4g/5 or 8 A1 5 [5] 2 EITHER: Resolve vertically: T cos 30° + R cos 30° = mg M1 A1 [T + R = 2mg/√3] Resolve horizontally: T sin 30° – R sin 30° = mu2/a sin 60° M1 A1 [T – R = 4mu2/a√3] OR: Any 2 other independent resolutions, e.g. Along BV (gives T): T cos 30° – mg cos 60° = (mu2/a sin 60°) sin 60° Normal to BV: R + T cos 60° – mg cos 30° = – (mu2/a sin 60°) cos 60° Along BA: T + R cos 60° – mg cos 30° = (mu2/a sin 60°) cos 60° Normal to BA: R cos 30° – mg cos 60° = – (mu2/a sin 60°) sin 60° Solve for T: T = (m/√3) (g + 2u2/a) A.G. A1 Solve for R: R = (m/√3) (g – 2u2/a) A1* Valid reason for deducing u2 Y ½ga (dep A1*): R [ 0 (A.E.F.) B1 6 1 [7] 3 Resolve vertically at equilibrium: 0⋅1g = λ 0⋅01/0⋅25 M1 Evaluate λ: λ = 25 A.G. A1 Use Newton’s Law at general point: 0⋅1 d2x/dt2 = 0⋅1g – λ(0⋅01 + x)/0⋅25 M1 Simplify: d2x/dt2 = – 1000x A.G. A1 Use SHM formula for x: x = 0⋅02 cos (t√1000) [or sin] M1 A1 Find reqd. time t: t = (1/√1000) cos-1(-0⋅01/0⋅02) or 2π/4√1000 + (1/√1000) sin-1(1/2) M1 A1 Evaluate: t = (1/√1000) 2π/3 = 0⋅0662 [s] A1 2 2 5 [9] 4 Find speed uQ of Q when striking plane: uQ = √(10 g) or 10 [m s-1] M1 Find speed vQ of Q when rebounding from plane: vQ = ½ uQ = 5 [m s-1] B1 Find magnitude of impulse: 0⋅04 (uQ + vQ) = 0⋅6 [N s] M1 A1 Find height risen by Q to collision in time t: hQ = vQ t – ½g t2 = 5t – 5t2 M1 EITHER: Find time for Q to fall to plane: uQ /g = 1 M1 Find height fallen by P in time (1 + t): hP = ½g(1 + t)2 = 5 + 10t + 5t2 M1 OR: State or imply P is at B when Q is at plane: [P’s speed at B is 10] (M1) Find height fallen by P in time t: h′P = 10t + ½gt2 = 10t + 5t2 (M1) Use hP + hQ = 10 or h′P + hQ = 5 to find t: 15t = 5, t = 1/3 M1 A1 [or relative motion can earn previous M1 M1 A1] Evaluate height above plane of collision: hQ = 10/9 or 1⋅11 [m] A1 4 6 [10]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 23 © UCLES 2010 Question Number Mark Scheme Details Part Mark Total 5 Consider speed components at Y normal to KN: v sin φ = ¾ u sin θ M1 Consider speed components at Y parallel to KN: v cos φ = u cos θ M1 Combine to eliminate u, v: tan φ = ¾ tan θ A.G. A1 Find XM in terms of a, θ: XM = 2a – a cot θ – a cot φ = 2a – (7/3) a cot θ A.G. M1 A1 Find alternative form of XM in terms of a, θ: XM = a cot MXN = a (4/3)2 cot θ M1 A1 Equate two forms of XM: (37/9) cot θ = 2 [tan θ = 37/18] M1 Evaluate θ: θ = 1⋅12 rad or 64⋅1° A1 EITHER: Combine speed components at N: w = u √(cos2 θ + (9/16)2 sin2 θ) M1 A1 OR: Consider speed components along KN: w cos MXN = u cos θ and MXN = cot-1 ((16/9) cot θ) (M1) √ on θ = 0⋅858 rad or 49⋅1° (A1√) Evaluate speed w at N: w = 0⋅669 u (allow 0⋅668 u) A1 3 6 3 [12] 6 Find confidence interval (allow z in place of t): 110⋅4 ± t √(50⋅42/25) M1 *A1 (using 24 in place of 25 loses A1) Use of correct tabular value: t24, 0.95 = 1⋅71[1] *B1 Evaluate C.I. correct to 3 sf (dep *A1, *B1): 110⋅4 ± 2⋅4 or [108⋅0, 112⋅8] A1 4 [4] 7 State (at least) null hypothesis (A.E.F.): H0: Data conforms to Law B1 Calculate expected values (to 1 dp): 30⋅10 17⋅61 12⋅49 9⋅69 7⋅92 22⋅18 M1 A1 Calculate value of χ2 : χ2 = 4⋅23 ± 0⋅02 M1 A1 Compare with consistent tabular value (to 2 dp): χ5, 0.9 2 = 9⋅236 B1 Valid method for reaching conclusion: Reject H0 if χ2 > tabular value M1 Correct conclusion (A.E.F., requires correct values): Data conforms to Benford’s Law A1 8 [8] 8 (a) (i) (ii) (b) Find prob. of 5 operating after 3 months: {exp (-3/2⋅5)}5 = {exp (-3 × 0⋅4)}5 = 0⋅30115 = 0⋅00248 (to 2 sf) M1 A1 Find prob. of 1 failing within one month: p1 = 1 – exp (-1/2⋅5) [= 0⋅3297] B1 Find prob. of 2 failing within one month: 5C2 p1 2 (1 – p1)3 = 0⋅327 M1 A1 Using 2⋅5 as