Cambridge A Level Mathematics - Further 9231 — 2024 Oct/Nov Paper 1 · Variant 1

This document has 16 pages. [Turn over Cambridge International AS & A Level DC (CE) 329672/2 © UCLES 2024 * 2 1 5 2 9 1 9 8 9 5 * FURTHER MATHEMATICS 9231/11 Paper 1 Further Pure Mathematics 1 October/November 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 [ ]. ,










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2 9231/11/O/N/24 © UCLES 2024 1 The matrix M represents the sequence of two transformations in the x-y plane given by a stretch parallel to the x-axis, scale factor k ( ) k 0 ! , followed by a shear, x-axis fixed, with (0, 1) mapped to (k, 1). (a) Show that k k M 0 1 = e o. [4] … … … … … … … … … … (b) The transformation represented by M has a line of invariant points. Find, in terms of k, the equation of this line. [3] … … … … … … … … … … … … … … * 0000800000002 * ,










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3 9231/11/O/N/24 © UCLES 2024 [Turn over The unit square S in the x-y plane is transformed by M onto the parallelogram P. (c) Find, in terms of k, a matrix which transforms P onto S. [1] … … … … … … (d) Given that the area of P is k 3 units 2 2, find the possible values of k. [2] … … … … … … … … … … … … … … … … … … … * 0000800000003 * ,










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4 9231/11/O/N/24 © UCLES 2024 2 Prove by mathematical induction that, for all positive integers n, ( ) tan x x x x P 1 d d n n n n 1 2 = + - - ` ` j j , where ( ) P x n is a polynomial of degree n 1 - . [6] … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000004 * ,










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6 9231/11/O/N/24 © UCLES 2024 3 The quartic equation x x 2 1 0 4 3 + - = has roots , , , a b c d. (a) Find a quartic equation whose roots are , , , 4 4 4 4 a b c d and state the value of 4 4 4 4 a b c d + + + . [5] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000006 * ,










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7 9231/11/O/N/24 © UCLES 2024 [Turn over (b) Find the value of 5 5 5 5 a b c d + + + . [3] … … … … … … … … … … … … … … (c) Find the value of 8 8 8 8 a b c d + + + . [2] … … … … … … … … … … … … * 0000800000007 * ,










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8 9231/11/O/N/24 © UCLES 2024 4 (a) Use the method of differences to find ( )( ) r k r k k 5 5 5 5 r n 1 + + + =/ in terms of n and k, where k is a positive constant. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000008 * ,










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9 9231/11/O/N/24 © UCLES 2024 [Turn over It is given that ( )( ) r k r k k 5 5 5 5 3 1 r 1 + + + = 3 =/ . (b) Find the value of k. [2] … … … … … … … … … (c) Hence find ( )( ) r k r k k 5 5 5 5 r n n2 + + + =/ in terms of n. [2] … … … … … … … … … … … … … … … * 0000800000009 * ,










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10 9231/11/O/N/24 © UCLES 2024 5 (a) Show that the curve with Cartesian equation x y xy 6 2 2 2 + = ` j has polar equation sin r 3 2 2 i = . [2] … … … … … … The curve C has polar equation sin r 3 2 2 i = , for r 0 2 1 G G i . (b) Sketch C and state the maximum distance of a point on C from the pole. [3] … * 0000800000010 * ,










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11 9231/11/O/N/24 © UCLES 2024 [Turn over (c) Find the area of the region enclosed by C. [2] … … … … … … … (d) Find the maximum distance of a point on C from the initial line. [6] … … … … … … … … … … … … … … … … … … … * 0000800000011 * ,









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12 9231/11/O/N/24 © UCLES 2024 6 The curve C has equation y x x x x 2 7 3 4 1 2 2 = - + + + . (a) Find the equations of the asymptotes of C. [2] … … … … … … (b) Find the coordinates of any stationary points on C. [4] … … … … … … … … … … … … … … … … … … … * 0000800000012 * ,









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13 9231/11/O/N/24 © UCLES 2024 [Turn over (c) Sketch C, stating the coordinates of any intersections with the axes. [5] … (d) Sketch the curve with equation y x x x x 2 7 3 4 1 2 2 = - + + + and state the set of values of k for which x x x x k 2 7 3 4 1 2 2 - + + + = has 4 distinct real solutions. [2] … * 0000800000013 * ,









