Cambridge A Level Mathematics - Further 9231 — 2017 Oct/Nov Paper 1 · Variant 3

*2188102920* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/13 Paper 1 October/November 2017 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. 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 23 printed pages and 1 blank page. JC17 11_9231_13/FP © UCLES 2017 [Turn over

2 1 Find n Ð r=1 4r −34r + 1, giving your answer in its simplest form. [4] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17

3 2 Find the general solution of the differential equation d2x dt2 + 2dx dt + 5x = 4 −5t2. [6] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17 [Turn over

4 3 (i) Show that dn+1 dxn+1 xn+1 ln x = dn dxn xn + n + 1xn ln x. [2] … … … … … (ii) Prove by mathematical induction that, for all positive integers n, dn dxn xn ln x = n! @ ln x + 1 + 1 2 + à + 1 n A . [5] … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17

5 4 The cubic equation 2x3 −3x2 + 4x −10 = 0 has roots !, " and '. (i) Find the value of ! + 1" + 1' + 1. [4] … … … … … … … … … … … (ii) Find the value of " + '' + !! + ". [4] … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17 [Turn over

6 5 The curve C has equation 2x3 + 3x2y −3y3 −16 = 0. (i) Find the coordinates of the point A on C at which dy dx = 0 and x ≠0. [5] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17

7 (ii) Find the value of d2y dx2 at A. [3] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17 [Turn over

8 6 The points A, B and C have position vectors 2i −j + k, 3i + 4j −k and −i + 2j + 4k respectively. (i) Find the area of the triangle ABC. [4] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17

9 (ii) Find the perpendicular distance of the point A from the line BC. [3] … … … … … … … … … … … … (iii) Find the cartesian equation of the plane through A, B and C. [2] … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17 [Turn over

10 7 The linear transformation T : >4 →>4 is represented by the matrix A, where A = € 1 −1 −2 3 5 −3 −4 25 6 −4 −6 28 7 −5 −8 31  . (i) Find the rank of A and a basis for the null space of T. [7] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17

11 … … … … … … … (ii) Find the matrix product A € −1 1 −1 1  and hence find the general solution of the equation Ax = € 3 21 24 27  . [3] … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17 [Turn over

12 8 Let In = Ó 1 40 0 secnx dx for n > 0. (i) Find the value of I2. [2] … … … (ii) Show that, for n > 2, n −1In = 2 1 2n−1 + n −2In−2. [5] … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17

13 (iii) The curve C has equation y = sec3x for 0 ≤x ≤1 40. The region R is bounded by C, the x-axis, the y-axis and the line x = 1 40. Find the volume of revolution generated when R is rotated through 20 radians about the x-axis. [4] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17 [Turn over

14 9 The curve C has equation y = 3x −9 x −2x + 1. (i) Find the equations of the asymptotes of C. [2] … … … … (ii) Show that there is no point on C for which 1 3 < y < 3. [4] … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17

15 (iii) Find the coordinates of the turning points of C. [3] … … … … … … … … … … … … (iv) Sketch C. [3] © UCLES 2017 9231/13/O/N/17 [Turn over

16 10 (i) Use de Moivre’s theorem to show that sin 51 = 5 sin 1 −20 sin31 + 16 sin51. [5] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17

17 (ii) Hence explain why the roots of the equation 16x4 −20x2 + 5 = 0 are x = ± sin 1 50 and x = ± sin 2 50. [3] … … … … … … … … … (iii) Without using a calculator, find the exact values of sin 1 50 sin 2 50 sin 3 50 sin 4 50 and sin21 50 + sin22 50. [4] … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17 [Turn over

18 11 Answer only one of the following two alternatives. EITHER (i) The vector e is an eigenvector of the matrix A, with corresponding eigenvalue ,, and is also an eigenvector of the matrix B, with corresponding eigenvalue -. Show that e is an eigenvector of the matrix AB with corresponding eigenvalue ,-. [3] … … … … … … … … … (ii) Find the eigenvalues and corresponding eigenvectors of the matrix A, where A = ` 0 1 3 3 2 −3 1 1 2 a . [6] … … … … … … … … … … © UCLES 2017 9231/13/O/N/17

19 … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17 [Turn over

20 (iii) The matrix B, where B = ` 3 6 1 1 −2 −1 6 6 −2 a , has eigenvectors ` 1 −1 0 a , ` 1 −1 1 a and ` 1 0 1 a . Find the eigenvalues of the matrix AB, and state corresponding eigenvectors. [4] … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17

