Cambridge A Level Mathematics - Further 9231 — 2015 May/June Paper 1 · Variant 3

*5151870499* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/13 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_13/RP © UCLES 2015 [Turn over

2 1 The quartic equation x4 −px2 + qx −r = 0, where p, q and r are real constants, has two pairs of equal roots. Show that p2 + 4r = 0 and state the value of q. [6] 2 The curve C has polar equation r = e41 for 0 ≤1 ≤!, where ! is measured in radians. The length of C is 2015. Find the value of !. [6] 3 Prove by mathematical induction that, for all positive integers n, n Ð r=1 1 2r2 −1 = n 2n + 1. [6] State the value of ∞ Ð r=1 1 2r2 −1. [1] 4 Use the formula for tanA −B in the List of Formulae (MF10) to show that tan−1x + 1 −tan−1x −1 = tan−1 @ 2 x2 A . 3 Deduce the sum to n terms of the series tan−1 @ 2 12 A + tan−1 @ 2 22 A + tan−1 @ 2 32 A + à . 4 5 Let In = Ô 1 20 0 sin 2n1 cos 1 d1, where n is a non-negative integer. (i) Use the identity sin P + sin Q  2 sin 1 2P + Q cos 1 2P −Q to show that In + In−1 = 2 2n −1, for all positive integers n. 5 (ii) Find the exact value of Ô 1 20 0 sin 81 cos 1 d1. [4] 6 Let Ï = cos 1 + i sin 1. Use the binomial expansion of 1 + Ïn, where n is a positive integer, to show that @ n 1 A cos 1 + @n 2 A cos 21 + à + @ n n A cosn1 = 2n cosn1 21 cos 1 2n1 −1. 7 Find @ n 1 A sin 1 + @n 2 A sin 21 + à + @ n n A sin n1. 2 7 The curve C has equation x2 + 2xy −4y2 + 20 = 0. Show that if the tangent to C at the point x, y is parallel to the x-axis then x + y = 0. [3] Hence find the coordinates of the stationary points on C, and determine their nature. [7] © UCLES 2015 9231/13/M/J/15

3 8 A line, passing through the point A 3, 0, 2, has vector equation r = 3i + 2k + ,2i + j −2k. It meets the plane , which has equation r.i + 2j + k = 3, at the point P. Find the coordinates of P. [3] Write down a vector n which is perpendicular to , and calculate the vector w, where w = n × 2i + j −2k. 3 The point Q lies in and is the foot of the perpendicular from A to . Use the vector w to determine an equation of the line PQ in the form r = u + -v. [4] 9 Find the particular solution of the differential equation d2x dt2 −3dx dt −10x = 2 sin t −3 cost, given that, when t = 0, x = 3.3 and dx dt = 0.9. [11] 10 The curve C has equation y = 4x2 −3x x2 + 1 . Verify that the equation of C may be written in the form y = −1 2 + 3x −12 2x2 + 1 and also in the form y = 9 2 −x + 32 2x2 + 1. [3] Hence show that −1 2 ≤y ≤9 2. [2] Without differentiating, write down the coordinates of the turning points of C. [2] State the equation of the asymptote of C. [1] Sketch the graph of C, stating the coordinates of the intersections with the coordinate axes and the asymptote. [3] [Question 11 is printed on the next page.] © UCLES 2015 9231/13/M/J/15 [Turn over

