Cambridge IGCSE Mathematics - Additional 0606 — 2024 Oct/Nov Paper 2 · Variant 2

This document has 16 pages. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/22 Paper 2 October/November 2024 2 hours You must answer on the question paper. No additional materials are needed. 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. ● 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 80. ● The number of marks for each question or part question is shown in brackets [ ]. * 7 3 3 8 2 9 6 7 8 6 * DC (PQ/SW) 336447/1 © UCLES 2024 ,










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2 0606/22/O/N/24 © UCLES 2024 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax bx c 0 2 + + = , x a b b ac 2 4 2 ! = - - Binomial Theorem ( ) a b a a b a b a b n n n r b 1 2 n n n n n r r n 1 2 2 f f + = + + + + + + - - - e e e o o o where n is a positive integer and ( )! ! ! n r n r r n = - e o Arithmetic series ( ) u a n d 1 n = + - ( ) { ( ) } S n a l n a n d 2 1 2 1 2 1 n = + = + - Geometric series u ar n n 1 = - ( ) ( ) S r a r r 1 1 1 n n ! = - - ( ) S r a r 1 1 1 = - 3 2. TRIGONOMETRY Identities sin cos A A 1 2 2 + = sec tan A A 1 2 2 = + ec cos cot A A 1 2 2 = + Formulae for ∆ABC sin sin sin A a B b C c = = cos a b c bc A 2 2 2 2 = + - sin bc A 2 1 T = * 0000800000002 * ,










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3 0606/22/O/N/24 © UCLES 2024 [Turn over 1 Solve the following simultaneous equations. x y 2 3 = x y 16 27 5 4 = [3] * 0000800000003 * ,










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4 0606/22/O/N/24 © UCLES 2024 2 Variables x and y are related by the equation y x x 1 2 = + . (a) Find x y d d . [3] (b) It is given that when y 12 = , x 4 = . Find the approximate change in x when y increases from 12 by the small amount 0.06. [3] (c) Find the x-coordinate of the stationary point on the curve y x x 1 2 = + . [2] * 0000800000004 * ,










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5 0606/22/O/N/24 © UCLES 2024 [Turn over 3 DO NOT USE A CALCULATOR IN THIS QUESTION. The polynomial p is defined by ( )x ax x x b 3 3 p 3 2 = - - + , where a and b are constants. (a) Given that x 2 = and x 1 =- are roots of the equation ( )x 0 p = , find a and b. [3] (b) Solve the equation ( )x 0 p = . [2] * 0000800000005 * ,










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6 0606/22/O/N/24 © UCLES 2024 4 Use a graphical method to solve the inequality x 2 8 4 2 - . [5] y x – 2 0 – 2 2 4 6 8 – 4 – 6 – 8 – 4 – 6 – 8 2 4 6 8 10 * 0000800000006 * ,










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7 0606/22/O/N/24 © UCLES 2024 [Turn over 5 Solve the following equations. (a) log log x x 18 2 2 16 + = [4] (b) 10 3 e e x x 2 1 2 1 - = + - - [4] * 0000800000007 * ,










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8 0606/22/O/N/24 © UCLES 2024 6 DO NOT USE A CALCULATOR IN THIS QUESTION. Write ( )( ) 5 3 6 2 2 - + - in the form a b 3 + , where a and b are constants. [5] * 0000800000008 * ,










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9 0606/22/O/N/24 © UCLES 2024 [Turn over 7 A class of 10 students includes Abby and Ben. (a) A group of 5 students is to be selected from the class. Find the number of possible groups in the following cases. (i) There are no restrictions. [1] (ii) The group includes both Abby and Ben. [2] (iii) The group includes either Abby or Ben, but not both. [2] (b) All 10 students are arranged in a line. How many arrangements are possible if there are exactly three students between Abby and Ben? [3] * 0000800000009 * ,










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10 0606/22/O/N/24 © UCLES 2024 8 Solve the equation cot cosec 2 3 2 9 2 i i + = for 90 90 ° ° G G i - . [6] * 0000800000010 * ,










