Cambridge A Level Mathematics - Further 9231 — 2022 May/June Paper 2 · Variant 3

This document has 20 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level DC (PQ/SG) 214267/1 © UCLES 2022 * 2 4 6 4 2 0 9 3 4 7 * FURTHER MATHEMATICS 9231/23 Paper 2 Further Pure Mathematics 2 May/June 2022 2 hours You must answer on the question paper. You will need: List of formulae (MF19) INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● If additional space is needed, you should use the lined page at the end of this booklet; the question number or numbers must be clearly shown. ● You should use a calculator where appropriate. ● You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator. ● Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. INFORMATION ● The total mark for this paper is 75. ● The number of marks for each question or part question is shown in brackets [ ].

2 9231/23/M/J/22 © UCLES 2022 1 Find the roots of the equation z 7 3 7i 3 = - , giving your answers in the form reii, where r 0 2 and r r 1 G i - . [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

3 9231/23/M/J/22 © UCLES 2022 [Turn over 2 (a) Find the coefficient of x2 in the Maclaurin’s series for lncosx - . [4] … … … … … … … … … … … … … (b) Find the length of the arc of the curve with equation lncos y x = - from the point where x 0 = to the point where r x 4 1 = . [4] … … … … … … … … … … … … …

4 9231/23/M/J/22 © UCLES 2022 3 The matrix A is given by A 6 5 1 9 8 1 5 5 2 = - - - f p. (a) Find the eigenvalues of A. [4] … … … … … … … … … … … … … … … … … … … … … … … …

5 9231/23/M/J/22 © UCLES 2022 [Turn over (b) Use the characteristic equation of A to show that p q A A I 1 2 = + - , where p and q are constants to be determined. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 9231/23/M/J/22 © UCLES 2022 4 It is given that tan x t t 1 = - + - and sinh y t t 1 = + - . (a) Show that x y t t t 1 1 d d 2 2 2 = - + + + . [4] … … … … … … … … … … … … … … … … … … … … … … … …

7 9231/23/M/J/22 © UCLES 2022 [Turn over (b) Find the value of x y d d 2 2 when t 4 3 = . [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 9231/23/M/J/22 © UCLES 2022 5 Find the solution of the differential equation ( ) x x x y y x 7 7 d d + + = for which y 7 = when x 1 = . Give your answer in the form ( ) y x f = . [9] … … … … … … … … … … … … … … … … … … … … … … … … …

9 9231/23/M/J/22 © UCLES 2022 [Turn over … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 9231/23/M/J/22 © UCLES 2022 6 n 1 n 2 n n 1 - x y 1 0 The diagram shows the curve with equation ( ) ln y x 1 = + for x 0 1 G G , together with a set of n rectangles each of width n 1. (a) By considering the sum of the areas of these rectangles, show that ( ) ln x x U 1 d n 0 1 1 + y , where ! ( )! ln ln U n n n n 1 2 n = - . [4] … … … … … … … … … … … … … …

11 9231/23/M/J/22 © UCLES 2022 [Turn over (b) Use a similar method to find, in terms of n, a lower bound Ln for ( ) ln x x 1 d 0 1 + y . [4] … … … … … … … … … … … … … … … … … (c) By simplifying U L n n - , show that ( ) lim U L 0 n n n - = " 3 . [2] … … … … … … … … …

12 9231/23/M/J/22 © UCLES 2022 7 The variables x and y are related by the differential equation x y y 4 3 d d 2 2 - = . It is given that, when x 0 = , y 3 =- and x y 2 d d = . (a) Find y in terms of x. [8] … … … … … … … … … … … … … … … … … … … … … … … …

13 9231/23/M/J/22 © UCLES 2022 [Turn over … … … … … … … … … … … … … … (b) Deduce the exact value of x for which y 0 = . Give your answer in logarithmic form. [3] … … … … … … … … … … … … …

14 9231/23/M/J/22 © UCLES 2022 8 (a) Find sin cos d n i i i y , where n 1 ! - . [2] … … … Let sin cos I d r , m n m n 0 2 1 i i i = y . (b) Show that, for m 2 H and n 0 H , I m n m I 1 , , m n m n 2 = + - - . [5] … … … … … … … … … … … … … … … … … … … … …

15 9231/23/M/J/22 © UCLES 2022 [Turn over (c) By considering the binomial expansion of z z 1 5 + b l , where cos sin z i i i = + , use de Moivre’s theorem to show that cos cos cos cos a b c 5 3 5i i i i = + + , where a, b and c are constants to be determined. [5] … … … … … … … … … … … … … … … … … … … … … … … …

16 9231/23/M/J/22 © UCLES 2022 (d) Using the results given in parts (b) and (c), find the exact value of I ,2 5. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

17 9231/23/M/J/22 © UCLES 2022 Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … … …

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9231/23/M/J/22 © UCLES 2022 20 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 Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge.