4 11 Answer only one of the following two alternatives. EITHER The linear transformation T : >4 →>4 is represented by the matrix M, where M = € 1 2 3 4 1 −1 2 3 1 −3 3 5 1 4 2 2  . The range space of T is denoted by V. (i) Determine the dimension of V. [3] (ii) Show that the vectors € 1 1 1 1  , € 2 −1 −3 4  , € 3 2 3 2  are a basis of V. [5] The set of elements of >4 which do not belong to V is denoted by W. (iii) State, with a reason, whether W is a vector space. [1] (iv) Show that if the vector € x y Ï t  belongs to W then x + y ≠Ï + t. [5] OR One of the eigenvalues of the matrix M, where M = ` 3 −4 2 −4 ! 6 2 6 −2 a , is −9. Find the value of !. [3] Find (i) the other two eigenvalues, ,1 and ,2, of M, where ,1 > ,2, [5] (ii) corresponding eigenvectors for all three eigenvalues of M. [3] It is given that x = ae1 + be2, where e1 and e2 are eigenvectors of M corresponding to the eigenvalues ,1 and ,2 respectively, and a and b are scalar constants. Show that Mx = pe1 + qe2, expressing p and q in terms of a and b. [3] 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 9231/13/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 9231/13 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 13 © 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 13 © 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 13 © Cambridge International Examinations 2015 Qn & Part Solution Marks 1 i 2 2 0 α β + = ii 2 2 4 p α αβ β + + = − iii 2 2 2 2 q α β αβ + = − iv 2 2 r α β = − All four correct (Any two correct) Use of β α = − in ii, iii or iv 2 4 2 2 and 4 0 p r p r α α ⇒ = = −⇒ + = (AG) and q = 0 (CAO) B2 (B1) M1 M1A1 B1 (6) Total 6 2 8 8 4 0 0 e 16e d 17e d s α α θ θ θ θ θ = + = ∫ ∫ = (AEF) (LNR) = 4 4 0 17 17 17 e e 2015 4 4 4 α θ α  ⇒ − =     4e 1955.837 1.89 α α ⇒ = ⇒ = K CAO M1A1 dM1A1 dM1A1 (6) Total 6 3 Hk : 2 1 1 (2 ) 1 2 1 k r k r k = ∑ = − + is true for some integer k. ⇒ + × = = − 1 1 2 1 3 1 1 2 1 2 H1 is true. )3 2 )( 1 2 ( 1 3 2 ) 3 2 )( 1 2 ( 1 1 2 1 ) 2 2 ( 1 1 2 2 2 + + + + = + + + + = − + + + k k k k k k k k k k k = 1 ]1 [2 1 ) 3 2 )( 1 2 ( )1 )( 1 2 ( + + + = + + + + k k k k k k 1 H H + ⇒ ∴ k k ∴(By Principle of Mathematical Induction) Hn is true for all positive integers n. (This mark requires all previous marks.) 2 1 1 1 (2 ) 1 2 r r ∞ = ∑ = − B1 B1 M1A1 A1 A1 (6) B1 (1) Total 7

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 13 © Cambridge International Examinations 2015 Qn & Part Solution Marks 4 )1 ( tan 1 + = − x A ; )1 )( 1 ( 1 )1 ( )1 ( ) tan( )1 ( tan 1 − + + − − + = − ⇒ − = − x x x x B A x B     = − ⇒ = − − 2 1 2 2 tan 2 ) tan( x B A x B A (AG) ]) 2 [ tan (tan )1 tan 3 (tan ) 0 tan 2 (tan LHS 1 1 1 1 1 -1 − − + + − + − = − − − − − n n K ]). 1 [ tan ]1 [ (tan 1 1 − − + + − − n n π 4 1 1 1 1 1 1 1 tan ]1 [ tan 0 tan 1 tan tan ]1 [ tan − + + = − − + + = − − − − − − n n n n M1A1 A1 (3) M1A1 A1A1 (4) Total 7 5 2 2 1 0 0 sin(2 ) sin(2 2) 2sin(2 1) cos d d cos cos n n n n n I I π π θ θ θ θ θ θ θ θ − + − − + = = ∫ ∫ (LR) 2 2 0 0 2cos(2 1) 2sin(2 1) d (2 1) n θ n n π π θ θ −   = − = −   −   ∫ (LR) 1 2 2 )1 2 ( 2 ] 0 [ − =     − − − = n n (AG) [ ] 2 2 2 1 0 0 0 sin 2 d 2sin d 2cos [0] [ 2] 2 cos I π π π θ θ θ θ θ θ = = = − = −− = ∫ ∫ 3 4 2 3 2 2 − = − = I 105 152 15 26 7 2 15 26 3 4 5 2 4 3 − = − = ⇒ = + = ⇒ I I M1A1 dM1A1 A1 (5) B1 M1A1 A1 (4) Total 9 6 n n z n n z n z n z 2 1 1 ) 1( 2       + +       +       + = + L Re θ θ θ n n n n n z n cos cos2 2 cos 1 1 } ) 1 {(       + +       +       + = + L Re [ ] [ ] [ ] ( ) 2 1 1 1 2 2 2 ([1 cos ] isin ) Re 2cos 2i sin cos n n θ θ θ θ θ + + = + = ( ) [ ] [ ] ( ) 1 1 1 2 2 2 2 cos Re cos i sin n n n θ θ θ + ( ) ( ) 1 cos cos 2 cos 2 cos 2 cos 1 2 1 2 1 − =       + +       +       ⇒ θ θ θ θ n θ n n n n n n n L (AG) Im{(1 ) } sin sin 2 sin 1 2 n n n n z n n θ θ θ    + = + + +       L       +       ⇒ 2 sin 1 n n θ θ 2 sin + +L       n n n n n cos 2 sin = θ ( ) 1 2θ sin( ) θ n 2 1 B1 M1A1 M1A1 A1 A1 (7) M1 A1 (2) Total 9

