Cambridge A Level Mathematics - Further 9231 — 2015 May/June Paper 1 · Variant 1

*4040221319* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/11 Paper 1 May/June 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 06_9231_11/RP © UCLES 2015 [Turn over

2 1 Use the List of Formulae (MF10) to show that 13 Ð r=1 3r2 −5r + 1 and 9 Ð r=0 r3 −1 have the same numerical value. [4] 2 Find the value of the constant k for which the system of equations 2x −3y + 4Ï = 1, 3x −y = 2, x + 2y + kÏ = 1, does not have a unique solution. [2] For this value of k, solve the system of equations. [4] 3 The sequence a1, a2, a3, à is such that a1 > 5 and an+1 = 4an 5 + 5 an for every positive integer n. Prove by mathematical induction that an > 5 for every positive integer n. [5] Prove also that an > an+1 for every positive integer n. [2] 4 The roots of the cubic equation x3 −7x2 + 2x −3 = 0 are !, " and '. Find the values of (i) 1 !""''!, (ii) 1 !" + 1 "' + 1 '!, (iii) 1 !2"' + 1 !"2' + 1 !"'2 . [6] Deduce a cubic equation, with integer coefficients, having roots 1 !", 1 "' and 1 '!. [2] 5 The curves C1 and C2 have polar equations C1 : r = 1 ï2, for 0 ≤1 < 20, C2 : r = sin 1 21, for 0 ≤1 ≤0. Find the polar coordinates of the point of intersection of C1 and C2. [2] Sketch C1 and C2 on the same diagram. [3] Find the exact value of the area of the region enclosed by C1, C2 and the half-line 1 = 0. [4] 6 A curve has equation x2 −6xy + 25y2 = 16. Show that dy dx = 0 at the point 3, 1. [4] By finding the value of d2y dx2 at the point 3, 1, determine the nature of this turning point. [5] © UCLES 2015 9231/11/M/J/15

3 7 Let In = Ó 1 20 0 xn sin x dx, where n is a non-negative integer. Show that In = n1 20n−1 −nn −1In−2, for n ≥2. 5 Find the exact value of I4. [4] 8 By considering n Ð r=1 Ï2r−1, where Ï = cos 1 + i sin 1, show that, if sin 1 ≠0, n Ð r=1 sin2r −11 = sin2n1 sin 1 . 7 Deduce that n Ð r=1 2r −1 cos B2r −10 2n C = −cosec 0 0 2n 1 cot 0 0 2n 1 . 4 9 The curve C has parametric equations x = 4t + 2t 3 2, y = 4t −2t 3 2, for 0 ≤t ≤4. Find the arc length of C, giving your answer correct to 3 significant figures. [6] Find the mean value of y with respect to x over the interval 0 ≤x ≤32. [5] 10 The matrix A is given by A = ` 2 2 −3 2 2 3 −3 3 3 a . The matrix A has an eigenvector ` 1 −1 1 a . Find the corresponding eigenvalue. [2] The matrix A also has eigenvalues 4 and 6. Find corresponding eigenvectors. [3] Hence find a matrix P such that A = PDP−1, where D is a diagonal matrix which is to be determined. [2] The matrix B is such that B = QAQ−1, where Q = ` 4 11 5 1 4 2 1 2 1 a . By using the expression PDP−1 for A, find the set of eigenvalues and a corresponding set of eigenvectors for B. [5] [Question 11 is printed on the next page.] © UCLES 2015 9231/11/M/J/15 [Turn over