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14 9231/11/O/N/24 © UCLES 2024 7 The lines l1 and l2 have equations ( ) r i j k i j k 3 2 2 m = + - + + + and ( ) r i j k i j k 2 9 4 2 n = + + - + - respectively. The plane 1 P contains l1 and is parallel to l2. (a) Find the equation of 1 P , giving your answer in the form ax by cz d + + = . [4] … … … … … … … … … … … The plane 2 P contains l2 and the point with coordinates ( , , ) 2 1 7 - . (b) Find the acute angle between 1 P and 2 P . [4] … … … … … … … … … … … … … * 0000800000014 * ,









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15 9231/11/O/N/24 © UCLES 2024 The point P on l1 and the point Q on l2 are such that PQ is perpendicular to both l1 and l2. (c) Find a vector equation for PQ. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000015 * ,









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16 9231/11/O/N/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. Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … * 0000800000016 * ,









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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 9231/11 Paper 1 Further Pure Mathematics 1 October/November 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 October/November 2024 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 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).

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 3 of 15 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. 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.

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 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.

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 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

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 6 of 15 Question Answer Marks Guidance 1(a) 0 0 1       k B1 [Stretch parallel to the x-axis, scale factor k ( 0)  k ]. (Allow without identification.) 1 0 1       k B1 [Shear, x-axis fixed, with ( ) 0,1 mapped to ( ) ,1 . k ] (Allow without identification.) 1 0 0 1 0 1 0 1      = =           k k k k M M1 A1 Correct order for M1, must have identified which matrix gives which transformation, AG. 4 1(b) 0 1 +      =          k k x kx ky y y B1 Transforms       x y to       X Y + = kx ky x M1 Sets    =       X x Y y 1− = k y x k oe A1 3 1(c) 1 1 1 0 − −   =     k k k M B1 (An alternative is possible.) 1

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 7 of 15 Question Answer Marks Guidance 1(d) 2 | | 3 = k k M1 Uses that det . = k M Without modulus is SC B1. 1 3 0   =  k k A1 2 Question Answer Marks Guidance 2 ( ) 1 2 d tan 1 d 1 − = + x x x so true when 1. = n B1 Differentiates once. Assume that ( ) ( )( ) 1 2 d tan P 1 , d − − = + k k k k x x x x where ( ) . d gP 1 e = − k x k B1 States inductive hypothesis. Must have ( ) . d gP 1 e = − k x k ( ) ( )( ) ( )( ) 1 1 1 2 2 1 d tan P ' 1 2 P 1 d + − − − + − = + − + k k k k k k x x x kx x x x M1 A1 Differentiates kth derivative using the product rule. ( )( ) ( ) ( )( ) 1 2 2 P ' 1 2 P 1 −− = + − + k k k x x kx x x so ( ) 1 degP + = k x k A1 Writes in the form ( )( ) 1 2 1 P 1 . −− + + k k x x So true when 1. = + n k By induction, true for all positive integers n. A1 Attempts to show degree of Pk+1 (x) is at most k (condone not showing coefficient of xk is non- zero) and states conclusion. 6

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 8 of 15 Question Answer Marks Guidance 3(a) 4 = y x B1 Uses correct substitution. ( ) 3 4 4 3 2 1 0 16 1 + −=  = − y y y y M1 Substitutes and obtains an equation not involving radicals. 3 2 3 4 16 1 4 6 4 = − + − + y y y y y M1 Uses binomial expansion. 2 4 3 20 6 4 1 0 − + − + = y y y y A1 Must be an equation. 4 4 4 4 20     + + + = B1 5 3(b) 2     + + + = − B1 ( ) ( ) 5 4 5 5 5 5 2 0 2 20 2     + − =  + + + = − + − x x x M1 Multiplies original equation by x and substitutes. 42 − A1 3 3(c) ( ) 8 8 8 8 2 20 2 6     + + + = − M1 Uses formula for sum of squares. Or alternative complete method eg another substitution. 388 A1 CAO. 2

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 9 of 15 Question Answer Marks Guidance 4(a) 5 1 1 (5 )(5 5 ) 5 5 5   = −   + + + + + +   k k r k r k r k r k M1 A1 Finds partial fractions. (Don’t allow a substitution of a value for k.) 1 0 5 1 1 1 1 1 0 1 (5 1 1 15 5 5 )(5 5 ) 5 5 =   = − + − + + −   + + + + +  + + +  + +  n r k k k k n k k k r k r k k n M1 Writes at least three terms, including last. (Allow any value of k.) 5 5 1 1 5  +   = −  +  +  n k k k A1 4 4(b) 1 5 3 5 5 3 2 =  = +  = + k k k k k M1 A1 2 4(c) 2 2 1 1 1 5 5 5 (5 )(5 5 ) (5 )(5 5 ) (5 )(5 5 ) − = = = = − + + + + + + + + +    n n n r n r r k k k r k r k r k r k r k r k M1 Or applies the method of differences again. 2 2 2 5 5 5 5 5 5 2 1 2 3 1 1 1 1 1 1 5 5 1 1       = − − − = −       + +     +   + + = + − + + + + n k k k k k k n k n k n k n n A1FT FT on their value of k (must be substituted in). 2

