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

*3084104856* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/12 Paper 1 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. 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 and 1 insert. JC16 06_9231_12/FP © UCLES 2016 [Turn over

2 1 The roots of the cubic equation 2x3 + x2 −7 = 0 are !, " and '. Using the substitution y = 1 + 1 x, or otherwise, find the cubic equation whose roots are 1 + 1 !, 1 + 1 " and 1 + 1 ' , giving your answer in the form ay3 + by2 + cy + d = 0, where a, b, c and d are constants to be found. [4] 2 Express 4 rr + 1r + 2 in partial fractions and hence find n Ð r=1 4 rr + 1r + 2. [5] Deduce the value of ∞ Ð r=1 4 rr + 1r + 2. [1] 3 Prove by mathematical induction that, for all positive integers n, 10n + 3 × 4n+2 + 5 is divisible by 9. [6] 4 A curve C has polar equation r2 = 8 cosec 21 for 0 < 1 < 1 20. Find a cartesian equation of C. [3] Sketch C. [2] Determine the exact area of the sector bounded by the arc of C between 1 = 1 60 and 1 = 1 30, the half-line 1 = 1 60 and the half-line 1 = 1 30. [3] [It is given that Ó cosecx dx = ln tan 1 2x + c.] 5 Let In = Ó 1 20 0 cosnx sin2x dx, for n ≥0. By differentiating cosn−1x sin3x with respect to x, prove that n + 2In = n −1In−2 for n ≥2. [5] Hence find the exact value of I4. [4] 6 Use de Moivre’s theorem to express cot 71 in terms of cot 1. [4] Use the equation cot 71 = 0 to show that the roots of the equation x6 −21x4 + 35x2 −7 = 0 are cot 1 14k0 for k = 1, 3, 5, 9, 11, 13, and deduce that cot2 1 140 cot2 3 140 cot2 5 140 = 7. [5] © UCLES 2016 9231/12/M/J/16

3 7 A curve C has equation y = x2 x −2. Find the equations of the asymptotes of C. [3] Show that there are no points on C for which 0 < y < 8. [4] Sketch C, giving the coordinates of the turning points. [3] 8 Find a cartesian equation of the plane 1 passing through the points with coordinates 2, −1, 3, 4, 2, −5 and −1, 3, −2. [4] The plane 2 has cartesian equation 3x −y + 2z = 5. Find the acute angle between 1 and 2. [3] Find a vector equation of the line of intersection of the planes 1 and 2. [4] 9 Find the value of the constant k such that y = kx2e2x is a particular integral of the differential equation d2y dx2 −4dy dx + 4y = 4e2x. * [4] Hence find the general solution of *. [3] Find the particular solution of * such that y = 3 and dy dx = −2 when x = 0. [4] 10 Write down the eigenvalues of the matrix A, where A = ` −2 1 −1 0 −1 2 0 0 1 a , and find corresponding eigenvectors. [4] Find a matrix P and a diagonal matrix D such that P−1AP = D, and hence find the matrix An, where n is a positive integer. [8] [Question 11 is printed on the next page.] © UCLES 2016 9231/12/M/J/16 [Turn over

4 11 Answer only one of the following two alternatives. EITHER A curve C has parametric equations x = e2t cos 2t, y = e2t sin 2t, for −1 20 ≤t ≤1 20. Find the arc length of C. [6] Find the area of the surface generated when C is rotated through 20 radians about the x-axis. [8] OR The linear transformation T : >4 →>4 is represented by the matrix M, where M = € 1 −2 3 −4 2 −4 7 −9 4 −8 14 −18 5 −10 17 −22  . Find the rank of M. [3] Obtain a basis for the null space K of T. [3] Evaluate M € 1 −2 2 −1  , and hence show that any solution of Mx = € 15 33 66 81  * has the form € 1 −2 2 −1  + ,e1 + -e2, where , and - are scalars and e1, e2 is a basis for K. [2] Hence obtain a solution x′ of * such that the sum of the components of x′ is 6 and the sum of the squares of the components of x′ is 26. [6] 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 9231/12/M/J/16

