Cambridge A Level Mathematics - Further 9231 — 2016 May/June Paper 2 · Variant 3

*0129980160* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 May/June 2016 3 hours Additional Materials: List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST An answer booklet is provided inside this question paper. You should follow the instructions on the front cover of the answer booklet. If you need additional answer paper ask the invigilator for a continuation booklet. 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, 3 blank pages and 1 insert. JC16 06_9231_23/2R © UCLES 2016 [Turn over

2 1 A particle P is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle moves in complete vertical circles with centre O. The tension in the string when P is at its lowest point is twice the tension in the string when P is at its highest point. Find, in terms of a and g, the greatest speed of P during the motion. [6] 2 2W 1 A B P A uniform rod AB of length 2a and weight W rests with its lower end A in contact with a rough horizontal surface. The rod rests on a smooth peg at the point P. A particle of weight 2W is suspended from the end B of the rod. The system is in limiting equilibrium with the angle between the rod and the horizontal surface equal to 1, where tan 1 = 4 3 (see diagram). The coefficient of friction between the rod and the surface is 2 3. Find the distance AP. [8] 3 A particle P of mass 0.2 kg is moving in a circle of radius 0.5 m. At time t s its velocity is t2 + ct + d m s−1, where c and d are constants. The magnitude of the resultant force acting on P when t = 3 is 2 5 17 N. The transverse component of the acceleration of P when t = 3 is 2 m s−2. Find the values of c and d. [8] 4 An object consists of a uniform rod AB, of mass 4m and length 6a, and a uniform disc, of mass m and radius 2a and with centre O. The end B of the rod is rigidly attached to a point on the circumference of the disc, so that ABO is a straight line. The object is free to rotate in a vertical plane about a fixed smooth horizontal axis through A, perpendicular to the plane of the disc. Show that the moment of inertia of the object about this axis is 114ma2. [4] The object is held with AB at an angle of 60Å with the downward vertical at A, and released from rest. Find, in terms of a and g, the speed of the point B when AB is vertical. [5] 5 Three small smooth spheres A, B and C, of masses 5m, m and km respectively, are at rest on a smooth horizontal surface. The spheres lie in a straight line with B between A and C. Sphere A is projected directly towards B with speed u. The coefficient of restitution between any pair of the spheres A, B and C is 4 5. Show that the speed of A after the first collision is 7 10u and find the speed of B. [3] Given that the speeds of A and C are equal after the second collision, (i) show that the value of k is 20 7 , [4] (ii) find the percentage loss in the kinetic energy of the system as a result of the two collisions. [5] © UCLES 2016 9231/23/M/J/16

3 6 Pia is practising for a high jump competition. In order to qualify for the competition, she needs to succeed in jumping a certain height. At each attempt the probability that she succeeds is 1 4. The number of attempts that Pia makes, up to and including her first success, is denoted by the random variable N. State the mean and variance of N. [2] Find the probability that Pia (i) succeeds in jumping the qualifying height in at most 4 attempts, [2] (ii) takes more than 6 attempts to succeed in jumping the qualifying height. [1] 7 The lifetime, in months, of a particular type of light bulb is a random variable T and its probability density function, f, is given by ft = T ,e−,t t ≥0, 0 otherwise, where , is a positive constant. The manager of a shopping mall uses this type of light bulb and he notes that, out of 1000 new light bulbs, 280 were still working after 12 months. (i) Find an estimate for the value of ,. [3] (ii) Hence estimate the mean of T. [1] An improved version of the light bulb is produced. The lifetime of the improved version also has a negative exponential distribution, with a mean 25% larger than the original mean. (iii) What percentage of the improved version of the light bulb fail within the first 12 months? [4] 8 The coach of a national athletics team carried out an investigation into the effect of high altitude training on the times of 400-metre runners. The times, in seconds, were recorded before and after a six-week period of high altitude training, for a random sample of 8 athletes. The results are given in the following table. Athlete A B C D E F G H Before 52.3 56.2 54.3 49.2 48.4 50.1 46.8 51.1 After 51.6 53.2 54.9 49.0 48.1 49.2 46.8 49.6 Stating any assumption that you make, test, at the 2.5% significance level, whether there is an improvement in athletes’ times after high altitude training. [8] 9 A company produces four different flavours of ice cream: mint, strawberry, chocolate and fudge. Customers were asked which of the four flavours they preferred. The responses from a random sample of 80 male customers and 70 female customers are given in the following table. Mint Strawberry Chocolate Fudge Male 25 12 30 13 Female 16 22 25 7 Test, at the 10% significance level, whether there is a difference between ice cream flavour preferences of male and female customers. [9] © UCLES 2016 9231/23/M/J/16 [Turn over

