Cambridge A Level Mathematics - Further 9231 — 2014 May/June Paper 2 · Variant 3

*5707588618* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 May/June 2014 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 an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. 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. JC14 06_9231_23/RP © UCLES 2014 [Turn over

2 1 Two small smooth spheres A and B have equal radii and have masses m and km respectively. They are moving in a straight line in the same direction on a smooth horizontal table. The speed of A is u and the speed of B is 2 3u. Sphere A collides directly with sphere B. The coefficient of restitution between the spheres is 4 5. (i) Show that the speed of A after the collision is u
2k + 5 5
k + 1 . [5] (ii) Given that the magnitude of the impulse experienced by B during the collision is 2 5mu, find the value of k. [2] 2 A particle P of mass m kg moves on an arc of a circle with centre O and radius a metres. At time t = 0 the particle is at the point A. At time t seconds, angle POA = sin2 2t. Show that the radial component of the acceleration of P at time t seconds has magnitude
4a sin2 4t m s−2. [2] Find (i) the value of t when the transverse component of the acceleration of P is first equal to zero, [2] (ii) the magnitude of the resultant force acting on P when t = 1 12
. [2] 3 A particle P of mass m 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. When the particle is hanging in equilibrium, with the string vertical, it is given a horizontal impulse of magnitude J. When the string makes an angle  with the downward vertical at O, the tension is T. (i) Show that T = J2 ma −mg
2 −3 cos . [5] (ii) It is given that J = m
kga, where k is a positive constant. Justifying your answers, describe the motion of P in each of the following cases: (a) k = 1, (b) k = 6. [4] © UCLES 2014 9231/23/M/J/14

3 4 B C 2a 2d A
A uniform rod AB has mass m and length 2d. The rod rests in equilibrium on a smooth peg C, with the end A resting on a rough horizontal plane. The distance AC is 2a and the angle between AB and the horizontal is , where cos  = 3 5. A particle of mass 1 2m is attached to the rod at B (see diagram). Find the normal reaction at A and deduce that d < 25 6 a. [6] The coefficient of friction between the rod and the plane is . Show that  ≥ 8d 25a −6d . [4] 5 O 5 2a A B C D 8a 6a A uniform rectangular lamina ABCD, in which AB = 8a and BC = 6a, has mass M. A uniform circular lamina of radius 5 2a has mass 1 3M. The two laminas are fixed together in the same plane with their centres coinciding at the point O (see diagram). A particle P of mass 1 2M is attached at C. The system is free to rotate about a fixed smooth horizontal axis through A and perpendicular to the plane ABCD. Show that the moment of inertia of the system about this axis is 2225 24 Ma2. [8] The system is released from rest with AC horizontal and D below AC. Find, in the form k g a  , the greatest angular speed in the subsequent motion, giving the value of k correct to 3 decimal places. [4] © UCLES 2014 9231/23/M/J/14 [Turn over

4 6 Employees at a particular company have been working seven hours each day, from 9 am to 4 pm. To try to reduce absence, the company decides to introduce ‘flexi-time’ and allow employees to work their seven hours each day at any time between 7 am and 9 pm. For a random sample of 10 employees, the numbers of hours of absence in the year before and the year after the introduction of flexi-time are given in the following table. Employee A B C D E F G H I J Before 42 35 96 74 20 5 78 45 146 0 After 34 32 100 72 31 2 61 35 140 0 Use a paired sample t-test to test, at the 10% significance level, whether the population mean number of hours of absence has decreased, following the introduction of flexi-time. [7] 7 James throws a discus repeatedly in an attempt to achieve a successful throw. A throw is counted as successful if the distance achieved is over 40 metres. For each throw, the probability that James is successful is 1 4, independently of all other throws. Find the probability that James takes (i) exactly 5 throws to achieve the first successful throw, [1] (ii) more than 8 throws to achieve the first successful throw. [2] In order to qualify for a competition, a discus-thrower must throw over 40 metres within at most six attempts. When a successful throw is achieved, no further throws are taken. Find the probability that James qualifies for the competition. [2] Colin is another discus-thrower. For each throw, the probability that he will achieve a throw over 40 metres is 1 3, independently of all other throws. Find the probability that exactly one of James and Colin qualifies for the competition. [3] 8 A random sample of 200 is taken from the adult population of a town and classified by age-group and preferred type of car. The results are given in the following table. Hatchback Estate Convertible Under 25 years 32 11 17 Between 25 and 50 years 45 24 6 Over 50 years 31 16 18 Test, at the 5% significance level, whether preferred type of car is independent of age-group. [8] 9 The continuous random variable X has distribution function F given by F
x =  0 x < 2, 1 8x −1 4 2 ≤x ≤10, 1 x > 10. Find the value of k for which P
X ≥k = 0.6. [3] The random variable Y is defined by Y = 2 ln X. Find the distribution function of Y. [3] Find the probability density function of Y and sketch its graph. [4] © UCLES 2014 9231/23/M/J/14

