Cambridge A Level Mathematics - Further 9231 — 2018 Oct/Nov Paper 1 · Variant 1

*5063558434* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/11 Paper 1 October/November 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 11_9231_11/2R © UCLES 2018 [Turn over

2 1 The vectors a, b, c and d in >3 are given by a = ` 1 2 1 a , b = ` 2 9 0 a , c = ` 3 3 4 a and d = ` 0 −8 3 a . (i) Show that a, b, c is a basis for >3. [3] … … … … … … … … … … … … … (ii) Express d in terms of a, b and c. [2] … … … … … … … … © UCLES 2018 9231/11/O/N/18

3 2 The roots of the equation x3 + px2 + qx + r = 0 are !, 2!, 4!, where p, q, r and ! are non-zero real constants. (i) Show that 2p! + q = 0. [4] … … … … … … … … … … … … … (ii) Show that p3r −q3 = 0. [2] … … … … … … … © UCLES 2018 9231/11/O/N/18 [Turn over

4 3 The sequence of positive numbers u1, u2, u3, à is such that u1 < 3 and, for n ≥1, un+1 = 4un + 9 un + 4 . (i) By considering 3 −un+1, or otherwise, prove by mathematical induction that un < 3 for all positive integers n. [5] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18

5 (ii) Show that un+1 > un for n ≥1. [3] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18 [Turn over

6 4 A curve is defined parametrically by x = t −1 2 sin 2t and y = sin2t. The arc of the curve joining the point where t = 0 to the point where t = 0 is rotated through one complete revolution about the x-axis. The area of the surface generated is denoted by S. (i) Show that S = a0 Ó 0 0 sin3t dt, where the constant a is to be found. [5] … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18

7 (ii) Using the result sin 3t = 3 sin t −4 sin3t, find the exact value of S. [3] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18 [Turn over

8 5 It is given that , is an eigenvalue of the matrix A with e as a corresponding eigenvector, and - is an eigenvalue of the matrix B for which e is also a corresponding eigenvector. (i) Show that , + - is an eigenvalue of the matrix A + B with e as a corresponding eigenvector. [2] … … … … … … … … The matrix A, given by A = ` 2 0 1 −1 2 3 1 0 2 a has ` 1 2 1 a , ` 1 4 −1 a and ` 0 1 0 a as eigenvectors. (ii) Find the corresponding eigenvalues. [3] … … … … … … … … … … © UCLES 2018 9231/11/O/N/18

9 … … … … … … … … … The matrix B has eigenvalues 4, 5 and 1 with corresponding eigenvectors ` 1 2 1 a , ` 1 4 −1 a and ` 0 1 0 a respectively. (iii) Find a matrix P and a diagonal matrix D such that A + B3 = PDP−1. [3] … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18 [Turn over

10 6 The curve C has equation y = x2 + ax −1 x + 1 , where a is constant and a > 1. (i) Find the equations of the asymptotes of C. [3] … … … … … … … … … … … … … … … (ii) Show that C intersects the x-axis twice. [1] … … … … … … © UCLES 2018 9231/11/O/N/18

11 (iii) Justifying your answer, find the number of stationary points on C. [2] … … … … … … … … … … … … (iv) Sketch C, stating the coordinates of its point of intersection with the y-axis. [3] © UCLES 2018 9231/11/O/N/18 [Turn over

12 7 (i) Use de Moivre’s theorem to show that sin 81 = 8 sin 1 cos 11 −10 sin21 + 24 sin41 −16 sin61. [6] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18

13 … … … … … … … (ii) Use the equation sin 81 sin 21 = 0 to find the roots of 16x6 −24x4 + 10x2 −1 = 0 in the form sin k0, where k is rational. [4] … … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18 [Turn over

14 8 The plane 1 has equation r = ` 5 1 0 a + s ` −4 1 3 a + t ` 0 1 2 a . (i) Find a cartesian equation of 1. [3] … … … … … … … … … … … The plane 2 has equation 3x + y −z = 3. (ii) Find the acute angle between 1 and 2, giving your answer in degrees. [2] … … … … … … … … … © UCLES 2018 9231/11/O/N/18

