Cambridge A Level Mathematics - Further 9231 — 2019 Oct/Nov Paper 1 · Variant 1

*5418419037* Cambridge Assessment International Education Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/11 Paper 1 October/November 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 22 printed pages and 2 blank pages. JC19 11_9231_11/FP © UCLES 2019 [Turn over

2 1 The curve C has equation y = xa for 0 ≤x ≤1, where a is a positive constant. Find, in terms of a, the coordinates of the centroid of the region enclosed by C, the line x = 1 and the x-axis. [6] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/11/O/N/19

3 2 It is given that y = lnax + 1, where a is a positive constant. Prove by mathematical induction that, for every positive integer n, dny dxn = −1n−1 n −1!an ax + 1n . [6] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/11/O/N/19 [Turn over

4 3 The integral In, where n is a positive integer, is defined by In = Ó 1 1 2 x−n sin 0x dx. (i) Show that nn + 1In+2 = 2n+1n + 0 −02In. [5] … … … … … … … … … … … … … (ii) Find I5 in terms of 0 and I1. [2] … … … … … … … © UCLES 2019 9231/11/O/N/19

5 4 The line y = 2x + 1 is an asymptote of the curve C with equation y = x2 + 1 ax + b. (i) Find the values of the constants a and b. [3] … … … … … … … (ii) State the equation of the other asymptote of C. [1] … … (iii) 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 2019 9231/11/O/N/19 [Turn over

6 5 Let SN = N Ð r=1 5r + 15r + 6 and TN = N Ð r=1 1 5r + 15r + 6. (i) Use standard results from the List of Formulae (MF10) to show that SN = 1 3N25N2 + 90N + 83. [3] … … … … … … … … … … … … … … … … (ii) Use the method of differences to express TN in terms of N. [4] … … … … … © UCLES 2019 9231/11/O/N/19

7 … … … … … … … … … … … … … … (iii) Find lim N→∞ N−3SNTN . [2] … … … … … … … … … … © UCLES 2019 9231/11/O/N/19 [Turn over

8 6 With O as the origin, the points A, B, C have position vectors i −j, 2i + j + 7k, i −j + k respectively. (i) Find the shortest distance between the lines OC and AB. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/11/O/N/19

9 (ii) Find the cartesian equation of the plane containing the line OC and the common perpendicular of the lines OC and AB. [4] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/11/O/N/19 [Turn over

10 7 The equation x3 + 2x2 + x + 7 = 0 has roots !, ", '. (i) Use the relation x2 = −7y to show that the equation 49y3 + 14y2 −27y + 7 = 0 has roots ! "' , " '! , ' !". [4] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/11/O/N/19

11 (ii) Show that !2 "2'2 + "2 '2!2 + '2 !2"2 = 58 49. [3] … … … … … … … … … … … … (iii) Find the exact value of !3 "3'3 + "3 '3!3 + '3 !3"3 . [2] … … … … … … … … … … © UCLES 2019 9231/11/O/N/19 [Turn over

12 8 The matrix M is defined by M = ` 2 m 1 0 m 7 0 0 1 a , where m ≠0, 1, 2. (i) Find a matrix P and a diagonal matrix D such that M = PDP−1. [7] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/11/O/N/19

13 … … … … … … … … … … … (ii) Find M7P. [3] … … … … … … … … … … … … … © UCLES 2019 9231/11/O/N/19 [Turn over

14 9 (i) Use de Moivre’s theorem to show that sec 61 = sec61 32 −48 sec21 + 18 sec41 −sec61. [6] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/11/O/N/19

15 (ii) Hence obtain the roots of the equation 3x6 −36x4 + 96x2 −64 = 0 in the form sec q0, where q is rational. [5] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/11/O/N/19 [Turn over

16 10 The matrix A is defined by A = ` 1 5 1 1 −2 −2 2 3 1 a . (i) (a) Find the rank of A when 1 ≠−1. [3] … … … … … … … … … … … (b) Find the rank of A when 1 = −1. [1] … … … Consider the system of equations x + 5y + z = −1, x −2y −2z = 0, 2x + 3y + 1z = 1. (ii) Solve the system of equations when 1 ≠−1. [3] … … … © UCLES 2019 9231/11/O/N/19