This document consists of 15 printed pages. © UCLES 2022 [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/23 Paper 2 Further Pure Mathematics 2 May/June 2022 MARK SCHEME Maximum Mark: 75 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 2022 series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 2 of 15 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.

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 3 of 15 Mathematics Specific Marking Principles 1 Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required then no marks will be awarded for a scale drawing. 2 Unless specified in the question, answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the degree of accuracy is not affected. 3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points. 4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). 5 Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or the method required, award all marks earned and deduct just 1 mark for the misread. 6 Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 4 of 15 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.  A or B marks are given for correct work only (not for results obtained from incorrect working) unless follow through is allowed (see abbreviation FT above).  For a numerical answer, allow the A or B mark if the answer is correct to 3 significant figures or would be correct to 3 significant figures if rounded (1 decimal place for angles in degrees).  The total number of marks available for each question is shown at the bottom of the Marks column.  Wrong or missing units in an answer should not result in loss of marks unless the guidance indicates otherwise.  Square brackets [ ] around text or numbers show extra information not needed for the mark to be awarded.

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 5 of 15 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

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 6 of 15 Question Answer Marks Guidance 1 1 6i π 3 7 3 7i 14e z     B1 Converts 7 3 7i  to exponential form. 1 1 3 18 i π 1 14 e z   M1 A1 Finds one root. 1 11 3 18 i π 2 14 e z  , 13 1 3 18 i π 3 14 e z   A1FT A1FT Finds other two roots. FT on modulus. Others given scores A1 A0. 5 Question Answer Marks Guidance 2(a)   d lncos tan d   x x x B1 Finds first derivative.   2 d tan sec d  x x x B1 Finds second derivative. ''(0) 1 2! 2  y M1 A1 Evaluates second derivative at zero. Accept 1 2 2. x 4

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 7 of 15 Question Answer Marks Guidance 2(b) 1 4 0 2 π 1 tan d x x   M1 Forms correct integral 1 4π 0 sec d x x  A1     1 4 π 1 1 2 4 0 ln ta π n x       M1 Integrates (answer given in List MF19). Accept ln(sec tan ).  x x   3 8 1 π ln tan 0.88  A1 Accept   ln 2 1 .  4 Question Answer Marks Guidance 3(a) 6 9 5 5 8 5 0 1 1 2          M1 Sets determinant equal to zero. 3 7 6 0      M1 Expands. 1, 2, 3   A1 A1 Award A1A0 for 2 correct solutions, A0A0 for 1 correct solution. 4

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 8 of 15 Question Answer Marks Guidance 3(b) 3 7 6 0    A A I B1 States that A satisfies its characteristic equation. 2 1 7 6 0     A I A M1 Multiplies through by 1  A . 1 2 7 1 6 6   A A I A1 3 Question Answer Marks Guidance 4(a) 2 d 1 1 d 1   y t t B1 2 d 1 1 d 1   x t t B1 2 2 2 2 d d dt 1 d 1 1 dt d 1          y y t x t t t x M1 Uses chain rule. 2 2 2 2 2 2 1 1 1 1 1                 t t t t t t A1 AG 4

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 9 of 15 Question Answer Marks Guidance 4(b)     1 2 2 2 2 2 4 2 1 2 1 1 d d d d                     t t t t t t t y t x t M1 A1 Applies product or quotient rule.        1 2 5 2 2 2 2 2 2 2 2 2 5 1 2 1 1 d 1 2 2 1 d 1               t t t y x t t t t t t M1 A1 Uses chain rule. Simplification not required for A1. 80 3  when 3 4.  t A1 5

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 10 of 15 Question Answer Marks Guidance 5 d 7 1 d ( 7) 7     y y x x x x B1 Divides through by ( 7)  x x . 7 1 1 ( 7) 7     x x x x or   49 4 2 7 2 7 7 ( 7)     x x x B1 Converts to partial fractions or completes the square. 1 1 ( 7) d ln ln( 7 7 ) e e 7          x x x x x x x M1 A1 Finds integrating factor. Accept non-zero multiple of . 7  x x     2 2 d 1 7 d 7 7 7 7             yx x x x x x x M1 Correct form on LHS and attempt to integrate RHS. RHS must be of the form   2 . 7 7    a b x x 1 ln( 7) 7( 7) ( ) 7        yx x x C x A1 7 7 8 8 ln8   C M1 Finds C, substitutes into their expression. 7 7 7 1 ln 8                 x y x x M1 A1 Divides their expression through by coefficient of y. 9

