Cambridge A Level Mathematics - Further 9231 — 2019 May/June Paper 1 · Variant 2

*7226382766* Cambridge Assessment International Education Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/12 Paper 1 May/June 2019 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 26 printed pages and 2 blank pages. JC19 06_9231_12/RP © UCLES 2019 [Turn over

2 1 A curve C has equation cosy = x, for −0 < x < 0. (i) Use implicit differentiation to show that d2y dx2 = −cot y @dy dx A2 . [4] … … … … … … … … … … … … … (ii) Hence find the exact value of d2y dx2 at the point 1 2, 1 30 ! on C. [2] … … … … … … … … © UCLES 2019 9231/12/M/J/19

3 2 Let un = 4 sinn −1 2 sin 1 2 cos2n −1 + cos 1. (i) Using the formulae for cos P ± cos Q given in the List of Formulae MF10, show that un = 1 cosn − 1 cosn −1. [2] … … … … … … (ii) Use the method of differences to find N Ð n=1 un. [2] … … … … … … … (iii) Explain why the infinite series u1 + u2 + u3 + à does not converge. [1] … … … … … © UCLES 2019 9231/12/M/J/19 [Turn over

4 3 The lines l1 and l2 have equations r = 6i + 2j + 7k + ,i + j and r = 4i + 4j + -−6j + k respectively. The point P on l1 and the point Q on l2 are such that PQ is perpendicular to both l1 and l2. Find the position vectors of P and Q. [8] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19

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6 4 It is given that, for n ≥0, In = Ó 1 0 xnex3 dx. (i) Show that I2 = 1 3e −1. [2] … … … … … … … (ii) Show that, for n ≥3, 3In = e −n −2In−3. [3] … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19

7 … … … … … … … … … … (iii) Hence find the exact value of I8. [3] … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19 [Turn over

8 5 A curve C is defined parametrically by x = 2 et + e−t and y = et −e−t et + e−t , for 0 ≤t ≤1. The area of the surface generated when C is rotated through 20 radians about the x-axis is denoted by S. (i) Show that S = 40 Ô 1 0 et −e−t et + e−t2 dt. [5] … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19

9 (ii) Using the substitution u = et + e−t, or otherwise, find S in terms of 0 and e. [3] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19 [Turn over

10 6 The equation x3 −x + 1 = 0 has roots !, ", '. (i) Use the relation x = y 1 3 to show that the equation y3 + 3y2 + 2y + 1 = 0 has roots !3, "3, '3. Hence write down the value of !3 + "3 + '3. [3] … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19

11 Let Sn = !n + "n + 'n. (ii) Find the value of S−3. [2] … … … … … … (iii) Show that S6 = 5 and find the value of S9. [4] … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19 [Turn over

12 7 Find the particular solution of the differential equation 10d2x dt2 + 3dx dt −x = t + 2, given that when t = 0, x = 0 and dx dt = 0. [10] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19

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14 8 (i) Prove by mathematical induction that, for z ≠1 and all positive integers n, 1 + z + z2 + à + zn−1 = zn −1 z −1 . [5] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19

15 (ii) By letting z = 1 2cos 1 + i sin 1, use de Moivre’s theorem to deduce that ∞ Ð m=1 1 2 m sin m1 = 2 sin 1 5 −4 cos 1. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19 [Turn over

16 9 It is given that e is an eigenvector of the matrix A, with corresponding eigenvalue ,. (i) Show that e is an eigenvector of A2, with corresponding eigenvalue ,2. [2] … … … … … … … … … The matrices A and B are given by A = ` n 1 3 0 2n 0 0 0 3n a and B = A + nI2, where I is the 3 × 3 identity matrix and n is a non-zero integer. (ii) Find, in terms of n, a non-singular matrix P and a diagonal matrix D such that B = PDP−1. [8] … … … … … … … … … … © UCLES 2019 9231/12/M/J/19

