Cambridge A Level Mathematics - Further 9231 — 2014 May/June Paper 1 · Variant 2

*2987069914* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/12 Paper 1 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. 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 4 printed pages. JC14 06_9231_12/FP © UCLES 2014 [Turn over

2 1 The equation x3 + px + q = 0, where p and q are constants, with q ≠0, has one root which is the reciprocal of another root. Prove that p + q2 = 1. [5] 2 Expand and simplify r + 14 −r4. [1] Use the method of differences together with the standard results for n Ð r=1 r and n Ð r=1 r2 to show that n Ð r=1 r3 = 1 4n2n + 12. 4 3 Prove by mathematical induction that, for all non-negative integers n, 112n + 25n + 22 is divisible by 24. [6] 4 Obtain the general solution of the differential equation d2x dt2 −6dx dt + 25x = 195 sin 2t. 6 5 The curve C has polar equation r = a1 + sin 1, where a is a positive constant and 0 ≤1 < 20. Draw a sketch of C. [2] Find the exact value of the area of the region enclosed by C and the half-lines 1 = 1 30 and 1 = 2 30. [4] 6 The linear transformation T : >4 →>4 is represented by the matrix M, where M = € 2 −1 1 3 2 0 0 5 6 −2 2 11 10 −3 3 19  . (i) Find the rank of M and state a basis for the range space of T. [4] (ii) Obtain a basis for the null space of T. [4] 7 Use de Moivre’s theorem to show that tan 51 = 5t −10t3 + t5 1 −10t2 + 5t4 , where t = tan 1. [4] Deduce that the roots of the equation t4 −10t2 + 5 = 0 are ± tan 1 50 and ± tan 2 50. [3] Hence show that tan 1 50 tan 2 50 = ï5. [2] © UCLES 2014 9231/12/M/J/14

3 8 The curve C has parametric equations x = t2, y = t −1 3t3, for 0 ≤t ≤1. Find (i) the arc length of C, [5] (ii) the surface area generated when C is rotated through 20 radians about the x-axis. [5] 9 The matrix M, where M = ` −2 2 2 2 1 2 −3 −6 −7 a , has an eigenvector ` 0 1 −1 a . Find the corresponding eigenvalue. [2] It is given that if the eigenvalues of a general 3 × 3 matrix A, where A = ` a b c d e f g h i a , are ,1, ,2 and ,3 then ,1 + ,2 + ,3 = a + e + i and the determinant of A has the value ,1,2,3. Use these results to find the other two eigenvalues of the matrix M, and find corresponding eigenvectors. [8] 10 It is given that In = Ô 1 40 0 sin2nx cosx dx, where n ≥0. Show that In −In+1 = 2− n+1 2 ! 2n + 1 . 5 Hence show that Ô 1 40 0 sin6x cosx dx = ln1 + ï2 −73 120ï2. [5] 11 The line l1 passes through the points A 2, 3, −5 and B 8, 7, −13. The line l2 passes through the points C −2, 1, 8 and D 3, −1, 4. Find the shortest distance between the lines l1 and l2. [5] The plane 1 passes through the points A, B and D. The plane 2 passes though the points A, C and D. Find the acute angle between 1 and 2, giving your answer in degrees. [6] © UCLES 2014 9231/12/M/J/14 [Turn over

4 12 Answer only one of the following two alternatives. EITHER The curve C has parametric equations x = t2, y = 2 −t 1 2, for 0 ≤t ≤2. Find (i) d2y dx2 in terms of t, [5] (ii) the mean value of y with respect to x over the interval 0 ≤x ≤4, [6] (iii) the y-coordinate of the centroid of the region enclosed by C, the x-axis and the y-axis. [3] OR The curve C has equation y = ax2 + bx + c x + d , where a, b, c and d are constants. The curve cuts the y-axis at 0, −2 and has asymptotes x = 2 and y = x + 1. (i) Write down the value of d. [1] (ii) Determine the values of a, b and c. [6] (iii) Show that, at all points on C, either y ≤3 −2ï6 or y ≥3 + 2ï6. [7] 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 9231/12/M/J/14

