Cambridge A Level Mathematics - Further 9231 — 2017 May/June Paper 1 · Variant 3

*8878049898* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/13 Paper 1 May/June 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 24 printed pages. JC17 06_9231_13/RP © UCLES 2017 [Turn over

2 1 The roots of the cubic equation x3 + 2x2 −3 = 0 are !, " and '. (i) By using the substitution y = 1 x2 , find the cubic equation with roots 1 !2 , 1 "2 and 1 '2 . [3] … … … … … … … … … … … (ii) Hence find the value of 1 !2 + 1 "2 + 1 '2 . [1] … … … … (iii) Find also the value of 1 !2"2 + 1 "2'2 + 1 '2!2 . [1] … … … … … © UCLES 2017 9231/13/M/J/17

3 2 (i) Verify that 2r + 1 rr + 1r + 2 = 1 2 D2r + 12r + 3 r + 1r + 2 −2r −12r + 1 rr + 1 E . [2] … … … … … … (ii) Hence show that n Ð r=1 2r + 1 rr + 1r + 2 = 1 2 D2n + 12n + 3 n + 1n + 2 −3 2 E . [2] … … … … … … … … … (iii) Deduce the value of ∞ Ð r=1 2r + 1 rr + 1r + 2. [2] … … … … … © UCLES 2017 9231/13/M/J/17 [Turn over

4 3 Prove, by mathematical induction, that n Ð r=1 r ln @r + 1 r A = ln @n + 1n n! A for all positive integers n. [6] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17

5 4 A curve C has equation x3 −3xy + y2 = 4. Find the value of d2y dx2 at the point 0, 2 of C. [7] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17 [Turn over

6 5 A curve C has parametric equations x = 2 5t 5 2 −2t 1 2, y = 4 3t 3 2, for 1 ≤t ≤4. (i) Find the exact value of the arc length of C. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17

7 (ii) Find also the exact value of the surface area generated when C is rotated through 20 radians about the x-axis. [3] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17 [Turn over

8 6 Let In denote Ó 2 0 4 + x2−n dx. (i) Find d dx x4 + x2−n! and hence show that 8nIn+1 = 2n −1In + 2 × 8−n. [5] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17

9 (ii) Use the result for integrating 1 x2 + a2 with respect to x, in the List of Formulae (MF10), to find the value of I1 and deduce that I3 = 3 10240 + 1 128. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17 [Turn over

10 7 (i) Use de Moivre’s theorem to prove that tan 41 = 4 tan 1 −4 tan31 1 −6 tan21 + tan41. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17

11 (ii) Hence find the solutions of the equation t4 −4t3 −6t2 + 4t + 1 = 0, giving your answers in the form tan k0, where k is a rational number. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17 [Turn over

12 8 Find the solution of the differential equation d2x dt2 + 6dx dt + 9x = 18t2 + 6t + 1, given that, when t = 0, x = 3 and dx dt = 0. [10] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17

13 … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17 [Turn over

14 9 The plane 1 passes through the points 1, 2, 1 and 5, −2, 9 and is parallel to the vector i + 2j + 3k. (i) Find the cartesian equation of 1. [4] … … … … … … … … … … … … … … … The plane 2 contains the lines r = 2i −3j + k + ,i −2j −k and r = 2i −3j + k + -2i + 3j −k. (ii) Find the cartesian equation of 2. [4] … … … … … … © UCLES 2017 9231/13/M/J/17

15 … … … … … … … … (iii) Find the acute angle between 1 and 2. [3] … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17 [Turn over

16 10 The matrix A is given by A = ` 6 −8 7 7 −9 7 6 −6 5 a . (i) Given that ` 1 1 0 a is an eigenvector of A, find the corresponding eigenvalue. [2] … … … … … … (ii) Given also that −1 is an eigenvalue of A, find a corresponding eigenvector. [2] … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17

17 (iii) It is given that the determinant of A is equal to the product of the eigenvalues of A. Use this result to find the third eigenvalue of A, and find also a corresponding eigenvector. [3] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17 [Turn over

18 (iv) Write down matrices P and D such that P−1AP = D, where D is a diagonal matrix, and hence find the matrix An in terms of n, where n is a positive integer. [6] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17

19 11 Answer only one of the following two alternatives. EITHER A curve C has polar equation r = 2a cos21 + 1 20 for 0 ≤1 < 20, where a is a positive constant. (i) Show that r = −2a sin 21 and sketch C. [4] … … … (ii) Deduce that the cartesian equation of C is x2 + y2 3 2 = −4axy. [2] … … … … … … … © UCLES 2017 9231/13/M/J/17 [Turn over

20 (iii) Find the area of one loop of C. [5] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17

