Cambridge IGCSE Mathematics - Additional 0606 — 2025 May/June Paper 2 · Variant 2

This document has 16 pages. [Turn over Cambridge IGCSE™ DC (CJ/CGW) 347693/1 © UCLES 2025 * 5 2 5 1 5 3 8 6 9 1 * ADDITIONAL MATHEMATICS 0606/22 Paper 2 May/June 2025 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 scientific calculator where appropriate. ● You must show all necessary working clearly. ● 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. ● For r, use either your calculator value or 3.142. INFORMATION ● The total mark for this paper is 80. ● The number of marks for each question or part question is shown in brackets [ ]. ,










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2 0606/22/M/J/25 © UCLES 2025 List of formulas Equation of a circle with centre (a, b) and radius r. (x – a)2 + (y – b)2 = r 2 Curved surface area, A, of cone of radius r, sloping edge l. r A rl = Surface area, A, of sphere of radius r. r A r 4 2 = Volume, V, of pyramid or cone, base area A, height h. V Ah 3 1 = Volume, V, of sphere of radius r. r V r 3 4 3 = Quadratic equation For the equation ax 2 + bx + c = 0, x a b b ac 2 4 2 ! = - - Binomial theorem ( ) … … a b a n a b n a b n r a b b 1 2 n n n n n r r n 1 2 2 + = + + + + + + - - - J L KK J L KK J L KK N P OO N P OO N P OO , where n is a positive integer and ( )! ! ! n r n r r n = - J L KK N P OO Arithmetic series un = a + (n – 1)d Sn = 2 1 n(a + l) = 2 1 n{2a + (n – 1)d} Geometric series un = arn – 1 Sn = ( ) r a r 1 1 n − − (r ≠ 1) S∞ = r a 1 − (|r| < 1) Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A Formulas for ∆ABC sin sin sin A a B b C c = = a2 = b2 + c2 – 2bc cos A ∆ = 2 1 ab sin C * 0000800000002 * ,










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3 0606/22/M/J/25 © UCLES 2025 [Turn over 1 Solve the inequality x 5 2 3 H + . [4] * 0000800000003 * ,










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4 0606/22/M/J/25 © UCLES 2025 2 In this question, all lengths are in metres and time is in seconds. A particle P moves in a straight line such that its displacement s from a fixed point O at time t is given by ( ) ( ) for s t t t 4 0 1 2 H = - - . (a) On the axes, sketch the displacement–time graph of P, stating the intercepts with the axes. [2] s t O (b) Find an expression for the velocity, v, of P. Give your answer in a factorised form. [2] * 0000800000004 * ,










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5 0606/22/M/J/25 © UCLES 2025 [Turn over (c) On the axes, sketch the velocity–time graph of P, stating the intercepts with the axes. [2] v t O (d) Find an expression for the acceleration, a, of P. [1] (e) On the axes, sketch the acceleration–time graph of P, stating the intercepts with the axes. [3] a t O * 0000800000005 * ,










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6 0606/22/M/J/25 © UCLES 2025 3 Functions f and g are such that ( ) f x for x x x x 4 3 0 2 2 = + + for x 2 2 = - ( ) g x . Solve the equation ( ) fg x 1 = . [4] * 0000800000006 * ,










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7 0606/22/M/J/25 © UCLES 2025 [Turn over 4 (a) Given that sin cos y x x 4 2 2 = , find the value of x y d d when r x 6 = . [4] (b) A curve has equation sin cos y x x 4 2 2 = . The normal to the curve at the point where r x 6 = meets the x-axis at the point P. Find the exact coordinates of P. [5] * 0000800000007 * ,










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8 0606/22/M/J/25 © UCLES 2025 5 (a) A 4-digit number is to be formed using the digits 0, 2, 4, 5, 6 and 8. The 4-digit number must not start with 0. Any digit may be used at most once in the 4-digit number. (i) Find how many 4-digit numbers can be formed. [1] (ii) Find how many even 4-digit numbers can be formed. [2] (iii) Find how many 4-digit numbers that are divisible by 5 can be formed. [2] (b) Solve the equation ( ) ( ) C C n n 1 33 10 n n 1 12 10 # # + = - + . [3] * 0000800000008 * ,










