Cambridge A Level Mathematics - Further 9231 — 2025 Oct/Nov Paper 1 · Variant 4

This document has 20 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level * 3 1 4 6 0 9 9 3 0 2 * FURTHER MATHEMATICS 9231/14 Paper 1 Further Pure Mathematics 1 October/November 2025 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 [ ]. DC (DE) 358922/2 © UCLES 2025 ,










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2 9231/14/O/N/25 © UCLES 2025 1 Prove by mathematical induction that n 1 n 5 6 5 1 H + b l for all positive integers n. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000002 * ,










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4 9231/14/O/N/25 © UCLES 2025 2 The cubic equation x bx cx d 0 3 2 + + + = , where d 0 ! , has roots a, b, c such that 1 c b = . (a) Show that d b 1 b b + = - . [3] … … … … … … … … … … … … (b) Show also that d c 1 1 b b + = - . [2] … … … … … … … … … … … … … * 0000800000004 * ,










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5 9231/14/O/N/25 © UCLES 2025 [Turn over (c) It is given that b = 3, c = -3 and d 0 2 . (i) Find the value of d. [2] … … … … … … … … … … … … (ii) Find the value of . 2 2 2 a b c + + [2] … … … … … … … … … … … … … * 0000800000005 * ,










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6 9231/14/O/N/25 © UCLES 2025 3 (a) Use standard results from the list of formulae (MF19) to show that ( )( )( ) r r r n n bn cn d 1 2 3 r n 3 2 1 4 1 + + + = + + + = ` j / , where b, c and d are integers to be determined. [3] … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000006 * ,










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7 9231/14/O/N/25 © UCLES 2025 [Turn over (b) Express ( )( )( ) r r r 1 2 3 2 + + + in partial fractions and hence use the method of differences to find ( )( )( ) r r r 1 2 3 2 r n 1 + + + =/ . [5] … … … … … … … … … … … … … … … … … … … … (c) State the value of ( )( )( ) r r r 1 2 3 2 r 1 + + + 3 =/ . [1] … … … … * 0000800000007 * ,










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8 9231/14/O/N/25 © UCLES 2025 4 The points A, B, C have position vectors i j k 3 5 5 + + , i j k 2 2 2 + + , i j k 2 + - , respectively, relative to the origin O. (a) Find the equation of the plane ABC, giving your answer in the form ax by cz d + + = . [5] … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000008 * ,










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9 9231/14/O/N/25 © UCLES 2025 [Turn over (b) The point D has position vector i j k 3 - + . Find the shortest distance between the lines AB and CD. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000009 * ,










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10 9231/14/O/N/25 © UCLES 2025 5 (a) (i) By writing r cos 12 1 b l as r r cos 4 1 6 1 - b l, show that r cos 6 2 12 1 4 1 = + b ` l j. [1] … … … … (ii) Show also that r sin 6 2 12 1 4 1 = - b ` l j. [1] … … … … The matrix M is such that M 6 2 6 2 2 6 6 2 1 0 0 1 4 1 = + - - + - f ep o. (b) The matrix M represents a sequence of two geometrical transformations in the x-y plane. Give full details of each transformation, and make clear the order in which they are applied. [4] … … … … … … … … … … … … … … * 0000800000010 * ,










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11 9231/14/O/N/25 © UCLES 2025 [Turn over (c) Write M 1 - as the product of two matrices, neither of which is I. [2] … … … … (d) Given that y = mx is an invariant line of the transformation represented by M, show that r r r sin cos sin m m 2 0 2 12 1 12 1 12 1 + - = b b b l l l and find the values of m in the form r r ec cot cos a b 12 1 12 1 + b b l l, where a and b are integers to be determined. [6] … … … … … … … … … … … … … … … … … … … * 0000800000011 * ,









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12 9231/14/O/N/25 © UCLES 2025 6 The curve C has polar equation cos r 2 1 i = , for r 0 G G i . (a) Sketch C. [2] In parts (b) and (c) you may use the identities sin sin cos 2 2 1 2 1 / i i i and cos cos 2 1 2 2 1 / i i - . (b) Find the exact value of the area of the region enclosed by C and the initial line. [4] … … … … … … … … … … … … … … … … … * 0000800000012 * ,









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13 9231/14/O/N/25 © UCLES 2025 [Turn over (c) Find the maximum distance of a point on C from the initial line. Give your answer in the form p q, where p and q are rational numbers to be determined. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000013 * ,









