Cambridge A Level Mathematics - Further 9231 — 2019 May/June Paper 1 · Variant 3

*1749707556* Cambridge Assessment International Education Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/13 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 23 printed pages and 1 blank page. JC19 06_9231_13/RP © UCLES 2019 [Turn over

2 BLANK PAGE © UCLES 2019 9231/13/M/J/19

3 1 Prove by mathematical induction that 33n −1 is divisible by 13 for every positive integer n. [5] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19 [Turn over

4 2 The curve C has polar equation r2 = ln1 + 1, for 0 ≤1 ≤20. (i) Sketch C. [2] (ii) Using the substitution u = 1 + 1, or otherwise, find the area of the region bounded by C and the initial line, leaving your answer in an exact form. [5] … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19

5 3 (i) Write down the fifth roots of unity. [2] … … … (ii) Find all the roots of the equation z10 + z5 + 1 = 0, giving each root in the form ei1. [5] … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19 [Turn over

6 4 (i) Use the method of differences to show that N Ð r=1 1 3r + 13r −2 = 1 3 − 1 33N + 1. [4] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19

7 (ii) Find the limit, as N →∞, of N2 Ð r=N+1 N 3r + 13r −2. [4] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19 [Turn over

8 5 The linear transformation T : >4 →>4 is represented by the matrix M, where M = € 1 2 0 4 5 2 1 −3 4 0 1 −7 −2 4 −1 !  . It is given that the rank of M is 2. (i) Find the value of ! and state a basis for the range space of T. [4] … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19

9 (ii) Obtain a basis for the null space of T. [4] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19 [Turn over

10 6 The curve C has equation y = x2 kx −1, where k is a positive constant. (i) Obtain the equations of the asymptotes of C. [3] … … … … … … … … … … … … (ii) Find the coordinates of the stationary points of C. [3] … … … … … … … … … © UCLES 2019 9231/13/M/J/19

11 … … … … … … … … … … … … (iii) Sketch C. [3] © UCLES 2019 9231/13/M/J/19 [Turn over

12 7 The line l1 passes through the points A −3, 1, 4 and B −1, 5, 9. The line l2 passes through the points C −2, 6, 5 and D −1, 7, 5. (i) Find the shortest distance between the lines l1 and l2. [5] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19

13 (ii) Find the acute angle between the line l2 and the plane containing A, B and D. [5] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19 [Turn over

14 8 Find the particular solution of the differential equation 9d2x dt2 + 6dx dt + x = 50 sin t, given that when t = 0, x = 0 and dx dt = 0. [10] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19

15 … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19 [Turn over

16 9 A cubic equation x3 + bx2 + cx + d = 0 has real roots !, " and ' such that 1 ! + 1 " + 1 ' = −5 12, !"' = −12, !3 + "3 + '3 = 90. (i) Find the values of c and d. [3] … … … … … … … (ii) Express !2 + "2 + '2 in terms of b. [2] … … … … (iii) Show that b3 −15b + 126 = 0. [4] … … … … … … … © UCLES 2019 9231/13/M/J/19

17 … … … … … … … … … … … … (iv) Given that 3 + iï12 is a root of y3 −15y + 126 = 0, deduce the value of b. [2] … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19 [Turn over

18 10 Let In = Ó 1 20 1 40 cotnx dx, where n ≥0. (i) By considering d dxcotn+1x, or otherwise, show that In+2 = 1 n + 1 −In. [5] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19

19 The curve C has equation y = cot x, for 1 40 ≤x ≤1 20. (ii) Find, in an exact form, the y-coordinate of the centroid of the region enclosed by C, the line x = 1 40 and the x-axis. [6] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19 [Turn over

20 11 Answer only one of the following two alternatives. EITHER A 3 × 3 matrix A has distinct eigenvalues 2, 1, 3, with corresponding eigenvectors ` 1 1 0 a , ` −1 0 b a , ` 0 1 −1 a respectively, where b is a positive constant. (i) Find A in terms of b. [9] … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19

