Cambridge A Level Mathematics - Further 9231 — 2014 Oct/Nov Paper 1 · Variant 1

*1234588211* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/11 Paper 1 October/November 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 11_9231_11/RP © UCLES 2014 [Turn over

2 1 Given that uk = 1
2k −1 − 1
2k + 1, express n
k=13 uk in terms of n. [4] Deduce the value of ∞
k=13 uk. [1] 2 A curve C has parametric equations x = et cost, y = et sin t, for 0 ≤t ≤1 2. Find the arc length of C. [6] 3 It is given that ur = r × r! for r = 1, 2, 3,  . Let Sn = u1 + u2 + u3 +  + un. Write down the values of 2! −S1, 3! −S2, 4! −S3, 5! −S4. 2 Conjecture a formula for Sn. [1] Prove, by mathematical induction, a formula for Sn, for all positive integers n. [4] 4 A curve C has equation y = 2x2 + x −1 x −1 . Find the equations of the asymptotes of C. [3] Show that there is no point on C for which 1 < y < 9. [4] 5 Find the value of a for which the system of equations x −y + 2 = 4, x + ay −3 = b, x −y + 7 = 13, where a and b are constants, has no unique solution. [3] Taking a as the value just found, (i) find the general solution in the case b = −5, [4] (ii) interpret the situation geometrically in the case b ≠−5. [1] 6 Use de Moivre’s theorem to show that cos 5 cos 16 sin4 −12 sin2 + 1. 5 By considering the equation cos 5 = 0, show that the exact value of sin2 1 10 is 3 − 5 8 . [4] © UCLES 2014 9231/11/O/N/14

3 7 Let In = 1 0 1 −xnex dx. Show that, for all positive integers n, In = nIn−1 −1. 3 Find the exact value of I4. [4] By considering the area of the region enclosed by the x-axis, the y-axis and the curve with equation y = 1 −x4ex in the interval 0 ≤x ≤1, show that 65 24 < e < 11 4 . 3 8 A circle has polar equation r = a, for 0 ≤ < 2, and a cardioid has polar equation r = a1 −cos , for 0 ≤ < 2, where a is a positive constant. Draw sketches of the circle and the cardioid on the same diagram. [3] Write down the polar coordinates of the points of intersection of the circle and the cardioid. [2] Show that the area of the region that is both inside the circle and inside the cardioid is 5 4 −2a2. 6 9 Given that xd2y dx2 + 2x + 2dy dx + 2 −3xy = 10e2x and that v = xy, show that d2v dx2 + 2dv dx −3v = 10e2x. 4 Find the general solution for y in terms of x. [7] 10 The line l1 is parallel to the vector i −2j −3k and passes through the point A, whose position vector is 3i + 3j −4k. The line l2 is parallel to the vector −2i + j + 3k and passes through the point B, whose position vector is −3i −j + 2k. The point P on l1 and the point Q on l2 are such that PQ is perpendicular to both l1 and l2. Find (i) the length PQ, [5] (ii) the cartesian equation of the plane containing PQ and l2, [4] (iii) the perpendicular distance of A from . [3] © UCLES 2014 9231/11/O/N/14 [Turn over

4 11 Answer only one of the following two alternatives. EITHER The roots of the quartic equation x4 + 4x3 + 2x2 −4x + 1 = 0 are , ,  and . Find the values of (i) +  +  + , [1] (ii) 2 + 2 + 2 + 2, [2] (iii) 1 + 1  + 1  + 1  , [2] (iv)  +   +   +   . [2] Using the substitution y = x + 1, find a quartic equation in y. Solve this quartic equation and hence find the roots of the equation x4 + 4x3 + 2x2 −4x + 1 = 0. [7] OR The square matrix A has  as an eigenvalue with e as a corresponding eigenvector. Show that if A is non-singular then (i)  ≠0, [2] (ii) the matrix A−1 has −1 as an eigenvalue with e as a corresponding eigenvector. [2] The 3 × 3 matrices A and B are given by A =  −2 2 −4 0 −1 5 0 0 3 and B = A + 3I−1, where I is the 3 × 3 identity matrix. Find a non-singular matrix P, and a diagonal matrix D, such that B = PDP−1. [10] 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 9231/11/O/N/14