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14 9231/14/O/N/25 © UCLES 2025 7 The curve C has equation y x x x 10 18 10 11 18 2 = - - - . (a) Find the equations of the asymptotes of C. [3] … … … … … … … … … … (b) Show that C has no stationary points. [4] … … … … … … … … … … … … … … … * 0000800000014 * ,









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15 9231/14/O/N/25 © UCLES 2025 [Turn over (c) Sketch C, stating the coordinates of the points of intersection with the axes. [3] … (d) Sketch the curve with equation y x x x 10 18 10 11 18 2 = - - - . [2] * 0000800000015 * ,









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16 9231/14/O/N/25 © UCLES 2025 (e) Given that x x x 10 18 10 11 18 4 2 1 - - - for x p 20 29 4441 1 1 - - and x q 20 29 4441 1 1 - + , find the values of p and q. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000016 * ,









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17 9231/14/O/N/25 © UCLES 2025 Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … … … * 0000800000017 * ,









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20 9231/14/O/N/25 © UCLES 2025 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 * 0000800000020 * ,










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This document consists of 20 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/14 Paper 1 Further Pure Mathematics 1 October/November 2025 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 2025 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 2 of 20 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level 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).

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 3 of 20 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.

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 4 of 20 Annotations guidance for centres Examiners use a system of annotations as a shorthand for communicating their marking decisions to one another. Examiners are trained during the standardisation process on how and when to use annotations. The purpose of annotations is to inform the standardisation and monitoring processes and guide the supervising examiners when they are checking the work of examiners within their team. The meaning of annotations and how they are used is specific to each component and is understood by all examiners who mark the component. We publish annotations in our mark schemes to help centres understand the annotations they may see on copies of scripts. Note that there may not be a direct correlation between the number of annotations on a script and the mark awarded. Similarly, the use of an annotation may not be an indication of the quality of the response. The annotations listed below were available to examiners marking this component in this series. Annotations Annotation Meaning More information required Accuracy mark awarded zero Accuracy mark awarded one Independent accuracy mark awarded zero Independent accuracy mark awarded one Independent accuracy mark awarded two Benefit of the doubt Blank Page Incorrect Dep Used to indicate DM0 or DM1

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 5 of 20 Annotation Meaning DM1 Dependent on the previous M1 mark(s) Follow through Indicate working that is right or wrong Highlighter Highlight a key point in the working Ignore subsequent work Judgement Judgement Method mark awarded zero Method mark awarded one Method mark awarded two Misread Omission or Other solution Off-page comment Allows comments to be entered at the bottom of the RM marking window and then displayed when the associated question item is navigated to. On-page comment Allows comments to be entered in speech bubbles on the candidate response. Judgment made by the PE Premature approximation Special case Indicates that work/page has been seen

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 6 of 20 Annotation Meaning Error in number of significant figures Correct Transcription error Correct answer from incorrect working

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 20 Mark Scheme Notes The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Mark for a correct result or statement independent of method marks. DM or DB When a part of a question has two or more ‘method’ steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly, when there are several B marks allocated. The notation DM or DB is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. FT Implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. • 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.

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 20 Abbreviations AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed) CWO Correct Working Only ISW Ignore Subsequent Working SOI Seen Or Implied SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) WWW Without Wrong Working AWRT Answer Which Rounds To

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 20 Question Answer Marks Guidance 1 ( ) 6 6 5 5 1 LHS = = 6 1 5 5 RHS 1 = + = B1 Checks base case. Assume that ( ) 6 5 1 5 1+ k k  for some positive integer k. B1 States inductive hypothesis. Then ( ) ( )( ) 6 6 1 5 1 5 5 1 + + k k  M1 Moving to the next term and using the inductive hypothesis. ( ) 6 6 6 1 1 5 25 5 5 5 1 1 = +  + = + + k k k A1 Convincing algebra to reach the required form. Hence, by induction, ( ) 6 5 1 5 1+ n n  is true for every positive integer .n A1 Must follow M1 A1. 5 Question Answer Marks Guidance 2(a) = −d B1 SOI 1    + + = −b M1 Uses    + + = −b and replaces 1 with .   OE 1   + = − d b A1 AG. 3