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 13 © Cambridge International Examinations 2015 Qn & Part Solution Marks 7 0 d d 8 d d 2 2 = −       + + x y y y x y x x (1st B1 for 1st and 3rd terms, 2nd B1 for rest including = 0). 0 d d when ) ( ) 4 ( d d = − = ⇒ + − = − ⇒ x y x y y x y x x y Substituting for 2 20 5 0 20 4 2 2 2 2 2 ± ⇒ = ⇒ = + − − ⇒ x x x x x y Coordinates are (2, –2) and (–2, 2) 2 2 d d d d ( 4 ) 1 4 1 d d d d y y y y x y x x x x   − + − = −−     (OE) Substituting either (2, –2) or (–2, 2) and 0 d d = x y At (2, –2) y″(2) = max. 10 1 ⇒ − At (–2, 2) y″(2) = 1 min. 10 ⇒ B1B1 B1 (3) M1 A1 M1A1 dM1 A1 A1 (7) Total 10 8 3 1 2 1 2 2 2 3 =                     − + . λ λ λ 1 3 2 2 2 2 3 − = ⇒ = − + + + λ λ λ λ P is (1, –1, 4) (Accept position vector.) n           = 1 2 1 w 5 1 2 1 4 2 1 2 3 −     = =     − −   i j k Direction of PQ is           −           − = − − 7 1 5 ~ 14 2 10 1 2 1 3 4 5 k j i (OE) Equation of PQ is r 1 5 1 1 4 7 µ         = − +         −     [N.B. For methods not using w at most three B1 marks.] M1 A1 A1 (3) B1 M1A1 (3) M1A1 dM1A1 (4) Total 10

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 13 © Cambridge International Examinations 2015 Qn & Part Solution Marks 9 t t B A x m m 2 5 e e : CF 0 ) 2 )( 5 ( − + = ⇒ = + − t q t p x t q t p x t q t p x cos sin sin cos cos sin : PI − − = ⇒ − = ⇒ + = && & t t t q t p t q t p t q t p cos 3 sin 2 cos 10 sin 10 sin 3 cos 3 cos sin − = − − + − − − ⇒ 11 3 2 and 3 11 3 0.1, 0.3 p q p q p q ⇒− + = + = ⇒ = − = 5 2 GS: e e 0.3 cos 0.1 sin t t x A B t t − = + + − CAO Initial conditions 3 3.3 3.0 = + ⇒ = + + ⇒ B A B A and from 5 2 5 e 2 e 0.3 sin 0.1 cos t t x A B t t − = − − − & 1 2 5 9.0 1.0 2 5 = − ⇒ = − − B A B A 5 2 1, 2 e 2e 0.3 cos 0.1 sin t t A B x t t − = = ⇒ = + + − CAO M1A1 M1 A1 dM1A1 A1 B1 M1 A1 A1 (11) Total 11 10 1 3 4 )1 ( 2 6 8 )1 ( 2 1 6 9 1 )1 ( 2 )1 3 ( 2 1 2 2 2 2 2 2 2 2 2 + − = + − = + + − + − − = + − + − x x x x x x x x x x x x 2 2 2 2 2 2 2 2 2 9 ( 3) 9 9 6 9 8 6 4 3 2 2( 1) 2( 1) 2( 1) 1 x x x x x x x x x x x x + + − − − − − − = = = + + + + 2 1 0 )1 ( 2 )1 3 ( 2 2 − ⇒ + − > > y x x 2 9 2 1 2 9 0 )1 ( 2 ) 3 ( 2 2 < < < > y y x x − ⇒ ⇒ + + (Not hence gets B1 only.) Turning points occur at equality, so 9 1 1 3, and , 2 3 2     − −         Asymptote is 4 = y . Intersection with axes at (0, 0) and 3 , 0 4       , Intersection with asymptote at 4 , 4 3   −     Correct graph. M1A1 A1 (3) B1 B1 (2) B1B1 (2) B1 (1) B1B1 B1 (3) Total 11

