Cambridge A Level Mathematics - Further 9231 — 2021 Oct/Nov Paper 1 · Variant 2

This document has 16 pages. Any blank pages are indicated. Cambridge International AS & A Level * 5 9 8 3 9 2 5 3 3 7 * DC (PQ/SG) 199016/2 © UCLES 2021 [Turn over FURTHER MATHEMATICS 9231/12 Paper 1 Further Pure Mathematics 1 October/November 2021 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 9231/12/O/N/21 © UCLES 2021 1 (a) Give full details of the geometrical transformation in the x-y plane represented by the matrix 6 0 0 6 e o. [1] … … Let 3 2 4 2 A = e o. (b) The triangle DEF in the x-y plane is transformed by A onto triangle PQR. Given that the area of triangle DEF is 13 cm2, find the area of triangle PQR. [2] … … (c) Find the matrix B such that 6 0 0 6 AB = e o. [2] … … … … … (d) Show that the origin is the only invariant point of the transformation in the x-y plane represented by A. [4] … … … … … … … … … …

3 9231/12/O/N/21 © UCLES 2021 [Turn over 2 It is given that y xeax = , where a is a constant. Prove by mathematical induction that, for all positive integers n, x y a x na d d e n n n n ax 1 = + - ` j . [6] … … … … … … … … … … … … … … … … … … … … … … … … …

4 9231/12/O/N/21 © UCLES 2021 3 Let ( ) ( ) ln S r r r 1 2 n r n 2 1 = + + =/ . (a) Using the method of differences, or otherwise, show that ( ) ln S n n 2 1 2 n = + + . [4] … … … … … … … … … … … … … … … … … … … … … … … … …

5 9231/12/O/N/21 © UCLES 2021 [Turn over Let ( ) ( ) ln S r r r 1 2 r 2 1 = + + 3 =/ . (b) Find the least value of n such that . S S 0 01 n 1 - . [3] … … … … … … … … … … … … … … … … … … … … … … … … …

6 9231/12/O/N/21 © UCLES 2021 4 The cubic equation x x x 2 3 3 0 3 2 + + + = has roots a, b, c. (a) Find the value of 2 2 2 a b c + + . [2] … … … … … … … … (b) Show that 1 3 3 3 a b c + + = . [2] … … … … … … … … … … … … … … … … …

7 9231/12/O/N/21 © UCLES 2021 [Turn over (c) Use standard results from the list of formulae (MF19) to show that ) ( ) ( ) ( ) ( r r r n n n an bn c 1 r n 4 1 2 1 3 3 3 a b c + + + = + + + + + + = ` ` j j / , where a, b and c are constants to be determined. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

8 9231/12/O/N/21 © UCLES 2021 5 The curve C has polar equation sin r 3 2 i = + , for 1 G r r i - . (a) The diagram shows part of C. Sketch the rest of C on the diagram. [1] i = 0 O The straight line l has polar equation sin r 2 i = . (b) Add l to the diagram in part (a) and find the polar coordinates of the points of intersection of C and l. [5] … … … … … … … … … … … … …

9 9231/12/O/N/21 © UCLES 2021 [Turn over (c) The region R is enclosed by C and l, and contains the pole. Find the area of R, giving your answer in exact form. [6] … … … … … … … … … … … … … … … … … … … … … … … … … …

10 9231/12/O/N/21 © UCLES 2021 6 The curve C has equation y x x 3 2 = - . (a) Find the equations of the asymptotes of C. [3] … … … … … … … (b) Show that there is no point on C for which y 0 12 1 1 . [4] … … … … … … … … … … … … … … … … … …

11 9231/12/O/N/21 © UCLES 2021 [Turn over (c) Sketch C. [2] (d) (i) Sketch the graphs of y x x 3 2 = - and y x 3 = - on a single diagram, stating the coordinates of the intersections with the axes. [4] (ii) Use your sketch to find the set of values of c for which x x x c 3 2 G - + has no solution. [1] … …

12 9231/12/O/N/21 © UCLES 2021 7 The points A, B, C have position vectors , 2 2 2 7 and i j j k i j k + - + + - respectively, relative to the origin O. (a) Find an equation of the plane OAB, giving your answer in the form p r.n = . [3] … … … … … … … … … … … … … … The plane P has equation x y z 3 2 1 - - = . (b) Find the perpendicular distance of P from the origin. [1] … … … … … … … …

