Cambridge A Level Mathematics - Further 9231 — 2011 Oct/Nov Paper 2 · Variant 3

*0026880427* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 October/November 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 11_9231_23/2R © UCLES 2011 [Turn over

2 1 A particle is moving in a circle of radius 2 m. At time t s its velocity is (t2 −12) m s−1. Find the magnitude of the resultant acceleration of the particle when t = 4. [4] 2 A particle P is moving in simple harmonic motion with centre O. When P is 5 m from O its speed is V m s−1, and when it is 9 m from O its speed is 3 5V m s−1. Show that the amplitude of the motion is 15 2 √2 m. [4] Given that the greatest speed of P is 3√2 m s−1, find V. [3] 3 A fixed hollow sphere with centre O has a smooth inner surface of radius a. A particle P of mass m is projected horizontally with speed 2 √(ag) from the lowest point of the inner surface of the sphere. The particle loses contact with the inner surface of the sphere when OP makes an angle θ with the upward vertical. (i) Show that cos θ = 2 3. [5] (ii) Find the greatest height that P reaches above the level of O. [4] 4 Two smooth spheres P and Q, of equal radius, have masses m and 3m respectively. They are moving in the same direction in the same straight line on a smooth horizontal table. Sphere P has speed u and collides directly with sphere Q which has speed ku, where 0 < k < 1. Sphere P is brought to rest by the collision. Show that the coefficient of restitution between P and Q is 3k + 1 3(1 −k). [6] One third of the total kinetic energy of the spheres is lost in the collision. Show that k = 1 3(2√3 −3). [5] 5 O C P 2a a A uniform solid sphere with centre C, radius 2a and mass 3M, is pivoted about a smooth horizontal axis and hangs at rest. The point O on the axis is vertically above C and OC = a. A particle P of mass M is attached to the sphere at its lowest point (see diagram). Show that the moment of inertia of the system about the axis through O is 84 5 Ma2. [4] The system is released from rest with OP making a small angle α with the downward vertical. Find (i) the period of small oscillations, [5] (ii) the time from release until OP makes an angle 1 2α with the downward vertical for the first time. [3] © UCLES 2011 9231/23/O/N/11

3 6 The continuous random variable X has probability density function f given by f(x) =   0 x < 1, 1 2 1 ≤x ≤3, 0 x > 3. Find the distribution function of X. [2] The random variable Y is defined by Y = X3. Find (i) the probability density function of Y, [3] (ii) the expected value and variance of Y. [3] 7 The lifetime, in hours, of a ‘Trulite’ light bulb is a random variable T. The probability density function f of T is given by f(t) =  0 t < 0, λe−λt t ≥0, where λ is a positive constant. Given that the mean lifetime of Trulite bulbs is 2000 hours, find the probability that a randomly chosen Trulite bulb has a lifetime of at least 1000 hours. [3] A particular light fitting has 6 randomly chosen Trulite bulbs. Find the probability that no more than one of these bulbs has a lifetime less than 1000 hours. [3] By using new technology, the proportion of Trulite bulbs with very short lifetimes is to be reduced. Find the least value of the new mean lifetime that will ensure that the probability that a randomly chosen Trulite bulb has a lifetime of no more than 4 hours is less than 0.001. [5] 8 A sample of 216 observations of the continuous random variable X was obtained and the results are summarised in the following table. Interval 0 ≤x < 1 1 ≤x < 2 2 ≤x < 3 3 ≤x < 4 4 ≤x < 5 5 ≤x < 6 Observed frequency 1 3 15 31 59 107 It is suggested that these results are consistent with a distribution having probability density function f given by f(x) =  kx2 0 ≤x < 6, 0 otherwise, where k is a positive constant. The relevant expected frequencies are given in the following table. Interval 0 ≤x < 1 1 ≤x < 2 2 ≤x < 3 3 ≤x < 4 4 ≤x < 5 5 ≤x < 6 Expected frequency 1 7 a b c 91 (i) Show that a = 19 and find the values of b and c. [4] (ii) Carry out a goodness of fit test at the 10% significance level. [7] © UCLES 2011 9231/23/O/N/11 [Turn over

