Cambridge A Level Mathematics - Further 9231 — 2016 May/June Paper 2 · Variant 1

*1085330974* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/21 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_21/2R © UCLES 2016 [Turn over

2 1 A bullet of mass 0.01 kg is fired horizontally into a fixed vertical barrier which exerts a constant resisting force of magnitude 1000 N. The bullet enters the barrier with speed 320 m s−1 and emerges with speed 20 m s−1. You may assume that the motion takes place in a horizontal straight line. Find (i) the magnitude of the impulse that acts on the bullet, [2] (ii) the thickness of the barrier, [2] (iii) the time taken for the bullet to pass through the barrier. [1] 2 A small smooth sphere A of mass m is moving with speed u on a smooth horizontal surface when it collides directly with an identical sphere B which is initially at rest on the surface. The coefficient of restitution between the spheres is e. Sphere B subsequently collides with a fixed vertical barrier which is perpendicular to the direction of motion of B. The coefficient of restitution between B and the barrier is 1 2. Given that 80% of the initial kinetic energy is lost as a result of the two collisions, find the value of e. [8] 3 A particle P is performing simple harmonic motion with amplitude 0.25 m. During each complete oscillation, P moves with a speed that is less than or equal to half of its maximum speed for 4 3 seconds. Find the period of the motion and the maximum speed of P. [8] 4 A particle P is at rest at the lowest point on the smooth inner surface of a hollow sphere with centre O and radius a. The particle is projected horizontally with speed u and begins to move in a vertical circle on the inner surface of the sphere. The particle loses contact with the sphere at the point A, where OA makes an angle 1 with the upward vertical through O. Given that the speed of P at A is 3 5ag, find u in terms of a and g. [5] Find, in terms of a, the greatest height above the level of O achieved by P in its subsequent motion. (You may assume that P achieves its greatest height before it makes any further contact with the sphere.) [5] © UCLES 2016 9231/21/M/J/16

3 5 O A B C D a 2a a 2a l A thin uniform rod AB has mass 3 4m and length 3a. The end A of the rod is rigidly attached to a point on the circumference of a uniform disc with centre C, mass m and radius a. The end B of the rod is rigidly attached to a point on the circumference of a uniform disc with centre D, mass 4m and radius 2a. The discs and the rod are in the same plane and CABD is a straight line. The mid-point of CD is O. The object consisting of the two discs and the rod is free to rotate about a fixed smooth horizontal axis l, through O in the plane of the object and perpendicular to the rod (see diagram). Show that the moment of inertia of the object about l is 50ma2. [6] The object hangs in equilibrium with D vertically below C. It is displaced through a small angle and released from rest, so that it makes small oscillations about the horizontal axis l. Show that it will move in approximate simple harmonic motion and state the period of the motion. [6] 6 The score when two fair dice are thrown is the sum of the two numbers on the upper faces. Two fair dice are thrown repeatedly until a score of 6 is obtained. The number of throws taken is denoted by the random variable X. Find the mean of X. [2] Find the least integer N such that the probability of obtaining a score of 6 in fewer than N throws is more than 0.95. [3] 7 A random sample of 9 observations of a normal variable X is taken. The results are summarised as follows. Σx = 24.6 Σx2 = 68.5 Test, at the 5% significance level, whether the population mean is greater than 2.5. [8] © UCLES 2016 9231/21/M/J/16 [Turn over

4 8 The random variable X has probability density function f given by fx = D 2e−2x x ≥0, 0 otherwise. (i) Find the distribution function of X. [2] (ii) Find the median value of X. [3] The random variable Y is defined by Y = eX. (iii) Find the probability density function of Y. [4] 9 Applicants for a national teacher training course are required to pass a mathematics test. Each year, the applicants are tested in groups of 6 and the number of successful applicants in each group is recorded. The overall proportion of successful applicants has remained constant over the years and is equal to 60% of the applicants. The results from 150 randomly chosen groups are shown in the following table. Number of successful applicants 0 1 2 3 4 5 6 Number of groups 1 3 25 51 38 30 2 Test, at the 5% significance level, the goodness of fit of the distribution B6, 0.6 for the number of successful applicants in a group. [10] 10 For a random sample of 6 observations of pairs of values x, y, where 0 < x < 21 and 0 < y < 14, the following results are obtained. Σx2 = 844.20 Σy2 = 481.50 Σxy = 625.59 It is also found that the variance of the x-values is 36.66 and the variance of the y-values is 9.69. (i) Find the product moment correlation coefficient for the sample. [5] (ii) Find the equations of the regression lines of y on x and x on y. [4] (iii) Use the appropriate regression line to estimate the value of x when y = 6.4 and comment on the reliability of your estimate. [2] © UCLES 2016 9231/21/M/J/16