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9 0606/22/M/J/25 © UCLES 2025 [Turn over 6 The volume, V, of a sphere is increasing at the constant rate of rcm s 2 3 1 - . Find the rate of change of the surface area, S, of this sphere when the volume of the sphere is rcm 36 3. [6] * 0000800000009 * ,










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10 0606/22/M/J/25 © UCLES 2025 7 The first three terms of an arithmetic progression can be written as , , ln ln ln x x x 2 5 2 3 2 7 ` ` ` j j j. (a) Given that x 1 2 , find the least number of terms for the sum of this progression to be greater than ln x 43 24 ` j. [6] * 0000800000010 * ,










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11 0606/22/M/J/25 © UCLES 2025 [Turn over (b) Given that the 25th term of this progression is equal to 408, find the exact value of x. [3] * 0000800000011 * ,









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12 0606/22/M/J/25 © UCLES 2025 8 y x B A O y x x 15 5 2 = - The diagram shows part of the curve y x 15 5 2 = - x . The curve meets the x-axis at the point A. The curve has a maximum at the point B. Find the area of the shaded region enclosed by the line AB and the curve. Give your answer in exact form. [11] * 0000800000012 * ,









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13 0606/22/M/J/25 © UCLES 2025 [Turn over Continuation of working space for Question 8. * 0000800000013 * ,









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14 0606/22/M/J/25 © UCLES 2025 9 (a) Solve the equation ° ° sec cosec for x x x 3 3 3 3 120 120 G G = - . [5] * 0000800000014 * ,









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15 0606/22/M/J/25 © UCLES 2025 [Turn over (b) Solve the equation r r r cos sin sin y y y 2 3 3 3 + + = + b b b l l l for r y 0 2 1 G . [5] Question 10 is printed on the next page. * 0000800000015 * ,









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16 0606/22/M/J/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. 10 The first three terms, in descending powers of x, in the expansion of x a x 3 1 1 n 2 2 2 - + ` e j o can be written as x x bx 729 972 12 10 8 + + , where a, b and n are constants. Find the values of a, b and n. [9] * 0000800000016 * ,









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This document consists of 18 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/22 Paper 2 May/June 2025 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 May/June 2025 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 May/June 2025 © Cambridge University Press & Assessment 2025 Page 2 of 18 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 May/June 2025 © Cambridge University Press & Assessment 2025 Page 3 of 18 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.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 4 of 18 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 Accuracy mark awarded two Accuracy mark awarded three Independent mark awarded zero Independent mark awarded one Independent mark awarded two Independent mark awarded three Benefit of the doubt Communication mark Incorrect Follow through Highlighter Highlight a key point in the working Ignore subsequent work Method mark awarded zero Method mark awarded one Method mark awarded two Method mark awarded three

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 18 Annotation Meaning Misread Omission 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. Premature rounding/approximation Special case Indicates that work/page has been seen Transcription error Correct Correct answer from incorrect working

