Cambridge A Level Mathematics - Further 9231 — 2024 May/June Paper 2 · Variant 3

This document has 20 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/23 Paper 2 Further Pure Mathematics 2 May/June 2024 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 [ ]. * 3 8 8 2 6 4 0 1 9 1 * DC (LK/CT) 329212/2 © UCLES 2024 ,

 
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3 9231/23/M/J/24 © UCLES 2024 [Turn over 1 Find the exact value of dx x x 1 4 1 2 2 2 7 - - y . [5] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655537603 * ,

 
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4 9231/23/M/J/24 © UCLES 2024 2 The curve C has parametric equations cosh x t = , sinh y t = , for t 0 5 3 1 G . The length of C is denoted by s. (a) Show that cosh d s t t 2 0 5 3 = y . [4] … … … … … … … … … … … … … … … … … … … … … … … … * 0019655537604 * ,

 
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5 9231/23/M/J/24 © UCLES 2024 [Turn over (b) By finding the Maclaurin’s series for cosh t2 up to and including the term in t2, deduce an approximation to s. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655537605 * ,

 
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6 9231/23/M/J/24 © UCLES 2024 3 The curve C has equation x xy y 2 8 12 3 3 + + = - . (a) Show that, at the point ( , ) 2 1 - - on , d d C x y 2 1 = - . [3] … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655537606 * ,

 
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7 9231/23/M/J/24 © UCLES 2024 [Turn over (b) Find the value of d d x y 2 2 at the point ( , ) 2 1 - - . [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655537607 * ,

 
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8 9231/23/M/J/24 © UCLES 2024 4 x y 0 1 2 3 4 N N 1 - The diagram shows the curve with equation y x 2 = - for x N 2 G G together with a set of ( ) N 2 - rectangles of unit width. (a) By considering the sum of the areas of these rectangles, show that r N N 1 2 3 1 1 r N 2 1 2 2 - + =/ . [5] … … … … … … … … … … … … … … … … … … … * 0019655537608 * ,

 
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9 9231/23/M/J/24 © UCLES 2024 [Turn over (b) Use a similar method to find, in terms of N, an upper bound for r 1 r N 2 1 =/ . [3] … … … … … … … … … … … … … … … … (c) Deduce lower and upper bounds for r 1 r 2 1 3 =/ . [2] … … … … … … … … … … * 0019655537609 * ,

 
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10 9231/23/M/J/24 © UCLES 2024 5 (a) Find the general solution of the differential equation d d d d sin t x t x x t 10 25 338 2 2 + + = . [7] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655537610 * ,

 
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11 9231/23/M/J/24 © UCLES 2024 [Turn over (b) Show that, for large positive values of t and for any initial conditions, ( ) sin x R t . z - , where the constants R and z are to be determined. [3] … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655537711 * ,

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12 9231/23/M/J/24 © UCLES 2024 6 (a) Show that z z z z z r n r n 4 2 2 4 2 2 1 = - - - + =/ , for z z 2 2 ! - . [2] … … … … … … … … … … … … … (b) By letting cos isin z i i = + , show that, if sin2 0 ! i , ( ) ( ) sin sin cos cos r n 4 2 2 2 4 2 r n 1 i i i i = - + =/ . [5] … … … … … … … … … … … … * 0019655537712 * ,

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13 9231/23/M/J/24 © UCLES 2024 [Turn over … … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655537713 * ,

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14 9231/23/M/J/24 © UCLES 2024 7 (a) Show that d d cosh x x x x x 2 9 2 9 3 9 2 1 2 - - = - - e o . [3] … … … … … … … … … … … (b) Find the solution of the differential equation d d x x y y x x 9 2 2 - = - , given that y 1 = when x 3 = . Give your answer in the form ( ) f y x = . [9] … … … … … … … … … … … … * 0019655537714 * ,

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16 9231/23/M/J/24 © UCLES 2024 8 The planes 1 P and 2 P do not intersect and are both perpendicular to i j k 2 3 + + . The line l intersects 1 P at the point ( , , ) 1 6 0 and intersects 2 P at the point ( , , ) 3 6 0 - . (a) Find Cartesian equations of 1 P and 2 P . [3] … … … … … … … … … … (b) Express the vector equation of l in the form a b x y z m = + f p , where a and b are vectors to be determined, and hence show that for points on l, x y 1 2 1 12 1 + = and z 0 = . [2] … … … … … … … … … … … … … * 0019655537716 * ,

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17 9231/23/M/J/24 © UCLES 2024 [Turn over The matrix A is given by A 1 1 2 2 3 3 0 2 1 12 1 = f p. (c) Show that the characteristic equation of A is 3 0 3 2 4 7 m m m - + = + and hence find the eigenvalues of A. [3] … … … … … … … … … … … … … … … … … … … … … … … … * 0019655537717 * ,

