Cambridge A Level Mathematics - Further 9231 — 2020 Oct/Nov Paper 1 · Variant 2

This document has 16 pages. Blank pages are indicated. DC (RW) 187466/1 © UCLES 2020 [Turn over * 9 1 1 7 0 4 4 3 9 4 * FURTHER MATHEMATICS 9231/12 Paper 1 Further Pure Mathematics 1 October/November 2020 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 [ ]. Cambridge International AS & A Level

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3 9231/12/O/N/20 © UCLES 2020 [Turn over 1 The cubic equation x bx cx d 0 3 2 + + + = , where b, c and d are constants, has roots a, b, c. It is given that 1 abc =- . (a) State the value of d. [1] … (b) Find a cubic equation, with coefficients in terms of b and c, whose roots are 1 a+ , 1 b+ , 1 c+ . [3] … … … … … … … … … … (c) Given also that 1 1 c a + =- - , deduce that ( )( ) c b b b c 2 3 3 - + - = - . [4] … … … … … … … … … … … …

4 9231/12/O/N/20 © UCLES 2020 2 Prove by mathematical induction that 7 1 n 2 - is divisible by 12 for every positive integer n. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 9231/12/O/N/20 © UCLES 2020 [Turn over 3 (a) By simplifying x x x x 1 1 n n n n 2 2 - + + + ` `j j, show that x x x x 1 1 1 n n n n 2 2 - + =- - + . [1] … … Let u x x x x 1 1 1 n n n n n 1 2 2 2 = + + + - + + + . (b) Use the method of differences to find u n N n 1 =/ in terms of N and x. [3] … … … … … … … … … (c) Deduce the set of values of x for which the infinite series u u u 1 2 3 f + + + is convergent and give the sum to infinity when this exists. [3] … … … … … … … … … …

6 9231/12/O/N/20 © UCLES 2020 4 The matrices A and B are given by A 0 1 1 0 = e o and B 3 3 2 1 2 1 2 1 2 1 = - f p. (a) Give full details of the geometrical transformation in the x-y plane represented by A. [1] … (b) Give full details of the geometrical transformation in the x-y plane represented by B. [2] … … … … … The triangle DEF in the x-y plane is transformed by AB onto triangle PQR. (c) Show that the triangles DEF and PQR have the same area. [3] … … … … … … … … … … … … … … …

7 9231/12/O/N/20 © UCLES 2020 [Turn over (d) Find the matrix which transforms triangle PQR onto triangle DEF. [2] … … … … … (e) Find the equations of the invariant lines, through the origin, of the transformation in the x-y plane represented by AB. [5] … … … … … … … … … … … … … … … … … … … … …

8 9231/12/O/N/20 © UCLES 2020 5 The curve C has polar equation r ( ) ln r 1 i = + - , for r 0 G G i . (a) Sketch C and state the polar coordinates of the point of C furthest from the pole. [3] … … (b) Using the substitution r u 1 i = + - , or otherwise, show that the area of the region enclosed by C and the initial line is r r r r ( ) ( ) ( ) ln ln 1 1 1 2 2 1 + + + - + ` j . [6] … … … … … … … … … … … … … … …

9 9231/12/O/N/20 © UCLES 2020 [Turn over … … … … … … … … (c) Show that, at the point of C furthest from the initial line, r r ( ) ( ) ln tan 1 1 0 i i i + - + - - = and verify that this equation has a root between 1.2 and 1.3. [5] … … … … … … … … … … … … … … … … …

10 9231/12/O/N/20 © UCLES 2020 6 Let a be a positive constant. (a) The curve C1 has equation y x a x a 2 = - - . [2] Sketch C1. The curve C2 has equation y x a x a 2 2 = - - b l . The curve C3 has equation y x a x a 2 = - - . (b) (i) Find the coordinates of any stationary points of C2. [3] … … … … … … … … … … … … … … …

11 9231/12/O/N/20 © UCLES 2020 [Turn over (ii) Find also the coordinates of any points of intersection of C2 and C3. [3] … … … … … … … … … … … … … (c) Sketch C2 and C3 on a single diagram, clearly identifying each curve. Hence find the set of values of x for which x a x a x a x a 2 2 2 G - - - - b l . [5]

12 9231/12/O/N/20 © UCLES 2020 7 The points A, B, C have position vectors i j k 2 2 - + - , i j k 2 2 - + + , 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] … … … … … … … … … … … … (b) Find the acute angle between the planes OBC and ABC. [4] … … … … … … … … … … …