21 (ii) Find A−1 ` 0 2 −2 a . [2] … … … … … … … … … … (iii) It is given that An ` 1 1 0 a = ` 4 4 0 a and An ` −1 0 b a = ` −1 0 b−1 a . Find the values of n and b. [3] … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19 [Turn over

22 OR The positive variables y and t are related by y = at, where a is a positive constant. (i) (a) By differentiating ln y with respect to t, show that dy dt = at ln a. [3] … … … … … (b) Write down d2y dt2 . [1] … … … (ii) Determine the set of values of a for which the infinite series y + dy dt + d2y dt2 + d3y dt3 + … is convergent. [3] … … … … … … … … © UCLES 2019 9231/13/M/J/19

23 A curve has parametric equations x = ta, y = at. (iii) Find d2y dx2 in terms of a and t, and show that, when t = 2, d2y dx2 = 21−2a1 −a + 2 ln a ln a. [7] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2019 9231/13/M/J/19

24 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. … … … … … … … … … … … … … … … … … … … … … 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 9231/13/M/J/19

This document consists of 22 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International Advanced Level FURTHER MATHEMATICS 9231/13 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.

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 2 of 22 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).

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 3 of 22 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.

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 4 of 22 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)

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 5 of 22 Question Answer Marks Guidance 1 33 1 26 −= is divisible by 13 B1 Checks base case. Assume that 33 1 − k is divisible by 13 for some positive integer k B1 States inductive hypothesis. Then 3 3 3 3 3 3 3 1 3 3 1 26 3 3 1 + −= −= ⋅ + − k k k k M1 Separates 33 1 − k is divisible by 13 A1 1 + ⇒ k k H H By induction, 33 1 − n is divisible by 13 for every positive integer n. A1 States conclusion. 5

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 6 of 22 Question Answer Marks Guidance 2(i) B1 Correct shape, domain and orientation. B1 The initial line is tangential to C at the pole. 2 2(ii) ( ) 2π 0 1 ln 1 d 2 A θ θ = + ∫ M1 States 2 1 d 2 θ ∫r with correct expression and limits. ( ) 2π 2π 1 0 1 ln 1 d ln d u u θ θ + + = ∫ ∫ M1 Applies given substitution correctly, changes their limits. [ ] [ ] 2π 1 2π 1 2π 1 1 1 1 ln d ln u u u u u + + + = − ∫ M1 A1 Integrates lnu (by parts or otherwise). ( ) 1 π ln 2π 1 π 2 A   = + + −     A1 AEF, must be exact. 5 0 θ =

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 7 of 22 Question Answer Marks Guidance 3(i) 2π exp i , 0, 1, 2 5 k k   = ± ±     B2 B1 for 1 correct fifth root. B1 for all 5 distinct, correct roots (AEF). SCB1 if only arguments are given. 3(ii) 5 1 3 i 2 2 = − ± z M1 Correctly solves quadratic in 5z 5 1 1 exp i2π or exp i2π 3 3 z k k         = + − +                 M1 A1 Writes in polar or exponential form and adds multiples of 2π 2π 4π 8π 2π 14π exp i , exp i ,exp i ,exp i ,exp i 15 15 15 3 15           ± ± ± ± ±                     A2 A1 for 5 distinct, correct roots. A1 for exactly 10 distinct, correct roots. Allow alternative exact values of θ such as 16π 4π 22π 26π 28π , , , , . 15 3 15 15 15 θ = Alternative method for 3(ii) 5 5 1 − + = − z z M1 Divides through by 5z 2cos5 1 θ = − M1 A1 Applies de Moivre’s theorem 2π 4π 8π 2π 14π exp i , exp i ,exp i ,exp i ,exp i 15 15 15 3 15           ± ± ± ± ±                     A2 A1 for 5 distinct, correct roots. A1 for exactly 10 distinct, correct roots. Allow alternative exact values of θ such as 16π 4π 22π 26π 28π , , , , . 15 3 15 15 15 θ = 5