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 10 of 15 Question Answer Marks Guidance 5(a) 4 2 c 6 os sin  = r r M1 Substitutes s co in s ,   = = y r x r and applies 2sin cos . sin2   = 2 3sin2 = r A1 AG. 2 5(b) B1 Correct position and symmetrical about 1 4 π = . B1 Single correct loop. 3 B1 States maximum distance or labels sketch. Allow ( ) 1 4 3, π but not ( ) 1 4 π, 3 . Allow 3sf. 3 5(c) π 2 2 π 3 1 0 0 3 2 2 2 c sin2 os2   = −      d M1 Forms 1 2 2 d .  r (Allow with wrong limits.) 3 2 A1 2 0 =

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 11 of 15 Question Answer Marks Guidance 5(d) 1 1 2 2 3 sin 2 sin   = y B1 1 1 2 2 0 s 2 in 2 cos sin 2 cos sin      − + = M1 A1 Sets d 0. d= y 2 2tan cos cos2 sin 0 tan2 tan tan 1 s t in2 an          + =  = −  = − − M1 Applies suitable trigonometric identity. Accept sin3 0. = 1 3 π = A1 5 4 3 2 3 1.40 2 = A1 AEF. 6 Question Answer Marks Guidance 6(a) 1 , 2 = x 3 = x B1 Vertical asymptotes. 2 = y B1 Horizontal asymptote. 2

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 12 of 15 Question Answer Marks Guidance 6(b) ( ) 2 2 2 2 d (2 7 3)(8 1) (4 1)(4 7) d 2 7 3 − + + − + + − = − + y x x x x x x x x x M1 Finds d . d y x Allow top line only for M1. 2 3 2 1 0 − + + = x x M1 Sets equal to 0 and forms equation. ( ) 1 1 3 5, , − ( ) 1, 3 − A1 A1 4 6(c) B1 Axes and asymptotes. Clear identification (label or clear intersection with axes at correct place). B1 3  x correctly approaching asymptotes, not too truncated. B1 1 2 3   x correct. B1 1 2  x correct. ( ) 1 3 0, B1 States coordinates of intersection with axis. May be seen on their graph. 5 yx

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 13 of 15 Question Answer Marks Guidance 6(d) B1FT FT from sketch in (c). At least two branches. 3  k B1 2 Question Answer Marks Guidance 7(a) 6 2 2 1 1 3 ~ 1 1 4 2 9 3 −         = −         − −     i j k M1 A1 Finds common perpendicular. Allow one error. 2(1) (3) 3( 2) 5 − + + − = − M1 Substitutes point on 1l . 2 3 5 − − = x y z A1 CAO. 4 yx

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 14 of 15 Question Answer Marks Guidance 7(b) 6 1 4 2 4 1 1 2 5   − =   −  i j k M1 A1 Finds the normal to 2 . 7 2 1 6 1 4 3 5 7 4 77 cos cos 14 7       −     −   − =  = M1 Uses dot product of normal vectors. 77.7 A1 No ISW. Accept 1.36 rad. 4

9231/11 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 15 of 15 Question Answer Marks Guidance 7(c) 1 2 1 3 , 2 1 2 4 5 4 2 9 1 2 2             + +             = + = −−  = −−             −+ +   − −     − + OP OQ PQ M1 A1 Finds PQ . 1 2 5 1 2 4 0 1 1 2           −− =       − − + − or 1 6 5 3 2 2 4 9 1               −− = −       − − + −  −     k M1 Uses that dot product of PQ with line direction is zero, or, alternatively, PQ is a multiple of the common perpendicular. 6 6 0  − + = A1 Deduces one equation. 2 1 5 4 4 0 0 2 2 21 42 11              −− − =  =      −   +    − − + A1 Deduces second equation. 3 1 4 1      =  =     −   OP or 1 2 6 5  −     = − =       OQ M1 Solves for  or  and substitutes into OP or OQ 1 6 6 3 5 9 −         = + −         −     t r A1 FT OE. FT using their common perpendicular. Must have" " = r . 7