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

*4111637505* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/11 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_11/FP © UCLES 2018 [Turn over

2 1 The curve C is defined parametrically by x = et −t, y = 4e 1 2t. Find the length of the arc of C from the point where t = 0 to the point where t = 3. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18

3 2 It is given that fn = 23n + 8n−1. By simplifying fk + fk + 1, or otherwise, prove by mathematical induction that fn is divisible by 9 for every positive integer n. [6] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18 [Turn over

4 3 The curve C has polar equation r = cos 21, for −1 40 ≤1 ≤1 40. (i) Sketch C. [2] (ii) Find the area of the region enclosed by C, showing full working. [3] … … … … … … … … © UCLES 2018 9231/11/M/J/18

5 … … … … … … … … … … (iii) Find a cartesian equation of C. [3] … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18 [Turn over

6 4 It is given that the equation x3 −21x2 + kx −216 = 0, where k is a constant, has real roots a, ar and ar−1. (i) Find the numerical values of the roots. [6] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18

7 … … … … … … … … … … … … … … … (ii) Deduce the value of k. [2] … … … … … … … … … © UCLES 2018 9231/11/M/J/18 [Turn over

8 5 Let Sn = n Ð r=1 −1r−1r2. (i) Use the standard result for n Ð r=1 r2 given in the List of Formulae (MF10) to show that S2n = −n2n + 1. [4] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18

9 (ii) State the value of lim n→∞ S2n n2 and find lim n→∞ S2n+1 n2 . [4] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18 [Turn over

10 6 The curve C has equation y = x2 + b x + b , where b is a positive constant. (i) Find the equations of the asymptotes of C. [3] … … … … … … … … … … … … … … … … (ii) Show that C does not intersect the x-axis. [1] … … … … … © UCLES 2018 9231/11/M/J/18

11 (iii) Justifying your answer, find the number of stationary points on C. [2] … … … … … … … … … … … … (iv) Sketch C. Your sketch should indicate the coordinates of any points of intersection with the y-axis. You do not need to find the coordinates of any stationary points. [3] © UCLES 2018 9231/11/M/J/18 [Turn over

12 7 Find the particular solution of the differential equation 49d2y dx2 + 14dy dx + y = 49x + 735, given that when x = 0, y = 0 and dy dx = 0. [10] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18

13 … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18 [Turn over

14 8 The linear transformation T : >4 →>3 is represented by the matrix M, where M = ` 1 2 ! −1 2 6 −3 −3 3 10 −6 −5 a and ! is a constant. When ! ≠0 the null space of T is denoted by K1. (i) Find a basis for K1. [5] … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18

15 When ! = 0 the null space of T is denoted by K2. (ii) Find a basis for K2. [3] … … … … … … … … … … … … … … … … (iii) Determine, justifying your answer, whether K1 is a subspace of K2. [2] … … … … … … … © UCLES 2018 9231/11/M/J/18 [Turn over

16 9 (i) Using the substitution u = tan x, or otherwise, find Ó sec2x tan2x dx. [2] … … … … … … … … … … … … It is given that, for n ≥0, In = Ó 1 40 0 secnx tan2x dx. (ii) Using the result that d dxsec x = tan x sec x, show that, for n ≥2, n + 1In = ï2n−2 + n −2In−2. [5] … … … … … … … © UCLES 2018 9231/11/M/J/18

17 … … … … … … … … … … … … (iii) Hence find the mean value of sec4x tan2x with respect to x over the interval 0 ≤x ≤1 40, giving your answer in exact form. [3] … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18 [Turn over

18 10 The line l1 is parallel to the vector ai −j + k, where a is a constant, and passes through the point whose position vector is 9j + 2k. The line l2 is parallel to the vector −ai + 2j + 4k and passes through the point whose position vector is −6i −5j + 10k. (i) It is given that l1 and l2 intersect. (a) Show that a = −6 13. [3] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18