15 (iii) Find an equation of the line of intersection of 1 and 2, giving your answer in the form r = a + ,b. [5] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18 [Turn over

16 9 The curve C has polar equation r = 5cot 1, where 0.01 ≤1 ≤1 20. (i) Find the area of the finite region bounded by C and the line 1 = 0.01, showing full working. Give your answer correct to 1 decimal place. [3] … … … … … … … … … … … Let P be the point on C where 1 = 0.01. (ii) Find the distance of P from the initial line, giving your answer correct to 1 decimal place. [2] … … … … … … … … … © UCLES 2018 9231/11/O/N/18

17 (iii) Find the maximum distance of C from the initial line. [3] … … … … … … … … … … … … … … … (iv) Sketch C. [2] © UCLES 2018 9231/11/O/N/18 [Turn over

18 10 (i) Find the particular solution of the differential equation d2x dt2 + 2dx dt + 10x = 37 sin 3t, given that x = 3 and dx dt = 0 when t = 0. [10] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18

19 … … … … … … … … … … … … (ii) Show that, for large positive values of t and for any initial conditions, x ≈37 sin3t −&, where the constant & is such that tan & = 6. [3] … … … … … … … … … … © UCLES 2018 9231/11/O/N/18 [Turn over

20 11 Answer only one of the following two alternatives. EITHER (i) By considering 2r + 12 −2r −12, use the method of differences to prove that n Ð r=1 r = 1 2nn + 1. [3] … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18

21 (ii) By considering 2r + 14 −2r −14, use the method of differences and the result given in part (i) to prove that n Ð r=1 r3 = 1 4n2n + 12. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18 [Turn over

22 The sums S and T are defined as follows: S = 13 + 23 + 33 + 43 + à + 2N3 + 2N + 13, T = 13 + 33 + 53 + 73 + à + 2N −13 + 2N + 13. (iii) Use the result given in part (ii) to show that S = 2N + 12N + 12. [1] … … … … (iv) Hence, or otherwise, find an expression in terms of N for T, factorising your answer as far as possible. [2] … … … … … … … … … … … (v) Deduce the value of S T as N →∞. [2] … … … … © UCLES 2018 9231/11/O/N/18

23 OR The curve C has equation x2 + 2xy = y3 −2. (i) Show that A −1, 1 is the only point on C with x-coordinate equal to −1. [2] … … … … … … … … For n ≥1, let An denote the value of dny dxn at the point A −1, 1. (ii) Show that A1 = 0. [3] … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18 [Turn over

24 (iii) Show that A2 = 2 5. [3] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18

25 … … … … … … … … … … Let In = Ô 0 −1 xn dny dxn dx. (iv) Show that for n ≥2, In = −1n+1An−1 −nIn−1. [3] … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18 [Turn over

26 … … … … … … … … … … (v) Deduce the value of I3 in terms of I1. [2] … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18

27 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. … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/O/N/18

28 BLANK PAGE 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 9231/11/O/N/18

This document consists of 21 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/11 Paper 1 October/November 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 October/November 2018 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 2 of 21 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.

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 3 of 21 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.

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 4 of 21 Abbreviations 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)

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 5 of 21 Question Answer Marks Guidance 1(i) EITHER: 1 2 3 2 9 3 36 10 3 9 1 0 1 0 4 = − −× = −≠ M1 A1 Calculates determinant OR: 2 3 0 2 9 3 0 4 0 α β γ α β γ α γ + + = + + = + = 2 0 β γ ⇒ − = 9 5 0 β γ − = (M1A1) Solves homogeneous system of equations. Eliminates one variable. 0 0 β α γ ⇒ = ⇒ = = Therefore , , a b c are linearly independent (and span 3 R ) so form a basis for 3 R . A1 States either that vectors are linearly independent or that vectors span 3 R . 3 1(ii) 1 2 3 0 2 9 3 8 1 0 4 3 l m n           + + = −                1 l m ⇒= = − and 1 n = . M1 Sets up system of equations. ⇒ = − − d c b a A1 2