4 9 A random sample of five metal rods produced by a machine is taken. Each rod is tested for hardness. The results, in suitable units, are as follows. 524 526 520 523 530 Assuming a normal distribution, calculate a 95% confidence interval for the population mean. [5] Some adjustments are made to the machine. Assume that a normal distribution is still appropriate and that the population variance remains unchanged. A second random sample, this time of ten metal rods, is now taken. The results for hardness are as follows. 525 520 522 524 518 520 519 525 527 516 Stating suitable hypotheses, test at the 10% significance level whether there is any difference between the population means before and after the adjustments. [8] © UCLES 2011 9231/23/O/N/11

5 10 Answer only one of the following two alternatives. EITHER A B q C P 2a h A uniform rod AB, of weight W and length 2a, rests with the end A on a rough horizontal plane. A light inextensible string BC is attached to the rod at B and passes over a small smooth fixed peg P, which is at a distance h vertically above A. A particle is attached at C and hangs vertically. The points A, B and C are all in the same vertical plane. In equilibrium the rod is inclined at an angle θ to the horizontal (see diagram). The coefficient of friction between the rod and the plane is µ. Show that µ ≥ 2a cos θ h + 2a sin θ . [7] Given that the particle attached at C has weight kW, angle ABP = 90◦and h = 3a, find (i) the value of k, [4] (ii) the horizontal component of the force on P, in terms of W. [3] OR The regression line of y on x obtained from a random sample of five pairs of values of x and y is y = 2.5x −1.5. The data is given in the following table. x 1 2 4 2 6 y 2 3 6 p q (i) Show that p + q = 19. [3] (ii) Find the values of p and q. [6] (iii) Determine the value of the product moment correlation coefficient for this sample. [3] (iv) It is later discovered that the values of x given in the table have each been divided by 10 (that is, the actual values are 10, 20, 40, 20, 60). Without any further calculation, state (a) the equation of the actual regression line of y on x, (b) the value of the actual product moment correlation coefficient. [2] © UCLES 2011 9231/23/O/N/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. 9231/23/O/N/11

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2011 question paper for the guidance of teachers 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 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 October/November 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 – October/November 2011 9231 23 © 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 – October/November 2011 9231 23 © 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 – October/November 2011 9231 23 © University of Cambridge International Examinations 2011 Question Number Mark Scheme Details Part Mark Total 1 Find tangential acceleration: 2t = 8 B1 Find radial acceleration: (42 – 12) 2/2 = 8 B1 Combine to give magnitude of acceleration: √(82 + 82) = 8√2 or 11⋅3 [ms–2] M1 A1 4 [4] 2 Apply v2 = ω2 (A2 – x2) at first point: V2 = ω2 (A2 – 52) B1 Apply v2 = ω2 (A2 – x2) at second point: (9/25)V2 = ω2 (A2 – 92) B1 Combine to find amplitude A: 25(A2 – 81) = 9(A2 – 25) A.G. 16 A2 = 25 × 72, A = 15√2/2 M1 A1 4 Find ω using vmax = ωA: ω = (3√2)/(15√2/2) = 2/5 M1 Find V using one of earlier eqns: V2 = (4/25) (225/2 – 25) = 14 V = √14 or 3⋅74 M1 A1 3 [7] 3 (i) Use conservation of energy: ½mv2 = ½mu2 – mga (1 + cos θ) B1 [v2 = 2ag (1 – cos θ)] Equate radial forces [may imply R = 0]: mv2/a = mg cos θ + R M1 Take R = 0 when contact lost: mv2/a = mg cos θ [v2 = ag cos θ] A1 Eliminate v2 and replace u2 by 4ag: 4mg – 2mg (1 + cos θ) = mg cos θ M1 Solve for cos θ: cos θ = 2/3 A.G. A1 5 (ii) Find further height h2 risen: h2 = v2 sin2 θ/2g M1 Substitute for v and θ: = (2ag/3) (5/9)/2g = 5a/27 M1 A1 Find total height risen above centre O: a cos θ + h2 = 23a/27 B1 4 [9] 4 Use conservation of momentum: 3mvQ = mu + 3kmu M1 A1 Use Newton’s law of restitution: vQ = e(u – ku) M1 A1 Eliminate vQ to find e: e = (3k + 1)/3(1 – k) A.G. M1 A1 6 Relate K.E. after and before collision: ½ 3mvQ 2 = ⅔ ½m (u2 + 3k2u2) M1 A1 Replace vQ by ⅓(1 + 3k)u and rearrange: (1 + 3k)2 = 2(1 + 3k2) 3k2 + 6k – 1 = 0 M1 A1 Find root k with 0 < k < 1: A.G. k = (–6 + √48)/6 = ⅓(2√3 – 3) A1 (Simply substituting given k earns M1 A0 A1) 5 [11] 5 Find MI of sphere about diameter: IC = (2/5) 3M (2a)2 [= 24Ma2/5] M1 Find MI of sphere about axis through O: IC + 3Ma2 [= 39Ma2/5] M1 Find MI of particle about axis through O: M (3a)2 [= 45Ma2/5] B1 Sum to find MI of system about O: I = 84 Ma2/5 A.G. A1 4 (i) State eqn of motion (A.E.F.): I d2θ /dt2 = – 3Mg a sin θ – Mg 3a sin θ M1 A1 Put sin θ ≈ θ (implied by using SHM): I d2θ /dt2 = – 6Mga θ M1 [ d2θ /dt2 = – (5g/14a) θ ] Find approx. period T from SHM formula: T = 2π /√(6Mga/(84 Ma2/5)) (A.E.F.) = 2π √(14a/5g) or 10⋅5√(a/g) M1 A1 5 (ii) Use appropriate SHM formula: θ = α cos ωt M1 Find time t to θ = ½α: t = (1/ω) cos–1 ½ = (1/ω) (π/3) = (π/3) √(14a/5g) M1 A1 3