19 (b) Find a cartesian equation of the plane containing l1 and l2. [4] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18 [Turn over

20 (ii) Given instead that the perpendicular distance between l1 and l2 is 330, find the value of a. [5] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18

21 11 Answer only one of the following two alternatives. EITHER (i) Show that if z = ei1 and z ≠−1 then z −1 z + 1 = i tan 1 21. [3] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18 [Turn over

22 (ii) Hence, or otherwise, show that if z is a cube root of unity then z3 −1 z3 + 1 + z2 −1 z2 + 1 + z −1 z + 1 = 0. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18

23 (iii) Hence write down three roots of the equation z3 −1z2 + 1z + 1 + z2 −1z3 + 1z + 1 + z −1z3 + 1z2 + 1 = 0 and find the other three roots. Give your answers in an exact form. [6] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18 [Turn over

24 OR It is given that e is an eigenvector of the matrix A, with corresponding eigenvalue ,. (i) Write down another eigenvector of A corresponding to ,. [1] … … … (ii) Write down an eigenvector and corresponding eigenvalue of An, where n is a positive integer. [2] … … … Let A = ` 3 0 0 2 7 0 4 8 1 a . (iii) Find a matrix P and a diagonal matrix D such that An = PDP−1. [7] … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18

25 … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/18 [Turn over

26 (iv) Determine the set of values of the real constant k such that ∞ Ð n=1 knAn −kAn+1 = kA. [4] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2018 9231/11/M/J/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/M/J/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/M/J/18

IGCSE™ is a registered trademark. 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 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.

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 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 May/June 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 May/June 2018 © UCLES 2018 Page 4 of 21 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.

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 5 of 21 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.

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 6 of 21 Question Answer Marks Guidance 1 d 1 d = − t x e t and 1 2 d 2 d = t y e t B1 ( ) 2 2 2 d d 1 d d t x y e t t     + = +         M1 A1 M1 for using 2 2 2 d d d d d d s x y t t t       = +             Arc length is 3 3 0 0 1 d t t e t e t   + = +   ∫ M1 M1 for good attempt at correct integral = 2 + ݁ଷ (or 22.1) A1 5 Question Answer Marks Guidance 2 We have that ( )1 9 = f is divisible by 9. B1 Checks base case. Assume that ( ) f k is divisible by 9. B1 Makes general statement. ( ) ( ) 3 3 3 1 1 2 8 2 8 + − + + = + + + = k k k k f k f k M1 Uses expansion of ( )1 . + f k 23.23k + 8.8k–1 + 23k + 8k–1 OE A1 Correct split of powers ( ) 3 1 9 2 8 − = + k k OE so ( )1 + f k is divisible by 9. A1 Alt method: f (k + 1) = 3 3 2 8 + + k k So if f (k) is divisible by 9, so is f (k +1), (and f (1) is divisible by 9), ( ) f n is divisible by 9 for every integer n ⩾1 A1 = 23.23k + 8.8k–1 M1 = 8 f(k) A1 6

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 7 of 21 Question Answer Marks Guidance 3(i) B1 Correct position including label of 1 on initial line, and symmetric about initial B1 Single correct loop. 2 3(ii) 2 4 4 1 2 2 π π θ θ −∫ cos d M1 Correct integral 4 4 1 cos4 1 4 π π θ θ − = + = ∫ d M1 Correct use of double angle formula 4 4 1 1 sin 4 4 4 π π θ θ −   +     = 8 π = 0.393 A1 3 1