4 10 A Physics teacher takes a random sample of size 5 from his students. Their marks in a written test (x) and a practical test (y) are given in the following table. Student A B C D E x 11 14 16 12 15 y 9 12 14 13 15 (i) Find the product moment correlation coefficient. [3] (ii) Test, at the 5% significance level, whether there is evidence of positive correlation. [4] The teacher takes a second random sample, this time of size 7, and the marks for this sample are summarised as follows. Σx = 96 Σx2 = 1366 Σy = 86 Σy2 = 1096 Σxy = 1220 (iii) Use the combined sample of size 12 to find the equation of the regression line of y on x, giving your answer in the form y = px + q, where p and q are constants. [4] One particular student scored 13 in the written test, but was absent for the practical test. (iv) Estimate this student’s mark in the practical test and comment on the reliability of your estimate. [2] © UCLES 2016 9231/23/M/J/16

5 11 Answer only one of the following two alternatives. EITHER The fixed points A and B are such that AB = 3.2 m and A is vertically above B. One end of a light elastic string, of natural length 0.8 m and modulus of elasticity 8 N, is attached to a particle P of mass 0.2 kg. The other end of the string is attached to A. One end of a second light elastic string, of natural length 1.2 m and modulus of elasticity 4 N, is attached to P and the other end is attached to B. (i) Find the length of AP when the particle is in equilibrium. [4] The particle is now pulled vertically down from its equilibrium position and released from rest at the point where AP = 1.65 m. (ii) Show that P performs simple harmonic motion and find the period of the motion. [6] (iii) Find the speed of P when it is at the mid-point of AB. Give your answer correct to 3 significant figures. [4] OR The annual salaries of workers at two factories, A and B, are to be compared. The salaries, in tens of thousands of dollars, at A and B, are denoted by x and y respectively. For a random sample of 40 workers in factory A and a random sample of 50 workers in factory B the results are as follows. Σx = 256.0 Σx2 = 1910.8 Σy = 382.9 Σy2 = 3148.8 The population mean salaries for A and B are denoted by -A and -B respectively. The population variances for salaries at A and B cannot be assumed to be equal. Test, at the 1% significance level, whether -B is greater than -A. [10] The width of an !% confidence interval for -B −-A is found to be 1.82. Find the value of !. [4] © UCLES 2016 9231/23/M/J/16

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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. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. 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 2016 9231/23/M/J/16