5 11 Answer only one of the following two alternatives. EITHER A B C 2a 1 2a 1 2a D The end A of a uniform rod AB, of length 2a and weight W, is freely hinged to a vertical wall. The end B of the rod is attached to a light elastic string of natural length 3 2a and modulus of elasticity 3W. The other end of the string is attached to the point C on the wall, where C is vertically above A and AC = 2a. A particle of weight 2W is attached to the rod at the point D, where DB = 1 2a. The angle ABC is equal to 1 (see diagram). Show that cos 1 = 3 4 and find the tension in the string in terms of W. [10] Find the magnitude of the reaction force at the hinge. [4] OR Petra is studying a particular species of bird. She takes a random sample of 12 birds from nature reserve A and measures the wing span, x cm, for each bird. She then calculates a 95% confidence interval for the population mean wing span, - cm, for birds of this species, assuming that wing spans are normally distributed. Later, she is not able to find the summary of the results for the sample, but she knows that the 95% confidence interval is 25.17 ≤- ≤26.83. Find the values of Σ x and Σ x2 for this sample. [6] Petra also measures the wing spans of a random sample of 7 birds from nature reserve B. Their wing spans, y cm, are as follows. 23.2 22.4 27.6 25.3 28.4 26.5 23.6 She believes that the mean wing span of birds found in nature reserve A is greater than the mean wing span of birds found in nature reserve B. Assuming that this second sample also comes from a normal distribution, with variance the same as the first distribution, test, at the 10% significance level, whether there is evidence to support Petra’s belief. [8] © UCLES 2016 9231/21/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 9231/21/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 9231/21 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 21 © 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 21 © 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 21 © Cambridge International Examinations 2016 Question Mark Scheme Details Part Mark Total 1 (i) Find magnitude of impulse by change in momentum: (±) 0⋅01 × (320 – 20) (if 320 + 20, can allow M1 but not MR) = (±) 3 [N s–1] M1 A1 [2] (ii) EITHER: Find thickness d by change in energy: 1000d = ½ × 0⋅01 × (3202 – 202) (M0 if ∆v2 = 3202 + 202) d = 0⋅51 [m] OR: [Find t from impulse and] use t to find d: d = ½ (320 + 20) t (M0 if 320 – 20) = 170 × 3/1000 = 0⋅51 [m] OR: Find d using F = ma, v2 = u2 + 2as a = 105, d = 0⋅51 [m] M1 A1 (M1 A1) (M1 A1) [2] (iii) EITHER: Find time t from constant deceln. formula: d = ½ (320 + 20) t, t = 0⋅003 [s] OR: Find time t from impulse: 1000 t = 3, t = 0⋅003 [s] B1 (B1) [1] 2 Use momentum and Newton’s law, e.g.: mvA + mvB = mu, vB – vA = eu (M0 for inconsistent signs; allow vA + vB = u) Combine to find vA and vB (may be implied): vA = ±(1 – e) u/2, vB = ±(1 + e) u/2 Relate speed wB of B after colln. with wall to vB, e.g.: wB = ±½ vB [= ±(1 + e) u/4] Relate KEs before and after (AEF): ½ mu2 / 5 = ½ mvA 2 + ½ mwB 2 (Treat error in 1/5 as M1 A0) Find e: 1/5 = (1 – e)2 /4 + (1 + e)2 /16 25e2 – 30e + 9 = (5e – 3)2 = 0 e = 3/5 or 0⋅6 M1 A1, A1 B1 M1 A1 M1 A1 [8] 3 EITHER: Find time during which speed is low e.g.: x = a cos ω t, v = (±) aω sin ω t ½ aω = aω sin ω t (working may be implied) tL = π/6ω or T/12 Find ω (implied?) or T from given time: 4tL = 4/3, ω = π/2 or T = 4 [s] Find other value (√ on first): T = 2π/ω = 4 [s] or ω = π/2 Find maximum speed vmax: vmax = 0⋅25ω = π/8 or 0⋅393 [m] OR: Find time during which speed is high e.g.: x = a sin ω t, v = aω cos ω t ½ aω = aω cos ω t (working may be implied) tH = π/3ω or T/6 Find ω (implied?) or T from given time, e.g.: 2π/ω – 4tL = 4/3, ω = π/2 or T = 4[s] Find other value (√ on first): T = 2π/ω = 4 [s] or ω = π/2 Find maximum speed vmax: vmax = 0⋅25ω = π/8 or 0⋅393 [m] OR: Find x when speed is ½ vmax: ω2 (a2 – x2) = ¼ ω2 a2 (AEF) x2 = ¾ a2, [±] x = √3a/2 or √3/8 Find time during which speed is low, e.g.: √3a/2 = a cos (ω t) , ωtL = π/6 or speed is high, e.g.: √3a/2 = a sin (ω t) , ωtH = π/3 Find ω, T and vmax as above M1 M1 A1 M1 A1 B1 M1 A1 (M1) (M1 A1) (M1 A1) B1 (M1 A1) (B1) (M1 A1) [8]