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 23 © University of Cambridge International Examinations 2011 Question Number Mark Scheme Details Part Mark Total 6 Integrate to find F(x) for 1 ≤ x ≤ 3: F(x) = ½ (x – 1) B1 State F(x) for other intervals of x: 0 (x < 1), 1 (x > 3) B1 2 (i) Relate dist. fn. G(y) of Y to X: G(y) = P(Y < y) = P(X 3 < y) (working may be omitted) = P(X < y1/3) = F(y1/3) = ½ (y1/3 – 1) M1 A1 Differentiate to find g(y): g(y) = y–2/3/6 (1 ≤ y ≤ 27) [= 0 otherwise] B1 3 (ii) Find expected value of Y (or X3): E(Y) = ∫1 27 y (y–2/3/6) dy = [y4/3/8] 1 27 or [x4/8] 1 3 = (81 – 1)/8 = 10 B1 Find variance of Y: E(Y 2) = ∫1 27 y2 (y–2/3/6) dy = [y7/3/14] 1 27 or [x7/14] 1 3 = (2187 – 1)/14 = 1093/7 Var (Y) = E(Y 2) – 102 = 393/7 or 56⋅1[4] M1 A1 3 [8] 7 State or find value of λ: λ = 1/2000 or 0⋅0005 B1 Find p = P(T ≥ 1000): 1 – ∫0 1000 λ e–λt dt = 1 + [e–λt]0 1000 = e–0⋅5 = 0⋅607 M1 A1 3 Find P(N = 1) where N of the 6 bulbs have T <1000: P(N = 1) = 6 p5 (1–p) [= 0⋅194] B1 Hence find P(N ≤ 1): P(N ≤ 1) = P(N = 1) + p6 = 0⋅244 M1 A1 3 Formulate inequality for new λ: 0⋅001 > ∫0 4 λ e–λt dt M1 = [–e–λt]0 4 = 1 – e–4λ A1 – 4λ > ln 0⋅999 A1 Find minimum mean from 1/λ: 1/ λ > –4/ln 0⋅999, min is 4000 M1 A1 5 [11] 8 (i) Find value of k by integrating f(x): [⅓kx3] 0 6 = 1, k = 3/63 = 1/72 B1 State and evaluate expression for a: A.G. a = 216 [⅓kx3]2 3 = 33 – 23 = 19 B1 Find b and c: b = 216 [⅓kx3]3 4 = 37, (MR: f(x) as distn. of table: max 3/4) c = 216 [⅓kx3]4 5 or 216–155 = 61 M1 A1 4 (ii) State (at least) null hypothesis: H0: f(x) fits data (A.E.F.) B1 Combine first 2 cells since exp. value < 5: O: 4 . . . E: 8 . . . B1 Calculate χ2 (to 2 dp ): χ2 = 6⋅69[4] M1 *A1 Compare consistent tabular value (to 2 dp): χ4, 0.9 2 = 7⋅779 [or if 3 or 0 cells combined: χ3, 0.9 2 = 6⋅251, χ5, 0.9 2 = 9⋅236] *B1 Valid method for reaching conclusion: Accept H0 if χ2 < tabular value M1 Conclusion (A.E.F., dep *A1, *B1): 6⋅69 < 7⋅78 so f(x) does fit A1 7 [11]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 23 © University of Cambridge International Examinations 2011 Question Number Mark Scheme Details Part Mark Total 9 Calculate sample mean: d = 2623/5 = 524⋅6 M1 Estimate population variance using 1st sample: s1 2 = (1376081 – 26232/5)/4 (allow biased here: 11⋅04 or 3⋅3232) [= 13⋅8 or 3⋅7152 ] M1 Find confidence interval (allow z in place of