Cambridge A Level Mathematics - Further 9231 — 2015 Oct/Nov Paper 1 · Variant 2

*6437283360* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/12 Paper 1 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. 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. JC15 11_9231_12/FP © UCLES 2015 [Turn over

2 1 The curve C is defined parametrically by x = 2 cos3t and y = 2 sin3t, for 0 < t < 1 20. Show that, at the point with parameter t, d2y dx2 = 1 6 sec4t cosect. 4 2 Find the general solution of the differential equation d2x dt2 + 4dx dt + 4x = 7 −2t2. 6 3 Given that a is a constant, prove by mathematical induction that, for every positive integer n, dn dxn xeax = nan−1eax + anxeax. 6 4 The sequence a1, a2, a3, … is such that, for all positive integers n, an = n + 5 n2 −n + 1 − n + 6 n2 + n + 1. The sum N Ð n=1 an is denoted by SN. Find (i) the value of S30 correct to 3 decimal places, [3] (ii) the least value of N for which SN > 4.9. [4] 5 The cubic equation x3 + px2 + qx + r = 0, where p, q and r are integers, has roots !, " and ', such that ! + " + ' = 15, !2 + "2 + '2 = 83. Write down the value of p and find the value of q. [3] Given that !, " and ' are all real and that !" + !' = 36, find ! and hence find the value of r. [5] 6 The matrix A, where A = ` 1 0 0 10 −7 10 7 −5 8 a , has eigenvalues 1 and 3. Find corresponding eigenvectors. [3] It is given that ` 0 2 1 a is an eigenvector of A. Find the corresponding eigenvalue. [2] Find a diagonal matrix D and matrices P and P−1 such that P−1AP = D. [5] © UCLES 2015 9231/12/O/N/15

3 7 The linear transformation T : >4 →>4 is represented by the matrix M, where M = € 1 −2 −3 1 3 −5 −7 7 5 −9 −13 9 7 −13 −19 11  . Find the rank of M and a basis for the null space of T. [6] The vector € 1 2 3 4  is denoted by e. Show that there is a solution of the equation Mx = Me of the form x = € a b −1 −1  , where the constants a and b are to be found. [4] 8 The curve C has equation y = 2x2 + kx x + 1 , where k is a constant. Find the set of values of k for which C has no stationary points. [5] For the case k = 4, find the equations of the asymptotes of C and sketch C, indicating the coordinates of the points where C intersects the coordinate axes. [6] 9 It is given that In = Ó e 1 ln xn dx for n ≥0. Show that In = n −1In−2 −In−1 for n ≥2. 6 Hence find, in an exact form, the mean value of ln x3 with respect to x over the interval 1 ≤x ≤e. [6] 10 Using de Moivre’s theorem, show that tan 51 = 5 tan 1 −10 tan31 + tan51 1 −10 tan21 + 5 tan41 . 5 Hence show that the equation x2 −10x + 5 = 0 has roots tan21 50 and tan22 50. [4] Deduce a quadratic equation, with integer coefficients, having roots sec21 50 and sec22 50. [3] [Question 11 is printed on the next page.] © UCLES 2015 9231/12/O/N/15 [Turn over

4 11 Answer only one of the following two alternatives. EITHER The points A, B and C have position vectors i, 2j and 4k respectively, relative to an origin O. The point N is the foot of the perpendicular from O to the plane ABC. The point P on the line-segment ON is such that OP = 3 4ON. The line AP meets the plane OBC at Q. Find a vector perpendicular to the plane ABC and show that the length of ON is 4 21. [4] Find the position vector of the point Q. [5] Show that the acute angle between the planes ABC and ABQ is cos−12 3 . [5] OR The curve C has polar equation r = a1 −cos 1 for 0 ≤1 < 20. Sketch C. [2] Find the area of the region enclosed by the arc of C for which 1 20 ≤1 ≤3 20, the half-line 1 = 1 20 and the half-line 1 = 3 20. [5] Show that @ ds d1 A2 = 4a2 sin21 21, where s denotes arc length, and find the length of the arc of C for which 1 20 ≤1 ≤3 20. [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. 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 9231/12/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 9231/12 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 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 12 © Cambridge International Examinations 2015 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 2015 9231 12 © Cambridge International Examinations 2015 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 2015 9231 12 © Cambridge International Examinations 2015 Qn & Part Solution Marks 1 t t x sin cos 6 2 − = & , t t y cos sin 6 2 = & t x y tan d d − = ⇒ (OE) t t t t t x y cosec sec 6 1 sin cos 6 1 sec d d 4 2 2 2 2 = − × − = AG B1 B1 M1A1 [4] Total 4 2 2 0 ) 2 ( 0 4 4 2 2 − = ⇒ = + ⇒ = + + m m m m CF: t t Bt A 2 2 e e − − + soi PI : p x q pt x r qt pt x 2 2 2 = ⇒ + = ⇒ + + = && & 2 2 2 7 4 4 4 4 8 2 t r qt pt q pt p − = + + + + + ⇒ 2 1 − = ⇒p , q = 1, r = 1 GS: 1 2 1 e e 2 2 2 + + − + = − − t t Bt A x t t M1 A1 M1 M1 A1 A1 [6] Total 6 3 n =1 in formula gives ax ax ax ax ax ax a e e e e 0 + = + ( ) 1 H e e e . 1 e e d d ⇒ + = + × = ax ax ax ax ax ax a x x x is true oe Assume Hk is true, i.e. ( ) ax k ax k ax k k x a e ka x x e e d d 1 + = − . ( ) ax k k x x e d d 1 1 + + ax k ax k ax k x a a ka e e e 1 + + + = ax k ax k x a a k e e )1 ( 1 + + + = 1 H + ⇒ k is true, hence by PMI Hn is true for all positive integers n. B1 B1 B1 M1 A1 A1 [6] Total 6 4 (i) 820 .4 931 36 6 931 36 871 35 7 8 3 7 3 7 1 6 = − =       − + +       − +       − K M1A1 A1 [3] (ii) ) 0 ( 79 . 34 79 . 10 21 .0 9.4 1 6 6 2 2 > − − ⇒ > + + + − n n n n n K 42 . 54 > ⇒n so 55 terms required. M1A1 dM1A1 [4] Total 7