21 (iv) Show that, at the points (other than the pole) at which a tangent to C is parallel to the initial line, 2 tan 1 = −tan 21. [3] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17 [Turn over

22 OR The matrix A, given by A = € 1 −1 0 2 3 −1 4 0 5 −8 −6 19 −2 3 2 −7  , represents a transformation from >4 to >4. (i) Find the rank of A and show that „€ 2 2 −1 0  , € 1 3 0 1  is a basis for the null space of the transformation. [6] … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/13/M/J/17

23 … … … … … … … … … … … … (ii) Show that if Ax = p € 1 3 5 −2  + q € −1 −1 −8 3  , where p and q are given real numbers, then x = € p + 2, + - q + 2, + 3- −, -  , where , and - are real numbers. [2] … … … … … … © UCLES 2017 9231/13/M/J/17 [Turn over

24 (iii) Find the values of p and q such that p € 1 3 5 −2  + q € −1 −1 −8 3  = € 3 7 18 −7  . [3] … … … … … … … (iv) Find the solution of the equation Ax = € 3 7 18 −7  of the form x = € 4 9 ! "  , where ! and " are positive integers to be found. [3] … … … … … … … … … … Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. © UCLES 2017 9231/13/M/J/17

® IGCSE is a registered trademark. This document consists of 17 printed pages. © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/13 Paper 1 May/June 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 will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 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 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 May/June 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 May/June 2017 © UCLES 2017 Page 4 of 17 Question Answer Marks Guidance 1(i) 2 1 1 = ⇒ = y x x y M1 Rearranging to make x the subject 1 2 1 2 3 0 3 + − = ⇒ = − y y y y y y M1 Substituting and squaring 3 2 9 12 4 1 0 ⇒ − + −= y y y SR B1 for finding cubic by manipulating roots A1 OE Total: 3 1(ii) 2 2 2 1 1 1 12 4 or 9 3 α β γ + + = B1FT Total: 1 1(iii) 2 2 2 2 2 2 1 1 1 4 9 α β β γ γ α + + = B1FT Total: 1

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 5 of 17 Question Answer Marks Guidance 2(i) RHS ( )( ) ( )( ) 3 2 2 4 8 3 4 1 2 1 2 1 2   + + − − +   =   + +     r r r r r r r r M1 ( ) ( )( ) 3 2 3 2 4 8 3 4 8 2 1 2 1 2   + + − + − −   =   + +     r r r r r r r r r ( ) ( )( ) ( ) ( )( ) 4 2 2 1 1 2 1 2 1 2   + +   = =   + + + +     r r r r r r r r A1 AG Total: 2 2(ii) Sum to n terms is: ( )( ) ( ) ( )( ) ( ) ( )( ) ( )( ) ( )( ) ( ) 2 1 2 1 2 3 2 1 2 1 2 3 2 1 2 1 1 3.5 1.3 5.7 3.5 2 2.3 1.2 3.4 2.3 1 1 1 2 1       − + − − + + − +       − + − +…+ − + −          + − + + +                 n n n n n n n n n n n n n n n n M1 ( )( ) ( )( ) 2 1 2 3 1 3 2 1 2 2   + +   = −   + +     n n n n A1 AG 2(iii) 1 3 1 4 1 2 4 4 ∞= × − = S . M1A1 Total: 4

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 6 of 17 Question Answer Marks Guidance 3 When n =1 1 1 2 1 ln2 ln2 ln (H 1!   × = = ⇒     is true ) B1 Assume, for some positive integer k, that [ ] 1 1 1 ln ln ! =   + +     =         ∑ k k r k r r r k B1 Hence ( ) ( ) 1 1 1 1 2 ln ln 1 ln ! 1 + =   + + +       = + +       +       ∑ k k r k r k r k r k k B1 ( ) ( ) ( ) 1 1 1 2 ln ! 1 + + + + = + k k k k k k k M1 ( ) ( ) ( ) ( ) 1 1 2 ln 1 ! 1 + + + = + + k k k k k k k ( ) ( ) 1 2 ln 1 ! + + = + k k k A1 Thus Hk ⇒ Hk+1 and hence by PMI Hn is true for all positive integers. A1 Total: 6

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 7 of 17 Question Answer Marks Guidance 4 3 2 d dy 3 3 2 0 d d   − + + =     y x y x y x x B1B1 SOI At (0,2): dy dy 3 6 4 0 d d 2 −+ = ⇒ = x x B1 2 2 dy dy d 6 3 3 3 d d d   − − +     y x x x x x M1A1 2 2 2 dy d 2 2 0 d d   + + =     y y x x A1 At (0,2): 2 2 2 2 9 9 9 d d 9 4 0 2 2 2 8 d d − − + + = ⇒ = y y x x A1 Total: 7