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2014 series 9231 FURTHER MATHEMATICS 9231/12 Paper 1, 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 12 © 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 12 © 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 12 © Cambridge International Examinations 2014 Qn & Part Solution Marks 1 Let roots be α, α–1 and β ⇒ α + α–1 + β = 0 (1) Any of three for M1 Product of roots ⇒ β = –q (2) A1 for another. Sum of products in pairs ⇒ 1+ β (α + α–1) = p (3) A1 for a third From (1) and (3) 1 – β2 = p Wrong sign in (2) scores Using (2) 1 – q2 = p or p + q2 = 1 M1A0A1M1A0 M1 A1 A1 M1 A1 [5] 2 (r + 1)4 – r4 = 4r3 + 6r2 + 4r + 1 (n + 1)4 – 14 = n r 1 4 = Σ r3 + n r 1 6 = Σ r2 + n r 1 4 = Σ r + n n4 +4n3 + 6n2 + 4n = n r 1 4 = Σ r3 + n(2n2 + 3n + 1) + 2n2 + 2n + n ⇒ … ⇒ n r 1 = Σ r3 = 4 1 n2(n + 1)2. (AG) B1 [1] M1 A1A1 A1 [4] 3 Hk: f(k) = 112k + 25k + 22 = 24λ f(0) = 1 + 1 + 22 = 24 = 1 × 24 ⇒H
is true. f(k + 1) – f(k) = 112k + 2 + 25k + 1 + 22 – (112k + 25k + 22) = 112k(121 – 1) + 25k(25 – 1) = 112k × 24 × 5 + 25k × 24 = 24µ Alternatively: f(k + 1) = 112k + 2 + 25k + 1 + 22 = 121.112k + 25.25k + 22 = (120 + 1)112k + (24 + 1)25k + 22 (OE) = 120.112k + 24.25k + 24λ = 24µ ⇒ f(k + 1) = 24µ + 24λ = 24(µ + λ) ⇒ Hk + 1 is true. Hence by PMI Hn is true for all non-negative integers. (Must see non-negative integers.) CSO: Final mark requires all previous marks. B1 B1 M1 A1 A1 (M1) (A1) (A1) A1 [6] 4 m2 – 6m + 25 = 0 ⇒ m = 3 ± 4i CF: x = e3t (A cos 4t + B sin 4t) PI: x – p sin 2t + q cos 2t ⇒x& = 2p cos 2t – 2q sin 2t ⇒x&& = –4p sin 2t – 4q cos 2t ⇒ 21p + 12q = 195 –12p + 21q = 0 ⇒ p = 7, q = 4 GS: x = e3t (A cos 4t + B sin 4t) + 7 sin 2t + 4 cos 2t M1 A1 M1 M1 A1 A1 [6]

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 12 © Cambridge International Examinations 2014 Qn & Part Solution Marks 5 Closed curve through (a, 0),       π 2 1 , 2a , (a, π),       π 2 3 ,0 . Completely correct shape for cardioid. Area = ∫ ∫ + + = + π π π π θ θ θ θ θ 3 2 3 1 2 3 2 3 1 2 2 2 d ) sin sin 2 1( 2 d ) sin 1( 2 a a (LNR) = ∫       − + π π θ θ θ 3 2 3 1 2 d 2 cos 2 1 sin 2 2 3 2 a (LR) = π π θ θ θ 3 2 3 1 2 2 sin 4 1 cos 2 2 3 2     − − a =       + + 3 8 1 1 4 1 2 π a (CAO) B1 B1 [2] M1 M1 M1 A1 [4] 6             − − − 19 3 3 10 11 2 2 6 5 0 0 2 3 1 1 2 →…→             − − 0 0 0 0 0 0 0 0 2 1 1 0 3 1 1 2 R(M) = 4 – 2 = 2 Basis for range space consists of any two independent column vectors from matrix M. 2x – y + z + 3t = 0 y – z + 2t = 0 t = λ, z = µ, y = µ – 2λ, x = – 2 5 λ (OE) Basis for null space is                                     − − 0 1 1 0 , 2 0 4 5 (OE) M1A1 A1 A1 [4] M1 M1 A1A1 [4]