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 20 Question Answer Marks Guidance 2(b) 1   − − + = d d c M1 Uses .    + + = c 1 1   − + = c d A1 AG. 2 2(c)(i) 2 1 3 3 3 4 0 + −=  − − = d d d d M1 Equates expressions given in (a) and (b), with correct use of coefficients. 4 = d A1 4 = d only for final answer. 2 2(c)(ii) 2 2 2 2 2    + + = − b c M1 2 2 2 2 ( 2 ) ( )          + + + + + + = − with correct use of coefficients. 15 A1 2

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 20 Question Answer Marks Guidance 3(a) ( ) 1 2 2 2 3 1 1 4 2 1 6 11 6 ( 1) ( 1)(2 1) ( 1) 6 = + + + = + + + + + + +  n r r r r n n n n n n n n M1 A1 Expands and substitutes formulae. ( ) 3 2 1 4 10 35 50 = + + + n n n n A1 3 3(b) 2 1 2 1 ( 1)( 2)( 3) 1 2 3 = − + + + + + + + r r r r r r M1 A1 Finds partial fractions. 1 2 (1) 2 (2) (3) ( 1)( 2)( 3) = = − + + + +  n r f f f r r r (2) 2 (3) (4) + − + f f f (3) 2 (4) (5) + − + f f f  ( 1) 2 ( ) ( 1) + − − + + f n f n f n ( ) 2 ( 1) ( 2) + − + + + f n f n f n M1 A1 Shows enough complete terms for cancellation to be clear and any terms that form part of the final answer. 1 ( ) .1 = + f r r 1 1 1 6 2 3 = − + + + n n A1 Accept 1 1 6 ( 2)( 3) − + + n n 5 3(c) 1 6 B1 FT their result of similar form 1

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 20 Question Answer Marks Guidance 4(a) 3 3 4 6 = −− − = −− −  AB AC i j k i j k B1 Finds direction vectors of two lines in the plane. 6 1 3 3 3 1 4 6 1     − − − = −     − − −   i j k M1 A1 Finds normal to the plane ABC. 6(3) 3(5) (5) 8 6 3 8 − + =  − + = x y z M1 A1 Substitutes point. CAO 5 4(b) 1 2 1 1 1 2 3 1 4 −             = − − = −             −        CD B1 18 1 3 3 7 1 2 4 1 −     − − − =     − − −   i j k M1 A1 Find common perpendicular. 1 18 1 4 4 7 0.207 374 374 6 1 − −       − = =       − −     M1 A1 Uses formula for shortest distance. 5

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 13 of 20 Question Answer Marks Guidance 5(a)(i) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 4 6 4 6 4 6 c π os cos co i π π π π π s s n sin = + − ( ) 1 1 1 4 4 4 2 3 2 6 2 = + = + B1 Uses cos( ) cos cos sin sin − = + A B A B A B , Every step must be seen because AG. 1 5(a)(ii) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 4 6 4 6 4 6 s π in sin co o π π π π π s c s sin = − − ( ) 1 1 1 4 4 4 2 3 2 6 2 = − = − B1 Uses sin( ) sin cos cos sin − = − A B A B A B , Every step must be seen because AG. 1 5(b) Reflection and rotation. B1 Correct order. B1 Reflection then rotation. Reflection in the x-axis. B1 Accept reflection in 0 = y . A rotation, 1 12 π [anticlockwise] about the origin. B1 Accept 15°. 4

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 14 of 20 Question Answer Marks Guidance 5(c) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 4 4 4 4 1 1 1 1 4 4 4 4 6 2 2 6 6 2 6 2 6 2 6 2 2 6 6 2 −     + − + −     =         − + − +     , 1 1 0 1 0 0 1 0 1 −     =     − −     B1 Both correct inverses required. ( ) ( ) ( ) ( ) 2 π π π 12 1 12 12 π cos sin sin cos       −   ( ) ( ) ( ) ( ) 1 1 4 1 4 1 1 4 4 6 2 6 2 1 0 0 1 2 6 6 2 −   + −    =    −    − +   M B1 Correct order of their inverses. ( ) ( ) ( ) ( ) 2 π π 1 12 12 12 1 π π cos sin 1 0 0 1 sin cos −      =    − −    M Or 4 4 1 1 1 0 6 2 6 2 0 2 6 6 2 −     + − =      − − +    M 2