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 6 of 21 Question Answer Marks Guidance 2(i) 2 4 p α α α + + = − B1 Sum of roots. 2 2 2 2 4 8 q α α α + + = B1 Sum of products in pairs. 2 14 7 q p α α = − M1 Combines equations. 2 0 p q α ⇒ + = A1 Verifies result (AG). 4 2(ii) 3 8 r α = − B1 Product of roots. 3 3 3 3 0 q r p r q p ⇒ = ⇒ − = B1 Verifies result (AG). 2

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 7 of 21 Question Answer Marks Guidance 3(i) ( ) 1 3 given u < B1 States base case. Assume that 3 ku < B1 States inductive hypothesis. Then 1 1 4 9 3 3 3 0 3 4 4 k k k k k k u u u u u u + + + − + − = − = > ⇒ < + + M1 A1 Hence, by induction, 3 n u < for all 1 n. . B1 States conclusion. 5 3(ii) 2 1 4 9 9 4 4 n n n n n n n u u u u u u u + + − + − = − = + + M1 A1 Considers 1 n n u u + − . So 1 3 0 n n n u u u + < ⇒ − > . B1 Uses 3 n u < 3

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 8 of 21 Question Answer Marks Guidance 4(i) d 1 cos2 d x t t = − d 2sin cos d y t t t = B1 B1 for d d x t and d d y t . 2 d 1 cos2 2sin d x t t t = − = M1 Uses an appropriate identity. 2 2 4 2 2 d d 4sin 4sin cos 2sin d d x y t t t t t t     + = + =         A1 ( )( ) 2 0 2 sin 2sin d 4 S t t t a π π = ⇒ = ∫ M1 A1 Uses the correct formula for S 5 4(ii) 3 0 0 4 sin d 3sin sin3 d S t t t t t π π π π = = − ∫ ∫ M1 Uses the result 3 4sin 3sin sin3 t t t = − 0 1 3cos cos3 3 t t π π   = − +     A1 1 1 16 3 3 3 3 3 π π     = − −−+ =         A1 Answer must be exact. 3

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 9 of 21 Question Answer Marks Guidance 5(i) = λ Ae e and = µ Be e ⇒ + = λ + µ Ae Be e e M1 Adds equations. ( ) ( ) ⇒ + = λ + µ A B e e A1 2 5(ii) 2 0 1 1 3 1 2 3 2 6 1 0 2 1 3       − =          M1 Multiples matrix with vector. 3 ⇒λ = k A1 Finds one eigenvalue. 1, 2 λ = A1 Finds other two. 3 5(iii) 1 1 0 2 4 1 1 1 0     =     −   P B1 3 3 3 7 0 0 343 0 0 0 6 0 0 216 0 0 0 27 0 0 3         = =               D M1 A1 Or correctly matched permutations of columns. 3

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 10 of 21 Question Answer Marks Guidance 6(i) Vertical asymptote is 1 x = −. B1 ( )( ) 2 1 1 1 x ax x x a a + −= + + − − or 2 1 1 1 x a x x ax + − + + − M1 By inspection or long division. Thus the oblique asymptote is 1 y x a = + − A1 3 6(ii) 2 4 0 a + > B1 1 6(iii) ( )( ) ( ) ( ) 2 2 2 1 2 1 0 2 1 0 1 x x a x ax x x a x + + − + − = ⇒ + + + = + Discriminant ( ) 4 4 1 0 a = − + < M1 Therefore there are no stationary points on C . A1 2

923 © U Qu 6 31/11 CLES 2018 uestion 6(iv) C Answe Cambridge Inter er rnational A Leve PUBLISHED Page 11 of 21 l – Mark Schem Marks B1 C B1 B1 E 3 me Correct y-intercep Each branch corre Octob Guidance pt and asymptotes ect ber/November 2 s drawn. 018

