Cambridge A Level Mathematics - Further 9231 — 2018 May/June Paper 1 · Variant 3

*5678150535* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/13 Paper 1 May/June 2018 3 hours Candidates answer on the Question Paper. Additional Materials: List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name in the spaces at the top of this page. 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 in the space provided. If additional space is required, you should use the lined page at the end of this booklet. The question number(s) must be clearly shown. 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 27 printed pages and 1 blank page. JC18 06_9231_13/2R © UCLES 2018 [Turn over

2 BLANK PAGE © UCLES 2018 9231/13/M/J/18

3 1 The variables x and y are such that y = −1 when x = 0 and @ x + dy dx A3 = y2 + x. (i) Find the value of dy dx when x = 0. [1] … … … (ii) Find also the value of d2y dx2 when x = 0. [4] … … … … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18 [Turn over

4 2 (i) Verify that ne −1 + e nn + 1en+1 = 1 nen − 1 n + 1en+1 . [1] … … … … … … … … Let SN = N Ð n=1 ne −1 + e nn + 1en+1 . (ii) Express SN in terms of N and e. [2] … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18

5 Let S = lim N→∞SN. (iii) Find the least value of N such that N + 1S −SN < 10−3. [3] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18 [Turn over

6 3 (i) Use de Moivre’s theorem to show that cos 41 = cos41 −6 cos21 sin21 + sin41. [3] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18

7 (ii) Hence find all the roots of the equation x4 −6x2 + 1 = 0 in the form tan q0, where q is a positive rational number. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18 [Turn over

8 4 The curve C has equation y = x2 + 7x + 6 x −2 . (i) Find the coordinates of the points of intersection of C with the axes. [2] … … … … … … … (ii) Find the equation of each of the asymptotes of C. [3] … … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18

9 (iii) Sketch C. [3] © UCLES 2018 9231/13/M/J/18 [Turn over

10 5 It is given that e is an eigenvector of the matrix A with corresponding eigenvalue ,. (i) Show that e is an eigenvector of A3 and state the corresponding eigenvalue. [3] … … … … … … … … … It is given that A = @ 2 0 −1 3 A . (ii) Find a matrix P and a diagonal matrix D such that A3 + I = PDP−1, where I is the 2 × 2 identity matrix. [5] … … … … … … … … … … © UCLES 2018 9231/13/M/J/18

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12 6 The equation 9x3 −9x2 + x −2 = 0 has roots !, ", '. (i) Use the substitution y = 3x −1 to show that 3! −1, 3" −1, 3' −1 are the roots of the equation y3 −2y −7 = 0. [2] … … … … … … … … … … The sum 3! −1n + 3" −1n + 3' −1n is denoted by Sn. (ii) Find the value of S3. [2] … … … … … … … … … © UCLES 2018 9231/13/M/J/18

13 (iii) Find the value of S−2. [4] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18 [Turn over

14 7 The lines l1 and l2 have vector equations r = ai + 9j + 13k + ,i + 2j + 3k and r = −3i + 7j −2k + -−i + 2j −3k respectively. It is given that l1 and l2 intersect. (i) Find the value of the constant a. [3] … … … … … … … … … … … … The point P has position vector 3i + j + 6k. (ii) Find the perpendicular distance from P to the plane containing l1 and l2. [4] … … … … … … … … © UCLES 2018 9231/13/M/J/18

15 … … … … … … … … … … (iii) Find the perpendicular distance from P to l2. [4] … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18 [Turn over

16 8 The curves C1 and C2 have polar equations, for 0 ≤1 ≤0, as follows: C1: r = a, C2: r = 2acos 1, where a is a positive constant. The curves intersect at the points P1 and P2. (i) Find the polar coordinates of P1 and P2. [2] … … … … … … … … … (ii) In a single diagram, sketch C1, C2 and their line of symmetry. [3] © UCLES 2018 9231/13/M/J/18

17 (iii) The region R enclosed by C1 and C2 is bounded by the arcs OP1, P1P2 and P2O, where O is the pole. Find the area of R, giving your answer in exact form. [5] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18 [Turn over

18 9 For the sequence u1, u2, u3, à, it is given that u1 = 8 and ur+1 = 5ur −3 4 for all r. (i) Prove by mathematical induction that un = 45 4 n + 3, for all positive integers n. [5] … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18