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18 9231/23/M/J/24 © UCLES 2024 (d) Find a matrix P and a diagonal matrix D such that A PDP n 1 = - , where n is a positive integer. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655537718 * ,

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19 9231/23/M/J/24 © UCLES 2024 Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … … … * 0019655537719 * ,

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20 9231/23/M/J/24 © UCLES 2024 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. BLANK PAGE * 0019655537720 * ,

 
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This document consists of 15 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/23 Paper 2 Further Pure Mathematics May/June 2024 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 2024 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 2024 © Cambridge University Press & Assessment 2024 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 descriptions 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 2024 © Cambridge University Press & Assessment 2024 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, non-integer 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 or sign 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 A or B 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 2024 © Cambridge University Press & Assessment 2024 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 2024 © Cambridge University Press & Assessment 2024 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 2024 © Cambridge University Press & Assessment 2024 Page 6 of 15 Question Answer Marks Guidance 1 2 2 4 1 3 ( 2) x x x      B1 Completes the square. 7 7 2 2 1 2 2 2 1 2 d sin 3 3 ( 2) x x x                   M1 A1 Applies formula. 7 7 2 2 2 1 1 2 2 2 sin cos 3 3 x x                                 1 2 3 1 1 1 π sin 3 sin 0     M1 A1 Uses limits. Answer must be in radians. 5 Question Answer Marks Guidance 2(a) sinh cosh x t y t     B1 Differentiates x and y with respect to t. 2 2 2 2 sinh cosh cosh 2 x y t t t       M1 A1 Applies 2 2 cosh sinh cosh2 t t t   . May use exponential form or 1 1 2 2 2 2 2 d cosh 1 d 1 sinh d . d sinh y t x t t x t                        3 5 0 cosh 2 d s t t  A1 AG 4

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 7 of 15 Question Answer Marks Guidance 2(b)   1 2 d cosh 2 sinh 2 sech 2 sinh 2 (cosh 2 ) d t t t t t t    B1 Finds first derivative.     3 2 2 2 2 2 d cosh2 2 cosh 2 tanh 2 cosh2 2 cosh2 sinh 2 cosh2 d t t t t t t t t      B1 Finds second derivative 3 3 5 5 1 2 3 3 0 0 1 d s t t t t          M1 A1 Applies Maclaurin’s series. 84 125 0.672   A1 CWO. 5 Question Answer Marks Guidance 3(a)   3 2 d 8 24 ' d y y y x  B1 Differentiates 3y correctly.     d 2 2 ' d xy xy y x   B1 Differentiates xy correctly. 2 2 3( 2) 2( 2) ' 2( 1) 24( 1) ' 0 10 20 ' 0 y y y            1 2 'y   B1 Substitutes ( 2, 1)   into 2 2 3 2 ' 2 24 ' 0, x xy y y y     AG. Must see intermediary step between substitution and the final answer. 3

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 8 of 15 Question Answer Marks Guidance 3(b) 6 2 '' 2 ' x xy y   B1 Differentiates 2 3 2 '. x xy  2 2 ' 24 '' 24 '(2 ) ' 0 y y y y y y     B1 B1 Differentiates 2 2 24 '. y y y  B1 for 2 2 ' 24 '' y y y  . B1 for   2 48 ' . y y (Allow alternative using quotient rule.)       2 2 1 2 2 1 0 ' 6 ' 48 ( 2) 2( 2) '' 4 24 1 0 1 20 '' 6 ( 2 ) y y y               M1 Substitutes ( 2, 1)   into their expression (must have derivatives 'y and '' y ). 13 10 '' y  A1 OE 5

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 9 of 15 Question Answer Marks Guidance 4(a) 2 1 2 2 2 1 2 ( 1) N r r N        B1 Forms sum of the areas of the rectangles. Need to see this sum written explicitly, may be in summation form, cannot be embedded in working. 2 2 2 1 1 1 d 2 N N x x x N           M1 A1 Evaluates integral, correct limits for M1. 2 2 2 2 1 1 d 1 N N r r x x N        M1 Compares with integral correctly (using >). 2 1 2 3 1 1 2 N r r N N      A1 AG 5 4(b) 2 2 2 3 4 N       M1* Forms the sum of the areas of appropriate rectangles. Allow addition of one extra rectangle or shift one left or right if limits on the integral are adjusted accordingly. 2 2 2 1 1 1 d 2 N N x x x N            DM1 Compares with integral, correct limits and correct inequality symbol (<). 2 1 7 1 4 N r r N      A1 3