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 12 of 21 Question Answer Marks Guidance 7(i) Write cos c θ = , sin s θ = . ( ) 8 cos8 sin8 i c is θ θ + = + M1 Uses binomial theorem. 7 5 3 3 5 7 sin8 8 56 56 8 c s c s c s cs θ ⇒ = − + − A1 ( ) 6 4 2 2 4 6 8 7 7 cs c c s c s s = − + − M1 Factorises. ( ) ( ) ( ) 3 2 6 4 2 2 4 6 2 2 2 2 4 6 7 7 1 7 1 7 1 c c s c s s s s s s s s − + − = − − − + − − M1 Uses 2 2 1 c s = − . ( ) ( ) ( ) 2 4 6 2 4 6 4 6 6 1 3 3 7 2 7 s s s s s s s s s = − + − − − + + − − 2 4 6 1 10 24 16 s s s = − + − A1 ( ) 2 4 6 sin8 8 1 10 24 16 cs s s s θ ⇒ = − + − A1 AG 6

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 13 of 21 Question Answer Marks Guidance 7(ii) ( ) 2 4 6 sin8 sin 2 2 4 1 10 24 16 sin 2 cs s s s θ θ θ = ⇒ = − + − M1 Uses sin2 2cs θ = to relate with equation in part (i) sin8 0 8 nπ θ θ = ⇒ = M1 Solves sin8 0 θ = sin 8 x π = A1 Gives one correct solution sin , 1, 2 3 1 ,2,3,9,10,11 8 n n or π = ± ± ± A1 Gives five other solutions. Allow different values of k as long as six distinct solutions are found. sin0 and sin 2 π must be excluded. 4

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 14 of 21 Question Answer Marks Guidance 8(i) 4 1 3 8 4 0 1 2 − = −+ − i j k i j k M1 Finds normal to 1 Π . 8 4 3 x y z −+ − = M1 A1 Uses point on plane to find cartesian equation, AEF. 3 8(ii) 1 3 8 1 9 11cos 4 1 θ −       ⋅ =       − −    M1 Uses scalar product. 72.5 θ ⇒ = ° A1 Accept 1.26 rad. 2 8(iii) 3 1 1 4 13 25 1 8 4 − = + + − − i j k i j k M1 A1 Finds direction of line of intersection. Point on both planes is, e.g. ( ) 1,1 ,1 . M1 A1 Finds point common to both planes. ( ) 4 13 25 = + + + λ + + r i j k i j k A1 States vector equation of line. 5

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 15 of 21 Question Answer Marks Guidance 9(i) 2 0.01 25 cot d 2 π θ θ ∫ M1 Uses 2 1 d 2 r θ ∫ . [ ]2 0.01 25 lnsin 2 π θ = A1 25 lnsin0.01 57.6 2 = − ≈ A1 3 9(ii) 1 1 1 2 2 2 5 5cos sin sin 2 2 y θ θ θ = = M1 Uses sin . y r θ = 0.01 0.5 y θ = ⇒ ≈ A1 2 9(iii) ( ) 1 2 d 5 sin 2 cos2 0 or max sin 2 1 d 2 y θ θ θ θ − = = = M1 A1 Sets d 0 d y θ = or considers max (AEF). ( ) 5 2 3.54 2 y ⇒ = = A1 3

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 16 of 21 Question Answer Marks Guidance 9(iv) B1 Intersecting the initial line only when x = 0 and y = 0. B1 Correct shape. 2

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 17 of 21 Question Answer Marks Guidance 10(i) 2 2 10 0 1 3 . m m m i + + = ⇒ = −± M1 Finds complementary function. ( ) cos3 sin3 t x e A t B t − = + A1 cos3 sin3 x p t q t = + M1 Finds particular integral. 3 sin3 3 cos3 9 cos3 9 sin3 x p t q t x p t q t ⇒ = − + ⇒ = − −   A1 ( ) ( ) 6 cos3 6 sin3 37sin3 p q t q p t t + + − = M1 6, 1 p q ⇒ = − = A1 ( ) cos3 sin3 6cos3 sin3 t x e A t B t t t − = + − + 3 x = when 0 t = 6 3 9 A A ⇒ − = ⇒ = B1 Uses initial conditions. ( ) ( ) 3 sin3 3 cos3 cos3 sin3 18sin3 3cos3 t t x e A t B t e A t B t t t − − = − + − + + +  M1 A1 0 x =  when 0 3 3 0 2 t B A B = ⇒ − + = ⇒ = ( ) 9cos3 2sin3 6cos3 sin3 t x e t t t t − = + − + A1 Obtains solution, AEF. 10 10(ii) As t →∞, 0 t e−→ 6cos3 sin3 x t t ⇒ ≈− + B1 Obtains limit. 1 6 sin3 6cos3 37 sin3 cos3 37 37 t t t t   − = −     M1 Converts to ( ) sin 3 R t φ − ( ) 1 37 sin 3 tan 6 . t − = − A1 AG 3