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 12 © Cambridge International Examinations 2014 Qn & Part Solution Marks 6 Alternative Solution Using Transpose Matrix MT =             − − − 19 11 5 3 3 2 0 1 3 2 0 1 10 6 2 2 →…→             0 0 0 0 0 0 0 0 4 2 2 0 10 6 2 2 R(M) = 4 – 2 = 2 Basis for range space consists of any two independent column vectors from matrix M. N.B. The following may appear in the first line, possibly in a 4 × 3 matrix. Using the same set of elementary row operations on the matrix:             d c b a             d c b a →…→             + − − + + d b a c b b a a 2 4 5 2 Basis for null space is                                     − − 0 1 1 0 , 2 0 4 5 (OE) M1A1 A1 A1 [4] M1 M1 A1A1 [4] 7 (cos θ + i sin θ)5 = cos 5θ + i sin 5θ = c5 + 5c4(is) + 10c3(is)4 + 10c2(is)3 + 5c(is)4 + (is)5 tan 5θ = θ θ 5 cos 5 sin = 4 2 3 5 5 3 2 4 5 10 10 5 cs s c c s s c s c + − + − =…= 4 2 5 3 5 10 1 10 5 t t t t t + − + − (AG) tan 5θ = 0 ⇒ t(t4 – 10t2 + 5) = 0 (Allow non-factorised form.) Rejecting t = 0, which implies θ = kπ, roots of t4 – 10t2 + 5 = 0 are: ±tan 5 1 π and ±tan 5 2 π, since 5θ = ±2π or 5θ = ±4π (AG) Product of roots = tan2 5 1 πtan2 5 2 π = 5 ⇒ tan 5 1 π tan 5 2 π = 5 (S.C. Award B1 if result deduced from information given in the question.) M1 A1 M1A1 [4] M1 A1 A1 [3] M1A1 [2]

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 12 © Cambridge International Examinations 2014 Qn & Part Solution Marks 8 (i) (ii) t x 2 = & 2 1 t y − = & 2 2 4 2 2 2 2 ) 1( 2 1 4 t t t t y x + = + − + = + & & ACF ∫ + = 1 0 2) 1( t s dt = 1 0 3 3      + t t = 3 4 or 1.33 LR ( ) ∫ ∫       − + = +      − = 1 0 1 0 5 3 2 3 d 3 1 3 2 2 d 1 3 1 2 t t t t t t t t S π π LNR 1 0 6 4 2 18 1 6 1 2 1 2     − + = t t t π LR π 9 11 = or 3.84 B1 M1A1 M1A1 [5] M1A1 M1A1 A1 [5] 9 ( ) 1 1 eigenvalue 1 1 0 1 1 0 λ = − = ⇒           − =           − M λ1 + λ2 + λ3 = –8 ⇒ λ2 + λ3 = –7 12 12 18 16 10 7 6 3 2 1 2 2 2 2 3 2 3 2 = ⇒ − = − + − = − − − − = − λ λ λ λ λ2 = –3 and λ3 = –4 or vice versa. (S.C. Award M1A1 for λ3 + 8λ2 + 19λ + 12 = 0 and A1 for both λ2 = –3 and λ3 = –4.)           − = ⇒ − = 0 1 2 3 2 2 e λ (OE)           − = ⇒ − = 1 0 1 4 3 3 e λ (OE) M1A1 [2] M1 M1A1 A1A1 M1A1 A1 [8]