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18 10 The curves C1 and C2 have equations y = ax x + 5 and y = x2 + a + 10x + 5a + 26 x + 5 respectively, where a is a constant and a > 2. (i) Find the equations of the asymptotes of C1. [2] … … … … … (ii) Find the equation of the oblique asymptote of C2. [2] … … … … … … (iii) Show that C1 and C2 do not intersect. [2] … … … … … … … … © UCLES 2019 9231/12/M/J/19

19 (iv) Find the coordinates of the stationary points of C2. [3] … … … … … … … … … … … (v) Sketch C1 and C2 on a single diagram. [You do not need to calculate the coordinates of any points where C2 crosses the axes.] [3] © UCLES 2019 9231/12/M/J/19 [Turn over

20 11 Answer only one of the following two alternatives. EITHER The curve C1 has polar equation r2 = 21, for 0 ≤1 ≤1 20. (i) The point on C1 furthest from the line 1 = 1 20 is denoted by P. Show that, at P, 21 tan 1 = 1 and verify that this equation has a root between 0.6 and 0.7. [5] … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19

21 The curve C2 has polar equation r2 = 1 sec21, for 0 ≤1 < 1 20. The curves C1 and C2 intersect at the pole, denoted by O, and at another point Q. (ii) Find the exact value of 1 at Q. [2] … … … … … … … … … … … (iii) The diagram below shows the curve C2. Sketch C1 on this diagram. [2] 1 = 0 O 1 = 1 20 C2 © UCLES 2019 9231/12/M/J/19 [Turn over

22 (iv) Find, in exact form, the area of the region OPQ enclosed by C1 and C2. [5] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19

23 OR The linear transformation T : >4 →>4 is represented by the matrix M = € −1 2 3 4 1 0 1 −1 1 −2 −3 a 1 2 5 2  . (i) For a ≠−4, the range space of T is denoted by V. (a) Find the dimension of V and show that € −1 1 1 1  , € 2 0 −2 2  and € 4 −1 a 2  form a basis for V. [5] … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19 [Turn over

24 (b) Show that if € x y z t  belongs to V then x + 2y = t. [4] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19

25 (ii) For a = −4, find the general solution of Mx = € −1 1 1 1  . [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/12/M/J/19

26 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 2019 9231/12/M/J/19

27 BLANK PAGE © UCLES 2019 9231/12/M/J/19

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 Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge. © UCLES 2019 9231/12/M/J/19

This document consists of 20 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/12 Paper 1 May/June 2019 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 2019 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 2 of 20 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 3 of 20 GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 4 of 20 Mark Scheme Notes The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark 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. Abbreviations 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 SC Special Case (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)

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 5 of 20 Question Answer Marks Guidance 1(i) ( ) 1 d d sin 1 sin d d − − = ⇒ = − y y y y x x M1 A1 Differentiates implicitly once. ( ) 2 2 2 2 d d d sin cos cot d d d −     = = −         y y y y y y x x x M1 A1 Differentiates again, AG. 4 1(ii) 1 d 2 sin d 3 3 π −   = − = −     y x B1 2 2 2 d 1 2 4 4 3 9 d 3 3 3 3   = − − = − = −     y x B1 AEF, must be exact. 2

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 6 of 20 Question Answer Marks Guidance 2(i) ( ) ( ) ( ) ( ) 1 1 4sin sin 2 cos 1 cos 2 2 cos 2 1 cos1 2cos cos 1   −   − −   = − + − n n n n n n M1 Uses formulae for c c s s . o o P Q ± ( ) 1 1 cos cos 1 n n − − A1 AG 2 2(ii) ( ) ( ) 1 1 1 1 4sin sin 1 1 2 2 cos 2 1 cos1 cos cos 1 N N n n n n n n = =   −     = − − + − ∑ ∑ ( ) 1 1 1 1 1 1 cos1 1 cos2 cos1 cos cos 1 = − + − +…+ − − N N M1 Applies (i), shows enough terms and cancelation. 1 1 cos N −+ A1 2 2(iii) cos N oscillates as →∞ N so does not converge. B1 States “oscillates” or refers to diverging values of cos N . 1 … + + + 3 2 1 u u u