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 18 of 21 Question Answer Marks Guidance 11E(i) ( ) ( ) 2 2 2 1 2 1 8 r r r + − − = B1 ( ) 2 2 1 8 1 2 1 n r r n = ⇒ = − + + ∑ M1 Sums both sides and uses method of differences. ( ) 1 1 1 2 n r r n n = ⇒ = + ∑ A1 AG. 3 11E(ii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4 4 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 r r r r r r + − − = + + − + − − M1 Uses difference of squares or expands. ( )( ) ( ) 2 3 8 2 8 16 4 r r r r = + = + A1 4 4 3 1 1 1 1 4 1 2 2 n n r r r r n = =     ⇒ + = − − + +         ∑ ∑ M1 Sums both sides and uses method of differences. ( ) 4 4 3 1 1 1 1 4 1 2 2 2 n r r n n n =     = + − − +         ∑ M1 Uses formula for r ∑ ( ) ( ) 2 2 2 2 2 2 1 1 1 1 1 1 1 2 2 2 2 2 n n n n n n            = + − + + − + = +                          A1 AG 5 11E(iii) ( ) ( ) ( ) ( ) 2 2 2 2 1 2 1 2 2 2 1 1 4 S N N N N = + + = + + B1 Uses formula for 3r ∑ . AG. 1

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 19 of 21 Question Answer Marks Guidance 11E(iv) ( ) ( ) ( ) ( ) 3 2 2 2 2 1 2 2 1 1 2 1 N r T S r N N N N = = − = + + − + ∑ M1 Eliminates even terms from S . ( ) ( ) 2 2 1 2 4 1 N N N + + + A1 Accept ( ) ( ) ( ) 2 N 1 2 2 1 . N N + + + 2 11E(v) ( ) 2 2 2 2 2 1 4 4 1 2 4 1 2 4 1 N S N N T N N N N + + + = = + + + + M1 Writes fraction as quadratic in N divided by quadratic in N . Converges to 2 as N →∞. A1 2 11O(i) ( )( ) 3 2 1 2 3 0 1 3 0 x y y y y y = −⇒ + − = ⇒ − + + = M1 Considers cubic polynomial in y . There is only one real root ( 1 = ). A1 2

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 20 of 21 Question Answer Marks Guidance 11O(ii) d 2 2 d y x x y x   + +     B1 Differentiates LHS correctly. 2 d 3 d y y x = B1 Differentiates RHS correctly. ( ) 2 d 3 2 2 2 d y y x x y x ⇒ − = + d 1, 1 0. d y x y x = − = ⇒ = B1 Substitutes ( ) 1,1 − , AG. 3 11O(iii) ( ) 2 2 2 d d d d 3 2 6 2 2 2 d d d d y y y y y x y x x x x   − + − = +     M1 M1 Differentiates equation again. 2 2 d d 2 1, 1, 0 d 5 d y y x y x x = − = = ⇒ = A1 Substitutes ( ) 1,1 − and d 0 d y x = , AG. 3 11O(iv) 0 0 0 1 1 1 1 1 1 1 1 d d d d d d d d n n n n n n n n n y y y x x x n x x x x x − − − − − − − −   = −     ∫ ∫ M1 A1 Integrates by parts. ( ) 1 1 1 1 . n n n A nI + − − = − − A1 AG. 3

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2018 © UCLES 2018 Page 21 of 21 Question Answer Marks Guidance 11O(v) ( ) ( ) 4 3 2 2 2 1 1 1 2 1 3 3 2 6 5 I A I A A I I = − − = − − − = + M1 A1 Uses reduction formulae. 2