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 8 of 22 Question Answer Marks Guidance 4(i) ( )( ) 1 1 1 1 3 1 3 2 3 3 2 3 1   = −   + − − +   r r r r M1 A1 Finds partial fractions. ( )( ) 1 1 1 1 1 1 1 1 1 3 1 3 2 3 1 4 4 7 3 2 3 1 N r r r N N =   = − + − +… −   + − − +   ∑ M1 At least 3 term including final term. 1 1 1 3 3 1   = −   +   N A1 AG 4 4(ii) ( )( ) ( )( ) ( )( ) 2 2 1 1 1 3 1 3 2 3 1 3 2 3 1 3 2 N N N r N r r N N N r r r r r r = + = = = − + − + − + − ∑ ∑ ∑ M1 Uses 2 2 1 1 1 = + = = = − ∑ ∑∑ N N N r N r r ( ) ( ) 2 3 3 3 3 1 3 3 1   = − − −     + +   N N N N N N M1 Applies (i) ( ) ( ) ( )( ) 3 2 2 2 3 3 1 3 3 1 3 1 3 1 − = − = + + + + N N N N N N N N A1 Allow simplification to common denominator. 1 as 9 → →∞ N B1 4

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 9 of 22 Question Answer Marks Guidance 5(i) 1 2 0 4 1 2 0 4 5 2 1 3 0 8 1 23 4 0 1 7 0 8 1 23 2 4 1 0 8 1 8 α α         − − −     →     − − −     − − − +     1 2 0 4 0 8 1 23 0 0 0 0 0 0 0 15 α     − −   →    −   M1 A1 Performs row operations correctly. ( ) 2 15 α = ⇒ = M r A1 CWO Basis for the range space is 1 2 5 2 , 4 0 2 4                       −     B1 Accept any two independent column vectors from M. 4 5(ii) 2 4 0 x y t + + = 8 23 0 y z t − + − = M1 Forms system of equations. 1 23 1 7 , , , 8 8 4 4 t z y x λ µ µ λ µ λ = = = − = − + M1 Uses two parameters. Basis for null space is 2 14 1 23 , 8 0 0 8   −         −                       A1 A1 AEF. Many alternatives are possible e.g. 2 4 4 0 1 0 0 2 , , , 8 23 23 7 0 1 1 1    − −                                  − −             − −          4

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 10 of 22 Question Answer Marks Guidance 6(i) 1 = x k B1 States vertical asymptote. ( )( ) 1 2 2 1 1 − − − − + + = − kx k x k k y kx M1 Finds oblique asymptote. Oblique asymptote is 1 2 − − = + y k x k A1 3 6(ii) ( ) ( ) 2 2 2 2 1 0 2 0 1 ′ − − = = ⇒ − = − x kx kx y kx x kx M1 Differentiates and equates to 0. 1 0, 2 x k − = A1 Finds x-coordinates. ( ) ( ) 1 2 0,0 , 2 , 4 − − k k A1 Finds y -coordinates 3

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 11 of 22 Question Answer Marks Guidance 6(iii) B1 Axes and asymptotes correct. B1 Upper branch correct. B1 Lower branch correct. Deduct at most 1 mark for poor forms at infinity. 3

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 12 of 22 Question Answer Marks Guidance 7(i) 2 1 4 , 1 5 0     = =       JJJG JJJG AB CD B1 Find the directions of the lines. 5 2 4 5 5 1 1 0 2 −     =     −   i j k M1 A1 Finds direction of common perpendicular. Allow any non- zero scalar multiple. 2 2 2 1 5 5 5 1 2 18 6 2.45 54 5 5 2 −     ⋅     −   = = = + + M1 A1 Uses formula for shortest distance. 5 7(ii) 26 13 2 4 5 8 ~ 4 2 6 1 4 2 − −         =             i j k M1 A1 Finds normal to the plane. 2 2 2 2 2 2 1 13 1 4 0 2 9 42 14 378 1 1 0 13 4 2 −     ⋅       = − = − + + + + M1 A1 FT Uses dot product of their normal with direction of line. ( ) 1 42 cos 90 27.6 or 0.481 rad 14 −  − − = °       A1 5