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 10 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/12 Paper 1 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 12 © 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 12 © 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 12 © Cambridge International Examinations 2016 Qu Solution Part Marks 1 1 1 1 1 y x x y = + ⇒ = − ( ) ( ) ( ) ( ) 3 3 2 2 1 7 0 2 1 7 1 0 1 1 y y y y + − = ⇒ + − − − = − − ( ) 3 2 3 2 7 3 3 1 1 2 0 7 21 20 8 0 y y y y y y y ⇒ − + − − + − = ⇒ − + − = ALT METHOD: ∑α, ∑αβ, αβγM1 A1, ∑(1+1/α) etc M1 A1 M1 A1 M1A1 [4] 2 2 4 2 1 2 r r r − + + + (Award B2 if written down by cover up rule.) 2 4 1 2 4 2 2 4 2 2 2 1 3 3 2 1 1 1 2 n n n n n n         − + + − + +…+ − + + − +         − + + +         = 2 2 1 1 2 n n − + + + (AEF) Sum to infinity = 1 M1A1 M1A1 A1 [5] B1 [1] 3 For n =1 10 +192 + 5 = 207= 1 9 23 H × ⇒ is true. Assume Hk is true for some positive integer k 2 10 3.4 5 9 n n α + ⇒ + + = Let f(n) = 2 10 3.4 5 n n+ + + Hence ( ) ( ) ( ) ( ) 2 f 1 f 1 0 10 1 3.4 4 1 n n n n + + − = − + − ( ) 2 9 10 4 n n+ = + 9β = Hence ( ) ( ) ( ) f 1 9 n β α + = + ⇒ Hk+1 is true 1 H is true and Hk ⇒ Hk+1 , hence by PMI Hn is true for all positive integers n. N.B. Or can show ( ) ( ) 2 f 1 9 10 2.4 5 n n α + + = − − for M1A1A1. (3rd ,4th&5th marks) B1 B1 M1 A1 A1 A1 [6]

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 12 © Cambridge International Examinations 2016 Qu Solution Part Marks 4 Using cos sin x r y r θ θ = = and 2 2 4 8 cosec 2 sin cos r r θ θ θ = ⇒ = cos . sin 4 4 r r xy θ θ ⇒ = ⇒ = (in simple form) Sketch: Curve in 1st quadrant with correct concavity, asymptotic to both axes. 1 1 3 3 1 1 6 6 1 8 cosec 2 d 2 ln tan 2 π π π π θ θ θ   =   ∫ 1 2 ln 3 ln 3   = −     =2 ln3orln9 B1 M1 A1 [3] B1B1 [2] M1A1 A1 [3] 5 ( ) ( ) 1 3 2 4 2 d cos sin 1 cos sin 3cos sin d n n n x x n x x x x x − − = − − + ( ) ( ) 1 1 2 1 3 2 2 2 2 0 0 cos sin 1 cos sin 1 cos d 3 n n n x x n x x x x I π π − −   ⇒ = − − − +   ∫ ( ) ( ) 2 0 1 1 3 n n n n I n I I − ⇒ = − − + − + ( ) ( ) 2 2 1 n n n I n I − ⇒ + = − (AG) ( ) 1 1 2 2 2 0 0 0 1 sin d 1 cos2 d 2 I x x x x π π = = − ∫ ∫ 1 2 0 sin2 2 4 x x π   = −     = 4 π 2 4 1 1 1 4 4 2 4 4 32 I I π π π = × = × × = M1A1 M1 M1 A1 [5] M1 A1 M1A1 [4]