17 … … … … … … … … … … (iii) Find the general solution when 1 = −1. [3] … … … … … … (iv) Show that if 1 = −1 and & ≠−1 then Ax = ` −1 0 & a has no solution. [2] … … … … … … © UCLES 2019 9231/11/O/N/19 [Turn over

18 11 Answer only one of the following two alternatives. EITHER It is given that w = cos y and tan y d2y dx2 + @dy dx A2 + 2 tan y dy dx = 1 + e−2x sec y. (i) Show that d2w dx2 + 2dw dx + w = −e−2x. [4] … … … … … … … … … (ii) Find the particular solution for y in terms of x, given that when x = 0, y = 1 30 and dy dx = 1 ï3. [10] … … … … … … … … © UCLES 2019 9231/11/O/N/19

19 … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/11/O/N/19 [Turn over

20 OR The curves C1 and C2 have polar equations, for 0 ≤1 ≤1 20, as follows: C1 : r = 2e1 + e−1, C2 : r = e21 −e−21. The curves intersect at the point P where 1 = !. (i) Show that e2! −2e! −1 = 0. Hence find the exact value of ! and show that the value of r at P is 4ï2. [6] … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/11/O/N/19

21 (ii) Sketch C1 and C2 on the same diagram. [3] (iii) Find the area of the region enclosed by C1, C2 and the initial line, giving your answer correct to 3 significant figures. [5] … … … … … … … … … … … … … … … … © UCLES 2019 9231/11/O/N/19

22 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 9231/11/O/N/19

23 BLANK PAGE © UCLES 2019 9231/11/O/N/19

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 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 9231/11/O/N/19

This document consists of 20 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/11 Paper 1 October/November 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 October/November 2019 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 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. GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 3 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. DM or DB 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 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. 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.

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 4 of 20 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 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) WWW Without Wrong Working AWRT Answer Which Rounds To

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 5 of 20 Question Answer Marks Guidance 1 1 1 1 0 0 1 d 1 1 +   = = =   + +   ∫ a a x A x x a a B1 Finds area of region. 1 1 1 2 1 0 0 0 1 d d 2 2 + +   = = = =   + +   ∫ ∫ a a x Ax xy x x x a a M1 A1 Finds 1 0 d ∫xy x . 1 1 1 2 1 2 2 0 0 0 1 2 d d 2 1 2 1 +   = = = =   + +   ∫ ∫ a a x Ay y x x x a a M1 A1 Finds 1 2 0 d ∫y x . ( ) ( ) 1 1 , , 2 2 2 1   + + =     + +   a a x y a a A1 Both coordinates correct. 6

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 6 of 20 Question Answer Marks Guidance 2 ( ) ( ) 1 0 1 d 0! 1 d 1 1 = = − + + y a a x ax ax so true for 1 = n . M1 A1 Proves base case. Assume that for some positive integer k. B1 States inductive hypothesis. Then so true for 1 = + n k . M1 A1 Differentiates k th derivative. By induction, true for every positive integer n . A1 States conclusion. 6 k k k k k ax a k x y )1 ( )! 1 ( )1 ( d d 1 + − − = − 1 1 1 1 1 1 )1 ( ! )1 ( )1 ( )! 1 ( )1 ( d d + + + − + + + − = + − − − = k k k k k k k k ax a k ax a k ka x y