4 11 Answer only one of the following two alternatives. EITHER Show that the substitution v = 1 y reduces the differential equation 2 y3 @dy dx A2 −1 y2 d2y dx2 −2 y2 dy dx + 5 y = 17 + 6x −5x2 to the differential equation d2v dx2 + 2dv dx + 5v = 17 + 6x −5x2. 4 Hence find y in terms of x, given that when x = 0, y = 1 2 and dy dx = −1. [10] OR The lines l1 and l2 have equations r = 8i + 2j + 3k + ,i −2j and r = 5i + 3j −14k + -2j −3k respectively. The point P on l1 and the point Q on l2 are such that PQ is perpendicular to both l1 and l2. Find the position vector of the point P and the position vector of the point Q. [8] The points with position vectors 8i + 2j + 3k and 5i + 3j −14k are denoted by A and B respectively. Find (i) −−→ AP × −−→ AQ and hence the area of the triangle APQ, (ii) the volume of the tetrahedron APQB. (You are given that the volume of a tetrahedron is 1 3 × area of base × perpendicular height.) [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 2015 9231/11/M/J/15

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the May/June 2015 series 9231 FURTHER MATHEMATICS 9231/11 Paper 1 (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 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 – May/June 2015 9231 11 © 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 – May/June 2015 9231 11 © 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 – May/June 2015 9231 11 © Cambridge International Examinations 2015 Qn & Part Solution Marks 1 2015 13 2 14 13 5 6 27 14 13 3 = × × − × × × + 2015 10 2 10 9 2 = −    × (Award M1 for subtracting 9 or 10 here.) M1A1 M1A1 (4) Total: 4 2 4 28 7 0 2 1 0 1 3 4 3 2 − = ⇒ − = ⇒ =           − − k k k Add 1st and 3rd ⇒ 3x – y =2 ( Same as 2nd) (OE) Set x = t (for example) (OE) 2 3 − = ⇒ t y , 7 5 4 4 z t = − (OE) – many forms M1A1 (2) M1 M1 A1A1 (4) Total: 6 3 a1 > 5 (given) ⇒ H1 is true. Assume Hk is true for some positive integer k, i.e. ak = 5 + δ, where δ > 0. 0 5 ) 5 )( 5 4 ( 5 25 25 4 5 5 25 4 5 2 2 1 > − − = − + = − + = − + k k k k k k k k k a a a a a a a a a , 5 1 > ⇒ + ka Or , 5 5 ) 5 ( 5 4 1 δ δ + + + = + k a ...) 25 5 1( 5 4 4 2 − + − + + = δ δ δ for 0 < δ < 5 ) ( 0 5 3 5 2 δ δ + + = [ ak+1 > 5, ( δ [ 5 is trivial). Hk ⇒ Hk +1 and H1 is true, hence by mathematical induction, the result is true for all=== n∈ + Z =(N.B. The minimum requirement is ‘true for all positive integers’.) k k k k a a a a 5 1 5 1 − = − + 1 5 < k a and 1 5 1 > k a k k k k a a a a < ⇒ < − ⇒ + + 1 1 0 B1 B1 M1A1 (M1) (A1) A1 (5) M1 A1 (2) Total: 7 4 (i) (ii) (iii) 7 α β γ + + = , 2 αβ βγ γα + + = , 3 αβγ = (Stated or implied by working.) 2 1 1 1 ( )( )( ) ( ) 9 αβ βγ γα αβγ = = 1 1 1 7 3 α β γ αβ βγ γα αβγ + + + + = = 2 2 2 1 1 1 α βγ αβ γ αβγ + + = 1 1 1 1 αβγ α β γ   + +     = 1 2 9 αβ βγ γα αβγ αβγ   + + =     ⇒ ⇒ = − + − 0 9 1 9 2 3 7 2 3 x x x 0 1 2 21 9 2 3 = − + − x x x B1 B1 M1A1 M1A1 (6) M1A1√ (2) Total: 8