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 8 of 17 Question Answer Marks Guidance 5(i) 2 2 3 1 3 1 2 2 2 2 2 2 4 − −     + = − + =…= +             & & x y t t t t t M1A1 SOI 4 3 1 2 2 1 d −   = +       ∫ s t t t M1 4 5 1 2 2 1 2 2 5   = +       t t M1 64 2 4 2 5 5     = + − +         72 5 = A1 Total: 5 5(ii) 4 4 3 3 1 3 2 2 2 1 1 4 8 2 d ( )d 3 3 π π −   = + = +       ∫ ∫ S t t t t t t t *M1 4 4 2 1 8 1 1 3 4 2 t t π    = +       DM1 [ ] 8 1 1 64 8 3 4 2 π     = + − +         1 90π = A1 Total: 3

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 9 of 17 Question Answer Marks Guidance 6(i) ( ) { } ( ) ( ) 1 2 2 2 2 d 4 4 2 4 d − − −− + = + − + n n n x x x nx x x M1A1 Integrate w.r.t. x: ( ) ( )( ) 2 2 1 2 2 2 0 0 4 2 4 4 4 d − −−   + = − + − +     ∫ n n n x x I n x x x M1 M1 ( ) 1 1 2.8 2 8 8 2 1 2.8 − − + + ⇒ = − + ⇒ = − + n n n n n n n I nI nI nI n I A1 AG Total: 5 6(ii) 2 2 1 1 2 0 0 1 1 1 d tan . 2 2 2 4 8 4 π π −     = = = =     +     ∫ x I x x M1A1 2 2 1 1 8 8 4 64 32 π π = + ⇒ = + I I M1A1FT 3 3 3 3 1 3 1 16 64 32 32 1024 128 π π = + + ⇒ = + I I A1 AG Total: 5

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 10 of 17 Question Answer Marks Guidance 7(i) ( ) ( ) ( ) ( ) ( ) 4 2 3 4 4 3 2 i 4 i 6 i 4 i i + = + + + + c s c c s c s c s s B1 SOI Equate real and imaginary parts M1 4 2 2 4 cos4 6 θ = − + c c s s A1 SOI 3 3 sin 4 4 4 θ = − c s cs A1 SOI ( ) ( ) 3 3 4 3 2 4 4 2 2 4 4 4 4 4tan 4tan tan 4 1 6tan tan 6 θ θ θ θ θ − ÷   − = =   − + − + ÷   c s cs c c c s s c A1 AG Total: 5 7(ii) 4 3 2 tan 4 1 4 6 4 1 0 θ = −⇒ − − + + = t t t t M1 M1 3 7 11 15 so 4 , , , 4 4 4 4 π π π π θ   =     M1 3 7 11 15 tan ,tan ,tan ,tan . 16 16 16 16 π π π π = t Allow ( 1 ) 0,1 , 2, 3 4 16 π − = k k oe A1 A1 Total: 5

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 11 of 17 Question Answer Marks Guidance 8 2 6 9 0 3 + + = ⇒ = − m m m M1 CF: 3 3 − − = + t t x Ae Bte A1 ¨ 2 2 and 2 = + + ⇒ = + = & x pt qt r x pt q x p M1 ( ) ( ) 2 2 2 6 2 9 18 6 1 ⇒ + + + + + = + + p pt q pt qt r t t M1 ( ) ( ) 2 2 9 9 12 2 6 9 18 6 1 ⇒ + + + + + = + + pt q p t p q r t t 2 , 2 , 1 ⇒ = = − = p q r whence PI: 2 2 2 1 − + t t A1 GS: 3 3 2 2 2 1 − − = + + − + t t x Ae Bte t t A1FT 3 = x when t = 0 ⇒ A = 2 B1 3 3 3 3 3 4 2 − − − = − − + + − & t t t x Ae Bte Be t M1 0 when 0 8 = = ⇒ = &x t B A1 Hence 3 3 2 2 8 2 2 1 − − = + + − + t t e te t t x A1 Total: 10

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 12 of 17 Question Answer Marks Guidance 9(i) 7 1 1 2 1 1 2 3 3 −     − = −       i j k M1A1 Cartesian equation of Π1 is 7 3 const + − = x y z M1 7 1 2 3 1 6 7 3 6 × + −× = ⇒ + − = x y z A1 Total: 4 9(ii) 5 1 2 1 1 2 3 1 7     − −= −     −   i j k M1A1 ( ) 5 2 3 7 1 20 5 7 20 × −− + × = ⇒ − + = x y z M1A1 Total: 4 9(iii) 7 5 1 . 1 3 7 cos 49 1 9 25 1 49 θ       −       −    = + + + + M1M1 13 cos 78.7 59 75 θ θ ⇒ = ⇒ = ° or 1.37 rad. A1 Total: 3