5 10 The lengths of a random sample of eight fish of a certain species are measured, in cm, as follows. 17.3 15.8 18.2 15.6 16.0 18.8 15.3 15.0 Assuming that lengths are normally distributed, (i) test, at the 10% significance level, whether the population mean length of fish of this species is greater than 15.8 cm, [7] (ii) calculate a 95% confidence interval for the population mean length of fish of this species. [4] 11 Answer only one of the following two alternatives. EITHER A particle P of mass m is suspended from a fixed point by a light elastic string of natural length l, and hangs in equilibrium. The particle is pulled vertically down to a position where the length of the string is 13 7 l. The particle is released from rest in this position and reaches its greatest height when the length of the string is 11 7 l. (i) Show that the modulus of elasticity of the string is 7 5mg. [2] (ii) Show that P moves in simple harmonic motion about the equilibrium position and state the period of the motion. [4] (iii) Find the time after release when the speed of P is first equal to half of its maximum value. [6] OR For a random sample of 12 observations of pairs of values
x, y, the equation of the regression line of y on x and the equation of the regression line of x on y are y = bx + 4.5 and x = ay + c respectively, where a, b and c are constants. The product moment correlation coefficient for the sample is 0.6. (i) Test, at the 5% significance level, whether there is evidence of positive correlation between the variables. [4] (ii) Given that b −a = 0.5, find the values of a and b. [3] (iii) Given that the sum of the x-values in the sample data is 66, find the value of c and sketch the two regression lines on the same diagram. [3] For each of the 12 pairs of values of
x, y in the sample, another variable  is considered, where  = 5y. (iv) State the coefficient of x in the equation of the regression line of  on x and find the value of the product moment correlation coefficient between x and , justifying your answer. [2] © UCLES 2014 9231/23/M/J/14

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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. 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 2014 9231/23/M/J/14

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2014 series 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 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 2014 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 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 – May/June 2014 9231 23 © Cambridge International Examinations 2014 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 – May/June 2014 9231 23 © Cambridge International Examinations 2014 Question Number Mark Scheme Details Part Mark Total 1 (i) Use conservation of momentum, e.g. mvA + kmvB R mu N ⅔kmu or vA N kvB R u (1 N ⅔k) B1 Use restitution (4 / 5 on wrong side is M0; vA – vB R – (4 / 5) (u – ⅔u) signs inconsistent with prev. eqn is A0): or vA – vB R – 4u / 15 M1 A1 Solve for vA (allow verification): (1 + k) vA = u(1 + ⅔k – 4k / 15) vA = u(2k + 5) / 5(k + 1) A.G. M1 A1 [vB = u(10k + 19) / 15(k + 1)] 5 (ii) Equate impulse to momentum change for A: mu – (2 / 5)mu = mvA 3 / 5 = (2k + 5) / 5(k + 1) , k = 2 M1 A1 OR B: ⅔kmu + (2 / 5)mu = kmvB ⅔k + (2 / 5) = k(10k + 19) / 15(k + 1) 10k2 + 16k + 6 = 10k2 + 19k, k = 2 (M1 A1) 2 7

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 2 Find radial acceleration from r (dθ / dt)2 [≡ rω2]: dθ / dt = (2 sin 2t) (2 cos 2t) = 2 sin 4t r (dθ / dt)2 = 4a sin2 4t A.G. M1 A1 2 (i) Find t by equating d2θ / dt2 to 0: d2θ / dt2 = 8 cos 4t = 0 t = π / 8 or 0⋅393 M1 A1 2 (ii) Find radial and transverse components of force: 4ma sin2 4π / 12 = 3ma and 8ma cos 4π / 12 = 4ma Combine to find magnitude of resultant force: √(32 + 42) ma = 5ma M1 A1 2 6