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 8 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 May/June 2016 MARK SCHEME Maximum Mark: 100 Published 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 May/June 2016 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 23 © Cambridge International Examinations 2016 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 Cambridge International A Level – May/June 2016 9231 23 © Cambridge International Examinations 2016 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 Cambridge International A Level – May/June 2016 9231 23 © Cambridge International Examinations 2016 Question Number Mark Scheme Details Part Mark Total 1 Relate u2, v2 at lowest and highest points by energy: ½ mv2 = ½ mu2 – 2mga [ v2 = u2 – 4ag ] Find tension T1 at lowest point by using F = ma: T1 = mu2/a + mg Find tension T2 at highest point by using F = ma: T2 = mv2/a – mg EITHER: Relate u2, v2 using T1 = 2T2 : 2v2 = u2 + 3ag Eliminate v2 to find u: u = √(11ag) OR: Relate T2, u2 by substituting for v2 : T2 = mu2/a – 5mg Use T1 = 2T2 to find u:u = √(11ag) [v = √(7ag)] B1 B1 B1 M1 M1 A1 (M1) (M1 A1) [6] 2 Take moments for rod about e.g. A: RP × AP = W × a cos θ + 2W × 2a cos θ [ = 5Wa cos θ = 3Wa ] or P: RA × AP cos θ – FA × AP sin θ = W × (AP – a) cos θ – 2W × (2a – AP) cos θ or B: RA × 2a cos θ – FA × 2a sin θ + RP × (2a – AP)= W × a cos θ or midpoint of AB: RP × (AP – a) – RA × a cos θ + FA × a sin θ = 2W × a cos θ or below B: RA × 2a cos θ + RP cos θ × (2a–AP) cos θ – RP sin θ × AP sin θ = W × a cos θ Formulate 2 more independent eqns. for forces, e.g.: Resolve forces on rod horizontally: FA = RP sin θ[= 4RP /5] Resolve forces on rod vertically: 3W – RA= RP cos θ [= 3RP /5] (or further independent moment eqn.) Eliminate θ from 3 independent eqns. for forces [sin θ = 4/5, cos θ = 3/5 ] Relate FA and RA : FA = ⅔ RA Find AP from eqns. in RP, RA and FA: [RP = 5W/3,FA = 4W/3, RA = 2W] AP = 9a/5 or 1⋅8a M1 A1 B1 B1 M1 M1 M1 A1 [8] 3 Find c by equating transverse acceln. to 2 at t = 3: 2 × 3 + c = 2,c = – 4 EITHER: Find radial force at t = 3: FR = 0⋅2 (32 + 3c + d)2 / 0⋅5 [= 0⋅4 (d – 3)2] Find transverse force at t = 3 by F = ma: Equate FT 2 + FR 2 to {(2√17)/5}2: 0⋅42 + 0⋅42 (32 + 3c + d)4 = 68/25 [FR 2 = 2⋅56, FR = 1⋅6] OR: Find radial acceln. at t = 3: aR =(32 + 3c + d)2 /0⋅5 [= 2 (d – 3)2] Valid use below of transverse acceln. at t = 3: aT = 2 Equate aT 2 + aR 2 to {(2√17)/(5 × 0⋅2)}2: 22 + 22 (32 [aR 2 = 64, aR = 8] Find d by solving with c = – 4, e.g.: (32 + 3c + d)4 = 17 – 1 = 16 (both values reqd. for A1) (d – 3)2 = 4,d = 1 or 5 (Omitting transverse comp. may earn M1 A1 M1 A1 B0 M0, max 4/8) M1 A1 M1 A1 B1 M1 A1 (M1 A1) (B1) (M1 A1) A1 [8]

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 23 © Cambridge International Examinations 2016 Question Number Mark Scheme Details Part Mark Total 4 Find or state MI of rod AB about axis at A: IAB= ⅓ 4m (3a)2 + 4m (3a)2 or (4/3) 4m (3a)2 [= 48 ma2] Find MI of disc about axis at A: Idisc = ½ m (2a)2 + m (8a)2 (M1 for either) [= 66 ma2] Find MI of object about axis at A: I = (48 + 66) ma2 = 114 ma2A.G. M1 A1 A1 A1 [4] Find angular speed ω when AB vertical by energy: ½ I ω2 = (4 × 3 + 8) mga (1 – cos 60º ) or (5mg × 4a) (1 – cos 60º ) (AEF) ω2 = 10mga /57 ma2 = 10g/57a Find speed of B from rω (AEF) :6a √(10g/57a) = √(120ag/19) or 2⋅51√(ag) [allow 7⋅95√a(g = 10) or 7⋅87√a(g = 9⋅8 or 9⋅81)] M1 A1 A1 M1 A1 [5] 5 For A & B use conservation of momentum, e.g.: 5mvA + mvB = 5mu (m may be omitted here and below) Use Newton’s law of restitution (consistent signs): vB – vA = (4/5) u Combine to find/verify vA and find vB , e.g.: 6vA = (5 – 4/5) u, vA = 7u/10 A.G. and vB = 3u/2 M1 M1 A1 [3] (i) For B & C use conservation of momentum, e.g.: mvB′ + kmvC = mvB Use Newton’s law of restitution (consistent signs): vC – vB′ = (4/5) vB Combine to find k using vC = vA , e.g.: (k + 1) vC = (1 + 4/5) vB k + 1 = 27/7, k = 20/7A.G. M1 M1 M1 A1 [4] (ii) Find vB′ (may be found earlier): vB′ = – u/2 EITHER: Find final KE, Efinal, of 3 particles: ½ 5mvA 2 + ½ mvB′ 2 + ½ kmvC 2 = ½ mu2 (49/20 + 1/4 + 7/5) = (41/20) mu2 or 2⋅05 mu2 OR: Find 1st loss/gain in KE (M1 for either): L1 = ½ 5mu 2 – ½ 5mvA 2 – ½ mvB 2 Find 2nd loss/gain in KE (A1 for both): L2 = ½ mvB 2 – ½ mvB′ 2 – ½ kmvC 2 [L1 = (3/20) mu2 , L2 = (3/10) mu2 ] Find loss in KE (ignore sign for M1) ½ 5mu2 – Efinalor L1 + L2 [= (9/20) mu2] Hence find percentage loss in KE: 100 (9/20) mu2 / (½ 5mu2)= 18% B1 M1 A1 (M1) (A1) M1 A1 [5] 6 State mean and variance of N: E(N) = 1/¼ = 4 Var(N) = ¾ / (¼)2 = 12 B1 B1 [2] (i) Find prob. of succeeding in at most 4 attempts: P(N ≤ 4) = 1 – (¾)4 or ¼ {1 + ¾ + (¾)2 + (¾)3} = 175/256or 0⋅684 M1 A1 [2] (ii) Find prob. of succeeding in more than 6 attempts: P(N > 6) = (¾)6 = 729/4096 or 0⋅178 B1 [1]

