Cambridge A Level Mathematics - Further 9231 — 2015 Oct/Nov Paper 2 · Variant 3

*6408093407* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/23 Paper 2 October/November 2015 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. JC15 11_9231_23/FP © UCLES 2015 [Turn over

2 1 ! 1 2a a A B C D A uniform ladder AB, of length 3a and weight W, rests with the end A in contact with smooth horizontal ground and the end B against a smooth vertical wall. One end of a light inextensible rope is attached to the ladder at the point C, where AC = a. The other end of the rope is fixed to the point D at the base of the wall and the rope DC is in the same vertical plane as the ladder AB. The ladder rests in equilibrium in a vertical plane perpendicular to the wall, with the ladder making an angle 1 with the horizontal and the rope making an angle ! with the horizontal (see diagram). It is given that tan 1 = 2 tan !. Find, in terms of W and !, the tension in the rope and the magnitudes of the forces acting on the ladder at A and at B. [9] 2 A small uniform sphere A, of mass 2m, is moving with speed u on a smooth horizontal surface when it collides directly with a small uniform sphere B, of mass m, which is at rest. The spheres have equal radii and the coefficient of restitution between them is e. Find expressions for the speeds of A and B immediately after the collision. [4] Subsequently B collides with a vertical wall which is perpendicular to the direction of motion of B. The coefficient of restitution between B and the wall is 0.4. After B has collided with the wall, the speeds of A and B are equal. Find e. [2] Initially B is at a distance d from the wall. Find the distance of B from the wall when it next collides with A. [4] 3 A and B are two fixed points on a smooth horizontal surface, with AB = 3a m. One end of a light elastic string, of natural length a m and modulus of elasticity mg N, is attached to the point A. The other end of this string is attached to a particle P of mass m kg. One end of a second light elastic string, of natural length ka m and modulus of elasticity 2mg N, is attached to B. The other end of this string is attached to P. Given that the system is in equilibrium when P is at M, the mid-point of AB, find the value of k. [3] The particle P is released from rest at a point between A and B where both strings are taut. Show that P performs simple harmonic motion and state the period of the motion. [5] In the case where P is released from rest at a distance 0.2a m from M, the speed of P is 0.7 m s−1 when P is 0.05a m from M. Find the value of a. [3] © UCLES 2015 9231/23/O/N/15

3 4 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 P is hanging at rest vertically below O, it is projected horizontally. In the subsequent motion P completes a vertical circle. The speed of P when it is at its highest point is u. Show that the least possible value of u is ag. [2] It is now given that u = ag. When P passes through the lowest point of its path, it collides with, and coalesces with, a stationary particle of mass 1 4m. Find the speed of the combined particle immediately after the collision. [4] In the subsequent motion, when OP makes an angle 1 with the upward vertical the tension in the string is T. Find an expression for T in terms of m, g and 1. [5] Find the value of cos 1 when the string becomes slack. [2] 5 A random sample of 10 observations of a normal variable X gave the following summarised data, where x is the sample mean. Σx = 222.8 Σx −x2 = 4.12 Find a 95% confidence interval for the population mean. [5] 6 A biased coin is tossed repeatedly until a head is obtained. The random variable X denotes the number of tosses required for a head to be obtained. The mean of X is equal to twice the variance of X. Show that the probability that a head is obtained when the coin is tossed once is 2 3. [2] Find (i) PX = 4, [1] (ii) PX > 4, [2] (iii) the least integer N such that PX ≤N > 0.999. [3] 7 The continuous random variable X has probability density function given by fx = T 1 21x2 1 ≤x ≤4, 0 otherwise. The random variable Y is defined by Y = X2. Show that Y has probability density function given by gy = T 1 42y 1 2 1 ≤y ≤16, 0 otherwise. [5] Find (i) the median value of Y, [2] (ii) the expected value of Y. [2] © UCLES 2015 9231/23/O/N/15 [Turn over

4 8 The number of goals scored by a certain football team was recorded for each of 100 matches, and the results are summarised in the following table. Number of goals 0 1 2 3 4 5 6 or more Frequency 12 16 31 25 13 3 0 Fit a Poisson distribution to the data, and test its goodness of fit at the 5% significance level. [10] 9 A random sample of 8 students is chosen from those sitting examinations in both Mathematics and French. Their marks in Mathematics, x, and in French, y, are summarised as follows. Σx = 472 Σx2 = 29 950 Σy = 400 Σy2 = 21 226 Σxy = 24 879 Another student scored 72 marks in the Mathematics examination but was unable to sit the French examination. Estimate the mark that this student would have obtained in the French examination. [5] Test, at the 5% significance level, whether there is non-zero correlation between marks in Mathematics and marks in French. [6] [Question 10 is printed on the next page.] © UCLES 2015 9231/23/O/N/15