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2015 9231 12 © Cambridge International Examinations 2015 Qn & Part Solution Marks 5 15 15 − = ⇒ = − = + + p p γ β α ( ) q 2 ) ( ) ( 2 2 2 2 2 = + + − + + = + + γ β α γ β α γα βγ αβ 71 ) 83 225 ( 2 1 = − = ⇒q α α − =15 36 ]) [ ( γ β + = 3 0 36 15 2 = ⇒ = + − ⇒ α α a , 12 ≠ α , e.g. since 122 > 83 or other reason 35 36 71 = − = βγ 105 35 3 − = × − = − = ⇒ αβγ r (extra answer penalised) B1 M1 A1 [3] M1 M1A1 B1 A1 [5] Total 8 6           −          − = − − = 1 0 1 ~ 6 0 6 7 5 7 10 8 10 : 1 k j i λ oe                     = − − = 1 1 0 ~ 20 20 0 10 10 10 0 0 2 : 3 k j i λ oe 2 1 2 0 2 2 4 0 1 2 0 8 5 7 10 7 10 0 0 1 − = ⇒           − =           − − =                     − − λ          − = 3 0 0 0 1 0 0 0 2 D ,           − = 1 1 1 1 0 2 0 1 0 P (or other multiples or permutations). Det P = –1 (or 1 depending on permutation). Adj P =           − − − 2 1 2 0 0 1 1 1 1 ⇒ P–1 =           − − − 2 1 2 0 0 1 1 1 1 ( or other permutations). M1A1 A1 [3] M1A1 [2] B1 B1 B1 M1A1 [5] Total 10

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2015 9231 12 © Cambridge International Examinations 2015 Qn & Part Solution Marks 7             − −             − −             − − − − − − − − 0 0 0 0 0 0 0 0 4 2 1 0 1 3 2 1 ~ 4 2 1 0 4 2 1 0 4 2 1 0 1 3 2 1 ~ 11 19 13 7 9 13 9 5 7 7 5 3 1 3 2 1 r(M) = 4 – 2 = 2 0 3 2 = + − − t z y x 0 4 2 = + + t z y E.g. Set z = λ and t = µ ⇒ µ λ 4 2 − − = y and µ λ 9 − − = x                         − −             − − ⇒ 1 0 4 9 , 0 1 2 1 is Basis             − − =             − − +             − − +             = 1 1 1 0 4 9 0 1 2 1 4 3 2 1 b a µ λ x Solving: 4 − = λ and 5 − = µ 50 = ⇒a , 30 = b . M1A1 A1 M1 M1 A1 [6] M1 M1 A1 A1 [4] Total 10 8 ( ) 0 1 2 ) 4 )( 1 ( 0 2 = × + − + + ⇒ = ′ kx x k x x y 0 4 2 0 2 ) 4 ( 4 2 2 2 = + + ⇒ = − − + + + ⇒ k x x kx x k x k x ⇒ < − 0 4 2 AC B no stationary points 0 8 16 < − ⇒ k ⇒ k > 2 for no stationary points. When k = 4: Vertical asymptote: x = –1 Oblique asymptote: ⇒ + − + = 1 2 2 2 x x y y = 2x + 2 Axes and asymptotes Each branch. M1 A1 M1A1 A1 [5] B1 M1A1 B1 B1B1 [6] Total 11