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 23 © Cambridge International Examinations 2016 Question Number Mark Scheme Details Part Mark Total 7 (i) Find F(t) by integration: F(t) = 1 – e–λt Estimate λ by equating 1 – F(12) to 280/1000: e–12λ= 0⋅28 λ = (– ln 0⋅28) / 12 = 0⋅106 B1 *M1 A1 [3] (ii) Estimate mean µ of T: µ = 1/λ = 9⋅43 B1 [1] (iii) Relate new exponential parameter λ′ to λ: λ′ =1/1⋅25µ = λ/1⋅25 or 0⋅8λ (B1 dependent on *M1) [= 0⋅08486 or 11⋅78–1] Find P(t ≤ 12): P(t ≤ 12) = 1 – e–12 λ′ = 1 – e–1⋅018 or1 – 0⋅280⋅8 Find percentage: = 63⋅9% B1 M1 A1 A1 [4] 8 State assumption, e.g.: Normal distribution(s) State hypotheses, e.g.: H0: µBefore – µAfter = 0, (AEF; B0 forx … or if ambiguous) H1: µBefore – µAfter > 0 Consider differences e.g. Before – After: 0⋅7 3⋅0 –0⋅6 0⋅2 0⋅3 0⋅9 0 1⋅5 Calculate sample mean d = 6 / 8 = 0⋅75 and and estimate population variance: s2 = (13⋅04 – 62/8) / 7 (allow biased here:1⋅0675 or 1⋅0332) = 61/50 or 1⋅22 or 1⋅1052 Calculate value of t: t = d/(s/√8) = 1⋅92 State or use correct tabular t-value: t7, 0.975 = 2⋅36[5] (or compared with t7, 0.975 s/√ 8 = 0⋅924) Consistent conclusion(AEF, √ on both t-values): [Accept H0:] No improvement in athletes’ times Wrong (hypothesis) test can earn only:B1 for assumption, B1 for hypotheses B1 B1 M1 M1 M1 A1 B1 B1 [8] 9 State (at least) null hypothesis (AEF): H0: No difference in preferences Find expected values (to 1 d.p.): 21⋅86718⋅13329⋅33310⋅667 (lose A1 if one error or rounded to integers) 19⋅13315⋅86725⋅6679⋅333 Calculate value of χ2 (to 3 s.f.): χ2 = 0⋅449 + 2⋅075 + 0⋅015 + 0⋅510 + 0⋅513 + 2⋅371 + 0⋅017 + 0⋅583 = 6⋅53 (allow 6⋅52) State or use correct tabular χ2 value (to 3 s.f.): χ3, 0.9 2 =6⋅25[1] State or imply valid method for conclusion e.g.: Reject H0 if χ2 ⩾ tabular value Correct conclusion, from correct values: (AEF) Difference in preferences B1 M1 A2 M1 A1 B1 M1 A1 [9]