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 12 © Cambridge International Examinations 2016 Qu Solution Part Marks 6 ( ) ( ) ( ) 7 7 7 6 i 7 i i c s c c s s + = + +…+ 7 5 2 3 4 6 6 4 3 2 5 7 cos7 21 35 7 sin7 7 35 21 c c s c s cs c s c s c s s θ θ − + − = − + − 7 5 3 6 4 2 cot 21cot 35cot 7cot cot7 7cot 35cot 21cot 1 θ θ θ θ θ θ θ θ − + − = − + − 6 4 2 cot 7 0 cot 0 21 35 7 0 , where cot x x x x θ θ θ = ≠ ⇒ − + − = = and and where 1, 3, 5, 9,1 1,1 3 14 k k π θ = = Product of roots ⇒ 3 5 9 11 13 cot cot cot cot cot cot 7 14 14 14 14 14 14 π π π π π π = − But 13 cot cot 14 14 π π = − , 3 11 cot cot 14 14 π π = − , 5 9 cot cot 14 14 π π = − Hence 2 2 2 1 3 5 cot cot cot 7. 14 14 14 π π π = (AG)(Penultimate line must be seen.) SC Award B1 if product of roots mentioned, without proper pairing seen. M1 A1 M1 A1 [4] M1 M1 A1 M1 A1 [5] 7 Vertical asymptote is x = 2. 4 2 2 y x x = + + ⇒ − Oblique asymptote is 2 . y x = + 2 2 2 0 2 x y x yx y x = ⇒ − + = − Quadratic has no real roots (i.e. no points on C) if ∆ < 0 ⇒ 2 8 0 y y − < ( ) 8 0 0 8 . y y y ⇒ − < ⇒ < < (AG) Correct inequality Axes and asymptotes. Each branch, showing (0,0) and (4,8). (Deduct at most 1 mark for poor forms at infinity and/or missing coordinates.) B1 M1A1 [3] B1 M1 M1A1 [4] B1 B1B1 [3]

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 12 © Cambridge International Examinations 2016 Qu Solution Part Marks 8 17 1 AB AC 2 3 8 34 ~ 2 3 4 5 17 1       × = − =       − −    i j k





 So ( ) 2 const const 2 2 3 3 using a point x y z + + = ⇒ = − + = ⇒ 2 3 x y z + + = 1 3 3 3 9 1 4 1 4 1 cos 2 . 1 cos 14 6 84 1 2 θ θ     + + + + = − ⇒ = =       70.9 θ ⇒ = ° or 1.24 radians Direction of line of intersection is 5 1 2 1 1 3 1 2 7     =     − −   i j k Finds point common to both planes is ( ) 1, 0 ,4 − or 13 4 , ,0 7 7       or ( 1 1 3 0, , ) 5 5 Equation of line of intersection is eg 1 5 0 1 4 7 t −         = +         −     r . M1 A1 M1 A1 [4] M1 M1 A1 [3] M1A1 M1 A1 [4] 9 2 2 2 2 e 2 e x x y kx kx = ′ + 2 2 2 2 2 e 8 e 4 e x x x y k kx kx + ′ = + ′ 2 2 2 2 2 2 2 2 2 2 2 e 8 e 4 e 8 e 8 e 4 e 4e x x x x x x x k kx kx kx kx kx + + − − + = 2 4 2 k k ⇒ = ⇒ = ( ) 2 2 4 4 0 2 0 2 m m m m − + = ⇒ − = ⇒ = CF: 2 2 e e x x y A Bx = + 2 2 2 2 e e 2 e x x x y A Bx x = + + 3 when 0 3 y x A = = ⇒ = ( ) ( ) 2 2 2 2 2 2 2 e e 2 e 4 e 4 e x x x x x y A B x x x = + + ′ + + 2 when 0 and A 3 8 y x B = − = = ⇒ = − ′ 2 2 2 2 3e 8 e 2 e x x x y x x ⇒ = − + B1 B1 M1 A1 [4] M1 A1 A1 [3] B1 M1 A1 A1 [4]