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 13 of 22 Question Answer Marks Guidance 8 ( ) 2 2 9 6 1 0 3 1 0 + + = ⇒ + = u u u M1 Auxiliary equation. CF: ( ) 1 3 e − = + t x At B A1 States CF. sin cos cos sin = + ⇒ = −  x p t q t x p t q t ¨ sin cos ⇒ = − − x p t q t M1 Forms PI and differentiates. 9 6 50 − − + = p q p 9 6 0 − + + = q p q M1 Substitutes. 3, 4. q p = − = − A1 ( ) 1 3 e 4sin 3cos − = + − − t x At B t t A1 States general solution. FT if correct form of CF/PI. ( ) 1 1 3 3 1 e e 4cos 3sin 3 − − = − + + − +  t t x At B A t t M1 Differentiates and forms equations using initial conditions. 3 = B B1 1 4 0 5 3 − + − = ⇒ = B A A A1 . ( ) 1 3 5 3 e 4sin 3cos − = + − − t x t t t A1 States PS. 10

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 14 of 22 Question Answer Marks Guidance 9(i) 1 1 1 5 αβ βγ αγ αβ βγ αγ α β γ αβγ + + + + = ⇒ = + + = c M1 A1 Uses 1 1 1 . αβ βγ αγ α β γ αβγ + + + + = 12 = d B1 3 9(ii) ( ) ( ) 2 2 2 2 2 α β γ α β γ αβ βγ γα + + = + + − + + M1 ( ) 2 2 10 10 b α β γ + + − = − A1 Alternative method for 9(ii) 2 2 5 126 90 5 36 0 − + − + = ⇒ = b bS b S b M1 A1 Sums over roots 2

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 15 of 22 Question Answer Marks Guidance 9(iii) 3 2 5 12 0 + + + = x bx x ( α β γ −= + + b ) M1 Formulates equation. ( ) 2 90 10 5 36 0 + − − + = b b b M1 A1 Sums over roots and uses 9(ii) 3 15 126 0 − + = b b A1 Alternative method 1 for 9(iii) ( ) ( )( ) 3 3 3 3 3 3 α β γ α β γ α β γ αβ βγ γα αβγ + + = + + + + + + + − M1 A1 Uses expansion of ( ) 3 α β γ + + 3 90 15 36 − = − + b b (M1) Substitutes known values. 3 15 126 0 − + = b b A1 Alternative method 2 for 9(iii) ( ) ( ) 2 2 2 2 2 2 10 α β γ α β γ αβ βγ γα + + = + + − + + = − b M1 A1 Finds sum of squares in terms of b. 2 3 5 126 10 15 126 0 − = − ⇒ − + = b b b b b M1 A1 Equates expressions. 4 9(iv) Real root is 6 = − b M1 A1 Gives the real root for b .

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 16 of 22 Question Answer Marks Guidance 10(i) ( ) ( ) n 1 n 2 d cot 1 cot csc d x n x x x + = − + M1 A1 Differentiates using chain rule. ( ) ( ) n 2 1 cot cot 1 = − + + n x x ( ) ( ) n 2 n 1 cot 1 cot + = − + − + n x n x M1 Uses 2 2 csc cot 1. = + x x OE ( )( ) 1 π n 1 2 2 1 π 4 cot 1 n n x n I I + +   = − + +   M1 Integrates both sides. ( )( ) 2 2 1 1 1 1 + + −= − + + ⇒ = − + n n n n n I I I I n A1 AG 10(i) Alternative method for 10(i) ( ) ( ) 2 2 1 1 2 2 1 sin cos 1 sin cos d cos d sin sin n n n n n n n n x x n x x x x x x + + + + +   − + − + =     M1 A1 Differentiates using quotient/chain rule. ( ) ( ) n n 2 1 cot 1 cot + − + − + n x n x M1 Separates ( )( ) 1 π n 1 2 2 1 π 4 cot 1 n n x n I I + +   = − + +   M1 Integrates both sides. ( )( ) 2 2 1 1 1 1 + + −= − + + ⇒ = − + n n n n n I I I I n A1 AG 5