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 7 of 20 Question Answer Marks Guidance 3(i) 1 1 1 1 2 1 1 2 2 sin π π cosπ d 1 1 n n n x x I x x x n n −− −− +   = −   −− −−   ∫ M1 A1 Integrates by parts. 1 1 1 1 1 2 2 2 π cosπ π sin π d 1 1 n n n x x x x x n n n n + − −       = + +     + + − −       ∫ M1 Integrates by parts again. 1 2 π 1 π 1 1 n nI n n n n +   = + −   + +   ( ) 1 2 1 π 1 2 π n n n n I I n n + +   ⇒ + = + −     M1 Uses nI . ( ) 1 2 2 1 2 π π n n n n n I n I + + ⇒ + = + − A1 AG 5 3(ii) 2 3 1 2 4 π π I I = + − ( ) 2 2 5 1 π 12 48 π 4 π π 2 I I = + − + − M1 Substitutes 3I into reduction formula. ( ) 2 3 4 5 1 1 4 2π 4π π π 24 I I ⇒ = + − − + A1 AEF, must be exact with fractions simplified. 2

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 8 of 20 Question Answer Marks Guidance 4(i) ( )( ) 2 1 2 1 + = + + + x ax b x c M1 Uses that 2 1 + x is the quotient. 1 1 , 2 4 ⇒ = = − a b A1 A1 3 4(ii) 1 2 = x B1 FT 1

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 9 of 20 Question Answer Marks Guidance 4(iii) B1 Intersection (0,-4) given and asymptotes drawn. B1 Left branch correct. B1 FT Right branch correct. Deduct at most one mark for poor forms at infinity. 3

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 10 of 20 Question Answer Marks Guidance 5(i) ( )( ) 2 1 1 1 5 1 5 6 25 35 6 = = = + + = + + ∑ ∑ ∑ N N N r r r r r r r N M1 Expands. ( )( ) ( ) 1 1 25 1 2 1 35 1 6 6 2     + + + + +         N N N N N N M1 Substitutes formulae for ∑r and 2 ∑r . ( ) ( ) 2 2 25 35 35 1 2 3 1 6 25 90 83 6 2 2 3   = + + + + + = + +     N N N N N N N A1 Simplifies to the given answer (AG). 3 5(ii) ( )( ) 1 1 1 1 5 1 5 6 5 5 1 5 6   = −   + + + +   r r r r M1 A1 Finds partial fractions. 1 1 1 1 1 1 1 5 6 11 11 16 5 1 5 6   = − + − + + −   + +   " N T N N M1 Expresses terms as differences. ( ) 1 1 1 1 1 5 6 5 6 30 5 5 6   − = −   + +   N N A1 At least 3 terms including last. 4 5(iii) 3 25 1 5 3 30 18 N N S T N → × = M1 A1 Divides N S by 3 N and takes limits as →∞ N 2

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 11 of 20 Question Answer Marks Guidance 6(i) 1 2 7   =    JJJG AB B1 9 3 1 1 1 6 2 1 2 7 3 1 −         × = − = − =         −     JJJG JJJG OC AB t i j k M1 A1 Finds direction of common perpendicular. 2 2 2 1 3 1 2 0 1 1 0.267 14 3 2 1       − ⋅       −   = = + + M1 A1 Uses formula for shortest distance. 5 6(ii) 1 1 1 1 4 3 2 1 5 t −     = − =     −   i j k n M1 A1 Finds normal to plane. ( ) ( ) ( ) 0 4 0 5 0 0 − + + = M1 Uses point on plane. 4 5 0 −+ + = x y z A1 AEF 4

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 12 of 20 Question Answer Marks Guidance 7(i) ( ) ( ) 7 7 2 7 7 7 0 − − + − + − + = y y y y ( ) ( ) ( ) 2 2 7 7 1 14 7 7 7 1 14 7 ⇒ − − + = − ⇒− − + = − y y y y y y M1 Uses given substitution and eliminates radical. 3 2 49 14 27 7 0 ⇒ + − + = y y y A1 AG 2 2 7 αβγ = = − x x y M1 Uses 7 αβγ = −. So roots are 2 2 2 , , α α β β γ γ αβγ βγ αβγ αγ αβγ αβ = = = A1 AG 4 7(ii) 2 2 2 2 1 1 1 27 , 7 49 α β γ βγ αγ αβ γ β α + + = − + + = − B1 States sum of roots and ' ' ' ' ' ' α β α γ β γ + + . 2 2 2 2 2 2 2 2 2 2 2 27 58 2 7 49 49 α β γ β γ α γ α β     + + = − − − =         M1 A1 Uses ( ) ( ) 2 2 2 2 ' ' ' ' ' ' 2 ' ' ' ' ' ' α β γ α β γ α β α γ β γ + + = + + − + + AG 3 7(iii) 3 3 3 3 3 3 3 3 3 58 2 49 14 27 21 49 7 α β γ β γ α γ α β       + + = − + − −             M1 Uses 3 2 49 ' 14 27 7 α α α = − + ′− ′ . 3 3 3 3 3 3 3 3 3 317 343 α β γ β γ α γ α β ⇒ + + = − A1 2