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 8 of 21 Question Answer Marks Guidance 3(iii) 2 2 cos sin θ θ = − r OE B1 Uses trig identity Thus ( ) 3 2 2 2 2 2 + = − x y x y M1 A1 Uses cosθ = x r or sin θ = y r orboth , AEF 3 Question Answer Marks Guidance 4(i) 3 216 6 αβγ = = ⇒ = a a M1 A1 Uses product of roots 1 21 − + + = a ar ar ( ) 1 6 1 21 − + + = r r M1 Uses sum of roots 2 2 5 2 0 2 − + = ⇒ = r r r or 0.5 = r M1 A1 Substitutes for a and solves quadratic Roots are 6, 12, 3 A1 6 4(ii) ( ) ( ) ( ) 6 12 6 3 12 3 126 αβ αγ βγ = + + = + + = k M1 A1 Or finds coefficient of x in ( )( )( ) 3 6 12 − − − x x x . Or substitutes root into equation 2

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 9 of 21 Question Answer Marks Guidance 5(i) S2n = 12 – 22 + 32 –42….. M1 Uses correct difference. so ( ) 2 2 2 2 2 2 2 1 1 1 1 2 2 8 n n n n r r n r r S r r r r = = = = = − = − ∑ ∑ ∑ ∑ A1 Alt method: Use ( ) ( ) 2 2 1 1 2 1 2 − − ∑ ∑ n n r r =A1 Thus ( )( )( ) ( )( ) 2 1 8 2 2 1 4 1 1 2 1 6 6 = + + − + + n S n n n n n n M1 ( ) ( ) 2 2 1 1 14 4 4 − + − ∑ ∑ ∑ n n n r r n r M1 Factorising, ( )( ) ( ) 2 1 2 1 4 1 4 4 2 1 3 = + + − − = − + n S n n n n n n A1 ( ) 2 1 = − + n n A1 AG 4 5(ii) 2 2 lim 2 →∞ = − n n S n B1 ( ) ( ) 2 2 2 1 2 1 2 1 + = + − + n n n S S n M1 So, ( ) ( ) ( )( ) 2 2 1 2 1 2 1 2 1 1 + = − + + + = + + n S n n n n n M1 Uses the result given in (i) or using 2 2 lim →∞ n n S n and correct sign Thus 2 1 2 lim 2 + →∞ = n n S n A1 Alt: Find limit from previous line directly 4

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 10 of 21 Question Answer Marks Guidance 6(i) Vertical asymptote is = − x b . B1 ( )( ) 2 2 + = + − + + x b x b x b b b or 2 0 − + + + x b x b x x b M1 By inspection or long division. Thus the oblique asymptote is = − y x b A1 3 6(ii) If 0 = y then 2 0 + = x b which has no real root. B1 Must refer to b > 0 OE 1 6(iii) ( ) ( ) ( ) 2 2 2 2 d 0 2 0 d + − + = = ⇒ + − = + x x b x b y x bx b x x b Or differentiating 2 + = − + + b b y x b x b and setting d 0 d = y x gives ( ) 2 2 1 0 + − = + b b x b . M1 Find d d y x and set = 0 2 0 + > b b Therefore there are two stationary points on C A1 Use discriminant or (x + b)2 to show two stationary points 2

923 © U Qu 31/11 CLES 2018 uestion 6(iv) C Answe Cambridge Inter r rnational A Leve PUBLISHED Page 11 of 21 l – Mark Schem Marks B1 B1 B1 3 me Intersection (0,1 Each branch cor Penalise at most Guidance 1) given and asym rrect t one mark for poo May/June 2 mptotes drawn or forms at infinit 018 ty

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 12 of 21 Question Answer Marks Guidance 7 The auxiliary equation is 2 1 1 0 7 7   + = ⇒ = −     n n . M1 CF: ( ) 1 7 − = + x y A Bx e . A1 PI: y px q = + , ' = y p and '' 0 = y . M1 Correct form of PI and derivatives 14 49 735 49, 49 p px q x p q + + = + ⇒ = = . M1 A1 Substitutes in equation correctly Thus ( ) ( ) 1 7 49 1 − = + + + x y A Bx e x . A1 FT FT only on correct form of CF AEF 0 = y when 0 = x gives 49 0 49 + = ⇒ = − A A . B1 ( ) 1 1 7 7 1 49 7 − − ′ = − + + + x x y A Bx e Be . M1 Differentiating their ݕ 0 ′ = y when 0 = x gives 1 49 0 56 7 − + + = ⇒ = − A B B A1 AEF Thus ( ) ( ) 1 7 49 56 49 1 . − = − + + + x y x e x A1 10