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 3 (i) Use conservation of energy: ½mv2 = ½mu2 – mg(a – a cos θ ) B1 Use F = ma radially: T – mg cos θ = mv2 / a B1 Relate u to impulse J: mu = J B1 Eliminate u and v to find T: T = J2 / ma – mg(2 – 3 cos θ) A.G. M1 A1 5 (ii)(a) Investigate v [and T] for k = 1 and describe motion: v2 = ga (2 cos θ – 1) [T = mg (3 cos θ – 1)] v = 0 [and T > 0] when cos θ = ½ (S.R. Award B1 for correct result based only on T) so P oscillates (A.E.F.) M1 A1 2 (b) Investigate v and T for k = 6 and describe motion: T = mg (3 cos θ + 4) [v2 = ga (2 cos θ + 4)] T > 0 for e.g. θ = π and v > 0 (S.R. Award B1 for correct result based only on T) so P does full circle(A.E.F.) M1 A1 2 9

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 4 EITHER: Resolve vertically: RA + RC cos α = 3mg / 2 B1 Take moments about A: RC 2a = mgd cos α + ½mg2d cos α M1 A1 Eliminate RC to find RA: RC = (mgd / a) cos α = 3mgd / 5a RA = 3mg / 2 – 9mgd / 25a or 3mg(25a – 6d) / 50a (A.E.F.) M1 A1 OR: Resolve along AB: RA sin α + FA cos α = mg sin α + ½mg sin α (B1) Take moments about C: (RA cos α) 2a – (FA sin α)2a= (mg cos α)(2a – d) – (½mg cos α)(2d – 2a) (M1 A1) Eliminate FA to find RA: RA = 3mg(25a – 6d) / 50a (A.E.F.) (M1 A1) Find limit on d from RA > 0: 25a – 6d > 0, d < 25a/6 A.G. B1 6 Find FA by e.g. horizontal resolution: FA = RC sin α = (3mgd / 5a)(4 / 5) = 12mgd / 25a M1 A1 Find inequality for µ from FA Y µRA (= loses A1): µ [ 8d / (25a – 6d) A.G. M1 A1 4 10

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 5 Find MI of rectangular lamina about O: I,O = ⅓M {(4a)2 + (3a)2} B1 [= 25 Ma2 / 3] Find MI of circular lamina about O: IΟ,O = ½ ⅓M (5a / 2)2 B1 [= 25 Ma2 / 24] EITHER: Find MI of combined laminas about O: IO = I,A + IΟ,A = 225 Ma2 / 24 M1 A1 Find MI of combined laminas about A: IA = IO + (4M / 3) 25a2 = 1025 Ma2 / 24 M1 A1 OR: Find MI of rectangle about A I,A = I,O + 25 Ma2 = 100 Ma2 / 3 and of circle about A: and IΟ,A = IΟ,O + ⅓M 25a2 =75 Ma2 / 8 (M1 A1) Find MI of combined IA = I,A + IΟ,A = 1025 Ma2 / 24 laminas about A: (M1 A1) Find MI of system about A: A.G. I = IA + 50 Ma2 = 2225 Ma2 / 24 M1 A1 8 State or imply that speed is max when AC vertical M1 Use energy when AC vertical (or at general point): ½ Iω2 = 4Mg / 3 × 5a + ½Mg × 10a or 11Mg / 6 × 70a / 11 [=35Mga / 3] M1 A1 Substitute for I and find k in max ang. speed ω: ω2 = (112 / 445)g / a, k = 0⋅502 A1 4 12