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 13 of 20 Question Answer Marks Guidance 8(i) Eigenvalues of (upper diagonal matrix) A are 2, m and 1. (Or from characteristic equation: ( )( )( ) λ 2 λ λ 1 0 m − − − = ) B1 λ 2: = 1 2 1 0 2 1 0 0 0 0 1 0 0 −       = − = =       −    m m t i j k e M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. λ : m = ( ) 2 7 2 1 7 2 2 0 0 7 0 0         = − = − = −             m m m m m t m i j k e A1 λ 1: = 3 6 1 6 1 1 1 7 7 0 1 7 1 1 + +         = = − = −         − − −     m m m t m m m i j k e A1 Thus 1 6 1 0 2 7 0 0 1 +     = − −     −   m m m m P and 2 0 0 0 0 0 0 1     =       m D M1 A1 FT Or correctly matched permutations of columns. No follow through on two or more zero eigenvectors. 7

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 14 of 20 Question Answer Marks Guidance 8(ii) = 7 7 1 6 1 2 0 0 0 2 7 0 0 0 0 1 0 0 1   +      − −        −    m m m m m M1 A1 FT Applies . 7 8 8 7 2 6 1 0 2 7 0 0 1   +   = − −     −   m m m m m A1 Order of columns might be swapped depending on P. 3 Question Answer Marks Guidance 9(i) Write cosθ = c , sinθ = s . ( ) 6 cos6 sin 6 θ θ + = + i c is M1 Uses binomial theorem. 6 4 2 2 4 6 cos6 15 15 θ ⇒ = − + − c c s c s s A1 ( ) ( ) ( ) 2 3 6 4 2 2 4 6 6 4 2 2 2 2 15 15 15 1 15 1 1 − + − = − − + − − − c c s c s s c c c c c c M1 Uses 2 2 1 = − c s . ( ) ( ) ( ) 6 4 2 2 2 4 2 4 6 15 1 15 1 2 1 3 3 = − − + − + − − + − c c c c c c c c c A1 6 4 2 32 48 18 1 = − + − c c c M1 Divides numerator and denominator by 6c . 6 6 4 2 2 4 6 1 sec sec6 32 48 18 1 32 48sec 18sec sec θ θ θ θ θ ⇒ = = − + − − + − c c c A1 AG 6 7 1 7 7 PD P P PD P M = = − 1 7 7 − = P PD M

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 15 of 20 Question Answer Marks Guidance 9(ii) ( ) 6 6 2 4 6 2 4 6 2 32 48 18 2 32 48 18 = − + − ⇒ = − + − x x x x x x x x M1 A1 Relates with equation in part (i). 1 sec6 2 cos6 2 θ θ = ⇒ = M1 Solves 1 cos6 2 θ = . sec18 π = x A1 Gives one correct solution. sec π = x q , 5 7 11 13 17 , , , , 18 18 18 18 18 = q A1 Gives five other solutions. Allow different values of q as long as all six solutions are found. 5