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2015 9231 12 © Cambridge International Examinations 2015 Qn & Part Solution Marks 9 x x x x x e − = ∫ ln d ln 1 In ∫ − = e 1 1 d ln . ) (ln x x x n [ ] x x x x x x n x x x x n n d ) ln ( 1 . ) (ln )1 ( ) ln ( ) (ln 2 e 1 e 1 1 − − − − = − − ∫ [ ] 1 2 e 1 2 )1 ( d )1 (ln ) )(ln 1 ( 0 − − − − − = − − − = ∫ n n n I I n x x x n (AG) Alternative for obtaining reduction formula: [ ] ∫ ∫ − − = × = e 1 1 1 e 1 d ) x (ln ) x (ln d 1 ) (ln e n n n n x n x x x I 1 e − − = ⇒ n n nI I Similarly 2 1 )1 ( e − − − − = n n I n I 2 1 1 )1 ( − − − − + = + ⇒ n n n n I n I nI I [ ] 1 2 )1 ( − −− − = ⇒ n n n I I n I (AG) [ ] 1 e 1 0 − = = e x I [ ] 1 ln e 1 1 = − = x x x I 2 e )1 1 e( 1 2 − = − − × = I ( ) e 2 6 ]) 2 e[ 1( 2 2 2 1 3 − = − − = − = I I I MV 1 2 6 1 e 3 − − = − = e e I B1 M1 M1A1 M1A1 [6] M1A1 A1 B1 M1 A1 [6] B1 B1 M1 A1 M1 A1 [6] Total 12

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2015 9231 12 © Cambridge International Examinations 2015 Qn & Part Solution Marks 10 θ θ θ θ 5 sin i 5 cos ) sin i (cos 5 + = + i c 5 i i 10 10 i 5 ) i ( 5 4 3 2 2 3 4 5 5 s s s c s c s c c s c + + − − + = + 4 2 3 5 5 3 2 4 5 10 10 5 5 tan cs s c c s s c s c + − + − = θ Divide numerator and denominator by c5 (stated or shown): θ θ θ θ θ θ 4 2 5 3 tan 5 tan 10 1 tan tan 10 tan 5 5 tan + − + − = ⇒ (AG) π π π π π θ θ , 5 4 , 5 3 , 5 2 , 5 1 0 5 tan = ⇒ = 0 5 10 3 5 = + − t t t has roots π π π π π tan , 5 4 tan , 5 3 tan , 5 2 tan , 5 1 tan                         0 5 10 2 4 = + − ⇒ t t has roots                         π π π π 5 4 tan , 5 3 tan , 5 2 tan , 5 1 tan . 0 5 2 tan 5 1 tan 2 2 2 2 =             −             − ⇒ π π t t since       − =       π π 5 4 tan 5 1 tan and       − =       π π 5 3 tan 5 2 tan . 0 5 10 2 = + − ⇒ x x has roots       π 5 1 tan2 and       π 5 2 tan2 . (AG) α α 2 2 tan 1 sec + = 0 5 )1 ( 10 )1 ( 1 1 2 = + − − − ⇒ − = ⇒ + = y y y x x y 0 16 12 2 = + − ⇒ y y B1 M1A1 M1 A1 [5] B1 B1 M1 A1 [4] M1 M1 A1 [3] Total 12

Page 9 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2015 9231 12 © Cambridge International Examinations 2015 Qn & Part Solution Marks 11 E E.g.                     = − − 1 2 4 ~ 2 4 8 4 0 1 0 2 1 k j i 21 4 1 2 4 1 2 4 . 0 0 1 2 2 2 = + +                     (AG) ) 2 4 ( 7 1 21 2 4 21 3 k j i k j i p + + =       + + = Line AP:          − +           = 1 2 3 0 0 1 t r For Q           = ⇒ = ⇒ = − 1 2 0 3 1 3 1 0 3 1 q t t E.g.          − = 0 2 1 AB ,           − = 1 4 0 3 1 BQ           = − − 4 1 2 1 4 0 0 2 1 k j i 3 2 cos 21 14 cos 21 4 2 8 cos 21 . 21 4 1 2 . 1 2 4 cos 1 1 1 1 − − − − = = + + =                     (AG) M1A1 M1A1 [4] B1 M1A1 M1A1 [5] B1 M1A1 M1A1 [5] Total 14

Page 10 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2015 9231 12 © Cambridge International Examinations 2015 Qn & Part Solution Marks 11O Closed curve through pole with correct orientation. Completely correct. θ θ θ π π d) cos cos 2 1( 2 1 2 2 1 2 2∫ + − × a ∫       + − = π π θ θ θ 2 1 2 d 2 cos 2 1 cos 2 2 3 a π π θ θ θ 2 1 2 2 sin 4 1 sin 2 2 3     + − = a       + = 2 4 3 2 π a 2 d d       θ s ) sin cos cos 2 1( 2 2 2 θ θ θ + + − = a θ θ θ 2 1 sin 4 2 1 sin 2. 2 ) cos 1( 2 2 2 2 2 2 a a a = = − = (AG) s ∫ × = π π θ θ 2 1 d 2 1 sin 2 2 a π π θ 2 1 2 1 cos 2 4    − = a a 2 4 = B1 B1 [2] M1M1 M1A1 A1 [5] B1 M1A1 M1 A1 M1A1 [7] Total 14