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 21 © Cambridge International Examinations 2016 Question Mark Scheme Details Part Mark Total 4 Find v2 at A from conservation of energy: ½mv2 = ½mu2 – mga(1 + cos θ ) Use F = ma radially at A with R = 0: mv2/a = mg cos θ Use v2 = 3ag/5 and eliminate cos θ to find u: cos θ = 3/5 u2 = v2 + 2ag (1 + cos θ ) = 19ag/5 u = √(19ag/5) or 1⋅95√(ag) (allow numerical value of g) or √(38a) or 6⋅16√(a) Find or imply vertical component of speed at A: V = v sin θ [= 4v/5] Find greatest height hA reached above A: hA = V2/2g = 24a/125 or 0⋅192a Find greatest height hO reached above O: hO= hA + a cos θ = 99a/125 or 0⋅792a M1 A1 B1 M1 A1 M1 M1 A1 M1 A1 [5] [5] 5 Find MI of rod about l: Irod = ⅓ ¾ m (3a/2)2 + ¾ m (½ a)2 [= ¾ ma2] Find MI of disc C about l using both theorems: Idisc C = ½ × ½ ma2 + m (3a)2 [= (37/4) ma2] Find MI of disc D about l using both theorems: Idisc D = ½ × ½ 4m(2a)2 + 4m (3a)2 [= 40 ma2] Sum to find MI of system about l: A.G. I = (¾ + 37/4 + 40) ma2 = 50 ma2 Use eqn of circular motion to find d2θ/dt2 where [–] I d2θ/dt2 = 4mg × 3a sin θ θ is angle of CD with vertical: – ¾mg × ½ a sin θ (A1 only for two correct terms on RHS; – mg × 3a sin θ A0 if cos θ used) [ = (69/8) mga sin θ ] Approximate sin θ by θ to show SHM: d2θ/dt2 = – (69g / 400a) θ (M0 if wrong sign or cos θ ≈ θ used) or – (0⋅1725 g / a) θ Find period T (AEF): T = 2π / √(69g/400a) = 40π √(a/69g) (allow g = 9⋅8 or 9⋅81) or 15⋅1√(a/g) or 4⋅78√a B1 M1 A1 M1 A1 A1 M1 A2 M1 A1 A1 [6] [6] 6 Find prob. p of score of 6 on one throw: p = 5/36 or 0⋅139 Find mean of X: 1/p = 36/5 or 7⋅2 Formulate condition for N (1 – qN is M0): 1 – qN – 1 > 0⋅95 Rearrange and take logs (any base) to give bound: (31/36) N – 1 < 0⋅05 N – 1 > log 0⋅05 / log 31/36 Find Nmin: N – 1 > 20⋅03, Nmin = 22 (N – 1 < 20⋅03 or N – 1 = 20⋅03 earns M1 M1 A0) B1 B1 M1 M1 A1 [2] [3] 7 Calculate sample mean: x = 24⋅6 / 9 or 41/15 or 2⋅73[3] Estimate population variance: s2 = (68⋅5 – 24⋅62 /9) / 8 (allow biased here: 0⋅14 or 0⋅37422) = 63/400 or 0⋅1575 or 0⋅39692 State hypotheses (AEF; B0 forx …): H0: µ = 2⋅5, H1: µ > 2⋅5 Calculate value of t (to 3 s.f.): t = (x – 2⋅5)/(s/√9) = 1⋅76 State or use correct tabular t-value (to 3 s.f.) : t8, 0.95 = 1⋅86[0] (or can comparex with 2⋅5 + 0⋅246 = 2⋅746 ) State or imply valid method for conclusion e.g.: Accept H0 if t < tabular value Conclusion (AEF, requires both values correct): 1⋅76 [ ±0⋅02] < 1⋅86 so popln. mean not greater than 2⋅5 B1 M1 B1 M1 A1 B1 M1 A1 [8]