19 (ii) Deduce the set of values of x for which the infinite series u1 −3x + u2 −3x2 + à + ur −3xr + à is convergent. [2] … … … … … … … … (iii) Use the result given in part (i) to find surds a and b such that N Ð n=1 lnun −3 = N2 ln a + N ln b. [3] … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18 [Turn over

20 10 It is given that t ≠0 and td2x dt2 + 2dx dt + 9tx = 3t2 + 1. (i) Show that if y = tx then d2y dt2 + 9y = 3t2 + 1. [3] … … … … … … … … … … … (ii) Find x in terms of t, given that x = 1 90 and dx dt = 2 3 when t = 1 30. [9] … … … … … … … … … © UCLES 2018 9231/13/M/J/18

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22 11 Answer only one of the following two alternatives. EITHER (i) Show that Ó 1 20 −1 20 ex cos x dx = 1 2 e 1 20 + e−1 20! . [4] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18

23 (ii) It is given that, for n ≥0, In = Ó 1 20 −1 20 e2x cosnx dx. Show that, for n ≥2, 4In = nn −1 Ó 1 20 −1 20 e2x sin2x cosn−2x dx −nIn, and deduce the reduction formula n2 + 4In = nn −1In−2. [6] … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18 [Turn over

24 (iii) Using the result in part (i) and the reduction formula in part (ii), find the y-coordinate of the centroid of the region bounded by the x-axis and the arc of the curve y = ex cos x from x = −1 20 to x = 1 20. Give your answer correct to 3 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18

25 OR Let V be the subspace of >4 spanned by v1 = € 1 2 0 2  , v2 = € −2 −5 5 6  , v3 = € 0 −3 15 18  and v4 = € 0 −2 10 8  . (i) Show that the dimension of V is 3. [3] … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/13/M/J/18 [Turn over

26 (ii) Express v4 as a linear combination of v1, v2 and v3. [2] … … … … … … … … (iii) Write down a basis for V. [1] … … … … Let M = € 1 −2 0 0 2 −5 −3 −2 0 5 15 10 2 6 18 8  . (iv) Find the general solution of Mx = v1 + v2. [6] … … … … … … … … © UCLES 2018 9231/13/M/J/18

27 … … … … … … … … … … … … … … … … The set of elements of >4 which are not solutions of Mx = v1 + v2 is denoted by W. (v) State, with a reason, whether W is a vector space. [2] … … … … … … … © UCLES 2018 9231/13/M/J/18 [Turn over

28 Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … 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 2018 9231/13/M/J/18

IGCSE™ is a registered trademark. This document consists of 20 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/13 Paper 1 May/June 2018 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 International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the May/June 2018 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 2 of 20 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 3 of 20 GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 4 of 20 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 FT 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.

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 5 of 20 The following abbreviations may be used in a mark scheme or used on the scripts: AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) 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 SOI Seen or implied 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.

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 6 of 20 Question Answer Marks Guidance 1(i) ( ) 3 2 d d 0 1 0 1 d d y y x x   + = − + ⇒ =     B1 1(ii) 2 2 2 d d 3 1 d d y y x x x     + +        M1 A1 Method mark for good attempt at implicit differentiation of LHS. d 2 1 d y y x = + B1 Note: may expand bracket before differentiation but M1 is still for implicit differentiation 2 2 2 2 d d 4 3 1 2 1 3 d d y y x x   ⇒ + = −+ ⇒ = −     A1 5 Question Answer Marks Guidance 2(i) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 e e 1 e 1 1 e 1 e 1 e 1 e n n n n n n n n n n n n n + + + + − − + − = = + + + B1 Verifies result (AG). 2(ii) ( ) 1 1 1 1 ( e 1 e N N n n n S n n + =   = − =     +   ∑ 1 1 e − 2e 2 1 + 2e 2 1 − 3e 3 + 1 . ………… e N N ( ) 1 1 ) SOI 1 eN N + − + = M1 Uses difference method to sum. ( ) 1 1 1 . e 1 eN N + − + A1