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 11 © Cambridge International Examinations 2015 Qn & Part Solution Marks 5 π θ θ 3 1 2 1 2 1 sin = ⇒ = ; Intersection at       π 3 1 , 2 1 (Accept 1.05 for π 3 1 .) C1: Circle centre at pole and radius 1/√2 C2: Curve, approx. correct orientation, from (0,0) to (1,π). Completely correct correct shape. 1 3 0 1 1 1 1 sin d 6 2 2 2 π π θ θ × × −∫ = π θ π 3 1 0 2 1 cos 2 2 1 12 1    − × 1 2 3 12 1 − + = π M1A1 (2) B1 B1 B1 (3) M1A1 M1A1 (4) Total: 9 6 d d 2 6 50 0 d d y y x x y y x x   − + + =     At (3,1) 0 ' 0 ' 50 6 ' 18 6 = ⇒ = + − − y y y (AG) 0 ]" )' [( 50 )' ' " ( 6 2 2 = + + + + − y y y y y y x At (3,1) 16 1 " 0 " 50 " 18 2 − = ⇒ = + − y y y ⇒ maximum. (All previous marks required for final mark, i.e. CSO) M1A1 A1 A1 (4) B1B1 M1A1 A1 (5) Total: 9 7 1 2 2 2 0 0 0 sin d [ cos ] cos d n n n nI x x x x x n x x x π π π − = = − + ∫ ∫ (LNR) 1 2 2 2 0 0 0 sin ( 1) sin d n n nx x n n x x x π π − −   = + − −   ∫ (LR) 2 1 )1 ( 2 − − − −       = n n I n n n π (AG) [ ] 2 2 0 0 0 sin d cos 1 n I x x x x π π = = − = ∫ 2 2 − = π I , using reduction formula. ( ) 24 12 2 1 2 12 2 4 3 3 4 + − = − −       × = π π π π I (If I2 found by integration M1A1 (if correct) then M1A1 for I4 from reduction formula.) M1A1 M1A1 A1 (5) B1 M1A1 A1 (4) Total: 9

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 11 © Cambridge International Examinations 2015 8 ( ) 2 1 2 1 2 2 1 2 1 1 ] [ 1 z z z z z z z n n r n r − − = − − = + = − ∑ ) sin (cos sin cos ) 2 sin 2 (cos 1 1 1 2 θ θ θ θ θ θ i i n i n z z z n + − − + − = − − = − θ θ θ sin 2 2 sin 2 cos 1 i n i n − − − = Equating imaginary parts: 2 2 1 2sin sin sin(2 1) 2 sin sin n r n n r θ θ θ θ θ = − = = ∑ (AG) Differentiating: θ θ θ θ θ θ θ cot cosec sin cosec n cos sin 2 )1 2 ( cos )1 2 ( sin 2 1 n n n r r n r − = − − ∑= Putting n 2 π θ = : 2 1 cos sin cos cosec sin cosec cot (2 1) (2 1) 2 2 2 2 2 2 2 n r r r n n n n n π π π π π π π = − − = −                         ∑             − =     − − ⇒∑= n n n r r n r 2 cot 2 cosec 2 )1 2 ( cos )1 2 ( 1 π π π . (AG) M1A1 M1A1 A1 M1A1 (7) M1A1 dM1 A1 (4) Total: 11 9 1 2 4 3 x t = + & 2 1 3 4 t y − = & t y x 18 32 2 2 + = + & & 4 1 3 4 2 2 0 0 1 (32 18 ) d (32 18 ) 27 s t t t   =∫ + = +     6. 32 32 104 27 1 2 3 2 3 =         − = (CAO) ( )( ) 3 32 4 4 4 2 2 0 0 0 0 d d d 4 2 4 3 d 16 4 6 d d x y x y t t t t t t t t t t   = = − + = + −     ∫ ∫ ∫ ∫ 4 0 3 2 5 2 2 5 8 8         − + = t t t ( = 51.2) MV = d 51.2 1.6 32 a b y x b a ∫ = = = − (CAO) B1 B1 M1A1 M1A1 (6) M1 M1A1 M1A1 (5) Total: 11