13 9231/12/O/N/21 © UCLES 2021 [Turn over (c) Find the acute angle between the planes OAB and P. [3] … … … … … … … … … … … … (d) Find an equation for the common perpendicular to the lines OC and AB. [10] … … … … … … … … … … … … … …

14 9231/12/O/N/21 © UCLES 2021 … … … … … … … … … … … … … … … … … … … … … … … … … … … …

15 9231/12/O/N/21 © UCLES 2021 Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … … …

16 9231/12/O/N/21 © UCLES 2021 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge. BLANK PAGE

This document consists of 16 printed pages. © UCLES 2021 [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/12 Paper 1 Further Pure Mathematics 1 October/November 2021 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 2021 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 2 of 16 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.

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 3 of 16 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.

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 4 of 16 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.

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 5 of 16 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

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 6 of 16 Question Answer Marks Guidance 1(a) Enlargement, scale factor 6. B1 1 1(b) det 6 8 2 = − = − A M1 Finds det . A Area 2 13 26 cm 2× = = A1 2 1(c) 1 2 1 2 4 2 3 − −   = −  −   A M1 Finds 1. − A 2 4 6 12 3 2 3 6 9 − −     = − =     − −     B A1 AEF Could be solved by equations. 2 1(d) 3 4 3 4 2 2 2 2 x x y y x y +      =      +      B1 Finds X Y       3 4 2 2 x y x x y y + =   + =  leading to 2 4 0 2 0 x y x y + =   + =  M1 Uses x X y Y     =         to form simultaneous equations. 2 y x = − leading to 6 0 x − = leading to 0, 0 x y = = M1 A1 Solves equations or states that 0 2 4 det 2 1      ≠  , AG. 4

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 7 of 16 Question Answer Marks Guidance 2 d e e ( 1)e d ax ax ax y ax ax x = + = + so true when 1. n = M1 A1 Differentiates once using the product rule. Assume that ( ) 1 d e . d k a k k k x y a x ka x − = + B1 States inductive hypothesis. ( ) ( ) ( ) 1 1 1 1 d e e ( 1) d k k ax ax k k k ax k k y a a x ka a a k a e x + − + + = + + = + + M1 A1 Differentiates kth derivative. So true when 1. n k = + By induction, true for all positive integers n. A1 States conclusion. 6 Question Answer Marks Guidance 3(a) ( ) 1 ln 2ln( 1) ln( 2) n r r r r = − + + +  B1 Separates logarithms into correct form using a difference. Or as logarithm of product. ln1 2ln2 ln3 ln2 2ln3 ln 4 ln3 2ln4 ln5 ln( 1) 2ln ln( 1) ln 2ln( 1) ln( 2) n n n n n n − + − + − + − − + + − + + +  M1 A1 Shows enough terms to make cancellation clear. 2 [ln1] ln 2 ln( 1) ln( 2) ln . 2( 1) n n n n + − − + + + = + A1 AG 4

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 8 of 16 Question Answer Marks Guidance 3(b) ln 2 S = − B1 States sum to infinity. AEF 2 ln 0.01 1 n n S S n +   − = <   +   leading to 0.01 2 e ( 1) n n + < + M1 Forms inequality Least value of n is 99 A1 CAO 3 Question Answer Marks Guidance 4(a) 2 ( 2) 2(3) − − M1 Uses formula for sum of squares. 2 − A1 2 4(b) 3 3 3 2( 2) 3( 2) 3(3) α β γ + + = − − − − − M1 Uses original equation or formula for sum of cubes. 1 A1 AG 2

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 9 of 16 Question Answer Marks Guidance 4(c) ( ) 3 3 2 2 3 3 3 r r r r α α α α + + = + + B1 Expands. ( ) ( ) ( ) ( ) ( ) 3 1 1 3 3 2 3 1 3( 2) 3( 2) 3 n n r r r r r r r r α β γ = = + + = + − + − + + + +   M1 A1 Collects like terms and uses results from parts (a) and (b). ( ) ( ) 1 2 2 2 2 3 1 2 6 4 3 4 6 ( 1) 6 ( 1)(2 1) ( 1) 3 ( 1) ( 1)(2 1) ( 1) n n n n n n n n n n n n n n n n − + − + + + + − + − + + + + M1 Applies formulae from MF19. ( ) ( ) 4 2 1 4 1 ( 1) 12 4(2 1) 3 ( 1) ( 1) 3 5 16 n n n n n n n n n n n + + − − + + + + + − − M1 A1 Simplifies. 6