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 7 of 20 Question Answer Marks Guidance 2(iii) S = 1 e B1 Finds S ( )( ) 1 N N S S + − 3 3 1 1 10 10 eN − − + < ⇒ < M1 Attempts to find difference between sum and sum to infinity. 1 3 e 10 N + ⇒ > ⇒ least such N is 6 . A1 6 Question Answer Marks Guidance 3(i) ( ) ( ) ( ) ( ) ( ) 4 3 4 4 3 2 2 4 6 4 c is c c is c s c is is + = + + − + + M1 Uses binomial theorem to expand ( ) 4 c is + . 4 2 2 4 cos4 6 c c s s θ ⇒ = − + M1 A1 Takes real part, AG. 3(ii) 4 2 4 cos4 tan 6 tan 1 cos θ θ θ θ = − + M1 A1 Divides through by 4 cos θ Let 4 2 tan , 6 1 0 cos4 0 x then x x θ θ = − + = ⇒ = 4 2 , 2 m m π θ π ⇒ = ± + ∈Z dM1 Following on from finding correct quartic Solves cos4 0 θ = . Roots are tanqπ where 1 3 5 7 , , , 8 8 8 8 q = . (q > 0) A1 A1 Alt methods: Solves tan(4θ) = (4t – t3 ) / (t4 – 6t2 + 1) and tan(4θ) = ∞. Solves equation in cot(θ) after dividing by sin4 θ 8

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 8 of 20 Question Answer Marks Guidance 4(i) ( ) ( ) 6,0 , 1,0 − − B1 States points of intersection with x -axis. ( ) 0, 3 − B1 States y -intercept 4(ii) One asymptote is 2 x = . B1 24 9 2 y x x = + + ⇒ − other asymptote is 9 y x = + . M1 A1 By inspection or long division. A0 if error in division 4(iii) B1 Sketches axes and asymptotes, labelled or to scale B1 Upper branch correctly located and orientated. B1 Lower branch correctly located and orientated. Penalise at most 1 mark for poor forms at infinity 8

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 9 of 20 Question Answer Marks Guidance 5(i) λ = Ae e SOI B1 ( ) ( ) ( ) 3 2 2 3 λ λ λ = = = = A e A Ae A Ae Ae e M1 Substitutes for Ae So eigenvalue is 3 λ . Special case: states eigenvalue is 3 B1 λ A1 5(ii) Eigenvalues of A are 2 and 3 . B1 Eigenvectors of A are 1 1    and 0 . 1    M1 AEF (allow any non-zero scalar multiple). So 1 0 1 1   =     P A1 Alt method: Find A3 + I = 9 0 19 28     −   B1 and 3 3 2 1 0 9 0 0 28 0 3 1   +   = =       +     D Columns of P and D can be permuted, but must match. M1 A1FT Eigenvalues (9, 28) and vectors 1 1    and 0 M1, A1 . 1    P and D FT on eigenvalues M1, A1FT 8

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 10 of 20 Question Answer Marks Guidance 6(i) Substitutes 1 3 y x + = M1 Accept substitution of 3 1 y x = − into given equation and derivation of equation in x . Obtains the given result A1 AG. 6(ii) 3 1 2 7 3 S S = + × M1 Uses 3 2 7 y y = + . Or uses formula for Σ (3α – 1)3 21 = A1 6(iii) ( )( ) ( )( ) ( )( ) ( )( )( ) 1 3 1 3 1 3 1 3 1 3 1 3 1 2 3 1 3 1 3 1 7 S α β α γ β γ α β γ − − − + − − + − − − = = − − − . M1 A1 Award M1A1 if 1 2 7 S−= − written down directly. 2 1 1 7 2 S S S − − = − M1 Uses 2 1 7 2 y y y − − = − . 2 4 49 s−= A1 Alt method: S–2 = 2 1 (3 1) α ∑ − = 2 2 2 2 2 (3 1) (3 1) (3 1) (3 1) (3 1) α β α β γ ∑ − − − − − = M1 A1 Alt method: Finds cubic with roots 1 3 1 α −, etc. M1 7z3 + 2z2 – 1 = 0 A1 Uses S2 = (S1)2 – 2xΣαβ M1 = 4 49 A1 ( )( ) ( )( )( ) ( ) ( ) 2 2 2 2 ( 3 1 3 1 ) 2 3 1 3 1 3 1 ( 3 1 ) 3 1 (3 1) (3 1) α β α β γ α α β γ ∑ − − − − − − ∑ − − − − M1 = 2 ( 2)2 – 2(7)(0) 7 − = 4 49 A1 8