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 17 of 22 Question Answer Marks Guidance 10(ii) 1π 2 2 2 1π 4 1 cot d 2 2 I y x x A A = = ∫ M1 Uses correct formula for .y Allow incorrect or missing limits. [ ] 1π 1 2 π 2 1π 1 4 π 4 cot d lnsin ln 2 A x x x = = = ∫ M1 A1 Integrates cot x to find area under the curve with correct limits. 2 0 1 1 1 π 4 I I = − = − M1 A1 Finds 2I 1 1 1 π ln 2 4 y   = −     A1 AEF. Must be exact. 6

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 18 of 22 Question Answer Marks Guidance 11E(i) 1 1 0 1 0 1 0 1 −     =     −   P b 2 0 0 0 1 0 0 0 3     =       D B1 B1 Writes P and D (accept correctly matched permutations of columns) det 1 = −− P b B1 T 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 − − −         = = −     + +     − − −     P b b b b b b b b M1 A1 Finds inverse of P. (Adj ÷ Det). 1 − = A PDP M1 Applies 1 − = A PDP 1 1 0 2 2 2 1 1 0 1 1 1 1 1 0 1 3 3 3 −       = −    +    − − −    A b b b b b M1 A1 Multiplies two adjacent matrices. 2 1 0 2 0 3 0 3 −     =     −   PD b 1 2 1 1 1 3 2 1 1 2 2 3 b b b b b b b +     − + −   +   − +   A1

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 19 of 22 Question Answer Marks Guidance 11E(i) Alternative method for 11E(i) 11 12 13 21 22 22 31 32 33 1 2 1 2 0 0       =           a a a a a a a a a 11 12 13 21 22 22 31 32 33 1 1 0 0 − −           =                 a a a a a a a a a b b 11 12 13 21 22 22 31 32 33 0 0 1 3 1 3           =           − −       a a a a a a a a a B3 Matches eigenvalues with eigenvectors to form systems of equations. B1 for each correctly matched pair. Three values found correctly from solving systems of equations. Another three values found correctly from solving systems of equations. Attempting to find final three values from systems of equations. M1 A1 M1 A1 M1 1 2 1 1 1 3 2 1 1 2 2 3 +     = − + −   +   − +   A b b b b b b b A1 Final answer given in matrix form. 9 11E(ii) 1 1 0 0 0 2 2 2 1 1 3 2 1 1 − −             = =             − − −       A A M1 A1 M1 for 1 1 0 0 2 2 1 . 2 1 − −         =         − −     A A 2

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 20 of 22 Question Answer Marks Guidance 11E(iii) B1 Uses M1 A1 M1 for stating 1 − = b b (since ). Must be only one answer ( 1 = b ) for A1. 3 2 0 1 1 2 0 1 1 2 = ⇒           =           n n A           =           0 1 1 2 0 1 1 n n A 1 0 1 0 1 1 1 = ⇒ = ⇒          − =          − − − b b b b b n A          − =          − b b n 0 1 0 1 A

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 21 of 22 Question Answer Marks Guidance 11O(i)(a) ln ln = y t a 1 d ln d = y a y t M1 A1 Differentiates both sides. d ln ln d t y y a a a t = = A1 AG 3 11O(i)(b) ( ) 2 2 2 d ln d = t y a a t B1 11O(ii) ln = r a B1 States common ratio. 1 1 ln 1 e e − −< < ⇒ < < a a M1 A1 Uses convergence condition for a geometric series. 3

9231/13 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 22 of 22 Question Answer Marks Guidance 11O(iii) 1 1 d d d ln ln d d d − −     = × = =        t t a a y y t a a a a x t x a at t M1 A1 Uses chain rule to find d d y x ( ) ( ) 1 2 2 1 ln 1 d d ln d d − − −   − −     =             a t a t a t a a a t a y a t x a t M1 A1 Uses quotient or product rule to find d d d d       y t x ( ) ( ) 1 2 2 2 2 3 1 ln 1 d ln d − − −   − −   =         a t a t a t a a a t a y a x a t ( ) ( ) 1 2 2 1 ln 1 ln − −   − −   =         t a a a t a a a t M1 A1 Uses chain rule to find 2 2 d d y x ( ) 2 1 2 2 d 2 2 ln 2ln 1 d − = ⇒ = − + a y t a a a x A1 Substitutes 2 = t , AG. 7