5 10 Answer only one of the following two alternatives. EITHER A B C D P Q 6a 3a 3a 3a 3a An object is formed by attaching a thin uniform rod PQ to a uniform rectangular lamina ABCD. The lamina has mass m, and AB = DC = 6a, BC = AD = 3a. The rod has mass M and length 3a. The end P of the rod is attached to the mid-point of AB. The rod is perpendicular to AB and in the plane of the lamina (see diagram). Show that the moment of inertia of the object about a smooth horizontal axis l1, through Q and perpendicular to the plane of the lamina, is 38m + Ma2. [4] Show that the moment of inertia of the object about a smooth horizontal axis l2, through the mid-point of PQ and perpendicular to the plane of the lamina, is 3 417m + Ma2. [2] Find expressions for the periods of small oscillations of the object about the axes l1 and l2, and verify that these periods are equal when m = M. [8] OR A farmer A grows two types of potato plants, Royal and Majestic. A random sample of 10 Royal plants is taken and the potatoes from each plant are weighed. The total mass of potatoes on a plant is x kg. The data are summarised as follows. Σx = 42.0 Σx2 = 180.0 A random sample of 12 Majestic plants is taken. The total mass of potatoes on a plant is y kg. The data are summarised as follows. Σy = 57.6 Σy2 = 281.5 Test, at the 5% significance level, whether the population mean mass of potatoes from Royal plants is the same as the population mean mass of potatoes from Majestic plants. You may assume that both distributions are normal and you should state any additional assumption that you make. [9] A neighbouring farmer B grows Crown potato plants. His plants produce 3.8 kg of potatoes per plant, on average. Farmer A claims that her Royal plants produce a higher mean mass of potatoes than Farmer B’s Crown plants. Test, at the 5% significance level, whether Farmer A’s claim is justified. [5] © UCLES 2015 9231/23/O/N/15

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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 2015 9231/23/O/N/15

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the October/November 2015 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 October/November 2015 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 – October/November 2015 9231 23 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 1 Find 3 independent equations for T, RA, RB : Resolve horizontally: RB = T cos α M1 A1 Resolve vertically: RA = W + T sin α M1 A1 Take moments about A: RB 3a sin θ = W (3a/2) cos θ (a may be omitted from moment eqns) + T a(sin α cos θ + cos α sin θ ) or + T a sin (α + θ ) or + T 3a cos θ sin α M1 A1 Take moments about B: RA3a cos θ = W (3a/2) cos θ + T 2a(sin α cos θ + cos α sin θ ) or + T 2a sin (α + θ ) or + T 3a sin θ cos α (M1 A1) Take moments about C: RA a cos θ + W (a/2) cos θ = RB 2a sin θ (M1 A1) Take moments about D: RA 3a cos θ – W (3a/2) cos θ = RB 3a sin θ (M1 A1) Solve for T, RA, RB (AEF in W and α): T = W / 2 sin α or ½W cosec α B1 RA = 3W / 2 B1 RB = W / 2 tan α or ½W cot α B1 9 9 2 For A & B use conservation of momentum, e.g.:2mvA + mvB = 2mu (allow 2vA + vB = 2u) M1 Use Newton’s law of restitution (consistent signs): vB – vA = eu M1 Combine to find vA and vB : vA = (2 – e) u/3, vB = 2(1 + e) u/3 A1, A1 Find e from vA = vB′ with vB′ = [–] 0⋅4 vB: (2 – e) = 0⋅8 (1 + e), e = 2/3 M1 A1 EITHER: Equate times in terms of reqd. distance x: (d – x)/ vA = d/vB + x/vB′ (AEF) M1 A1 [speeds need not be found: vA = vB′ = 4u/9, vB = 10u/9] Use vA = vB′ = 0⋅4 vB to solve for x: d – x = 0⋅4 d + x, x = 0⋅3 d M1 A1 OR: Find dist. moved by A when B reaches wall: dA = (d/vB) vA = 0⋅4 d (M1 A1) Find reqd. distance x: x = ½ (d – dA) = 0⋅3 d (M1 A1) 4 2 4 10