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the October/November 2014 series 9231 MATHEMATICS 9231/11 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 October/November 2014 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 2014 9231 11 © 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 Cambridge International A Level – October/November 2014 9231 11 © 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 Cambridge International A Level – October/November 2014 9231 11 © Cambridge International Examinations 2014 Qn & Part Solution Marks 1 ( ) ( ) ( ) 1 2 1 1 2 1 29 1 27 1 27 1 25 1 ... + −− + + − + − n n ∑= + − = n r n k u 13 1 2 1 5 1 ∑ ∞ = = 13 5 1 r k u M1A1 M1A1 (4) B1 (1) [5] 2 t t x t t sin e cos e − = & t t y t t cos e sin e + = & ( ) t t t t y x 2 2 2 2 2 2 e 2 sin cos e 2 = + = + & & ∫ = π 2 1 0 d e 2 t s t [ ] ( )1 2 e 2 2 1 2 1 0 − = = π π e t ( = 5.39) M1A1 B1 M1 A1A1 (6) [6] 3 2! – S1 = 1 , 3! – S2 = 1 , 4! – S3 = 1 , 5! – S4 = 1 (Two correct B1, all four correct B2) Sn = (n +1)! – 1 2! – 1 = 2 – 1 = 1 ⇒ H1 is true. Hk: Sk = (k +1)! – 1 (k +1)! – 1 + (k + 1) × (k + 1)! = (k + 1)!(1 + k +1) – 1 = ([k +1] + 1)! – 1 Hence Hk ⇒ Hk+1 So result holds for all positive integers (by PMI). B2,1,0 (2) B1 (1) B1 B1 M1 A1 (4) [7] 4 Vertical asymptote is x = 1 ( ) 3 2 1 2 3 2 1 + = ⇒ − + + = − x y x x y is the oblique asymptote. ( ) ( ) 0 1 1 2 2 = − + − + y x y x has real roots ( ) ( )1 8 1 2 − − − ⇔ y y > 0 9 10 2 + − ⇔ y y > 0 ( )( ) 9 1 − − ⇔ y y > 0 Hence ( )( ) ⇒ < − − 0 9 1 y y no real roots. i.e. ⇒ < < 9 1 y no points on C. (AG) Thumbnail sketch, or similar, required. B1 M1A1 (3) M1A1 M1 A1 (4) [7]

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 11 © Cambridge International Examinations 2014 5 0 7 1 1 3 1 1 2 1 1 = − − − − 1 0 5 5 − = ⇒ = + ⇒ a a           − → →           − − − − − 0 9 4 | | | 0 0 0 5 0 0 2 1 1 ... 13 5 4 | | | 7 1 1 3 1 1 2 1 1 , or by elimination methods, x – y + 2z = 4 5z = 9           +           =           ⇒ 0 1 1 8.1 0 4.0 t z y x (OE) Planes form a prism, or words to that effect. M1 A1A1 (3) M1 A1 M1A1 (4) B1 (1) [8] 6 ( ) ( ) ( ) ( ) ( )5 4 3 2 2 3 4 5 5 i i 5 i 10 i 10 i 5 i s s s c s c s c c s c + + + + + = + 4 2 3 5 5 10 5 cos cs s c c + − = θ ( ) ( )     + − − − = 4 2 2 2 2 5 1 10 1 s s s s c ( ) ( )1 12 16 5 10 10 2 1 2 4 4 4 2 4 2 + − = + + − + − = s s c s s s s s c (AG) Alternative: using ( ) θ n z z n cos 2 1 = + − ( ) θ θ θ θ cos 20 3 cos 10 5 cos 2 cos 2 5 + + = θ θ θ θ cos 10 3 cos 5 5 cos cos 16 5 + + = ⇒ But θ θ θ cos 3 cos 4 3 cos 3 − = (Can be quoted, if known.) ( ) 5 cos 20 cos 16 cos 5 cos 2 4 + − = ⇒ θ θ θ θ Uses θ θ 2 2 sin 1 cos − = to obtain cos ( )1 12 16 5 2 4 + − = s s c θ (AG) π π π π π θ θ 10 9 10 7 2 1 10 3 10 1 , , , , 0 5 cos = ⇒ = 8 5 3 32 64 144 12 2 ± = − ± = s Since ( ) ( ) ( ) 8 5 3 10 1 2 10 3 2 10 1 2 sin , sin sin − = < π π π (Or by any other valid method.) (AG) Justification required for final mark. B1 M1A1 M1 A1 (5) (M1) (B1) (A1) (M1A1) B1 M1A1 A1 (4) [9]

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 11 © Cambridge International Examinations 2014 7 ( ) [ ] ( ) ∫ − − + − = 1 0 1 1 0 d e 1 e 1 x x n x I x n x n n = 0 – 1 + nIn–1 = nIn–1 – 1 [ ] 1 e d e 1 0 1 0 0 − = = = ∫ x x e x I 41 24 17 24 5 12 1 4 0 1 2 3 4 − = − = − = − = I I I I I 65 e 24 − = Or for the last 3 marks: I1 = e – 2, I2 = 2e – 5 I3 = 6e – 16 and I4 = 24e – 65 Draws appropriate sketch, or explains clearly. (Must be seen.) 1 65 e 24 0 1 graph below Area 0 < − < ⇒ < < Hence 4 11 24 65 24 66 24 65 e e < < ⇒ < < M1A1 A1 (3) B1 M1A1 A1 (M1A1) (A1) (4) B1 M1 A1 (3) [10] 8 Circle sketched Cardioid – correct location and orientation – correct indentation near pole. ( ) 2 , π a and ( ) 2 3, π a (B1 for reverse, or ( ) 2 3 2 , cos 1 π π θ θ = ⇒ − = a a seen.) Area ( ) θ θ π π d cos 1 2 2 2 0 2 2 1 2 2 1 − × + = ∫ a a (Half circle + Area of sector) ( ) ∫ + − + = 2 0 2 2 2 2 1 d cos cos 2 1 π θ θ θ π a a ( ) ∫ + − + = 2 0 2 1 2 3 2 2 2 1 d 2 cos cos 2 π θ θ θ π a a (Use of double angle formula.) [ ] 2 0 4 1 2 3 2 2 2 1 2 sin sin 2 π θ θ π θ + − + = a a (Integration) ( ) ( ) 2 4 5 4 3 2 2 2 1 2 2 a a a − = − + = π π π (AG) B1 B1B1 (3) B1B1 (2) B1M1 A1 M1 M1 A1 (6) [11]