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 10 of 15 Question Answer Marks Guidance 4(c) 1 2 3 2 r r     B1 Needs to clear (using > or words) is that value is lower bound. 1 2 7 4 r r     B1 FT Needs to clear (using < or words) is that value is upper bound. Must have scored final A mark in part (b). FT on fully correct alternative to part 4(b). 2 Question Answer Marks Guidance 5(a) 2 10 25 0 5 m m m      M1 Auxiliary equation.   5 e t x At B    A1 Complementary function. Allow ‘ x ’ missing. sin cos ' cos sin '' sin cos x p t q t x p t q t x p t q t         M1 A1 Particular integral and its derivatives. For M1, both derivatives must be present and of the correct form (only allow sign errors). 10 25 338 p q p     10 25 0 q p q    M1 Substitutes and equates coefficients. 12 p  5 q  A1   5 e 12sin 5cos t x At B t t      A1 Must have ‘ x ’. 7

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 11 of 15 Question Answer Marks Guidance 5(b)   5 12 13 13 12sin 5cos 13 sin cos x t t t t     M1 Considering PI only. 12sin 5cos 13sin( 0.395...) t t t    A1 A1 A1 for R and A1 for . Condone 22.6 . Withhold final A1 mark if working with incorrect PI. CWO. 3 Question Answer Marks Guidance 6(a)   4 4 4 4 4 4 4 1 4 1 ( ) ( ) 1 n n z z z z z z z z         M1 Uses sum of geometric series.   4 2 4 2 2 4 4 2 2 2 1 1 n n z z z z z z z z z           A1 Divides numerator and denominator by 2z , AG. 2 6(b) 2 2 2isin2 z z    B1 Simplifies denominator. 4 2 2 2 2 cos( c i 4 2) is n(4 2) os i 2 n 2 s isi 2 in 2 n z z z n n z                M1 A1 Applies de Moivre’s theorem to numerator. (Used at least once for M1.) Accept 2 2 2 2isin cos sin . cos z          1 c cos(4 2) sin 4 2sin os2 2 n r n r           M1 Equates imaginary parts.   1 2 cos 2 2 cos(4 ) sin 4 sin 2 n r n r          A1 AG. Must see evidence of eliminating i from the denominator. 5

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 12 of 15 Question Answer Marks Guidance 7(a) 2 1 d 9 9 cosh d 2 2 3 x x x x          M1 Applies product rule and 2 1 d 1 cosh d 3 9 x x x         2 2 2 2 1 9 9 2 2 9 2 9 x x x x      A1 OE 2 2 2 2 9 1 9 9 2 2 9 x x x x       A1 Puts over common denominator, AG. 3

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 13 of 15 Question Answer Marks Guidance 7(b) 2 d 9 d x x y y x x    B1 Divides through by .x 1 1 ln d e e x x x         M1 A1 Finds integrating factor.   1 2 d 9 d x y x x    M1 Correct form on LHS,   d d Iy x for their integrating factor I. 2 1 1 9 9 cosh 2 2 3 x x x y x C       M1 A1 Integrates RHS. For M1, RHS must be of the form 2 9, c x  where c is a non-zero constant. 1 3 C  M1 Substitutes initial conditions. 1 2 9 1 9 cosh 2 2 3 3 x x y x x            M1 A1 Divides through by their integrating factor. Accept 2 2 9 1 1 9 ln 9 . 2 2 3 3 3 x x y x x x                   9 Question Answer Marks Guidance 8(a)   1 1 2 6 3 0 d    or     2 3 2 6 3 0 d     M1 Substitutes point into 2 3 x y z   . 2 3 13 x y z    A1 2 3 9 x y z    A1 3

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 14 of 15 Question Answer Marks Guidance 8(b) 1 2 12 0 6 0 x y z                                        or 0 1 6 0 1 6 x y z                                         M1 Forms vector equation of line. 6 1 , 0 6 y x z      1 1 2 12 1, 0 x y z     A1 AG. CWO. 2 8(c) 1 1 2 12 2 1 2 3 1 3       M1 Uses ) det(  I A (attempt an expansion).           3 1 1 1 4 2 12 2 ( 0 3 1 ) 2 2 2                 3 2 3 1 1 4 4 2 7 4 2 3 3 0 3                2 3 7 4 3 0        A1 Expands determinant, expansion of brackets shown, AG. CWO. (Must have ‘ 0  ’ here or above.) 7 1 2 2 0, ,   B1 3

9231/23 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 15 of 15 Question Answer Marks Guidance 8(d) 1 4 3 2 1 1 11 2 12 12 3 0: 1 2 3 18 0 11                           i j k  M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. 21 2 7 5 21 2 2 2 3 7 2 4 6 : 2 3 ~ 6 1 3 1                    i j k 3 2 3 3 1 2 2 2 5 7 2 4 6 : 2 3 ~ 6 1 3 7                           i j k A1 A1 Thus 3 6 6 18 6 6 11 1 7              P and    7 2 1 2 0 0 0 0 0 0 0 n n              D M1 A1 Or correctly matched permutations of columns. Do not accept 7 2 1 2 0 0 0 0 0 . 0 0 n            D M1 is for matching their eigenvectors correctly with their eigenvalues. M0 if a column of zeros appears in P. 6