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 23 © Cambridge International Examinations 2016 Question Number Mark Scheme Details Part Mark Total 10 (i) Find correlation coefficient r : r = Sxy / √(Sxx Syy) with e.g. Sxy = 872 – 68 × 63/5 = 15⋅2 (or 3⋅04) Sxx = 942 – 682/5= 17⋅2 (or 3⋅44) Syy = 815 – 632/5= 21⋅2 (or 4⋅24) (working may be implied) r = 0⋅796 M1 A1 *A1 [3] (ii) State both hypotheses (B0 for r …): H0: ρ = 0, H1: ρ > 0 State or use correct tabular one-tail r-value: r5, 5% = 0⋅805 State or imply valid method for conclusion e.g.: Accept H0 if r < tab. value (AEF) Correct conclusion(AEF, dep *A1, *B1): No positive correlation B1 *B1 M1 A1 [4] (iii) Calculate reqd. values for combined sample: Sxy = 2092 – 164 × 149/12 = 55⋅67 Sxx = 2308 – 1642/12= 66⋅67 [Syy = 1911 – 1492/12= 60⋅92 ] Calculate gradient p in y –y = p (x –x) : p =Sxy / Sxx= 0⋅835 and find q: q = 149/12– p × 164/12 = 12⋅42 – 0⋅835 × 13⋅67 =1⋅00[5] M1 A1 M1 A1 [4] (iv) Estimate mark from regression line when x = 13: y = 11⋅9so mark = 12 Consistent valid comment on reliability, e.g.: Reliable since 13 (or 12) in range (√ on r′ ) or Reliable since r′ = 0⋅874 ≈ 1 or Reliable since r′ > r12, 5% = 0⋅497 B1 B1 [2] 11 (a) (i) Find AP0 by equating equilibrium tensions and wt.: 8 (AP0 – 0⋅8)/0⋅8 (Lose A1 for one incorrect term) = 4 (2 – AP0)/1⋅2 + 0⋅2g or if e = AP0 – 0⋅8: 8e/0⋅8 = 4 (1⋅2 – e)/1⋅2 + 0⋅2g [e = 9/20 or 0⋅45] AP0 = 5/4 or 1⋅25 [m] M1 A2 A1 [4] (ii) Apply Newton’s law at general point, e.g.: ± 0⋅2 d2x/dt2 = – 8 (AP0 – 0⋅8 + x)/0⋅8 (lose A1 for each incorrect term) + 4 (2 – AP0 – x)/1⋅2 + 0⋅2g or ± 0⋅2 d2y/dt2 = 8 (AP0 – 0⋅8 – y)/0⋅8 – 4 (2 – AP0 + y)/1⋅2 + 0⋅2g Simplify to give standard SHM eqn, e.g.: d2x/dt2 = – (10x +10x/3) / 0⋅2 SC: B1 if no derivation (max 3/6) =– 200x / 3 State or find period using 2π/ω with ω = √(200/3): T = 2π√(3/200) = π√(3/50) or 0⋅245 π or 0⋅770 [s] M1 A2 A1 M1 A1 [6] (iii) Find or state displacement x at midpoint: x = 1⋅6 – AP0[= 0⋅35] Find or state amplitude a: a = 1⋅65 – AP0 [= 0⋅4] Find speed v from v2 = ω2 (a2 – x2): v2 = (200/3) (0⋅42 – 0⋅352) v = √2⋅5,v = 1⋅58 [m s–1] B1 B1 M1 A1 [4]

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 23 © Cambridge International Examinations 2016 Question Number Mark Scheme Details Part Mark Total 11 (b) State hypotheses (B0 forxA …), e.g.: H0: µB = µA ,H1: µB > µA Estimate both popn. variances: sA 2 = (1910⋅8 – 2562/40) / 39 and sB 2 = (3148⋅8 – 382⋅92/50) / 49 (allow biased here: 6⋅81 or 2⋅6102 sA 2 = 6⋅985 or 454/65 or 2⋅6432 and 4⋅331 or 2⋅0812) and sB 2 = 4⋅419 or 2⋅1022 Estimate combined variance (may be implied): s2 = sA 2/40 + sB 2/50 (to 3 s.f) =0⋅263 or 0⋅5132 Calculate value of z (or –z): z = (xB – xA) / s = (7⋅658 – 6⋅4) / s (to 3 s.f.) = 1⋅258/0⋅5128 = 2⋅45 State or use correct tabular z value:(to 3 s.f.) z 0.99 = 2⋅326(SC: allow 2⋅358) (or can comparexB –xA = 1⋅258 with 1⋅19) Correct conclusion (AEF, √ on z, dep *B1) :z > tabular value so µB is greater than µA Note: Basing test on equal variances can earn first 4 marks only B1 M1 A1 A1 M1 A1 M1 A1 *B1 B1 [10] Calculate value of z using width of interval: z = 1⋅82 / 2s = 0⋅91 / 0⋅5128 (allow M1 for z = 1⋅82 / s) = 1⋅77[46] Find Φ(z) and value of α : Φ(z) = 0⋅962 = 1 – ½ (1 – α/100) (overlooking % is misread; α = 100 {1 – 2 (1 – 0⋅962)} M0 for α = 96⋅2 or finding 100 – α) = 92⋅4 M1 A1 M1 A1 [4]