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 7 of 20 Question Answer Marks Guidance 3 6 4 2 2 , 4 6 2 6 7 7 λ λ λ µ λ µ µ µ + −−             = + = − ⇒ = − −             −+       JJJG JJJG JJJG OP OQ PQ M1 A1 Finds JJJG PQ . ˙ 2 1 2 6 1 0 7 0 λ λ µ µ −−     − − =     −+   Or 2 1 2 6 1 1 0 1 7 0 6 1 6 λ λ µ µ −−         − − = = −         −+ − −     k k i j k M1 Uses that dot product of JJJG PQ with line directions is 0 . Or, alternatively, JJJG PQ is multiple of common perpendicular. 2 6 0 λ µ − − = A1 Deduces one equation. CWO. ˙ 2 0 2 6 6 0 6 37 19 7 1 λ λ µ λ µ µ −−       − − − = ⇒ + =       −+    A1 Deduces second equation. CWO. 3, 1 λ µ = − = M1 A1 Solves simultaneous equations. 3 4 1 , 2 7 1 OP OQ         = − = −             JJJG JJJG A1 States JJJG OP and JJJG OQ . 8

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 8 of 20 Question Answer Marks Guidance 4(i) ( ) 3 3 1 1 2 0 0 1 1 d 1 3 3   = = −   ∫ x x x e x e e M1 A1 Must show working, AG. 2 4(ii) 3 3 3 1 1 1 2 2 2 3 0 0 0 1 2 d d 3 3 − − − −   = = −     ∫ ∫ n x n x n x n n I x x e x x e x e x M1 A1 Integrates by parts. ( ) 3 3 2 3 2 3 3 n n n e n I I e n I − − − − ⇒ = − − A1 AG 3 4(iii) ( ) ( ) 5 2 1 1 1 3 1 3 3 3 = − = −+ = I e I e e M1 A1 Applies reduction formula once and uses (i). ( ) ( ) 8 5 1 1 6 2 . 3 3 = − = − I e I e A1 Must be exact. 3

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 9 of 20 Question Answer Marks Guidance 5(i) ( ) ( ) 2 2 e e d d e e − − − − = + t t t t x t B1 ( ) ( ) ( ) ( )( ) ( ) ( ) 2 2 2 2 2 e e e e 2e 2e d 4 d e e e e e e − − − − − − + − − = = = + + + t t t t t t t t t t t t y t B1 Differentiates and simplifies. Accept 2 e e 1 . e e t t t t − −   − −  +   ( ) ( ) ( ) ( ) 2 2 2 2 4 4 4 e e 16 4 e e d d d d e e e e − − − − − + +     + = =         + + t t t t t t t t x y t t M1 A1 Attempt at writing 2 2 d d d d     +         x y t t as a square. ( ) 1 1 2 0 0 e e 2 e e 2π d 4π d e e e e e e t t t t t t t t t t S t t − − − − −   − −   = =    + +     + ∫ ∫ A1 Uses correct formula, simplifies to AG. 5 5(ii) 1 1 e e 2 1 2 2 4π d 4π e e S u u u − − + + − −   = = −   ∫ M1 A1 Applies given substitution. 1 1 1 4π . 2 e e−   = −   +   A1 AEF, must be exact. 3

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 10 of 20 Question Answer Marks Guidance 6(i) ( ) 3 1 = + y y M1 Obtains an equation in y not involving radicals. 3 2 3 2 3 3 1 3 2 1 0 y y y y y y y = + + + ⇒ + + + = A1 AG 3 3 = − S B1 3 6(ii) 3 3 3 3 3 3 3 3 3 3 2 2 1 α β β γ α γ α β γ − + + = = = − − S M1 A1 2 6(iii) ( ) ( ) 2 6 3 2 2 5 = − − = S M1 A1 Uses ( ) 2 2 2 α β α α αβ ≠ ∑ = ∑ + ∑ . AG. ( ) ( ) 9 6 3 3 2 3 3 5 2 3 3 12 = − − − = − − − − = − S S S M1 A1 Sums 3 2 3 2 1 0 − + −= y y y for 3 3 3 , , α β γ = y . 4