Page 3 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2015 9231 23 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 3 Find k by equating equilibrium tensions: mg (a/2)/a = 2mg (3a/2 – ka) / ka M1 A1 (vertical motion can earn M1 only) ½ = 3/k – 2, k = 6/5 or 1⋅2 A1 Apply Newton’s law at general point, e.g.: m d2x/dt2 = – mg (a/2 + x)/a (lose A1 for each incorrect term) + 2mg (3a/2 – ka – x) / ka or m d2y/dt2 = + mg (a/2 – y)/a – 2mg (3a/2 – ka + y) / ka M1 A2 Simplify to give standard SHM eqn, e.g.: d2x/dt2 = – (1 + 2/k)gx / a S.R.: B1 if no derivation (max 2/5) = – 8gx/3a A1 State or find period using 2π/ω with ω = √(8g/3a): T = 2π√(3a/8g) or π√(3a/2g) (√ on ω) or 3⋅85√(a/g) or 1⋅22√a [s] B1 Substitute values in v2 = ω2 (x0 2 – x2): 0⋅72 = (8g/3a){(0⋅2a)2 – (0⋅05a)2} M1 A1 Solve to find numerical value of a: 0⋅49 = (8g/3) × 0⋅0375a, a = 0⋅49 A1 3 5 3 11

Page 4 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2015 9231 23 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 4 EITHER: Find tension at top from F = ma vertically: T = mu2/a – mg B1 OR: Use energy at e.g. θ to upward vertical: ½ mv2 = ½ mu2 + mga (1 – cos θ ) Find tension T by using F = ma radially: T′ = mv2/a – mg cos θ Eliminate v2 : = mu2/a + mg (2 – 3 cos θ ) Find T at top by taking θ = 0: T = mu2/a – mg (B1) Find umin by requiring T ⩾ 0 at top [or T > 0]: u2/a – g ⩾ 0 so umin = √ag A.G. B1 Find v at bottom from conservation of energy: ½mv2 = ½mu2 + mg × 2a M1 v2 = ag + 4ag, v = √(5ag) A1 Find new speed V from conservation of momentum: m′V = mv with m′ = m + ¼m M1 V = 4v/5 = 4√(ag/5) or (4/5) √(5ag) AEF A1 Find w2 at angle θ from conservation of energy: ½ m′w2 = ½ m′V2 (condone m instead of m′ here since cancels out) – m′ga (1 + cos θ ) M1 A1 [ w2 = ag (6/5 – 2 cos θ )] S.R. Invalid energy method (max 2/5): ½ m′w2 = ½ mu2 [gives T ′ = (5mg/4)(2 – 3 cos θ ) ] + mga (1 – cos θ ) – ¼mga (1 + cos θ ) (B1) Find tension T ′ there by using F = ma radially: T ′ = m′w2/a – m′g cos θ B1 Eliminate w2 : = m′V2/a – m′g (2 + 3 cos θ ) A1 Substitute for m′ and V: = (5mg/4)(6/5 – 3 cos θ ) AEF or 3mg/2 – (15/4) mg cos θ A1 Find cos θ when string becomes slack from T ′ = 0: cos θ = ⅓ × 6/5 = 2/5 or 0⋅4 M1 A1 S.R. Allow if found from T ′ = mg (6/5 – 3 cos θ ) 2 4 5 2 13 5 Find or use sample mean x = 222⋅8 / 10 = 22⋅28 and estimate population variance: s2 = 4⋅12 / 9 (allow biased here: 0⋅412 or 0⋅6422) = 0⋅458 or 103/225 or 0⋅6772 M1 Find confidence interval (allow z in place of t) e.g.: 22⋅28 ± t √(0⋅458 / 10) M1 A1 Use of correct tabular value: t9, 0.975 = 2⋅26[2] A1 Evaluate C.I. correct to 3 s.f.: 22⋅3 ± 0⋅48[4] or [21⋅8, 22⋅8] A1 5 5