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 16 of 20 Question Answer Marks Guidance 10(i) 1 5 1 1 5 1 1 2 2 0 7 3 2 3 0 0 1 θ θ         − − → − −         +     M1 A1 Reduces to echelon form. At least one row operation for M1. ( ) 3 = r A if 1 θ ≠− A1 ( ) 2 = r A if 1 θ = − B1 4 10(ii) x 5y + z + = 1 − 7y − 3z − = 1 ( 1)z θ + = ( 1) θ + M1 Uses reduced form of augmented matrix or eliminates variables from scratch. 4 6 1, , 7 7 = = − = z y x A1 A1 One correct. All three correct. 3 10(iii) x 5y + z + = 1 − 7y − 3z − = 1 ( 1)z θ + = ( 1) θ + = z t M1 Uses parameter. 3 1 8 2 , 7 7 + − = − = t t y x A1 A1 3

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 17 of 20 Question Answer Marks Guidance 10(iv) 5 1 + + = − x y z , 7 3 1 − − = y z , ( ) 1 1 θ φ + = + z M1 Uses reduced form of augmented matrix or eliminates variables from scratch 1 1 θ φ = −⇒ = − so no solution (inconsistent). A1 2 Question Answer Marks Guidance 11E(i) d d cos sin d d = ⇒ = − w y w y y x x B1 2 2 2 2 2 d d d sin cos d d d   = − −     w y y y y x x x B1 2 2 2 2 2 d d d d d 2 sin cos 2sin cos d d d d d   + + = − − − +     w w y y y w y y y y x x x x x M1 Uses substitution to obtain w - x equation, AG. ( ) 2 2 cos e sec e − − = − = − x x y y A1 4

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 18 of 20 Question Answer Marks Guidance 11E(ii) 2 2 1 0 1 + + = ⇒ = − m m m M1 Finds CF. CF: ( )e− = + x w Ax B A1 PI: 2 2 2 e 2 e 4 e − − − = ⇒ = − = ′ ′ ⇒ ′ x x x w k w k w k M1 Forms PI and differentiates. 4 4 1 1 − + = −⇒ = − k k k k A1 ( ) 2 e e − − = + − x x w Ax B A1 States general solution. 0 = x , , ⇒ B1 Uses initial conditions to find constants. ( ) 2 e e 2e − − − ′ + = − + + x x x w Ax B A M1 Differentiates general solution. 0 = x , , , M1 A1 Substitutes initial conditions. 1 2 3 cos e e 2 − − −     = − −         x x y x A1 States particular solution for y in terms of x . 10 π 3 1 = y 2 1 = w 2 3 = B π 3 1 = y 3 3 '= y 1 2 ' 2 3 2 1 2 1 − = ⇒ + + − = − ⇒ − = A A w

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 19 of 20 Question Answer Marks Guidance 11O(i) ( ) 2 2 e e 2 e e e e 2 α α α α α α − − − − = + ⇒ − = M1 Sets equations equal and divides by e e α α − + . 2e 2e 1 0 e 1 2 α α α − −= ⇒ = + M1 A1 Forms quadratic in eα , AG. ( ) ln 1 2 α = + A1 Must be exact. ( ) 2 1 2 2 1 4 2 = + + − = r M1 A1 Substitutes to find r . 6 11O(ii) B1 1 C has correct shape. B1 2 C has correct shape. B1 Intersection points positioned correctly. 3

9231/11 Cambridge International A Level – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 20 of 20 Question Answer Marks Guidance 11O(iii) ( ) ( ) ( ) ( ) ln 1 2 ln 1 2 2 2 2 2 0 0 1 2 e e d e e d 2 θ θ θ θ θ θ +√ +√ − − + − − ∫ ∫ ( ) ln 1 2 2 2 4 4 0 1 1 5 2e 2e e e d 2 2 θ θ θ θ θ +√ − − = + + − − ∫ M1 A1 Uses 2 1 d 2 r θ ∫ to formulate correct area. ( ) ln 1 2 2 2 4 4 0 1 1 5 e e e e 8 8 θ θ θ θ θ +√ − −   = + − − +     M1 A1 Expands and integrates. ( ) ( ) ( ) ( ) 2 2 4 4 1 5ln(1 2) 1 2 1 2 1 2 1 2 5.82 8 − −   = + √ + + − + − + − + =     A1 5