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 11 of 20 Question Answer Marks Guidance 7 ( )( ) 2 10 3 1 0 2 1 5 1 0 + −= ⇒ + − = u u u u M1 Axillary equation CF: 1 1 5 2 e e − = + t t x A B A1 PI: ¨ 0 = + ⇒ = ⇒ =  x p qt x q x M1 Forms PI and differentiates. 3 2 1, 5. − − = + ⇒ = − = − q p qt t q p M1 A1 Substitutes. GS: 1 1 5 2 e e 5 − = + −− t t x A B t A1 States general solution. 1 1 5 2 1 1 e e 1 2 5 − = − + −  t t x A B M1 Differentiates. 5 + = A B 1 1 1 2 5 − + = A B M1 Forms simultaneous equations. 0, 5 A B = = A1 1 5 5e 5 = −− t x t A1 States PS. 10

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 12 of 20 Question Answer Marks Guidance 8(i) 1 1 1 1 − = − z z So true when 1 = n . B1 Shows base case. Assume that 1 1 1 1 − − + +…+ = − k k z z z z B1 States inductive hypothesis. Then ( ) 1 1 1 1 1 1 1 , 1 1 1 + − −+ − − − + +…+ + = + = = − − − k k k k k k k z z z z z z z z z z z z so true when 1 = + n k M1 A1 Combines fractions. 1 + → k k H H Hence, by induction, true for all positive integers. A1 States conclusion. 5

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 13 of 20 Question Answer Marks Guidance 8(ii) Since 1 < z , 0 1 1 ∞ = − = − ∑ m m z z B1 States 1 < z and uses formula for sum to infinity of geometric progression. 1 0 1 2 sin Im Im 1 1 cos i sin 1 2 2 θ θ θ ∞ ∞ − = =       − = =         + −   ∑ ∑ m m m m m z M1 A1 Uses de Moivre’s theorem. 2 2 1 1 cos 1 i sin 2 2 Im 1 1 cos cos 1 sin 4 4 θ θ θ θ θ     − −−           − + +     M1 Multiply numerator and denominator by conjugate. 1 sin 2sin 2 5 5 4cos cos 4 θ θ θ θ = − − A1 States imaginary part, AG. 5

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 14 of 20 Question Answer Marks Guidance 9(i) ( ) 2 2 λ λ = = = A e A Ae Ae e M1 A1 AG 2 9(ii) Eigenvalues of A are , 2 n n and 3n. B1 : λ = n 3 1 0 1 3 0 0 0 0 0 0 n t n −       = = =          1 i j k e M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. 2 : λ = n 2 1 1 3 0 0 0 0 n n n t n n       = − = =          2 i j k e A1 3 : λ = n 2 3 3 2 1 3 0 0 0 0 2 2 n n t n n n         = − = =         −     3 i j k e A1 Eigenvalues of + n A I are 2 , 3 n n and 4n B1 Thus 1 1 3 0 0 0 0 2     =       n n P and ( ) ( ) ( ) 2 2 2 2 0 0 0 3 0 0 0 4       =       n n n D M1 A1 Or correctly matched permutations of columns. 8

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 15 of 20 Question Answer Marks Guidance 10(i) 5 and = − = x y a B1 B1 2 10(ii) ( ) ( )( ) 2 x 10 5 26 5 5 1 + + + + = + + + + a x a x x a M1 By inspection or long division. oblique asymptote is 5 y x a = + + A1 2 10(iii) 2 10 5 26 0 + + + = x x a M1 Puts y –values equal and forms quadratic equation. ( ) 2 10 4 5 26 4 20 0 − + = −− < a a so no intersection point A1 Correct discriminant and conclusion. 2 10(iv) ( )( ) 2 5 2 10 10 5 26 0 + + + − − − − − = x x a x ax x a M1 Differentiates and forms quadratic equation. 2 10 24 0 x x + + = A1 Stationary points are ( ) 4, 2 − + a and ( ) 6, 2 − − a A1 Must have both points. 3