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 13 of 21 Question Answer Marks Guidance 8(i) Reduces to row echelon form 1 2 α 1 1 2 α 1 2 6 3 3 ~ 0 2 3 2α 1 3 10 6 5 0 4 6 3α 2 − −         − − −− −         − − −− −     M1 Good attempt at REF 1 2 α 1 ~ 0 2 3 2α 1 0 0 α 0 −     −− −       A1 Solves system of equations 2 αz t 0 + + −= x y M1 Forms system of equations from row echelon matrix ( ) 2 3 2 0 α − + −= y z t 0 α = z Since 0 α ≠ , 0 = z and A1 Three correct equations 1 0 1 ( ) 0 2             =               K A1 AEF 5

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 14 of 21 Question Answer Marks Guidance 8(ii) If 0 α = then 2 0 + −= x y t 2 3 0 − −= y z t M1 Forms system of 2 equations 2 6 0 3 3 1 0 2 , 2 0 1 0 2 3 −                ⇒ =      −           K any of or A1 A1 AE 3 8(iii) A basis vector for ܭଵ forms part of a basis for ܭଶ or a linear combination of basis vectors from K2 form basis for K1. M1 Credit should be given for a correct conclusion using their bases. Therefore ܭଵ is a subspace of ܭଶ. A1 FT 2

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 15 of 21 Question Answer Marks Guidance 9(i) 2 d sec d u x x = so integral becomes 2 ∫u du M1 Uses substitution correctly 3 1 tan 3 = + x C A1 2 9(ii) 2 2 2 4 0 sec sec tan π − =∫ n nI x x x B1 Separates into correct structure = n 2 3 2 4 4 4 0 0 1 2 sec tan sec tan 3 3 π π − − −   −   ∫ n n x x x xdx M1 Uses integration by parts correctly = 2 2 2 3 − n – ( ) 2 2 2 4 0 2 sec tan sec 1 3 π − − − ∫ n n x x x dx M1 Uses 2 2 tan sec 1 = − x x = 2 2 2 2 2 2 3 3 3 − − − − − + n n n n n I I A1 Thus ( ) ( ) ( ) 2 2 3 2 2 2 − − + − = + − n n n n I n I . A1 AG 5

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 16 of 21 Question Answer Marks Guidance 9(iii) 3 4 2 0 tan 1 3 3 π   = =     x I B1 FT Attempts to find I2 4 2 5 2 2 = + I I M1 Uses reduction formula Mean value 4 4 32 15 π π = = I A1 3 Question Answer Marks Guidance 10(i)(a) Point on 1l is 9 2 λ λ λ     −     +   a and on 2l is 6 5 2 10 4 µ µ µ −−     −+     +   a B1 Point of intersection: 6 9 5 2 2 10 4 λ µ λ µ λ µ −−         − = −+         + +     a a M1 Equates coordinates of points 6 1, 12, 13 a ⇒ = = = − µ λ A1 AG 3

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 17 of 21 Question Answer Marks Guidance 10(i)(b) Normal to the plane: 6 30 6 1 1 6 ~ 13 5 13 13 13 6 2 4 13 − − = − + − − + − i j k i j k i j k M1 A1 Uses cross product to find normal to the plane AEF Using point on plane: e.g. ( ) 5 9 2 43 − = M1 Substitutes a point Equation of plane: 13 5 43 − + − = x y z A1 AEF 4