13 9231/12/O/N/20 © UCLES 2020 The point D has position vector ti j - . (c) Given that the shortest distance between the lines AB and CD is 10, find the value of t. [6] … … … … … … … … … … … … … … … … … … … … … … … … … …

14 9231/12/O/N/20 © UCLES 2020 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. … … … … … … … … … … … … … … … … … … … … … … … … … …

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16 9231/12/O/N/20 © UCLES 2020 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 2020 [Turn over Cambridge International AS & A Level FURTHER MATHEMATICS 9231/12 Paper 1 Further Pure Mathematics 1 October/November 2020 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 2020 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 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 2020 © UCLES 2020 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 2020 © UCLES 2020 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 2020 © UCLES 2020 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 2020 © UCLES 2020 Page 6 of 16 Question Answer Marks Guidance 1(a) 1 d = B1 1 1(b) 1 1 y x x y = +  = − B1 Uses correct substitution. 3 2 ( 3) ( 2 3) 0 y b y c b y b c + − + − + + − = M1 A1 Substitutes and expands. Alternative method for question 1(b) 1)( 1) 1)( 1) ( 1)( 1) 2 3 ( ( c b α β α γ β γ + + + + + + + + = − + B1 ( 1 1 1) 3 ,b α β γ + + + + + = − ) ( 1)( 1 ( 1) c b α β γ + + + = − M1 Using these relationships. 3 2 ( 3) ( 2 3) 0) y b y c b y b c + − + − + + − = A1 3 1(c) ( ) 1 3 b β + = − − B1 Uses sum of roots. 3 1)( 1 ( 2 ) c b α α + − − + + = B1 Uses sum of products in pairs. ( )( )( ) ( ) 1 1 1 b c α β α + + = − + − − M1 Applies product of roots. ( )( ) 2 3 3 c b b b c  − + − = − A1 AG 4

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 7 of 16 Question Answer Marks Guidance 2 2 7 1 48 −= is divisible by 12. B1 Checks base case. Assume that 2 7 1 k − is divisible by 12 for some positive integer .k B1 States inductive hypothesis. Then 2 2 2 2 2 7 1 49 7 1 48 7 7 1 k k k k + −= ⋅ −= ⋅ + − Or 2 2 2 2 (7 1) (7 1) 48 7 k k k + − − − = ⋅ M1 Separates 2 7 1. k − 2 2 7 1 k+ − is divisible by 12. A1 So if is true for n k = (could be written earlier) it is also true for 1 n k = + . Hence, by induction, true for every positive integer .n A1 Depends on previous M1 and A1. 5

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 8 of 16 Question Answer Marks Guidance 3(a) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 1 1 1 n n n n n n x x x x x x −√ + + √ + = − + = − B1 Uses the difference of two squares and obtains the given answer. AG 1 3(b) ( ) ( ) 2 2 2 1 1 1 n n n n n u x x x x + + = + √ + − −√ + B1 Uses the identity given in part (a). ( ) ( ) ( ) ( ) 2 4 2 3 6 2 4 1 1 1 1 1 N n n u x x x x x x x x = = + √ + − −√ + + + √ + − −√ +  ( ) ( ) 2 2 2 1 1 1 N N N N x x x x + + + + + √ + − −√ +  M1 Writes at least three complete correct terms, including the first and last. ( ) ( ) 2 2 1 2 1 1 N N x x x x + + = + √ + − −√ + A1 (Don’t allow in terms of n .) 3 3(c) 1 1 x −< < B1 ( ) 1 2 2 1 3 0 as 1 1 N x n u u u x x + →  + + + = − −√ →∞ +  M1 A1 Finds sum to infinity. 3

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 9 of 16 Question Answer Marks Guidance 4(a) Reflection in the line . y x = B1 Only mention reflection (and no other transformation). 1 4(b) Rotation B1 Writes ‘rotation’ or ‘rotate’ (and no other transformation) 1 3 π anticlockwise about the origin. B1 Or 60° 2 4(c) 1 1 2 2 1 1 2 2 3 3   √ =     −√   AB or det det det = AB A B M1 Finds AB or uses product of determinants. Full marks here may be obtained by arguing that both reflection and rotation preserve [absolute] value of area and so also does their combination. det 1 = − AB B1 Area of PQR = |-1| Area of DEF A1 3 4(d) ( ) 1 1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 3 3 3 3 −     −√ − √ = − =         − √ −√     AB M1 A1 or using (AB)–1 = B–1A–1 2