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 13 © Cambridge International Examinations 2015 Qn & Part Solution Marks 11 E (i)             → →             − − 0 0 0 0 8 5 0 0 1 1 3 0 4 3 2 1 2 2 4 1 5 3 3 1 3 2 1 1 4 3 2 1 L ⇒Dim (V) = 3 M1A1 A1 (3) (ii)             =             +             − − +             0 0 0 0 2 3 2 3 4 3 1 2 1 1 1 1 c b a i 0 3 2 = + + c b a ii 0 2 = + − c b a iii 0 3 3 = + − c b a i – iii ⇒ b = 0 ⇒ a + 3c = 0 and a + 2c = 0 ⇒ c = 0 and a = 0. So vectors are linearly independent. Since                                     − −             2 3 2 3 , 4 3 1 2 , 1 1 1 1 has three linearly independent vectors, it forms a basis for V. M1 M1A1 A1 B1 (5) (iii) W is not a vector space as it does not contain the zero vector. B1 (1)

Page 9 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 13 © Cambridge International Examinations 2015 Qn & Part Solution Marks (iv)             + + − − + − − → →             − − t z y x z y x y x x t z y x 3 5 2 0 0 0 5 0 0 1 3 0 3 2 1 2 4 1 3 3 1 2 1 1 3 2 1 L Alternatively: 1 2 3 1 1 2 1 3 3 1 4 2 x y z t α β γ             −       = + +       −             (i.e. in V) 2 3 2 5 2 3 3 2 5 4 2 x x y y z z t t α β γ α β γ α β γ α β γ α β γ α β γ = + + ⇒ + = + +  = − +  = − + ⇒ + = + +  = + +  0 ) ( ) ( = + − + ⇒ t z y x Vector belongs to V iff .t z y x + = + (OE) So belongs to W iff .t z y x + ≠ + (AG) M1A1 (M1A1) A1 M1 A1 (5) Total 14 11 O 12 4 2 4 9 6 0 80 80 0 1 2 6 7 α α α − − + = ⇒ + = ⇒ = − M1A1 A1 (3) (i) 0 2 6 2 6 1 4 2 4 3 = − − − − − − − λ λ λ 0 162 63 3 = + − ⇒ λ λ 3. and 6 0 ) 6 )( 3 )( 9 ( 2 1 = = ⇒ = − − + ⇒ λ λ λ λ λ M1 M1A1 M1A1 (5)

Page 10 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9231 13 © Cambridge International Examinations 2015 Qn & Part Solution Marks (ii)           − = ⇒ − = 2 2 1 9 e λ (Award M1A1 for any one 1 1 2 2 2 2 6 2 and 3 1 1 2 λ λ −       = ⇒ = = ⇒ =          e e and A1 for the other two). 2 1 2 2 1 1 2 1 2 1 3 6 ) ( e e e e Me Me e e M Mx b a b a b a b a + = + = + = + = λ λ M1A1 A1 (3) B1 B1 B1 (3) Total 14