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 16 of 20 Question Answer Marks Guidance 10(v) B1 Asymptotes drawn, intersection correct. B1 1 C correct. B1 2 C correct. 3

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 17 of 20 Question Answer Marks Guidance 11E(i) 1 2 2 cos θ θ = x M1 Uses cosθ = x r . 1 1 1 2 2 2 d 1 2 cos 2 sin cos 0 d 2 θ θ θ θ θ θ θ −     = − + =             M1 A1 Sets derivative of cosθ r equal to zero. 1 1 2 2 1 sin cos 0 cos 2 sin 2 tan 1 2 θ θ θ θ θ θ θ θ θ − − + = ⇒ = ⇒ = A1 AG ( ) ( ) ( ) ( ) 2 0.6 tan 0.6 1 0.179 and 2 0.7 tan 0.7 1 0.179 −= − −= B1 Shows sign change. 5

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 18 of 20 Question Answer Marks Guidance 11E(ii) ( ) 1 2 2 π 2 sec sec 2 0 4 θ θ θ θ θ θ = ⇒ − = ⇒ = M1 A1 Finds value of θ . 2 11E(iii) B1 Correct shape. B1 Intersection correct. 2 ߠൌ0

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 19 of 20 Question Answer Marks Guidance 11E(iv) ( ) π π π 4 4 4 2 2 0 0 0 1 1 1 2 d sec d 2 sec d 2 2 2 θ θ θ θ θ θ θ θ − = − ∫ ∫ ∫ M1 Forms correct integral. ( ) ( ) π 4 π 4 0 0 1 1 2 tan 2 tan d 2 2 θ θ θ θ θ θ  − − −   ∫ M1 A1 Integrates by parts. ( ) π π 2 4 4 0 0 1 1 2 tan lncos 2 2 θ θ θ θ θ    −  − +     A1 2 π π 1 π 1 1 π π 1 ln2 ln2 1 . 8 2 2 4 2 4 8 4         − − + = + −                   A1 AEF, must be exact. 5 11O(i)(a) 1 2 3 4 1 2 3 4 1 0 1 1 0 2 4 3 1 2 3 0 0 0 4 1 2 5 2 0 0 0 0 − −         −     → →     − − +         " a a 1 2 3 4 v v v v 1 2 3 4 ' ' ' ' v v v v M1A1 Reduces M or T M to echelon form. dim = V rank = 3 . A1 1 1 2 2 3 4 1 2 3 ' ' ' 0 0 + + = ⇒ = = = c v c v c v c c c M1 Shows 1 2 4 ', ', ' v v v are linearly independent. Thus 1 2 4 , , v v v are linearly independent (and so form a basis for V ). A1 5

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 20 of 20 Question Answer Marks Guidance 11O(i)(b) 1 2 4 1 2 4 1 0 1 0 2 3 1 2 0 0 4 1 2 2 0 4 6 − −         − +     → →     − + +     +     " x x y y x a z a z x t t x M1 A1 2 4 α β γ = − + + x α γ = − y Uses row operations or 2 α β γ = − + z a 2 2 α β γ = + + t System is consistent when ( ) 2 2 + = + ⇒ + = t x y x x y t Or ( ) ( ) 2 4 2 2 2 α β γ α γ α β γ − + + + − = + + = t M1 A1 AG 4 11O(ii) 2 3 4 0 −+ + + = x y z t 2 4 3 0 + + = y z t M1 Finds basis for null space. 3 , , 2 , 2 t z y x µ λ λ µ λ µ = = = − − = − + A1 A basis is { } 1 2 1 2 2 3 , , 1 0 0 2   −         − −      =                  e e A1 AEF 1 1 0 1 0 1 0 1 −          =          M so particular solution is 1 0 0 0    =    e B1 Finds particular solution. General solution is 1 2 λ µ = + + x e e e A1 FT Accept their basis. Must have correct particular solution. 5