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11 0606/22/O/N/24 © UCLES 2024 [Turn over 9 In this question time is measured in seconds. (a) A particle is moving in a straight line with constant velocity of ms 6 1 - . At time t 0 = , it passes a fixed point A. At time t 5 = it suddenly changes direction and moves with a different constant velocity along the same straight line. It passes the point A again at time t 15 = . Sketch the velocity−time graph for the motion. [3] v t 5 10 15 6 −6 0 (b) Another particle is moving in a straight line with constant acceleration. At time t 0 = it passes a fixed point B with velocity ms 8 1 - - . It passes the point B again at time t 20 = . Sketch the velocity−time graph for the motion. [3] v t 5 10 15 20 0 * 0000800000011 * ,









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12 0606/22/O/N/24 © UCLES 2024 10 The diagram shows part of the curve y x x 4 2 = - and the line y 4 =- . The curve and the line intersect at the point A. y O x A y x x 4 2 = - y 4 =- h (a) The maximum point on the curve is at a perpendicular distance h from the line y 4 =- . Find the value of h. [4] * 0000800000012 * ,









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13 0606/22/O/N/24 © UCLES 2024 [Turn over (b) Find the exact x-coordinate of A. [3] (c) Find the acute angle between the tangent to the curve at A and the line y 4 =- . [4] * 0000800000013 * ,









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14 0606/22/O/N/24 © UCLES 2024 11 In this question i is a unit vector in the positive x-direction and j is a unit vector in the positive y-direction. Time is in seconds and distances are in metres. The diagram shows the initial positions and velocities of two particles, A and B, that move in the x-y plane. y O Particle A Particle B x 10 ms–1 ms–1 60° 3 5 ( , ) 2 3 9 Particle A starts from the origin O at time t 0 = . It moves with constant speed ms 10 1 - in the direction 60° above the x-axis. (a) Find the exact values of the components of the velocity of particle A in the x-direction and the y-direction. [2] (b) Find, in terms of t, the position vector of particle A at time t. [1] * 0000800000014 * ,









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15 0606/22/O/N/24 © UCLES 2024 [Turn over Particle B starts from the point ( , ) 2 3 9 at time t 0 = . It moves with constant speed ms 3 5 1 - parallel to the positive x-axis. (c) Find, in terms of t, the position vector of particle B at time t. [2] (d) Hence show that the particles collide. [4] Question 12 is printed on the next page. * 0000800000015 * ,









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16 0606/22/O/N/24 © UCLES 2024 12 A metal tank is in the shape of a cuboid with a square base of side x m and an open top. The tank has a volume of m 5 3. Given that x can vary, and that the area of the metal used to make the tank is a minimum, find the dimensions of the tank. [6] 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. * 0000800000016 * ,