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 12 © Cambridge International Examinations 2014 Qn & Part Solution Marks 10 ∫ + + = π x x x 4 1 0 2 n 2 1 n d cos sin I (LR) ∫ − = + π 4 1 0 2 2 1 d cos sin ) cos 1( x x x x I n n (LR) ∫ − = π 4 1 0 2 d sin cos x x x I n n π 4 1 0 1 2 1 2 sin       + − = + n x I n n . 1 2 2 1 2 1 1 + = − ⇒      + − + n I n n n (AG) [ ] ( ) 2 1 ln tan sec ln d sec 4 1 0 0 0 4 1 + = + = =∫ π π x x x x I ( ) 2 1 2 1 ln 1 − + = I ( ) 2 6 1 2 1 2 1 ln 2 − − + = I ( ) ( ) 120 2 73 2 1 ln 2 20 1 2 6 1 2 1 2 1 ln 3 − + = − − − + = I (AG) B1 M1 A1 M1 A1 [5] M1A1 M1 A1 A1 [5]

Page 9 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 12 © Cambridge International Examinations 2014 Qn & Part Solution Marks 11           − − =           − = 4 2 5 8 4 6 CD AB Common perpendicular is                     − − − = − − − 2 1 2 ~ 16 8 16 4 2 5 4 2 3 k j i ⇒           − = 9 4 1 AD shortest distance = 3 16 4 1 4 2 1 2 9 4 1 = + +           ⋅           − or 5.33 Normal to Π1 is          − = − − 14 31 2 4 2 3 9 4 1 k j i Normal to Π2 is           = − − − 18 49 34 4 2 5 9 4 1 k j i 2 2 2 2 2 2 18 49 34 14 31 2 18 14 49 31 34 2 cos + + + + × + × + × − = θ ⇒ θ = 36.7° (CAO) B1 M1A1 M1A1 [5] M1A1 A1 M1A1 A1 [6]

Page 10 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2014 9231 12 © Cambridge International Examinations 2014 Qn & Part Solution Marks 12E (i) (ii) (iii) . ) 2 ( 4 1 d d ) 2 ( 2 1 , 2 2 1 2 1 t t x y t y t x − − = ⇒ − − = = − & & t t t t t t x y 2 1 ) 2 ( 2 4 ) 2 ( 4 ) 2 ( 4 d d 2 1 2 1 2 2 1 2 2 ×       − − −       − = − (OE) ) 3 4 ( ) 2 ( 16 1 2 3 3 t t t − − = (Any correct simplified form.) ∫ ∫ − = 4 0 2 0 2 1 d ) 2 ( 2 d t t t x y       − +         − − = ∫ 2 0 2 3 2 0 2 3 d ) 2 ( 3 2 ) 2 ( 3 2 2 t t t t (LNR) 2 15 32 2 5 ) 2 ( 15 4 0 2 2 0 =           − − + = t (LR) MV 2 15 8 2 15 32 4 1 0 4 d 4 0 = × = − = ∫ x y or 0.754. ∫ ∫ =     − = − = 4 0 2 0 2 0 3 2 2 3 4 3 1 d ) 2 ( 2 2 1 d 2 1 t t t t t x y or 1.33. 2 16 5 2 8 5 2 32 15 3 4 d d 2 1 4 0 4 0 2 = = × = = ∫ ∫ x y x y y or 0.442. M1A1 M1A1 A1 [5] M1A1 M1A1 M1A1 [6] M1A1 A1 [3] 12O (i) (ii) (iii) d = –2 x = 0 ⇒ y = –2 = d c ⇒ c = 4 y = ax + (b + 2a) + 2 2 4 4 − + + x b a Oblique asymptote y = x + 1 ⇒ a = 1, b + 2a = 1 ⇒ b = –1 0 4 2 )1 ( 2 4 2 2 = + + + − ⇒ − + − = y x y x x x x y For real x B2 – 4AC > 0 ⇒ (y + 1)2 – 4(2y + 4) > 0 ⇒ y2 – 6y – 15 > 0 ⇒ (y – 3)2 > 24 (or for solving y2 – 6y – 15 = 0) ⇒ y – 3 < – 6 2 or y – 3 > – 6 2 (or for thumbnail sketch) ⇒ y < 3 – 6 2 or y > 3 + 6 2 (AG) (Note: if done by differentiation, max and min must be demonstrated for full marks.) B1 [1] M1A1 M1A1 A1A1 [6] M1 M1 A1 M1A1 M1 A1 [7]