Page 9 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 6 Consider differences e.g. before – after: 8 3 –4 2 –11 3 17 10 6 0 M1 Calculate sample mean d = 34 / 10 = 3⋅4 and and estimate population variance: s2 = (648 – 342 / 10) / 9 (allow biased here: 53⋅24 or 7⋅2972) = 2662 / 45 or 59⋅16 or 7⋅6912 M1 State hypotheses (A.E.F.; B0 forx), e.g.: H0: µB – µA = 0, H1: µB – µA > 0 B1 Calculate value of t: t = d / (s/√10) = 1⋅398 or 1⋅4 M1 A1 State or use correct tabular t-value: t9, 0.9 = 1⋅38[3] B1 (or can compared with 3⋅36[4]) Consistent conclusion (A.E.F, on both t–values): [Reject H0:] Hours of absence have decreased B1 7 7 Wrong test can earn only B1 for hypotheses and B1 for conclusion

Page 10 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 7 (i) Find P(N = 5): P(N = 5) = (¾)4 × ¼ = 81 / 1024 or 0⋅079[1] B1 1 (ii) Find P(N > 8): P(N > 8) = (¾)8 = 6561 / 65536 or 0⋅1[00] M1 A1 2 Find prob. PJ that James qualifies: PJ = P(N Y 6) = 1 – (¾)6 or {1 + ¾ + (¾)2 + (¾)3 + (¾)4 + (¾)5}¼ = 3367 / 4096 or 0⋅822 M1 A1 2 Find prob. PC that Colin qualifies: PC = 1 – (⅔)6 [= 0⋅9122] B1 Find prob. that exactly one qualifies: PJ (1 – PC) + PC (1 – PJ) = (3367 / 4096) (64 / 729) + (665 / 729) (729 / 4096) = 0⋅0722 + 0⋅1624 = 0⋅235 (allow 0⋅234) M1 A1 3 8

Page 11 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 8 Find expected values (to 1 d.p.): 32⋅4 15⋅3 12⋅3 40⋅5 19⋅125 15⋅375 35⋅1 16⋅575 13⋅325 M1 A1 State (at least) null hypothesis (A.E.F.): H0: Car type is independent B1 Calculate χ2 (to 1 d.p.): χ2 = 0⋅005 + 1⋅208 + 1⋅796 + 0⋅500 + 1⋅243 + 5⋅716 + 0⋅479 + 0⋅020 + 1⋅640 = 12⋅6 (allow 12⋅7) M1 A1 State or use correct tabular χ2 value (to 3 s.f.): χ4, 0.95 2 = 9⋅488 OR 9⋅49 B1 Valid method for reaching conclusion: Reject H0 if χ2 > tabular value M1 Conclusion consistent with correct values (A.E.F): Car type is dependent on age A1 8 8

Page 12 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 9 Find k for which P(X [ k) = 0⋅6: 0⋅6 = 1 – F(k) M1 = 1 – (k / 8 – ¼) M1 k = 26 / 5 or 5⋅2 A1 3 Find G(y) from Y = 2 ln X for 2 Y x Y 10: G(y) = P(Y < y) = P(2 ln X < y) (allow < or Y throughout) = P(X < ey / 2) = F(ey / 2) (result may be stated) = ey / 2 / 8 – ¼ (2 ln 2 Y y Y 2 ln 10) or (ln 4 Y y Y ln 100) or (1⋅39 Y y Y 4⋅61) M1 A1 State G(y) for other values of x: 0 (y < 2 ln 2) and 1 (y > 2 ln 10) B1 3 Find g(y) for 2 ln 2 Y y Y 2 ln 10: g(y) = ey / 2 / 16 M1 A1 Sketch positive exponential for 2 ln 2 Y y Y 2 ln 10 B1 Show g(y) = 0 on either side of this interval B1 4 10

Page 13 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 10 (i) Calculate sample mean: x = 132 / 8 = 16⋅5 M1 Estimate population variance: s2 = (2192⋅06 – 1322 / 8) / 7 (allow biased here: 1⋅757 or 1⋅3262) = 703 / 350 or 2⋅009 or 1⋅4172 M1 State hypotheses (A.E.F.; B0 forx): H0: µ = 15⋅8, H1: µ > 15⋅8 B1 Calculate value of t (to 3 s.f.): t = (x – 15⋅8) / (s / √8) = 1⋅4[0] M1 A1 State or use correct tabular t–value (to 3 s.f.): t7, 0.9 = 1⋅41[5] B1 (or can comparex with 15⋅8 + 0⋅709 = 16⋅51) Consistent conclusion (A.E.F, on both t–values): [Accept H0:] Popn. mean not greater than 15⋅8 B1 7 (ii) Find confidence interval (allow z in place of t) e.g.: 16⋅5 ± t √(2⋅009 / 8) M1 A1 (Use of 15⋅8 does not lose M1) Use of correct tabular value: t7, 0.975 = 2⋅36[5] A1 Evaluate C.I. correct to 3 s.f.: 16⋅5 ± 1⋅18[5] or [15⋅3, 17⋅7 A1 4 11