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 11 of 20 Question Answer Marks Guidance 7(i) Solve two equations 9 2 7 2 λ µ + = + 13 3 2 3 λ µ + = −− M1 to obtain 3 λ = − and 2 µ = −. A1 Use third equation to obtain 2 a = . A1 7(ii) Normal to the plane is 12 4 = − + n i k M1 A1 Perpendicular distance 6 12 1 6 0 10 3.16 160 8 4 − −       = = =       −    ⋅ Alt method: Finds equation of plane M1, A1: Finds foot of perpendicular from P to plane M1 Hence length A1 Alt method: Find equation of plane 3x – z + 7 = 0 M1, A1 Use formula 3 3 0 1 6 7 9 1 × + × − + + = 10 M1, A1 M1 A1 7(iii) Cross product of direction of P to 2l with direction of 2l 6 1 2 6 2 10 8 3 6 − − −           × = −           − − −      M1 A1 Alt method: Find N (foot of perpendicular) in terms of parameter and uses scalar product with n to find parameter M1, A1 so PN = 3 0 1         −   Perpendicular distance from P to 2l is 2 10 6 10 3.16 2 3 + + = = −+ − i j k i j k M1 A1 Find length PN M1, A1 11

923 © U Qu 8 8 31/13 CLES 2018 uestion 8(i) 2 a a = , 3 a π       8(ii) cos cos θ θ ⇒ = ± and 2 , 3 a π       C Answ 1 2 ± Cambridge Inter wer rnational A Leve PUBLISHED Page 12 of 20 l – Mark Schem Marks M1 A1 B1 B1 B1 me Eliminates r . Both points need Semicircle for C Half of C2 inclu Other half of C2 Guidance ded for A1. C1 including r = a. uding r = 2a. 2 and line of symm May/June 2 . metry. 018

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 13 of 20 Question Answer Marks Guidance 8(iii) 2 2 2 3 4 cos d a π π θ θ ∫ M1 Finds area of segment OP1 and OP2 of C2 2 2 3 2 cos2 1 d a π π θ θ = + ∫ M1 Uses ( ) 2 1 cos cos2 1 2 θ θ = + 2 2 3 1 2 sin 2 2 a π π θ θ   = +     = 2 3 2 2 4 3 a π π     − +             = 2 3 3 2 a π   −       A1 Integrates correctly. Area 2 2 2 2 3 3 6 3 2 6 2 a a a a π π π   = − − = − +       M1 A1FT M1 for subtracting ‘their’ OP1P2 from 2 6 a π 10

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 14 of 20 Question Answer Marks Guidance 9(i) 5 : 4 3 4 n n n P u   = +     Let 1 n = then 1 5 4 3 8 4 P  + = ⇒     true. B1 States proposition. Proves base case. Assume kP is true for some k . Then B1 States inductive hypothesis. 1 1 5 5 4 3 3 4 4 k ku correct step +         = + − =               M1 Proves inductive step. 1 5 4 3, 4 k+   = +     A1 So Pk ⇒ Pk+1. Therefore, by induction, nP is true for all positive integers. A1 States conclusion. 9(ii) ( ) 5 5 3 4 4 4 4 n n n n n x u x x     − = =         so r = 5 4 x       M1 So series is convergent for 5 4 4 1 1 4 5 5 x x −< < ⇒− < < A1

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 15 of 20 Question Answer Marks Guidance 9(iii) 1 1 5 ln( 3) ln 4 4 n N N n n n u = =     − =           ∑ ∑ 1 1 5 ln ln4 4 N N n n n = =   = +    ∑ ∑ M1 Alt method: 1 5 ln( 3) 4 4 n N n n u ln =   − = ∏    ∑ M1 = ln 4N 1 5 4 n N       ∏ = Nln4+ ln 5 4 n ∑       ( ) 1 5 1 ln ln4 2 4 N N N = + + Use ( ) 1 1 1 2 N n n N N = = + ∑ . M1 = Nln4 + ( ) 1 ln( 2 N N + 5 4 ) M1 ( ) 2 5 5 ln ln 2 5 , 2 5 2 2 N N a b = + ⇒ = = oe Alt method: Writes series as an AP M1, uses summation formula M1 Correct answer A1 A1 ( ) 2 5 5 ln ln 2 5 , 2 5 2 2 N N a b = + ⇒ = = A1 10 Question Answer Marks Guidance 10(i) ' y x x t ′ = + B1 '' y x x tx ′ + + ′ = ′ ′ B1 Substitute correctly B1 AG 10(ii) Auxiliary equation: 2 9 0 3 . m m i + = ⇒ = ± M1 Correct auxiliary CF cos3 sin3 . A t B t = + A1