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 11 © Cambridge International Examinations 2014 9 2 2 2 2 d d d d 2 d d d d d d x y x x y x v x y x y x v xy v + = ⇒ + = ⇒ = ( ) ( ) x y x x y x x y x 2 2 2 e 10 3 2 d d 2 2 d d = − + + + x x v x v x v e xy x y x y x y x x y 2 2 2 2 2 2 e 10 3 d d 2 d d 10 3 d d 2 d d d d 2 = − + ⇒ = −       + + + ⇒ ( )( ) 1 ,3 0 1 3 0 3 2 2 − = ⇒ = − + ⇒ = − + m m m m m CF: x x B A e e 3 + − x x x k v k v k v 2 2 2 e 4 e 2 e = ′′ ⇒ = ′ ⇒ = x x x x x k k k k k 2 2 2 2 2 e 2 : PI 2 10 5 e 10 e 3 e 4 e 4 ⇒ = ⇒ = ⇒ = − + ⇒ GS: ( ) x x x x x x B A x y B A v 2 3 2 3 e 2 e e 1 e 2 e e + + = ⇒ + + = − − B1B1 M1A1 (4) M1 A1 M1 M1A1 M1A1 (7) [11]

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 11 © Cambridge International Examinations 2014 10 (i) (ii) (iii)           −           − − = − − − 1 1 1 k j i ~ 3 3 3 3 1 2 3 2 1 k j i 6 4 6 − + = BA Shortest distance 2 2 2 6 1 4 . 1 6 1 1 1 1 4 3       −       −    + + = ( = 2.31) Alternative 16 14 13 0 3 2 1 . 3 3 6 2 4 2 6 = − − ⇒ =           − −           + + + + − − − − µ λ µ λ µ λ µ λ 26 113 14 0 3 1 2 . 3 3 6 2 4 2 6 − = + ⇒ =          −           + + + + − − − − µ λ µ λ µ λ µ λ 52 38 9 9 , λ µ ⇒ = − = Shortest distance 4 3 = . (= 2.31)                     − − − = − − 1 5 4 ~ 1 5 4 3 1 2 1 1 1 k j i Cartesian equation of Π: 15 2 5 12 5 4 − = + − − = + + z y x Distance of A from Π: 2 2 2 1 5 4 1 5 4 . 6 4 6 + +                     − 38 42 = ( = 5.86) M1A1 B1 M1A1 (5) (M1) (A1) (M1A1) (A1) M1A1 M1A1 (4) M1A1 A1 (3) [12]

Page 9 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2014 9231 11 © Cambridge International Examinations 2014 11E (i) (ii) (iii) (iv) 4 − = + + + δ γ β α ( ) 12 2 2 4 2 2 2 2 2 = × − − = + + + δ γ β α ( ) 4 1 4 1 1 1 1 = − − = + + + δ γ β α 12 1 12 2 2 2 2 = = + + + = + + + αβγδ δ γ β α αβγ δ αβδ γ αγδ β βγδ α 1 1 − = ⇒ + = y x x y ( ) ( ) 3 8 6 1 4 1 2 4 3 4 − + − = − + − y y y y y ( ) ( ) 7 8 2 1 1 4 1 2 2 2 + − = + − − − y y y y 0 4 4 1 4 2 4 2 4 2 3 4 = + − = + − + + ⇒ y y x x x x ( ) 2 0 2 2 2 ± = ⇒ = − y y (twice). 1 2 − ± = ⇒x (twice). (Some indication of four roots for final mark.) B1 (1) M1A1 (2) M1A1 (2) M1A1 (2) M1A1 A1 A1 A1 M1A1 (7) [14] 11O (i) (ii) Ae = λe; since A is non-singular Ae ≠ 0 ⇒ λ ≠ 0 (e ≠ 0) . Ae = λe ⇒ A–1Ae = A–1λe ⇒ e = λA–1e ⇒ e e A λ 1 1 = − Eigenvalues of A are –2 , –1 , 3 (1 mark for any one, 2 marks for all three.) Corresponding eigenvectors are:          −                     20 25 6 , 0 1 2 , 0 0 1 (M1A1 for one, A1 for each other.) (N.B. May come from using eigenvalues of A + 3I.)           − = 20 0 0 25 1 0 6 2 1 P Eigenvalues of A + 3I are 1 , 2 , 6 (Award B3 if obtained from A + 3I.) Eigenvalues of (A + 3I)–1 are 6 1 2 1 , ,1           = 6 1 2 1 0 0 0 0 0 0 1 D (CAO) M1A1 (2) M1 A1 (2) B2,1,0 M1A1 A1 A1 B1 B1 B1 B1 (10) [14]