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This document consists of 12 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/22 Paper 2 October/November 2024 MARK SCHEME Maximum Mark: 80 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.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 2 of 12 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors. GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 3 of 12 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. 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 marks, awarded for a valid method applied to the problem. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. For accuracy marks to be given, the associated Method mark must be earned or implied. B Mark for a correct result or statement independent of Method marks. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’ is used to indicate that a particular M or B mark is dependent on an earlier mark in the scheme. Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working nfww not from wrong working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 4 of 12 Question Answer Marks Partial marks 1 Correct elimination of one unknown e.g. 4 3 1 2 x       = 27 16 or 4 5 3 1 2 x x       = 27 16 or 4 5 27 16 2 3 y y =       or 4 5 81 16 27 16 x x  =     or 4 5 32 16 27 243 y y   =     oe M1 x = 3 and y = 9 2 oe and no other solutions A2 A1 for x = 3 or y = 9 2 oe 2(a) ( ) ( ) 1 1 2 2 d 1 1 2 1 2 or 1 2 2 d 2 x x x x − − + = +  +  oe B2 B1 for ( ) 1 2 1 2 k x − + where k is a positive constant, k ≠ 1 1 1 2 2 1 (1 2 ) 2 [1](1 2 ) 2 x their x x −      +  + +   oe, isw B1 FT their d 1 2 d x x + Alternative ( ) ( ) 1 2 3 2 2 1 2 2 6 2 x x x x − +  + or ( ) ( ) 1 2 2 2 1 (1 2 ) 2 2 (1 2 ) 2 x x x x x − +  + + (B3) B2 for ( ) ( ) 1 2 3 2 2 1 2 2 x x ax bx − +  + or ( ) ( ) 1 2 2 2 1 (1 2 ) 2 x x ax bx − +  + where a and b are constants or B1 for ( ) 1 2 3 2 2 k x x − + or ( ) 1 2 2 (1 2 ) k x x − + where k is a positive constant, soi 2(b) Uses their 1 1 2 2 4(1 2(4)) (1 2(4)) − + + + in an attempt at a small changes relationship M1 FT x = 4 substituted into their derivative 0.06 δx = 1 1 2 2 4(1 2(4)) (1 2(4)) their −     + + +     oe M1 dep previous M1 FT their 13 3 0.0138 or 0.01384[6…] or 9 650 oe nfww A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 5 of 12 Question Answer Marks Partial marks 2(c) 1 2 1 2 x x x + + + = 0 oe and solves as far as x = …. M1 FT a derivative of the form 1 2 1 2 ax b x x + + + or 1 2 1 2 a b x x + + + or 1 2 ax b x + + oe x = 1 3 − oe A1 3(a) 8a − 12 − 6 + b = 0 oe and –a − 3 + 3 + b = 0 oe and a = 2, b = 2 B3 B1 for 8a − 12 − 6 + b = 0 oe B1 for – a − 3 + 3 + b = 0 oe 3(b) [2x3 − 3x2 − 3x + 2 =] (x − 2), (x + 1), (2x − 1) or (x2 – x – 2), (2x − 1) and x = 2, 1 2 , −1 OR [2x3 − 3x2 − 3x + 2 =] (x + 1), (2x2 – 5x + 2) or (x – 2), (2x2 + x – 1) and correct factorisation or method of solution of the quadratic and x = 2, 1 2 , −1 OR 2 – 1 + x = 3 2 −   −    or 2 – 1 + x = 3 2 or for 2  – 1  x = 2 2 −or 2  – 1  x = –1 and x = 2, 1 2 , −1 B2 B1 for [2x3 − 3x2 − 3x + 2 =] (x − 2) and (x + 1) seen or (x2 – x – 2) seen or (x + 1), (2x2 – 5x + 2) or (x – 2), (2x2 + x – 1) B1 for 2 – 1 + x = 3 2 −   −    or 3 2 or for 2  – 1  x = 2 2 −or –1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 6 of 12 Question Answer Marks Partial marks 4 Correct intersecting graphs -2 2 4 6 8 2 4 6 8 10 x y 0 3 y = |2x – 8|: M1 for an attempt to draw the sections from (0, 8) to (2, 4) and (6, 4) to (8, 8) with at least one side accurate or for  shape with vertex at (4, 0) A1 for correct graph B1 for y = 4 drawn critical values: 2, 6 M1 dep on 3 marks awarded for intersecting graphs x < 2, x > 6 mark final answer A1 5(a) Correctly derives correct equation free of logarithms e.g. 9 18 16 x = or x = 162 oe or 9 72 2 x = or x = 28 oe or 9 18 4 2 x = oe M3 M2 for correctly changing to consistent bases and correct use of one other log law or correct use of log 1 a a = in a correct equation e.g. 2 16 16 log log 18 1 4 x x + = oe or 2 2 2 log log 18 4 x x + = oe or 2 log log 18 log 2 4log 2 x x x x x x + = or M1 for correctly changing to consistent bases or correct use of one other log law or correct use of log 1 a a = in a correct equation x = 256 nfww A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 7 of 12 Question Answer Marks Partial marks 5(b) 4 2 2 1 e 3e 10 x x + + − − [= 0] or ( ) 2 2 1 2 1 e 3e 10 x x + + − − [= 0] B1 Solves or factorises their 3-term quadratic in 2 1 e x+ M1 FT their 3-term quadratic in 2 1 e x+ 2 1 e 5 x+ = nfww A1 x = 1 ln5 2 −+ oe, isw or 0.305 or 0.3047[18…] isw A1 and no other solution Alternative ( )   2 2 2 1 e e 3e 10e 0 x x − − − = oe or ( ) ( )   2 2 2 2 e e 3e e 10 0 x x − − = oe or ( ) ( )   4 2 2 e e 3e e 10 0 x x − − = oe (B1) Solves or factorises their 3-term quadratic in 2 e x oe (M1) FT their 3-term quadratic in 2 e x 2 5 e e x = or 1.839[…] or 5 e e x = or 1.356[…] nfww (A1) x = 1 5 ln 2 e oe, isw or 0.305 or 0.3047[18…] isw (A1) and no other solution 6 ( ) 2 5 3 6 2 − + soi B1 ( ) 2 6 2 6 2 2 12 + = + + or 8 + 2 12 B1 5 3 8 4 3 8 4 3 8 4 3 − −  + − or ( ) 5 3 2 3 2 3 4 2 3 − −  − + oe M1 FT ( ) 5 3 a b c d − + where a, b, and c are non-zero constants and d is an integer 2 2 40 20 3 8 3 12 40 20 3 8 3 12 or 64 48 8 (4 3) − − + − − + − − or 2 2 10 5 3 2 3 3 10 5 3 2 3 3 or 4 4(2 ( 3) ) − − + − − + − A1 13 7 3 4 4 − oe, nfww A1 dep on all previous marks awarded