Page 14 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 11 A (i) EITHER: Find extension e at equilibrium point: e = 5l / 7 B1 Resolve forces at eq. pt. to find modulus λ: λe / l = mg , λ = 7mg / 5 A.G. B1 OR: Use conservation of energy to find λ: ½λ{(6l / 7)2 – (4l / 7)2} / l =mg2l / 7 λ = 7mg / 5 A.G. (M1 A1) 2 (ii) Use Newton’s Law at general point: m d2x / dt2 = mg – λ (e+x) / l [ or – mg + λ (e–x) / l ] M1 A1 Simplify to give standard SHM eqn: d2x / dt2 = – λx / lm or – 7gx / 5l A1 S.R.: Stating this without derivation (max 2 / 4): (B1) Find period T = 2π / ω with ω2 = 7g / 5l: T = 2π / √(7g / 5l) or 2π√(5l / 7g) B1 4

Page 15 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 (iii) EITHER: Equate P’s speed to one-half its max. speed: ω2(l2 / 72 – x2) = ¼ ω2 l2 / 72 , M1 A1 Find x: (A.E.F.) x2 = ¾ l2 / 72 , x = (√3 / 14) l A1 either: Find t from x = x0 cos ωt: t = ω–1 cos–1 {(√3 / 14)l / (l / 7)} M1 A1 = ω–1 cos–1 (√3/2) = π / 6ω = (π / 6) √(5l / 7g) A1 or: Find t from x = x0 sin ωt: t = ¼T – ω–1sin–1 {(√3/14)l/(l / 7)} (M1 A1) = ¼T – ω–1 sin–1 (√3 / 2) = (π/2 – π / 3) √(5l / 7g) = (π / 6) √(5l / 7g) (A1) OR: Equate P’s speed to one-half its max. speed: ωx0 sin ωt or ωx0 cos ωt = ½ωx0 (M1 A1; A1) Find first value of t: t = ω–1 sin–1 (1 / 2) or ¼T – ω–1 cos–1 (1 / 2) (M1 A1) = (π / 6) √(5l / 7g) (A1) 6 12

Page 16 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 23 © Cambridge International Examinations 2014 11 (b) (i) State both hypotheses (B0 for r …): H0: ρ = 0, H1: ρ > 0 B1 State or use correct tabular one-tail r-value: r12, 5% = 0⋅497 *B1 Valid method for reaching conclusion: Reject H0 if 0⋅6 > tabular value M1 Consistent conclusion (A.E.F, on 0⋅497): There is positive correlation A1 4 (ii) Use r2 = ab to eliminate a or b: 0⋅5 = 0⋅62 / a – a or b – 0⋅62 / b M1 Solve and select correct values: a2 + 0⋅5a – 0⋅36 = 0 or b2 – 0⋅5b – 0⋅36 = 0 (a + 0⋅9)(a – 0⋅4) = 0 or (b + 0⋅4)(b – 0⋅9) = 0 (A0 if a = –0⋅9, b = –0⋅4 not rejected) a = 0⋅4 and b = 0⋅9 M1 A1 3 (iii) Usex [= 5⋅5] in 1st regression eqn to findy: y = (66 / 12) b + 4⋅5 [= 9⋅45] M1 Usex andy in 2nd regression eqn to find c: c = x – ay = 1⋅72 B1 Sketch both regression lines on one diagram y = 0⋅9x + 4⋅5 and x = 0⋅4y + 1⋅72 ⇒ y = 2⋅5x – 4⋅3 B1 3 (iv) State coefficient of x in eqn of z on x ( on b): 5 × b = 4⋅5 B1 State value of r with valid justification, e.g.: r = √{(4⋅5 × (0⋅4 / 5)} = 0⋅6 or r is unchanged by scaling so 0⋅6 B1 2 12