21 OR The polar equation of a curve C is r = a1 + cos 1 for 0 ≤1 < 20, where a is a positive constant. (i) Sketch C. [2] (ii) Show that the cartesian equation of C is x2 + y2 = ax + x2 + y2. [2] … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17 [Turn over

22 (iii) Find the area of the sector of C between 1 = 0 and 1 = 1 30. [4] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17

23 (iv) Find the arc length of C between the point where 1 = 0 and the point where 1 = 1 30. [5] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/O/N/17

24 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 2017 9231/13/O/N/17

® IGCSE is a registered trademark. This document consists of 17 printed pages. © UCLES 2017 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/13 Paper 1 October/November 2017 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 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 2 of 17 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 October/November 2017 © UCLES 2017 Page 3 of 17 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 October/November 2017 © UCLES 2017 Page 4 of 17 Question Answer Marks Guidance 1 2 1 1 1 16 8 3 = = = = − − ∑ ∑ ∑ n n n r r r r u r r n M1A1 M1 for split into 3 parts ( )( ) ( ) 1 2 1 1 16 8 3 6 2 + + + = − − n n n n n n M1 For using formulae correctly in their expression ( ) 2 16 12 13 3 = = + − … n n n (3 terms) A1 OE 4 Question Answer Marks Guidance 2 CF: m2 + 2m + 5 = 0 ⇒ m = –1 ± 2i M1 e–t(A cos 2t + B sin 2t) A1 PI: x = pt2 + qt + r ⇒ x = 2pt + q ⇒ x = 2p M1 2p +4pt + 2q + 5pt2 + 5qt+5r = 4 – 5t2 M1 ⇒ p = –1, q = 4 5 , r = 22 25 A1 GS: x = e–t (A cos 2t + B sin 2t) + 2 22 4 25 5 + − t t A1FT 6

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 5 of 17 Question Answer Marks Guidance 3(i) ( ) ( ) ( ) ( ) 1 1 1 1 d d 1 ln . 1 lnx d d d 1 lnx d + + + +   = + + =     + + n n n n n n n n n n n x x x n x x x x x n x x M1A1 AG 2 3(ii) Assume Hk is true ⇒ ( ) d 1 1 ln ! ln 1 2 d   = + + +…+     k k k x x k x k x B1 Statement of Hk seen ( ) [ ] ( ) 1 1 1 d d ln 1 lnx d d + + + = + + k k k k k k k x x x k x x x M1 [ ] 1 1 ! 1 ! ln 1 2   = + + + + +…+     k k k x k A1 ( ) 1 1 1 ! ln 1 2 1   = + + + +…+ ⇒   +   k x k Hk+1 is true A1 Check H1 is true and Hk is true ⇒ H௞ାଵis true; hence, by PMI, Hn is true for all positive integers n. A1 5

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 6 of 17 Question Answer Marks Guidance 4(i) 3 2 5 2 3 2 5 2 α β γ αβ βγ γα αβγ β γ αβ βγ γα αβγ + + = + + = = + + = + + = = B1 (Can be awarded in (ii) if not seen here) SOI ( )( )( ) ( ) ( ) 1 1 1 1 α β γ αβγ αβ βγ γα α β γ + + + = + + + + + + + M1A1 Multiply out and group for M1 1 1 5 2 1 1 9 2 2 = + + + = A1FT Alt method: Let x= y–1 M1 Sub and expand 2y3 – 9y2 16y – 19 = 0 M1, A1 Product of roots = 19/2 A1 4 4(ii) ( )( )( ) 1 1 1 1 1 1 2 2 2 β γ γ α α β α β γ     + + + = − − −         M1 Alt methods: 2 3 ( ) ( ) or 2 α αβ αβγ α α αβγ α = ∑ ∑ − ∑ ∑ + −∑ ( ) ( ) 27 9 3 8 4 2 α β γ αβ βγ γα αβγ = − + + + + + − A1 27 9 3 3 2 5 2 8 4 2 2 = − × + × − = − M1A1 4

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 7 of 17 Question Answer Marks Guidance 5(i) ( ) 2 2 2 2 2 6 6 3 ' 9 0 * 2 ( ) (3 ) ′ + + − = ⇒ − ′ + = x xy x y y y x x y y x y M1A1 0and 0 ′ = ≠ ⇒ = − y x x y M1A1 3 3 3 2 3 3 16 x x x ⇒ − + = ⇒ A is (2, –2) A1 5 5(ii) ( ) 2 2 2 12 6 6 6 3 18 9 0 ′ ′ ′′   + + + + − + ′ ′′ =   x xy y xy x y y y y y *M1 2 2 0 8 4 4 12 0 = = − = ⇒− + ′ ′ ′ − = ′ ′ x y y y y DM1 1 2 ′′ ⇒ = y A1 3