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 11 of 15 Question Answer Marks Guidance 6(a)           1 1 1 1 2 1 0 ln 1 d ln 1 ln 1 ln 1           n n n n n n n x x M1 A1 Forms the sum of the areas of the rectangles.           2 ! 1 1 1 ! ln ln 1 ln 2 n l 2 n l ln           n n n n n n n n n n M1 A1 Applies laws of logarithms, AG. 4 6(b)           1 1 1 1 1 2 1 0 ln 1 d ln 1 ln 1 ln 1            n n n n n n n x x M1 A1 Forms the sum of the areas of appropriate rectangles.         1 1 ( 1)ln ln 1 l n 2 n 2 l 1           n n n n n n M1 Applies laws of logarithms.     2 1 ! 1 1 ! ln ln    n n n n n n A1 Equivalent answers: (2 )! 1 1 ! ln ln ln 2;   n n n n n (2 1)! 1 ( 1)! ln ln .    n n n n 4 6(c) ln 2   n n U L n M1 Simplifies their  n n U L to . c n ln2 0  n as  n A1 AG 2

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 12 of 15 Question Answer Marks Guidance 7(a) 1 2 2 4 1 0 leading to , m m   M1 Auxiliary equation. 1 1 2 2 e e    x x y A B A1 Complementary function. Allow " "  y missing. 3 leading to 3 k k   M1 A1 Substitutes .  y k 1 1 2 2 e e 3     x x y A B A1 FT General solution. Must have  y or sensible equivalent, such as .  GS y Don’t allow .  GS 1 1 2 2 1 1 2 2 ' e e    x x y A B B1 Derivative of general solution. 1 1 2 2 0 2 leading to 2, 2 A B A B A B        M1 Substitutes initial conditions into general solution, solves simultaneous equations. 1 1 2 2 2e 2e 3     x x y A1 Particular solution. Must have  y or sensible equivalent, such as .  PS y Don’t allow .  PS 8 7(b) 3 1 1 2 2 4 1 0 leading to 4sinh 3 leading to sinh y x x     M1 Makes x the subject. y must be of the form 1 2 .    bx bx a e a e c   3 9 4 16 1 3 4 sinh ln 1 ln 2      M1 Uses logarithmic form. y must be of the form 1 2 .    bx bx a e a e c 2ln 2 ln 4   x A1 3

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 13 of 15 Question Answer Marks Guidance 8(a) 1 cos sin ( ) cos d 1         n n C n M1 A1 Allow with " " C missing. 2 8(b) 1 2 1 1 2 2 π 1 1 π 0 0 0 π 1 2 2 sin 1 s cos sin cos d si in cos 1 1 n d m n m n n m m n n                          M1 A1 Integrates by parts using part (a). Allow with limits missing. 1 2π 2 0 2 sin (1 sin )d 1 cos 1 n m m n          M1 Applies 2 2 . c i o 1 s s n    , 2, , 1 1 1 1        m n m n m n m m I I I n n M1 Combines with LHS. , 2, , 2, 1 1 leading to . 1 1 m n m n m n m n n m m m I I I I n n m n           A1 AG.

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 14 of 15 Question Answer Marks Guidance 8(b) Alternative method for question 8(b)   1 1 2 2 π π 2 2 2 0 0 2 s sin co d in o si s d 1 c s s n co m n m n             M1 Applies 2 2 sin 1 cos    1 2 1 2 π π 2 2 0 2 0 cos d sin c si d n os m n m n             1 2π 2 1 2, 0 sin cos cos d m n m n I          1 1 2 2 1 0 0 π π 1 2, cos si 1 1 sin cos 1 1 n d m n n m m n n I m m                      M1 A1 1 2 1 cos ' ( 1)cos sin 1 ' sin cos sin 1                  n m n m u n v v m u , , , 2, 2, 1 1 1 1 1                 m n m n m n m n m n n n I I I I m m I M1 A1 Combines with LHS. AG 5 8(c) 1 2cos    z z B1         5 3 3 1 5 1 5 5 10            z z z z z z z z M1 A1 Expands and groups. A0 if no grouping shown.     5 5 5 2 c cos s 2 3 co 2 s5 2 o 10 cos        M1 Substitutes 2cos .     n n z z n 5 5 5 1 16 16 8 c s os co co 3 co 5 s s        A1 5

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 15 of 15 Question Answer Marks Guidance 8(d) 1 2,5 0,5 7  I I M1 Applies reduction formula in part (b). 1 1 2 2 π π 5 5 5 1 16 16 8 1 1 1 0,5 7 7 7 0 0 cos c d cos3 c s os o d 5 I          M1 Applies part (c). 1 2π 5 1 7 0 5 1 80 48 8 sin sin n 5 3 si           A1 8 105  A1 4