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 8 of 12 Question Answer Marks Partial marks 7(a)(i) 252 B1 7(a)(ii) 56 B2 B1 for 8C3 or 8! 5! 3!  oe 7(a)(iii) 140 B2 B1 for 8C4 × 2 oe or 10C5 – 8C5 – 8C3 oe 7(b) 483 840 B3 B2 for 8! × 6 [× 2] or 241 920 oe or B1 for 8! or 40 320 or 8! × 2 or 80 640 8 cosec2 2θ + 3cosec 2θ − 10 [= 0] or 2 10sin 2 3sin2 1 [ 0]   − − = B2 B1 for correctly writing the equation in terms of one trigonometric function e.g. cosec2 2θ – 1 + 3 cosec 2θ = 9 or 2 2 1 sin 2 3 9 sin 2 sin 2    − + = Solves or factorises their 3-term quadratic in cosec2 or sin2 e.g. (cosec 2θ − 2) (cosec 2θ + 5) [= 0] or (2sin 2θ − 1)(5sin 2θ + 1) [= 0] M1 FT their 3-term quadratic in cosec2 or sin2 [sin 2θ = 1 2 sin 2θ = −1 5  = ] 15 75 −5.8 or –5.76 to –5.77 −84.2 or –84.23 to –84.232 and no other angles in range; nfww A3 A2 for any 2 correct, ignoring extras in range; nfww or A1 for one correct angle or one correct double angle; nfww

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 9 of 12 Question Answer Marks Partial marks 9(a) Horizontal line, v = 6 for 0 ⩽ t ⩽ 5 B1 Horizontal line, v = −3 for 5 ⩽ t ⩽ 15 B2 B1 for horizontal line for 5 ⩽ t ⩽ 15 with v = k where k < 0 or v = 3 for 5 ⩽ t ⩽ 15 or v = −3 for t > 5 and at least 7 ⩽ t ⩽ 13 If 0 scored, award: SC2 for Horizontal line, v = –6 for 0 ⩽ t ⩽ 5 and Horizontal line, v = 3 for 5 ⩽ t ⩽ 15 OR SC1 for Horizontal line, v = –6 for 0 ⩽ t ⩽ 5 and Horizontal line for 5 ⩽ t ⩽ 15 with v = k where k > 0 9(b) Single line with positive gradient passing through (0, –8), (10, 0) and (20, 8) B3 B2 for a single line with positive gradient • passing through a point indicated as (0, –8) or (20, 8) and the point (10, 0) • or passing through points indicated as (0, –8) and (20, 8) which does not pass through (10, 0) or B1 for a single line with positive gradient • passing through a point indicated as (0, –8) or (20, 8) to or through any point on the t-axis • or passing through (10, 0) but with both endpoints incorrect or unlabelled If 0 scored, award SC1 for a single line with negative gradient passing through (0, 8), (10, 0) and (20, –8)