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 8 of 17 Question Answer Marks Guidance 6(i) 5 2 4 2 5 3 3 3 = + − = − − + = − + + JJJG JJ JJJG JG AB BC AC i j k i j k i j k B1 2 correct required 21 3 18 (*) × = + + JJJG JJJG AB BC i j k M1A1 OE Area of triangle ABC 2 2 2 1 3 21 3 18 13.9 86 2 2   = + + =     A1 Alt method: Use scalar product to find angle (M1A1 Find area using Area = ½ ab sin C or equivalent M1A1) 4 6(ii) 2 2 2 2 2 2 21 3 18 d 4 2 5 × + + = = + + JJJG JJJG JJJG AB BC BC M1A1 Alt method: Find angle at C 1 4.15 430 5   =     A1 Area triangle = sin C × |AC| Alt method: Use equation of BC to find D (foot of perpendicular) in terms of parameter and scalar product to find parameter , λ= 8/15. Find length (M1A1) 3 6(iii) From (*) Cartesian equation is 7ݔ+ ݕ+ 6ݖ= const. M1 Through ( 2, –1 , 1) Hence 7ݔ+ ݕ+ 6ݖ= 19 A1 2

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 9 of 17 Question Answer Marks Guidance 7(i) 1 1 2 3 1 1 2 3 5 3 4 25 0 1 3 5 6 4 6 28 0 0 0 0 7 5 8 31 0 0 0 0 − − − −         − −     →…→     − −     − −     M1A1 r(A) = 4 – 2 = 2 A1 x – y – 2z + 3t = 0 x – y + 3z + 5t = 0 B1 , 8 , 3 5 λ µ λ µ λ µ = = ⇒ = − − = − − z t x y M1 Basis for null space is 1 8 3 5 , 1 0 0 1 λ µ   − −           − −                           19 0 0 19 , 5 8 3 1             −       A1 A1 OE 7 7(ii) 1 3 1 21 1 24 1 27 −             =     −         A B1 1 1 8 1 3 5 1 1 0 1 0 1 λ µ − − −             − −       = + +       −             x M1A1FT OE 3

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 10 of 17 Question Answer Marks Guidance 8(i) [ ] 1 1 4 2 4 2 0 0 sec d tan 1 π π = = = ∫ I x x x M1A1 2 8(ii) 1 4 2 2 0 sec .sec d π − = ∫ n nI x x x M1 ( ) 1 1 4 2 3 4 0 0 sec tan 2 sec (sec tan )tan d π π − −   = − −   ∫ n n x x n x x x x x M1A1 ( ) ( ) 1 1 4 2 2 2 4 0 0 sec tan 2 sec sec 1 d π π − −   = − − −   ∫ n n x x n x x x M1A1 ( ) ( ) 1 1 2 2 1 2 2 − − ⇒ − = + − n n n n I n I AG 5

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 11 of 17 Question Answer Marks Guidance 8(iii) Volume of revolution 1 4 2 6 0 d sec d y x x x π π π = ∫ = ∫ M1 4 4 4 3 2 2 1 3 I I = + × ⇒ = M1 6 6 4 28 5 4 4 3 15 I I = + × ⇒ = M1 Volume of revolution 28 15 π = A1 4

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 12 of 17 Question Answer Marks Guidance 9(i) Degree of numerator < degree of denominator ⇒ݕ= 0 is horizontal asymptote. B1 ( )( ) 1 2 0 + − = x x 1 ⇒ = − x and 2 ⇒ = x are vertical asymptotes. B1 2 9(ii) ( ) 2 3 9 2 0 − + + − = yx y x y M1 No points on C if ( ) ( ) 2 3 4 9 2 0 + − − < y y y M1 2 9 30 9 0 ⇒ − + < y y 2 3 10 3 0 ⇒ − + < y y A1 ( )( ) 1 3 1 3 0 3 3 ⇒ − − < ⇒ < < y y y A1 AG 4 9(iii) ( ) ( )( ) 2 d 0 3 2 3 9 2 1 0 dx = ⇒ − − − − − = y x x x x B1 ( )( ) 1 5 0 x x ⇒ ⇒ − − = … B1 ⇒ Turning points are (1,3) and 1 5, 3       . B1 3