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 10 of 16 Question Answer Marks Guidance 4(e) 1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 3 3 3 3 x y x y x y   √+   √     =        −√ − √       B1 Transforms x y       to X Y       . Accept 1 t m       instead of . x y       2 1 1 1 1 2 2 2 2 3 3 m m m − √ = √+ M1A1 Uses Y mX = . OE. Allow working with y mx c = + as long as 0 c = is stated explicitly. 2 2 1 3 3 2 3 1 0 2 3 m m m m m m − √ = √+  + √−=  = ± −√ A1 (2 y x = −√3) and (2 0 y x + + √3) = A1 OE 5

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 11 of 16 Question Answer Marks Guidance 5(a) B1 Correct shape, r decreasing. B1 Section 1 2 π π θ ≤ ≤ correct, becoming tangential at π. θ = ( ) ln(1 ,0 π) + B1 May be seen on their diagram. Allow (1.42, 0) 3 5(b) π 1 π 2 2 1 1 2 2 0 1 ln π )d d l 1 n ( u u θ θ + − + =   M1 Uses correct formula with correct limits. π 1 1 π 2 1 2 1 1 l l n n u u du u + + =    −   M1 A1 Integrates by parts once. = [ ] π 1 π 1 π 2 1 2 1 1 1 1 ln 1 l d n u u u u u + + +   − −        M1 Integrates by parts again (or uses known result for integral of ln u) π 1 1 2 2 1 ln ln u u u u u +   = − +   A1 1 2 2 π) π) π) π) π (1 ln (1 (1 ln(1 + + − + + + = 1 2 π) π)(ln(1 π) 2) π (1 ln(1 + − + + + A1 AG 6 0 θ =

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 12 of 16 Question Answer Marks Guidance 5(c) ) ln(1 π sin y θ θ − = + B1 Uses sin y r θ = sin π )cos 0 d l 1 π n(1 d y θ θ θ θ θ − = − = + − + M1 A1 Sets derivative equal to zero. 0 (1 π )ln(1 π ) tan 0 cosθ θ θ θ ≠  + − + − − = A1 AG (1 π 1.2)ln(1 π 1.2) tan1.2 0.602 + − + − − = and (1 π 1.3)ln(1 π 1.3) tan1.3 0.634 + − + − − = − B1 Shows sign change. Values correct to 2sf should be shown. 5

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 13 of 16 Question Answer Marks Guidance 6(a) B1 Asymptotes labelled. B1 Correct position and shape. Not too truncated. 2 6(b)(i) ( ) ( ) 3 2 d 0 d 2 a x a y x x a − = − = − M1 Sets d 0. d y x = Solves for .x M1 Since 2 , x a ≠ stationary point is( ) ,0 . a A1 3 2a x 1 y

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 14 of 16 Question Answer Marks Guidance 6(b)(ii) ( ) 2 ( )( 2 ) x a x a x a − = − − or ( ) 2 ( )( 2 ) x a x a x a − = − − − M1 Finds a critical point. x a = and 3 2 a or (a, 0) or ( 3 2 a , 1) A1 Both x values or one correct point. (a, 0) and ( 3 2 a , 1) A1 3 6(c) B1 Axes and asymptotes. B1 FT 3 C FT from their sketch in part (a). Clearly identified. B1 Correct shape of 2. C Clearly identified. B1 Relative positions and intersections correct. x ⩽ 3 2 a B1 Accept algebraic method. 5 x y 1 2a 2 C 3 C 2 C 3 C 2 C 3 C

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 15 of 16 Question Answer Marks Guidance 7(a) 3 2 4 2 AB AC = −+ = − + j k i j k   B1 Finds direction vectors of two lines in the plane. 2 3 BC = − − i j k  5 0 1 3 3 1 2 1 1   − =   −  i j k M1 A1 Finds normal to the plane ABC. OE 5(0) 3( 2) 1(1) 5 5 3 5 x y z + − + = − + + = − M1 A1 Substitutes point. OE 5 7(b) 5 2 1 2 2 0 2 1 4   − =   −  i j k M1 A1 Finds normal to the plane OBC. 5 35 4 5 3 5 s 1 co 2 4 θ  =     M1 Uses dot product of normal vectors. 28.1° A1 0.49(0) rad 4

9231/12 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 16 of 16 Question Answer Marks Guidance 7(c) 2 0 1 3 3 1 1 t t t −     − =     −   i j k M1 A1 Finds common perpendicular. 2 2 2 2 1 4 10 4 3 4 10 4 10 2 t t t t t −    −−    − =    + +        M1 A1 Uses formula for perpendicular distance. ( ) 2 2 2 4 10 10 (4 10 ) 10 4 10 4 10 t t t t −− =  + = + + M1 Sets equal to 10 and solves for t. 3 10 t = A1 6