t) e.g.: 524⋅6 ± t √(13⋅8/5) M1 (inconsistent use of 4 or 5 loses M1) Use of correct tabular value: t4, 0.975 = 2⋅776 (2 d.p.) A1 Evaluate C.I. correct to 3 s.f. (needs correct s, t): 524⋅6 ± 4⋅6[1] or [520⋅0, 529⋅2] A1 5 State hypotheses: H0: µb = µa , H1: µb ≠ µa B1 Estimate population variance using 2nd sample: s2 2 = (2720780 – 52162/10)/9 (allow biased here: 11⋅44 or 3⋅3822) [= 572/45 or 12⋅711 or 3⋅5652] M1 Estimate population variance for combined sample: s2 = (4×13⋅8 + 9×12⋅71)/13 = 848/65 or 13⋅05 M1 A1 Calculate value of t (to 2 dp): t = (524⋅6 – 521⋅6)/( s√(5–1 + 10–1)) = 1⋅52 M1 *A1 Compare with correct tabular t value: t13, 0.95 = 1⋅77[1] *B1 Correct conclusion (AEF, dep *A1, *B1): No difference in means B1 8 [13] 10a Take moments about P for system [i.e. rod]: FA h = Wa cos θ M1 A1 Take moments about B for rod: FA 2a sin θ + Wa cos θ = RA 2a cos θ M1 A1 Eliminate FA to give RA : RA = ½W + (Wa sin θ)/h M1 Find inequality for µ: µ ≥ FA/RA µ ≥ 2a cos θ/(h + 2a sin θ) A.G. M1 A1 7 (i) Use kW = T to express in terms of FA or RA : kW = FA /sin θ or (W – RA)/cos θ M1 Substitute for FA or RA : k = (a/h) cot θ or (½ – (a/h) sin θ)/cos θ M1 A1 Substitute for h and θ : k = √5/6 or 0⋅373 A1 4 (ii) Find horizontal component NP : NP = T sin θ or kW sin θ M1 Substitute for k and θ : (√5/6)(2/3)W = √5W/9 or 0⋅248W M1 A1 3 [14]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 23 © University of Cambridge International Examinations 2011 Question Number Mark Scheme Details Part Mark Total 10b (i) Use regression line or 1st normal eqn, e.g.: Σy/5 = 2⋅5 Σx/5 – 1⋅5 B1 Use data to substitute for Σx and Σy: 11 + p + q = 2⋅5 × 15 – 5 × 1⋅5 p + q = 37⋅5 – 7⋅5 – 11 = 19 A.G. M1 A1 3 (ii) Use formula for b or 2nd normal eqn: 2⋅5 = (32+2p+6q – 15×30/5)/(61 – 152/5) or 32+2p+6q = 2⋅5×61 – 1⋅5×15 M2 A1 (A.E.F.) p + 3q = 49 (or 3q – p = 41) A1 Solve any two simultaneous eqns for p, q: p = 4, q = 15 M1 A1 6 (iii) Find correlation coefficient r: r = (32+2p+6q – 15×30/5)/√{(61 – 152/5) (49 + p2 + q2 – (11 + p + q)2/5)} M1 = (130 – 15 × 30/5)/√{(61 – 152/5) (290 – 302/5)} A1 or 2⋅5√{(61 – 152/5)/ (49 + p2 + q2 – (11 + p + q)2/5)} (M1) = 2⋅5 √{(61 – 152/5)/ (290 – 302/5)} (A1) = 40/√(16×110) or 2⋅5√(16/110) = 0⋅953 A1 3 (iv) (a) State eqn of actual regression line: y = 0⋅25x – 1⋅5 B1 (b) State new value of r or say unchanged: Same value as found in (iii) B1 2 [14]