923 © U Q Q 31/13 CLES 2017 Question 9(iv) Question 10(i) ( 5 sin5θ Im c sin5θ ( ) ( 5 5 4 3 i 5 i 10 i θ = + = + + Im c s c s c s 4 2 3 5 10 θ = − + c s c s ( 2 2 5 1   = − −   s s 5sin 2 θ =…= − C Answer Axes, asymp (0, 4.5) (3,0) RH branch ; Answer ) ( ) 2 3 2 10 i 5 + + s c s 5 + s 2 2 4 10 1   − − +   s s s 3 5 20sin 16sin θ θ + Cambridge Inter ptotes and points o . Other two branch ( ) ( ) ) 4 5 5 i i + c s s ) 4 rnational A Leve PUBLISHED Page 13 of 17 Mar on axes hes B1 Mar M1 M l – Mark Schem ks B1 1B1 3 rks B1 SOI 1A1 M1 A1 AG 5 me G G Guidance Guidance October/Novem 2 mber 017

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 14 of 17 Question Answer Marks Guidance 10(ii) If 1 2 0 , , 5 5 θ π π = ± ± then sin5 0 θ = B1 5 3 16 20 5 0 ⇒ − + = s s s , where sinθ = s , ( ) 4 2 16 20 5 0 ⇒ − + = s s s B1 0 0 θ = ⇒ = s B1 Hence roots of 4 3 16 20 5 0 − + = s s are 1 2 sin , sin 5 5 π π ± ± AG 3 10(iii) Since 4 1 3 2 sin sin and sin sin 5 5 5 5 π π π π     = − − = − −         B1 4 3 2 1 sin sin sin sin 5 5 5 5 1 2 1 2 5 sin sin sin sin 5 5 5 5 16 π π π π π π π π        =                         − − =                 M1A1 ( ) 2 2 20 1 2 5 sin sin 5 5 16 4 π π − + = − = A1 4

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 15 of 17 Question Answer Marks Guidance 11E(i) λ = Ae e and B µ = e e M1A1 µ µ µλ λµ = = = = ABe A e Ae e e M1 AG 3 11E(ii) ( )( ) 2 1 5 6 0 λ λ λ + − + = A1 ( )( )( ) 1 2 3 0 λ λ λ + − − = A1 1, 2, 3. λ = − M1 Eigenvectors are 1 1 1 1 , 1 and 0 0 1 1           − −                respectively. A1A1 Uses either vector product or equations to find eigenvectors 6 11E(iii) 1 3 6 1 1 3 1 2 1 1 3 3 6 6 2 0 0 µ −           − − − = ⇒ = −           −      M1 Similarly, other two eigenvalues of B are – 2 and 4. A1 Eigenvalues of AB are 3 , – 4 and 12 A1 Corresponding eigenvectors are 1 1 1 1 , 1 and 0 . 0 1 1           − −                A1 4

923 © U Q 1 31/13 CLES 2017 Question 11OR 1OR(ii) ( = r a 2 + x y ( ) 2 1 cosθ + ⇒ x 2 2 2 ( ( = + + y a x x y C Answer 2 2 1   + = +  y a x 2)) Cambridge Inter 2 2   +  x y rnational A Leve PUBLISHED Page 16 of 17 Mar M l – Mark Schem ks B1 Closed curv location. B1 Cardioid wi 2 M1 A1 Substitutes 2 me G ve starting and end ith indication of c for r and cos(θ) Guidance ding at pole, in ap orrect scale. October/Novem 2 pproximately corre mber 017 ect

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 17 of 17 Question Answer Marks Guidance 11OR(iii) Sector area ( ) 1 3 2 2 0 1 2cos cos d 2 π θ θ θ = + + ∫ a 1 3 2 0 3 cos2 2cos d 2 2 2 π θ θ θ   = + +     ∫ a M1A1 1 2 3 0 3 sin 2 2sin 2 2 4 π θ θ θ   = + +     a M1 ( ) 2 4 9 3 16 π = + a A1 4 11OR(iv) Arc length ( ) ( ) 1 3 2 2 2 2 0 1 2cos cos sin d π θ θ θ θ = + + + − ∫ a a M1A1 1 3 0 2 2cos d π θ θ = + ∫ a A1 1 3 0 4sin 2 2 π θ   = =     a a M1A1 5