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2015 9231 23 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 6 (i) (ii) (iii) Find prob. p of head from mean = 2 × variance: 1/p = 2 × (1 – p)/p2, p = ⅔ A.G. M1 A1 Find P(X = 4) (denoting 1 – p by q [= ⅓]): P(X = 4) = q3 × p = 2/81 or 0⋅0247 B1 Find or state P(X > 4): P(X > 4) [= 1 – (1 + q + q2 + q3) × p = 1 – (1 – q4) ] = q4 = 1/81 or 0⋅0123 M1 A1 Formulate condition for N: 1 – qN > 0⋅999, [ (⅓)N < 0⋅001 ] M1 Take logs (any base) to give bound for N: N > log 0⋅001 / log ⅓ M1 Find Nmin: N > 6⋅29, Nmin = 7 A1 (N < 6⋅29 or N = 6⋅29 earns M2 A0) 2 1 2 3 8 7 (i) (ii) Find F(x) for 1 ⩽ x ⩽ 4: F(x) = (x3 – 1)/63 B1 Find G(y) from Y = X2 for 1 ⩽ x ⩽ 4: G(y) = P(Y < y) = P(X2 < y) = P(X < y1/2) = F(y1/2) (result may be stated) = (y3/2 – 1)/63 M1 A1 Find g(y) for corresponding range of y: g(y) = y1/2 /42 A.G. A1 Find or state corresponding range of y : 1 ⩽ y ⩽ 16 A.G. B1 Find median value m of Y: (m3/2 – 1)/63 = ½ m = 32⋅5 2/3 = 10⋅2 M1 A1 Find E(Y) [or equivalently E(X2)]: E(Y) = ∫ y g(y) dy = ∫ y3/2 dy /42 = [y5/2]1 16 /105 = 1023/105 = 341/35 or 9⋅74 M1 A1 5 2 2 9

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2015 9231 23 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 8 Find mean of sample data [for use in Poisson distn.]: λ = 220/100 = 2⋅2 B1 State (at least) null hypothesis (AEF): H0: Poisson distn. fits data or λ = 2⋅2 B1 Find expected values 100λre-λ/r! (to 1 d.p.): 11⋅080 24⋅377 26⋅814 19⋅664 (ignore incorrect final value here for M1) 10⋅8151 4⋅759 2⋅491 M1 A1 Combine last two cells so that exp. value ⩾ 5: Oi : 3 Ei : 7⋅25 M1* Calculate value of χ2 (to 2 d.p.; A1 dep M1*): χ2 = 0⋅076 + 2⋅879 + 0⋅653 + 1⋅448 + 0⋅441 + 2⋅491 (allow 7⋅95 if 1 d.p. exp.values used) = 7⋅99 M1 A1 State or use consistent tabular value (to 3 s.f.): 5 cells: χ3 ,0.95 2 = 7⋅815 6 cells: χ4, 0.95 2 = 9⋅488 (correct) 7 cells: χ5, 0.95 2 = 11⋅07 B1 State or imply valid method for conclusion e.g.: Accept H0 if χ2 < tabular value M1 Conclusion (AEF, requires both values correct): Distn fits or λ = 2⋅2 A1 Not combining cells [so χ2 = 8⋅64] can earn B1 B1 M1 A1 M0 M1 B1 M1 (max 7) 10 10 9 Calculate gradient b1 in y –y = b1(x –x) : Sxy = 24 879 – 472 × 400/8 = 1 279 Sxx = 29 950 – 4722/8 = 2 102 b1 = Sxy / Sxx = 0⋅6085 (3 s.f.) M1 A1 Find regression line of y on x: y = 400/8 + b1 (x – 472/8) M1 A1 = 50 + 0⋅6085 (x – 59) = 0⋅6085x + 14⋅1 Find y when x = 72: = 57⋅9 or 58 Allow use of regression line of x on y (since neither variable clearly independent): Syy = 21 226 – 4002/8 = 1 226 b2 = Sxy / Syy = 1⋅043 (M1 A1) x = 472/8 + b2 (y – 400/8) (M1 A1) = 1⋅043 y + 6⋅85 Y = 62⋅5 or 62 (A1) 5