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 10 of 16 Question Answer Marks Guidance 5(a) B1 Correct symmetrical shape, closed loop. 1 5(b) Line l parallel to initial line and correct side of pole. B1 2 n 2 3sin 2si θ θ = + M1 Forms quadratic in sin . θ Or in r 4 3 r r = + 1 2 sinθ = M1 Solves for sin . θ ( ) 1 6 4, π ( ) 5 6 4, π A1 A1 SC1 For finding both angles correctly 5

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 11 of 16 Question Answer Marks Guidance 5(c) ( ) π 6 π 2 2 1 2 3 2s 2 n d i θ θ − × +  M1 Finds the part of the required area enclosed by the curved outer edge and two line segments from the pole. Limits must be correct. ( ) π 6 π 2 c 9 2 12sin d 2 1 os θ θ θ − + − +  M1 Uses double angle formula and integrates. [ ] π 6 2 π 13 22 3 2 3 1 12cos sin2 π 1θ θ θ − − − = − A1 A1 13 π 22 3 2 6 π 3 (4cos ) 2 − + × M1 Adds area of triangle. 5 22 3 2 π 3 − A1 6

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 12 of 16 Question Answer Marks Guidance 6(a) 3 x = B1 States vertical asymptote. 9 3 3 y x x = + + − leading to 3 y x = + M1 A1 Finds oblique asymptote. 3 6(b) 2 3 yx y x − = leading to 2 3 0 x yx y − + = M1 A1 Forms quadratic in x. 2 0 4(3 ) y y − < leading to 2 12 0 y y − < M1 Uses that discriminant is negative. 0 12 y < < A1 AG 4

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 13 of 16 Question Answer Marks Guidance 6(c) B1 Axes and asymptotes. B1 Branches correct. 2 𝑦ൌ 𝑥ଶ 𝑥െ3 𝑥ൌ3

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 14 of 16 Question Answer Marks Guidance 6(d)(i) B1 FT FT from sketch in (c). B1 Correct shape at infinity. B1 Correct shape of 3. y x = − B1 Correct intercepts with axes (may be seen on graph). 4 6(d)(ii) c ⩽ –3 B1 1 x y 𝑦ൌቤ𝑥ଶ 𝑥െ3ቤ 𝑥ൌ3

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 15 of 16 Question Answer Marks Guidance 7(a) 2 1 2 2 0 2 ~ 1 0 1 1 1 2         = = − −         − − −     i j k n M1 A1 Finds common perpendicular. 1 1 0 1     − =     −   r. A1 3 7(b) 2 2 2 1 1 14 1 3 2 = + + B1 Divides by magnitude of the normal to Π 0.267 1 7(c) 1 1 1 3 3 14 cos 1 2 α       − − =       − −     leading to cos 3 14 6 α = M1 A1 FT Takes dot product of normal vectors. 22.2° A1 Accept 0.388 radians. Mark final answer. 3

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 16 of 16 Question Answer Marks Guidance 7(d) 20 5 2 1 7 12 ~ 3 2 3 1 1 4 −         − = −         − − −     i j k M1 A1 Finds direction of common perpendicular. 2 2 2 1 , 2 3 7 OP OQ μ λ μ μ −         = = −         −       leading to 2 2 2 7 2 3 PQ μ λ μ λ μ λ  − −     = −     − +  M1 A1 Finds PQ  . 2 2 2 2 3 3 0 7 1 2 μ λ μ λ μ λ − −       − − =      − − +      or 2 2 5 2 3 3 2 7 1 k μ λ μ λ μ λ −         − = −        − − +      M1 Uses that dot product of PQ  with line direction is zero, or, alternatively, PQ  is a multiple of the common perpendicular (parameter k not 1). 14 10 14μ λ + = A1 Deduces one equation. 6 2 2 2 5 2 3 1 0 2 4 7 14 7 μ λ μ λ μ λ μ λ − −       − =        −   − + +  =  A1 Deduces second equation. 1 10 λ = − leading to 2 1 1 10 7 OP     = −     −    M1 A1 Solves for λ and substitutes into OP  . 0.2 5 0.1 3 1 0.7 k −         = − + −             r B1 FT FT using their common perpendicular. 10