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 15 of 20 Question Answer Marks Guidance 5(d) ( ) ( ) ( ) ( ) 2 π π π 1 π 12 12 1 2 cos sin sin cos     =   −   M B1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 π π π π π 12 12 1 π π π 12 12 12 12 2 1 cos sin cos sin sin cos sin cos     +      =      − −      y x y x y x B1 FT their M Transforms       x y to       X Y ( ) ( ) ( ) ( ) ( ) π 12 12 12 π π 12 π sin cos cos sin − = + x m x m x mx M1 Uses = y mx and = Y mX . ( ) ( ) ( ) 12 12 1 π 2 2 π π sin 2 cos sin 0 + − = m m A1 AG. ( ) ( ) ( ) 2 6 2 2 6 2 6 2 0 − + + − − = m m ( ) ( ) ( ) ( ) 12 12 12 12 2 2 π π π π 2cos 4cos 4sin 2sin −  + = m M1 Applies quadratic formula and ( ) ( ) π 12 2 2 π 12 cos sin 1. + = OE. May see ( ) ( ). 2 3 6 2 = − +  + m ( ) ( ) 1 π π 12 2 cosec cot  = − m A1 Or 1, 1 = − =  a b 6

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 16 of 20 Question Answer Marks Guidance 6(a) B1* Initial line drawn and ( ) 1,0 joined to O by a single arc lying in both of the first two quadrants and no others. DB1 First B1 awarded and also correct shape at extremities and r decreasing. 2 6(b) π 1 1 0 2 2 2 cos d   M1 Uses 1 2 2 d r with correct limits.   π 0 4 0 π 1 1 4 1d o s c s in     = + = +  M1 A1 Applies given identity and integrates. 1 4 π = A1 4

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 17 of 20 Question Answer Marks Guidance 6(c) 1 2 cos sin   = y B1 Uses sin = y r . 1 1 1 2 2 2 cos cos sin sin 0     − = M1* Differentiates. A1 Correct and equal to zero. ( ) 2 2 1 1 2 2 1 1 2 2 cos 2cos 1 sin cos 0     − − = 2 2 1 2 1 1 2 2 cos 0 2cos 1 sin 0      −− = ( ) 2 1 1 2 2 1 2 2 2 2cos 1 1 cos 0 3cos 2 0    −− − =  − = DM1 Forms an equation in one trig ratio using correct formulae. (May be in terms of sin or tan) 2 2 1 2 cos 3 = A1 Solving equation to find one of 2 2 1 1 sin 3 = 2 2 1 1 2 tan = 2 2 sin 3 = 1 cos 3 = ( ) 2 1 1 2 2 4 1 2 2 3 3 3 3 2cos sin 2 1   = = − = y A1 Accept 4 9 3 . 6 Question Answer Marks Guidance 7(a) 1.8 = x B1 ( )( ) 5 27 5 7 1 10 7 27 0 0 10 18 10 18 1 18 = + − − − + − = − x x y x x x M1 Must be an attempt to find the constant term. 0.7 = + y x A1 3

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 18 of 20 Question Answer Marks Guidance 7(b) ( ) 2 2 d 100 360 378 d 10 18 − + = − y x x x x M1 Differentiates. 2 50 180 189 0 − + = x x ( ) 2 d 54 or 1 d 10 18     = +   −   y x x A1 Forms quadratic equation if using discriminant method. (Or writes d d y x in a form to show it is positive) 2 180 4(50)(189) 5400 0 − = −  (or ' 0  y )  No stationary points M1 A1 Consideration of discriminant or sign of 'y with correct conclusion. 4

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 19 of 20 Question Answer Marks Guidance 7(c) B1 Axes and asymptotes labelled. B1 Branches correct. (0,1), ( 0.9,0), − ( ) 2,0 B1 May be seen on their diagram. 3 x y

9231/14 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 20 of 20 Question Answer Marks Guidance 7(d) B1 FT from their attempt in (c). B1 Everything correct with sharp changes of direction at intersections with x axis. 2 7(e) 2 10 11 18 29 4441 4 10 18 20     − − − = − =  −   x x x x 2 10 11 18 4 10 18 − − = − x x x M1 Forms an equation to find required critical values for x . 2 10 51 54 0 − + = x x A1 3 2 , = p 18 5 = q A1 3 18 2 5 , = = x x A1 ,p q clearly identified in correct order. 4 x y