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 18 of 21 Question Answer Marks Guidance 10(ii) 1 1 6 5 2 4 − = − − + − a i aj ak a i j k 2 6 5 36 26 −     − = +       a a a B1 Finds the magnitude of the cross product of the direction vectors of the lines 6 6 5 9 5 36 78 10 2 a a a − −       −− ⋅− = +       −    M1 A1 Takes dot product of the correct vectors 2 3 30 36 26 36 78 + = + a a M1 Puts distance equal to 3 30 ( ) ( ) 2 2 15 18 13 6 13 ⇒ + = + a a ( ) 2 26 3 0 ⇒ − = a a 3 ⇒ = A1 5

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 19 of 21 Question Answer Marks Guidance 11E(i) If 1 ≠− z then ( ) 1 1 2 2 1 1 2 2 1 2 sin or 1 2 cos 1 θ θ − −   − − = =     + +   + z z z i z z z M1 A1 Using de Moivre’s theorem or multiplying numerator and denominator by cos 1 sin θ θ + −i 1 2 sin 1 2 tan 1 2 2cos 2 θ θ θ = = i i A1 3 11E(ii) If 1 = z then 1 0 −= rz so the given sum is zero. B1 If 1 ≠ z then 2 3 π = i z e or 2 3 π − = i z e , so B1 Must consider all three roots of unity 4 2 3 2 3 3 3 2 4 2 3 3 1 1 1 1 1 0 1 1 1 1 1 π π π π ± ± ± ± − − − − − + + = + + + + + + + i i i i z z z e e z z z e e M1 A1 AEF two of three terms correct for M1 Using (ii), or by direct calculation, 4 2 3 3 4 2 3 3 1 1 1 4 1 2 tan tan 0 2 3 2 3 1 1 i i i i e e i i e e ± ± ± ± − −     + = + =         + + π π π π π π A1 AG 5

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 20 of 21 Question Answer Marks Guidance 11E(iii) 2 3 1, , π ± = i z e ( )( )( ) ( )( )( ) ( )( )( ) 3 2 2 3 3 2 1 1 1 1 1 1 1 1 1 0 − + + + − + + + − + + = z z z z z z z z z B1 Writes down the cube roots of unity. AEF, must be exact 6 5 4 2 3 3 0 ⇒ + + − − − = z z z z z M1 Expands ( )( )( ) 6 5 4 2 3 2 3 3 1 1 3 2 3 + + − − − = + − − + z z z z z z z z z M1 A1 Factorises 2 32 1 2 1, 2 6 3 3 ± ⇒ = − = ± i z i M1 A1 Finds other three roots. AEF, must be exact 6 Question Answer Marks Guidance 11O(i) e.g. 2e B1 Allow any scalar multiple µe where 0,1 µ ≠ 1 11O(ii) Eigenvector: e, Eigenvalue: λ n B1 B1 Allow any scalar multiple µe where 0 µ ≠ Note: An e = λ n e SCB1 2

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 21 of 21 Question Answer Marks Guidance 11O(iii) Eigenvalues of (diagonal matrix) A : 3, 7,1 λ = (Or from characteristic equation: ( )( )( ) 3 7 1 0 λ λ λ − − − = ) B1 3: λ = 1 8 2 2 4 0 4 1 4 8 2 0 0 − −         = = =         −     t i j k e M1 A1 Uses vector product (or equations) to find corresponding eigenvectors 7: λ = 2 0 0 4 0 0 24 3 4 8 6 32 4       = − = − =       − −    t i j k e A1 1: λ = 3 0 0 2 6 0 0 0 4 8 0 8 1       = = =       −    t i j k e A1 Thus 2 0 0 1 3 0 0 4 1 −     =       P and 3 0 0 0 7 0 0 0 1     =         n n D . B1 FT B1 FT Or correctly matched permutations of columns. FT on non-zero and distinct eigenvalues and vectors 7 11O(iv) ( ) 1 1 1 1 1 + + + + = − = − ∑ N n n n n N N n k A k A kA k A M1 A1 Method of differences 1 1 0 + + → N N k A as →∞ N for M1 1 1 . 7 7 − < < k A1 4