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 10 of 12 Question Answer Marks Partial marks 10(a) Maximum point at (2, 1) oe, nfww and h = 5 nfww B4 B3 for maximum point at (2, 1) oe or B2 for maximum point when x = 2 or B1 for a correct method which could be used to find the maximum point ( ) 1 0 when 0 and 4 x x x   − = =     x = 4 or [differentiating and equating to 0:] 2 1 0 4 x − = or [completing the square to find:] ( )2 1 1 2 4 x − − If 0 scored, SC1 for h = 5 with incorrect or no method shown 10(b) x2 − 4x − 16 [= 0] oe or ( )2 1 2 4 1 4 x − − = −− oe B1 FT their attempt to complete the square, if already seen and of the form ( )2 a b x c + + where a, b and c are constants and b is negative Solves their 3-term quadratic using the formula or completing the square M1 FT their rearrangement of 2 4 4 x x −= − oe x = 2 2 5 + oe A1 10(c) First derivative 1 − 2 x and substitution of their 2 2 5 + or 6.47 M1 FT their attempt to differentiate 2 4 x x − , if already seen, and their 2 2 5 + 1 – 2 2 5 2 4 +  or 5 − or awrt –2.24 soi A1 dep on correct derivative and correct x-coordinate of A Correct method to find the angle e.g. 1 2 1 – 2 5 tan 2 4 −  + −      soi or 1 tan 5 − or – ( ) 1 tan 5 − − soi M1 dep on previous M1 A1 Degrees: awrt 65.9 or Radians: awrt 1.15 A1 11(a) x = 5 y = 5 3 B2 B1 for either component correct

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 11 of 12 Question Answer Marks Partial marks 11(b) 5 0 0 5 3 t   +         oe, isw B1 FT their 5 5 3       , which must be a vector with at least one non-zero component 11(c) 5 2 3 3 9 0 t       +         oe, isw B2 B1 for x component 5 2 3 3t + seen or y component 9 seen or 5 2 3 3 9 0 t       +         with at most one error but must include t 11(d) 5 3 t = 9 or 5t = 5 2 3 3t + M1 FT their position vector of A and their position vector of B providing • both are in terms of t and • at least one is of form a c t b d    +       where a, b, c and d are constants ( ) 5 3 t = 9 and 5t = 5 2 3 3t + oe A1 Shows the exact times to be the same e.g. t = 9 3 3 5 5 3 = oe and 10 3 3 2 3 3 5 t t = →= or 6 3 3 3 10 5 t t = →= or 5 3 3 3 3 5 t t = →= oe OR Finds a correct value for t, as above, and shows this satisfies the other equation OR Finds a correct value for t, as above, and shows both particles are at 9 3 9         oe at this time A2 A1 for t = 9 5 3 and 2 3 10 3 t = oe

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 12 of 12 Question Answer Marks Partial marks 12 Correct use of x2h = 5 to find an expression that can be used to eliminate h M2 M1 for x2h = 5 soi Surface area: x2 + ( ) 2 5 4 x x oe B1 Derivative of the surface area: 2 2 20 x x− − oe B1 FT their surface area of form 2 b ax x + Equates their 2 2 20 x x− − to 0 and solves to find a value of x M1 FT their derivative of form 2 b ax x + oe x = 2.15 or 2.154[…] h = 1.08 or 1.077[…] nfww A1 dep on all previous marks awarded Alternative Correct use of 2 5 x h = to find an expression that can be used to eliminate x (M2) M1 for 2 5 x h = soi Surface area: 2 5 h       + 5 4h h  oe (B1) Derivative of the surface area: 1 2 2 1 5 4 5 2 h h −  − oe (B1) FT their surface area of form a b h h + Equates their 1 2 2 1 5 4 5 2 h h −  − to 0 and solves to find a value of h (M1) FT their derivative of form 2 a b h h + oe h = 1.08 or 1.077[…] x = 2.15 or 2.154[…] nfww (A1) dep on all previous marks awarded