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 13 of 17 Question Answer Marks Guidance 10(i) 1 6 8 7 1 2 7 9 7 1 2 2 6 6 5 0 0 λ − −         − = − ⇒ =−         −     M1A1 Total: 2 10(ii) 2 2 1 1 0 1 λ     = −⇒ =     −   e M1A1 OE Total: 2 10(iii) Det A = 10 3 5 λ ⇒ = 3 1 1 1   ⇒ =    e B1B1B1 OE Total: 3

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 14 of 17 Question Answer Marks Guidance 10(iv) 1 1 1 1 0 1 0 1 1     =     −   P 2 0 0 0 1 0 0 0 5 −     = −       D (Check for consistency.) B1FT B1FT Det P = –1 (From working) Adj P = 1 2 1 1 1 0 1 1 1 −     −     − −   1 1 2 1 1 1 0 1 1 1 − − −     ⇒ = −     −   P M1A1 Alternative Method: Reduced row echelon form M1, A1 ( ) ( ) 2 0 0 1 1 1 1 2 1 1 0 1 0 1 0 1 1 0 0 1 1 1 1 1 0 0 5   − − −             = − −           − −         n n n n A M1 Allow An = P Dn P-1 =… ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 5 1 2 2. 2 1 5 5 2 5 2 2. 2 5 5 2 5 1 1 5 5 +         + − −− − + − − −−                 = −− − − −−                   −− − −         n n n n n n n n n n n n n n n n n n n A1 Total: 6

923 © U Ques 11E 11E 11E 31/13 CLES 2017 stion E(i) 2 c  =  r a E(ii) 4 s = − r a E(iii) Area of ( 2 1 2 1 π π = ∫ a 2 θ  =  a 1 cos2 cos sin 2 θ π − sin cos 4 . θ θ = − y a r one loop is 1 2 1 2 π π∫ ) 1 cos4 d θ θ − 1 2 1 sin 4 4 π π θ  −  = C 1 n2 sin 2 2 θ π =  a 3 . 4 ⇒ = − y x r axy r r 2 2 2 4 sin 2 d θ θ = a a 2 1 2π = a Cambridge Inter Answe ( ) 0 sin2 .1 2 θ − = − ( ) 3 2 2 2 ( ⇒ + = y x y 2 2 1 2 2sin 2 d π π θ θ ∫ rnational A Leve PUBLISHED Page 15 of 17 er 2 sin 2θ a 4 ) = −axy l – Mark Scheme To To To Marks M1A1 B1 B1 otal: 4 M1A1 otal: 2 M1 M1A1 M1A1 otal: 5 May/June 2 Guidance AG Loops in 2nd an quadrant Symmetry and s correct AG OE 2017 e nd 4th shape

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 16 of 17 Question Answer Marks Guidance 11E(iv) sin 2 sin 2 sin θ θ θ = = − y r a B1 ( ) d 2 2cos2 sin sin2 cos 0 d θ θ θ θ θ = − + = y a M1 2cos2 sin sin 2 cos θ θ θ θ = − ( 2tan tan 2 ) θ θ ⇒ = − A1 AG Total: 3 11O(i) 1 1 0 2 1 1 0 2 3 1 4 0 0 1 2 3 5 8 6 1 9 0 0 0 0 2 3 2 7 0 0 0 0 − −         − −     ⇒…⇒     − −     − −     M1A1 r(A) = 4 – 2 = 2 A1FT 2 0 − + = x y t 2 3 0 + − = y z t M1 ' and ' 2 ' ' , 2 ' 3 ' λ µ λ µ λ µ = = ⇒ = − + = − + z t x y ⇒ Basis of null space is 2 1 2 3 , 1 0 0 1                     −             A1A1 AG Total: 6

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 17 of 17 Question Answer Marks Guidance 11O(ii) 2 1 2 2 3 2 3 0 0 0 1 0 0 0 0 1 λ µ λ µ λ µ λ µ   + +                     + +            − = ⇒ = + + =            − −                         p p p q q q A x x M1A1 Total: 2 11O(iii) 1 1 3 3 1 7 3 and 3 7 2 , 1 5 8 18 2 3 7 p q p q p q p q −             −       + = ⇒ − = − = ⇒ = = −       −       − −       M1A1 A1 Total: 3 11O(iv) 2 2 4 2 2 λ µ λ µ + + = ⇒ + = 1 2 3 9 2 3 10 λ µ λ µ −+ + = ⇒ + = M1 1 , 4 λ µ ⇒ = − = ⇒ Solution of 3 7 18 7       =     −   Ax is 4 9 1 4       =       x A1A1 Total: 3