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2015 9231 23 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total Find product moment correlation coefficient r: r = 1 279 / √(2 102 × 1 226) or √(0⋅6085 × 1⋅043) = 0⋅797 M1 A1* State both hypotheses (B0 for r …): H0: ρ = 0, H1: ρ ≠ 0 B1 State or use correct tabular two-tail r-value: r8, 5% = 0⋅707 B1* State or imply valid method for conclusion e.g.: Reject H0 if |r| > tab. value (AEF) M1 Correct conclusion (AEF, dep A1*, B1*): There is non-zero correlation A1 6 11 10A Find MI of lamina about Q: Ilamina = ⅓m{(3a)2 + (3a/2)2} + m(9a/2)2 M1 A1 [= (15/4 + 81/4) ma2 = 24 ma2] State or find MI of rod about Q: Irod = (⅓ + 1) M (3a/2)2 [= 3Ma2] B1 Sum to find MI of object about Q: I1 = 24 ma2 + 3 Ma2 = 3 (8m + M) a2 A.G. A1 Find MI of object about mid-point of PQ: I2 = (15/4 + 32) ma2 + ⅓ M (3a/2)2 = (51/4) ma2 + ¾ Ma2 = ¾ (17m + M) a2 A.G. M1 A1 Use eqn of circular motion to find d2θ/dt2 for axis l1: [–]I1 d2θ/dt2 = mg × (9a/2) sin θ + Mg × (3a/2) sin θ M1 A1 [ = (9m/2 + 3M/2) ga sin θ ] [Approximate sin θ by θ and] find ω1 2 in SHM eqn: ω1 2 = (3m + M)g / 2(8m + M) a M1 Find period T1 for axis l1 from 2π/ω1: (AEF) T1 = 2π√{2(8m + M) a / (3m + M)g} A1 Use eqn of circular motion to find d2θ/dt2 for axis l2: [–]I2 d2θ/dt2 = mg × 3a sin θ M1 [Approximate sin θ by θ and] find ω2 2 in SHM eqn: ω2 2 = 4mg / (17m + M) a M1 Find period T2 for axis l2 from 2π/ω2: (AEF) T2 = 2π√{(17m + M) a / 4mg} A1 Verify that T1 = T2 when m = M: (AEF) T1 = 2π√(18 a / 4g) = T2 B1 [Taking m = M throughout 2nd part can earn: M1 A1 M1 A0 M1 M1 A0 B1 (max 6/8)] 4 2 8 14

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2015 9231 23 © Cambridge International Examinations 2015 Question Number Mark Scheme Details Part Mark Total 10B State hypotheses (B0 forx …), e.g.: H0: µX = µY , H1: µX ≠ µY B1 State assumption (AEF): Distributions have equal variances B1 Find sample means and estimate popln. variances: x = 4⋅2, y = 4⋅8 sX 2 = (180 – 422/10) / 9 (allow biased here: 0⋅36 or 0⋅62) = 0⋅4 or 0⋅63252 sY 2 = (281⋅5 – 57⋅62/12) / 11 (allow biased here: 0⋅4183 or 0⋅64682) = 0⋅4564 or 251/550 or 0⋅67552 M1 Estimate (pooled) common variance: s2 = (9 sX 2 + 11 sY 2) / 20 (AEF) (note sX 2 and sY 2 not needed explicitly) or (180 – 422/10 + 281⋅5 – 57⋅62/12) / 20 = 0⋅431 or 0⋅65652 M1 A1 Calculate value of t (to 3 s.f.): [-] t = (y –x) / s √(1/10 + 1/12) = 2⋅13 M1 A1 State or use correct tabular t value: t20, 0.975 = 2⋅086 [allow 2⋅09] B1* (or can comparey –x = 0⋅6 with 0⋅586) Correct conclusion (AEF, √ on t, dep *B1): t > 2⋅09 so mean masses not same B1 S.R. Implicitly taking sX 2, sY 2 as popln. variances: z = ( y – x ) / √(sX 2/10 + sY 2/12) (may also earn first B1 B1 M1) = 0⋅6 / √(0⋅078 = 2⋅15 Comparison with z0.975 and conclusion: 2⋅15 > 1⋅96 (can earn at most 5/9) so mean masses not same (B1) State hypotheses (B0 forx …), e.g.: H0: µX = 3⋅8 , H1: µX > 3⋅8 or H0: µX = µZ, H1: µX > µZ B1 Calculate value of t using sX from above: t = (4⋅2 – 3⋅8) / (sX /√10) = 2⋅0 M1 A1 State or use correct tabular t value: t9, 0.95 = 1⋅833 [allow 1⋅83] B1* (or can compare 0⋅4 with 0⋅367) Correct conclusion (A.E.F., √ on t, dep *B1): t > 1⋅833, so claim is justified or mean mass of Royals > mean mass of Crowns B1 9 5 14