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 11 © Cambridge International Examinations 2015 10 ⇒           − − =           −           − − 3 3 3 1 1 1 3 3 3 3 2 2 3 2 2 eigenvalue is – 3. λ = 4 : 7 1 2 2 3 7 ~ 1 3 3 1 0 0     − − =     − −   i j k (Or by equations) λ = 6 : 3 1 4 2 3 3 ~ 1 3 3 3 6 2         − − = − −         − − − −     i j k P =           − − − 2 0 1 1 1 1 1 1 1 D =          − 6 0 0 0 4 0 0 0 3 B = QAQ –1 = QPAP–1Q–1; = QPA (QP) –1 QP =           − − − − − 3 3 0 7 5 1 17 15 2 (CAO) Eigenvalues are –3, 4 and 6 (same as those for A). (CAO) Eigenvectors are:           0 1 2 ,           3 5 15 ,           3 7 17 . M1A1 (2) M1A1 A1 (3) B1√B1√ (2) M1;A1 B1 B1 B1 (5) Total: 12 11 (e) 2 d 1 d d d v y x y x = − 2 2 2 2 2 2 2 2 2 2 2 2 3 d 1 d d d 1 d 2 d d d d d d d v y y v y y x y x x y y x y x     = − + = − +         2 2 2 2 2 3 2 2 2 2 2 d 1 d 2 d 5 d d 17 6 5 ~ 2 5 17 6 5 d d d d d y y y v v x x v x x y x y x y x y x x  − − + = + − + + = + −     (AG) 0 5 2 2 = + + m m ) 2 sin 2 cos ( : 2 1 x B x A e CF i m x + ⇒ ± − = ⇒ − PI : p v q px v r qx px v 2 " 2 ' 2 = ⇒ + = ⇒ + + = Equate coefficients: 17 5 2 2 6 5 4 5 5 = + + = + − = r q p q p p 3 , 2 , 1 = = − = r q p GS : 2 2 3 ) 2 sin 2 cos ( x x x B x A e v x − + + + = − When x = 0 2 2 1 = ⇒ = v y and d d 1 4 d d y v x x = − ⇒ = B1 M1A1 A1 (4) M1A1 M1 M1 A1 A1

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 11 © Cambridge International Examinations 2015 1 3 2 − = ⇒ + = ⇒ A A ) 2 sin 2 cos ( ) 2 cos 2 2 sin 2 ( ' x B x A e x B x A e v x x + − + − = − − 2 1 1 2 2 1 2 4 = ⇒ = ⇒ + + = ⇒ B B B 1 2 2 3 2 cos 2 sin 2 1 − −     − + +       − = ⇒ x x x x e y x B1 M1 A1 A1 (10) Total: 14 110 (i) (ii) (a) (b) p = , 3 2 2 8           − + λ λ q =           + − − − + = ⇒           − − + µ µ λ λ µ µ 3 17 2 2 1 3 3 14 2 3 5 QP 5 4 5 0 3 17 2 2 1 3 . 0 2 1 − = + ⇒ =           + − − − +           − µ λ µ µ λ λ 53 13 4 0 3 17 2 2 1 3 . 3 2 0 = − − ⇒ =           + − − − +           − µ λ µ µ λ λ Solving : λ = 3, µ = – 5 Whence: p =           − 3 4 11 , q =           − 1 7 5 .           − − − =           − = 2 9 3 , 0 6 3 AQ AP 12 3 6 0 6 3 9 2 45     − =     − − − −   i j k (CAO) ; Area = 2205 2 1 (=23.5) 3 12 1 . 6 3 17 45 1 1 735 AB 1 Volume 2205 ( 122.5) 3 2 6 2205 17 −          −      − −      = ⇒ = × = =     −   uuur M1A1 M1A1 A1 M1A1 A1 (8) B1 M1A1 A1 M1A1 (6) Total: 14

Page 9 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 11 © Cambridge International Examinations 2015 Alternative: for marks 3,4,5,6 and 7 in part (i) 6 1 2 0 3 0 2 3 2   − =   −  i j k λ − k 6 = 3 1 2 2 3 − = + + λ k 17 3 2 = −µ k Solving; k = 1 , λ = 3 and µ = – 5 (If k missing, or assumed to be 1, deduct 1 mark.) (M1A1) (A1) (M1A1)