parameter, not mean, can earn M1s only in (a) Find prob. of n operating after c months: {exp (-cλ)}n [λ = 0⋅4] M1 A1 Show same as 1 operating after nc months: = exp (-ncλ) M1 A1 2 3 4 [9]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 23 © UCLES 2010 Question Number Mark Scheme Details Part Mark Total 9 (a) (i) (ii) (iii) (b) (i) (ii) State valid comment on each diagram (A.E.F.): Str. line with negative gradient B1 (Almost) no linear relation B1 Close to str. line with +ve gradient B1 Calculate correlation coefficient: r = (13040 – 700 × 275/7) / √{(149000 – 7002/7)(17351 – 2752/7)} M1 A1 = – 14460 /√(79000 × 6547⋅4285 = – 0⋅636 A1 State hypotheses: H0: ρ = 0, H1: ρ < 0 B1 Compare with consistent tabular value: r7, 5% = 0⋅669 B1 Valid method for reaching conclusion: Reject H0 if r < – tabular value M1 Correct conclusion (needs correct values): No negative correlation (A.E.F.) A1 3 3 4 [10] 10 State valid assumption (A.E.F.): Popln. of diffs. has Normal distn. B1 State hypotheses: H0: µ2 – µ1 = 0, H1: µ2 – µ1 > 0 B1 Consider differences eg: 4⋅0 5⋅1 2⋅3 3⋅1 0 1⋅1 0⋅9 1⋅2 M1 Calculate sample mean: x = 17⋅7/8 [= 2⋅2125] M1 Estimate population variance: s2 = (60⋅37 – 17⋅72/8) / 7 (allow biased: 2⋅6511 or 1⋅6282) [= 3⋅0298 or 1⋅7412 ] M1 Calculate value of t (to 2 dp): t = x /(s/√8) = 3⋅60 [or 3⋅59] M1 *A1 Compare with correct tabular t value: t7, 0.99 = 2⋅998 (to 2 dp) *B1 Correct conclusion (AEF, dep *A1, *B1): There is an increase B1 Formulate inequality, with any tabular t value: ( x – w)/(s/√8) > t7, 0.9 M1 Use correct tabular value (to 2 dp): t7, 0.9 = 1⋅415 A1 Evaluate inequality for w: w < 1⋅34 [or ≤] A1 S.R. Allow M1 A1 A0 if = or < used in inequality 9 3 [12] 11 EITHER EITHER: Find 2 indep. eqns for RB, FB only: M1 Moments for BA about A: FB 2a sin β – RB 2a cos β = Wa sin β M1 A1 Moments for system about C: FB 6a sin β + RB 2a cos β = 9Wa sin β M1 A1 Add eqns to find FB : FB = 5W/4 A.G. M1 A1 OR: If RC, FC introduced, resolve vertically: FB + FC = 3W (B1) Any 2 moment eqns indep. of above resln. e.g.: For BA about A: FB 2a sin β – RB 2a cos β = Wa sin β For CA about A: FC 4a sin β – RC 4a cos β = 4Wa sin β For system about B: FC 6a sin β – RC 2a cos β = 9Wa sin β (2 system eqns are equiv. to vert. resolution) (2 × M1 A1) Solve eqns using RB = RC to find FB : FB = 5W/4 A.G. (M1 A1) Find FC by eg vertical resolution for rods: FC = 7W/4 B1 Find RB [or RC] from a moment eqn: RB [= RC ] = ¾ W tan β B1 Find one of FB /RB , FC /RC eg: FB /RB = 5 / (3 tan β) M1 A1 Find other eg: FC /RC = 7 / (3 tan β) A1 Find set of possible values of µ: µ > 7 / (3 tan β) (allow [) M1 A1 7 7 [14]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 23 © UCLES 2010 Question Number Mark Scheme Details Part Mark Total 11 OR For time to pack 50 items, find E(T50) and Var(T50): E(T50) = 50 × 1⋅5 [= 75] and Var(T50) = 50 × 0⋅42 [= 8] B1 State valid justification (A.E.F.) e.g.: By central limit theorem or Since n [or 50] is large B1 Find Normal approximation to P(T50 < 70): Φ((70 – 75)/√8) M1 = 1 – Φ(1⋅768) = 0⋅0385 ± 0⋅0001 A1 For time Tn to pack n items, find E(Tn) and Var(Tn): E(Tn) = 1⋅5n and Var(Tn) = 0⋅42 n B1 Use Normal approximation for P(Tn < 70) > 0⋅9: Φ((70 – 1⋅5n) / (0⋅4√n)) [ 0⋅9 M1 Invert function: (70 – 1⋅5n)/ 0⋅4√n [ 1⋅282 A1 Rearrange as quadratic expression in √n: (√n)2 + 0⋅3419√n – 140/3 [Y 0] M1 Find positive root when expression is zero: 44⋅4 A1 Find greatest integer value of n: nmax = 44 A1 For difference in times, find E(A–H ) and Var(A–H): E(A–H) = 75 – 65 = 10 and Var(A–H) = 8 + 12⋅5 = 20⋅5 M1 A1 Find Normal approximation to P(A–H > 0): Φ((10)/√20⋅5) = Φ(2⋅209) = 0⋅986 M1 A1 4 6 4 [14]