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 21 © Cambridge International Examinations 2016 Question Mark Scheme Details Part Mark Total 8 (i) Find or state distribution function F(x) for x ⩾ 0: F(x) = ∫ f(x) dx = – e-2x + c Use F(0) = 0 to find F(x): F(x) = 0 (x < 0), 1 – e-2x (x ⩾ 0) M1 A1 [2] (ii) Find median value m from F(m) = ½: 1 – e–2m = ½ , e2m = 2 m = ½ ln 2 or 0⋅347 M1 M1 A1 [3] (iii) Find or state G(y) from Y = eX for x ⩾ 0: G(y) = P(Y < y) = P(eX < y) (allow < or ⩽ throughout) = P(X < ln y) = F(ln y) = 1 – e–2 ln y [= 1 – 1/y2 ] Find g(y) by differentiation: g(y) = 2 / y3 State corresponding range of y: for y ⩾ 1 [g(y) = 0 for y < 1] M1 A1 A1 A1 [4] 9 State (at least) null hypothesis in full: (AEF) H0: Given distribution fits data Find exp. values using 150 6Ci q6–i pi with p = 0⋅6, q = 0⋅4 (to 3 s.f.): 0⋅6144 5⋅5296 20⋅736 41⋅472 (allow A1 if only one error 46⋅656 27⋅9936 6⋅9984 or if all values correct to 2 s.f.) Combine 0⋅6144 & 5⋅5296 since 1st exp. value < 5: Oi: 4 . . . Ei: 6⋅14[4] . . . Calculate χ2 (result correct to 3 s.f.): χ2 = 0⋅748 + 0⋅877 + 2⋅189 + 1⋅606 + 0⋅144 + 3⋅570 = 9⋅13 State or use consistent tabular value (to 3 s.f.): 6 cells: χ5,0.95 2 = 11⋅07 [ or if no or more cells combined: 7 cells: χ6, 0.95 2 = 12⋅59 5 cells: χ4, 0.95 2 = 9⋅488 4 cells: χ3, 0.95 2 = 7⋅815 3 cells: χ2, 0.95 2 = 5⋅991 ] State or imply valid method for conclusion e.g.: Accept H0 if χ2 < tabular value Conclusion (AEF, requires both values correct): 9⋅13 [± 0⋅01] < 11⋅1 so distn. fits [data] (Allow A1 here for e.g. “It is a good fit”) S.C. 150 6Ci qi p6– i can earn B1 M1 B1 M1 B1 M1 (max 6/10) B1 M1 A2 B1 M1 A1 B1 M1 A1 [10] 10 (i) Find ∑ x and ∑ y (M1 for either): (∑ x)2 = 6 (844⋅20 – 6 × 36⋅66) = 3775⋅44 SC: Allow M1 if 5 used instead of 6 (max 4/11), ∑ x = 61⋅2 or x = 10⋅2 giving 62⋅97, 50⋅97 and r = 0⋅961[5] (∑ y)2 = 6 (481⋅5 – 6 × 9⋅69) = 2540⋅16 ∑ y = 50⋅4 or y = 8⋅4 Find correlation coefficient r, e.g.: Sxy = 625⋅59 – 61⋅2 × 50⋅4/6 = 111⋅51 Sxx = 844⋅20 – 61⋅22/6 or 6 × 36⋅66 = 219⋅96 Syy = 481⋅50 – 50⋅42/6 or 6 × 9⋅69 = 58⋅14 r = Sxy / √(Sxx Syy) = 0⋅986 M1 A1 A1 M1 A1 [5]