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 16 of 20 Question Answer Marks Guidance PI 2 a +b +c y t t = so y ’ = 2at + b and y ’’ = 2a M1 Differentiate twice and substitute 1 1 a , b 0, c . 3 27 = = = A1 2 1 1 cos3 sin3 3 27 y A t B t t = + + + A1FT Their CF + their PI both in correct form 9 x π = when 3 t π = gives 1 27 A = . B1 2 3 sin3 3 cos3 3 y A t B t t = − + + ′ M1 Differentiating their y of equivalent difficulty 2 3 x′ = when 3 t π = gives 27 B π = − . A1 2 cos3 sin3 9 1 27 t t t x t π − + + = A1 AEF 12

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 17 of 20 Question Answer Marks Guidance 11E(i) ( ) 2 2 2 2 2 2 e cos d e cos e sin d x x x I x x x x x π π π π π π − − −   = = − −   ∫ ∫ M1 A1 The parts can be either way round. 2 2 e sin x x I π π −   = −   M1 Integrates by parts again. 2 2 2 e e AG I π π − ⇒ = + A1 Alt method: Integrate Re(eix)ex by parts M1, M1, A1, A1

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 18 of 20 Question Answer Marks Guidance 11E(ii) 2 2 2 2 1 2 2 e cos e cos sin d 2 2 x n x n n n I x x x x π π π π − − −   = +     ∫ M1 Integrates by parts once. ( ) ( ) 2 2 2 2 2 0 e cos 1 sin cos d 4 x n n n x n x x x π π − − = − − − ∫ M1 Integrates by parts again. ( ) 2 2 2 2 2 4 1 e sin cos d x n n n I n n x x x nI π π − − ⇒ = − − ∫ A1 Simplifies to answer. (AG.) ( ) ( ) ( ) 2 2 2 2 2 4 1 e 1 cos cos d x n n n I n n x x x π π − − + = − − ∫ M1 For splitting sin2x ( ) ( ) ( ) 2 4 1 1 n n n n I n n I n n I − ⇒ + = − − − M1 ( ) ( ) 2 2 4 1 n n n I n n I − ⇒ + = − A1 AG

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 19 of 20 Question Answer Marks Guidance 11E(iii) ( ( ) 2 2 2 2 2 0 2 2 1 1 e 1 d ) e e 4 4 2 8 x y x I I π π π π π π − − −   = = = = −     ∫ M1 A1 Uses reduction formula to find 2I . ( ) 2 2 2 1 e e 0.575 2 8 e e I y I π π π π − −   −   = = =   +   M1 A1 Uses correct formula for y . 14 Question Answer Marks Guidance 11O(i) 1 2 0 0 1 2 0 0 2 5 3 2 0 1 3 2 0 5 15 10 0 0 1 1 2 6 18 8 0 0 0 0 − −         − − −     → →             " M1 A1 Reduces to echelon or reduced row echelon form. At least 2 manipulations…. Or finds 4 columns dependent, 3 columns independent dimV = rank = 3 . 1 2 3 4 c c c c A1 11O(ii) 4 3 2 1 4 3 2 1 2 2 c c c c = − − ⇒ = − − v v v v M1 A1 OE 11O(iii) { } 1 2 3 , , v v v B1 OE

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 20 of 20 Question Answer Marks Guidance 11O(iv) 1 2 1 1 0 0    + =    v v M so particular solution is 1 1 0 0       M1 A1 Finds particular solution. 2 0 x y − = 3 2 0 y z t + + = 0 z t + = M1 Finds basis for null space. So basis for null space is 2 1 1 1                 −           A1 AEF 1 2 1 1 . 0 1 0 1 λ          = +    −       x M1 A1 Finds general solution. Alt method: Set up equations M1,A1: Solve equations M1, A1: x = etc. M1, A1 or using augmented matrix. 11O(v) Not closed under addition 1 0 1 0 1 1 0 0 0 0 0 0                + =                  M1 Accept any valid reason with evidence. so not a vector space. A1 14