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 12 © Cambridge International Examinations 2016 Qu Solution Part Marks 10 Eigenvalues are:–2, –1, 1 Eigenvectors are; 1 1 0 0 , 1 , 1 0 0 1      (oe) 1 1 0 2 0 0 0 1 1 , 0 1 0 0 0 1 0 0 1 −         = = −             P D or equivalent (in correct order) 1 1 1 1 0 1 1 0 0 1 − −     = −       p ( ) 1 1 n n n − − = = P AP P A P D ( ) ( ) 2 0 0 1 1 0 1 1 1 0 1 1 0 1 0 0 1 1 0 0 1 0 0 1 0 0 1 n n n   − −             ⇒ = − −                   A Accept 1 n − PD P here. ( ) ( ) ( ) ( ) ( ) 1 2 2 2 1 1 0 0 1 1 0 1 1 0 0 1 0 0 1 n n n n n n +   − −− −       = − −          A = ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 2 1 2 1 0 1 1 1 0 0 1 n n n n n n n + +   − −− + − − + −     − − +       B1 M1A1 A1 [4] B1 B1 M1A1 M1A1 M1 A1 [8]

Page 9 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 12 © Cambridge International Examinations 2016 Qu Solution Part Marks 11 (e) 2 2 2 2 2e cos2 2e sin2 2e cos2 2e sin2 t t t t x t ty t t = − = +

( ) 2 2 4 2 2 2 2 4e cos 2 2cos2 sin 2 sin 2 cos 2 2cos2 sin 2 sin 2 t x y t t t t t t t t + = − + + + +

4 8e t = 1 2 2 1 2 2 2e d t s t π π − = ∫ ( ) 1 2 2 1 2 2 e 2 e e t π π π π − −   = = −   or 2 2 sinhπ or 32.7 1 1 2 2 2 2 4 1 1 2 2 2 e sin 2 .2 2e d 4 2 e sin2 d t t t S t t t t π π π π π π − − = = ∫ ∫ Let 1 2 4 1 2 e sin2 d t I t t π π − = ∫ 1 1 2 2 4 4 1 1 2 2 cos2 e 2e cos2 d 2 t t t t t π π π π − −   = − +     ∫ 1 1 2 2 2 2 4 4 1 1 2 2 e e sin 2 2 e 2e sin 2 2 2 2 t t t t π π π π π π − − −           = − + −                   ∫ 2 2 e e 0 4 2 I π π − − = + − 2 2 e e 10 I π π − − ⇒ = ( ) 2 2 2 2 e e 2 2 4 2 . e e 10 5 S π π π π π π − − − ⇒ = = − or 4 2 sinh 2 5 π π or 952 (3sf) B1 M1 A1 M1 M1A1 [6] M1 M1A1 A1A1 M1 A1 A1 [8]

Page 10 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 12 © Cambridge International Examinations 2016 Qu Solution Part Marks 11 (o) 1 2 3 4 1 2 3 4 2 4 7 9 0 0 1 1 . 4 8 14 18 0 0 0 0 5 1017 22 0 00 0 − − − −         − − −     ⇒…⇒     − −     − −     r(M) = 4 – 2 = 2 2 3 4 0 x y z t − + − = 0 z t −= and 2 t z y x λ µ µ λ = = = ⇒ = + Basis for K is 1 2 0 1 , 1 0 1 0                     1 2 3 4 1 1 4 6 4 15 2 4 7 9 2 2 8 14 9 33 4 8 14 18 2 4 16 28 18 66 5 1017 22 1 5 20 34 22 81 − − + + +               − − − + + +        = =        − − + + +        − − − + + +        1 1 2 2 0 1 2 1 0 1 1 0 λ µ         −     = + +         −     x since M 1 15 2 33 2 66 1 81         −     =         −     and M 1 2 0 1 0 1 0 1 0 λ µ             + =               . Sum of components = 6 ⇒ ( ) 3 3 6 2 λ µ µ λ + = ⇒ = − Sum of squares of components = 26 ⇒ 2 2 5 4 4 3 10 26 µ λµ λ λ + + + + = ⇒ ( ) 2 2 4 8 4 0 1 0 λ λ λ − + = ⇒ − = ⇒ 1, 1 λ µ = = 4 1 ' 3 0     −   =       x M1A1 A1 [3] M1 M1 A1 [3] B1 B1 [2] B1 M1A1 M1 A1 A1 [6]