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 21 © Cambridge International Examinations 2016 Question Mark Scheme Details Part Mark Total (ii) Calculate gradients in both lines : b1 = Sxy / Sxx = 0⋅507 and b2 = Sxy / Syy or r2/ b1 = 1⋅918 Find one regression line, e.g. of y on x: y = 50⋅4/6 + 0⋅507 (x – 61⋅2/6) = 0⋅507 x + 3⋅23 Find other regression line, e.g. of x on y: x = 61⋅2/6 + 1⋅918 (y – 50⋅4/6) = 1⋅92 y – 5⋅91 SC: Use PA –1 if results only correct to 2 s.f. B1 M1 A1 A1 [4] (iii) Find x when y = 6⋅4: x = 6⋅36 (allow 6⋅37 or 6⋅38) Valid comment on reliability, e.g.: Reliable since r ≈ 1 [√ on r] or 6⋅4 within range of y B1 B1 [2] 11A Denoting angle between AB and CA produced by Φ and denoting mid-point of BC by E, say. Take moments for rod about A (AEF): W × a sin Φ + 2W × (3a/2) sin Φ = T × AE Substituting Φ = 2θ and AE = 2a sin θ: W × a sin 2θ + 2W × (3a/2) sin 2θ = T × 2a sin θ Hence find tension T in terms of θ : T = 2W sin 2θ / sin θ [ = 4W cos θ ] Find T in terms of θ using Hooke’s Law: T = 3W (BC – 3a/2) / (3a/2) = W (8 cos θ – 3) Equate to find cos θ : 4 cos θ = 8 cos θ – 3 cos θ = ¾ A.G. Hence find T (allow assumption of cos θ = ¾): T = 3W Find horizontal force X at A (ignore sign): X = T sin θ [= 3W√7 / 4] Find vertical force Y at A (ignore sign): Y = 3W – T cos θ [= 3W / 4] Find magnitude of force at A: √(X2 + Y2) = (3/√2) W or (3√2 / 2) W or 2⋅12 W M1 A1 A1 A1 M1 A1 A1 M1 A1 B1 B1 B1 M1 A1 [10] [4]

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 21 © Cambridge International Examinations 2016 Question Mark Scheme Details Part Mark Total 11B Find Σ x via sample meanx: Σ x = 12x = 12 × ½ (25⋅17 + 26⋅83) = 12 × 26 = 312 EITHER: Find estimated s.d. sA, or sA 2: t sA / √12 = ½ (26⋅83 – 25⋅17) with t = t11, 0.975 = 2⋅201 sA = 0⋅83 √12 / 2⋅201 (to 3 s.f.) = 1⋅306 or sA 2 = 1⋅706 Find Σ x2 from sA or sA 2: sA 2 = { Σ x2 – (Σ x)2/12 } / 11 (M0 for sA 2 = {…}/12) Σ x2 = 11 sA 2 + (Σ x)2/12 = 8130[⋅8] OR: Find estimated s.d. σA, or σA 2: t σA / √11 = ½ (26⋅83 – 25⋅17) with t = t11, 0.975 = 2⋅201 σA = 0⋅83 √11 / 2⋅201 (to 3 s.f.) = 1⋅25 or σA 2 = 1⋅564 Find Σ x2 from σA or σA 2: σA 2 = { Σ x2 – (Σ x)2/12 } / 12 (M0 for σA 2 = {…}/11) Σ x2 = 12 σA 2 + (Σ x)2/12 = 8130[⋅8] State hypotheses (B0 forxA …), e.g.: H0: µA = µB , H1: µA > µB Estimate B’s popln. variance (to 3 d.p..): sB 2 = (4507⋅62 – 1772/7) / 6 (allow biased here: 4⋅578) [ = 5⋅341 ] Find pooled estimate of common variance: s2 = (11 sA 2 + 6 sB 2) / 17 (note sB 2 may be implied, earning M1 above) = (18⋅77 + 32⋅05) / 17 = 2⋅989 or 1⋅7292 (to 3 s.f.) Calculate value of t (or –t): t = (26 – 177/7) / s √(1/12 + 1/7) = (26 – 25⋅29) / s √(1/12 + 1/7) (to 3 s.f.) = 0⋅7143 / 0⋅8222 = 0⋅869 State or use correct tabular t value: (to 3 s.f.) t17, 0.9 = 1⋅333 (or can comparexA –xB = 0⋅7143 with 1⋅536) Consistent conclusion (AEF, √ on t, dep *B1): t < tabular value so Petra’s belief not supported or wing span of A not greater M1 A1 M1 A1 M1 A1 (M1 A1 M1 A1) B1 M1 M1 A1 M1 A1 *B1 B1 [6] [8]