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 6 of 18 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 May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 18 Question Answer Marks Partial marks 1 5 2*3 + x and 5 2* 3 + − x oe M1 where * could be = or any inequality sign Critical values: 1 5 , –1 soi A2 A1 for one correct value x ⩽ –1 x ⩾1 5 oe mark final answer A1 Alternative method ( ) 2 2 5 2 *3 + x (M1) where * could be = or any inequality sign Critical values: 1 5 , –1 soi (A2) A1 for  2 25 20 5 *0 + − x x x ⩽ –1 x ⩾1 5 oe mark final answer (A1) 2 (a) Fully correct graph with all intercepts stated O t s −   2 B1 for correct cubic shape with maximum in the first quadrant, a point of tangency to the t-axis at the minimum point; must meet the s-axis B1 for correct intercepts; must have attempted the correct shape 2(b) ( )( ) ( ) 2 : 2 4 1 4 (1) − − + − v t t t or 3 2 : 9 24 16 − + − s t t t and 2 : 3 18 24 − + v t t M1 : 3( 4)( 2) − − v t t A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 18 Question Answer Marks Partial marks 2(c) Fully correct graph with all intercepts stated O t v    B2 B1 for correct quadratic shape with minimum in the fourth quadrant; must meet the v-axis B1 for correct intercepts; must have attempted the correct shape 2(d) 6 18 − t or 3(t – 4) + (3t – 6) oe B1 FT their 3-term quadratic expression oe for v Correct linear expression for a from their v which must be either a 3-term quadratic or an expression that would simplify to a 3-term quadratic 2(e) Fully correct ruled graph with all intercepts stated O t a −  3 B1 for a single ruled line with positive gradient in fourth and first quadrants; must meet the a-axis and cross the t-axis B1 FT for a line with t-intercept 3 FT –thei r(–18) their 6 providing their a is a linear expression of the form mt + c B1 FT for a line with a-intercept –18 FT their (–18) providing their a is a linear expression of the form mt + c

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 18 Question Answer Marks Partial marks 3 3 2 1 2 4 + = + + x x soi OR ( ) 1 g f (1) − = x and 1 4 f ( ) 3 − = − x x x oe 2 M1 for   f( 2) = 1 + x soi or for ( ) 1 g f (1) − = x soi 2 = x nfww 2 M1 FT for correct equation without an algebraic fraction e.g. 2 2 + = x FT their 3 2 1 2 4 + = + + x x providing M1 previously awarded and fg(x) is an algebraic fraction Alternative method Solves 3 1 4 = + x x to find x = 2 and states ( ) g 2 = x (2) M1 for solving ( ) f 1 = x to find x = k and states ( ) g = x k , where k is a constant 2 = x nfww (2) M1 FT for 2 + = x k FT ( ) g = x k their providing M1 previously awarded

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 18 Question Answer Marks Partial marks 4(a) d d   =     y x ( ) 4 cos2 (2cos2 ) sin2 ( 2sin2 ) + − x x x x oe M1 Use of the product rule with correct structure ( ) d 4 cos2 (2cos2 ) sin 2 ( 2sin 2 ) d   = + −     y x x x x x oe, isw A1 2 2 π 6 d 2π 2π 4 2cos 2sin d 6 6 =      = −        x y x oe M1 dep on previous M1 FT their d d y x providing it is of the form 2 2 cos 2 sin 2 − a x b x or equivalent where a, b are positive integers π 6 d 4 d =     = −     x y x A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 18 Question Answer Marks Partial marks 4(b) π When 3 6   = =     x y B1 Gradient of normal = 1 4 M1 FT π 6 1 d from part d = − x y their x (a) Equation of normal: 1 π 3 4 24 = + − y x or 1 1 π and 3 4 4 6   = + = +     y x c c or 1 π 3 4 6   − = −     y x oe M1 FT π 6 1 d d = − x y their x and their y , providing their y  0 1 π 0 3 4 6   − = −     x M1 FT their equation of normal providing gradient is π 6 1 d d = − x y their x π 4 3, 0 6   −     or exact equivalent; mark final answer A1 dep on all previous marks awarded and a fully correct solution to (a) or a correct derivative and gradient of tangent at x = π 6 stated in this part 5(a)(i) 300 B1 5(a)(ii) 252 2 M1 for 21  4  3  1 oe OR Starts with 5 OR ends with 0: 1  4  3  5 oe or 60 or Starts with 2, 4, 6 or 8 OR ends with 2, 4, 6 or 8: 4  4  3  4 oe or 192 OR [All possible –] ends with 5: 4  4  3  1 oe or 48

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 12 of 18 Question Answer Marks Partial marks 5(a)(iii) 108 2 M1 for 9  4  3  1 oe OR Ends with 5: 4  4  3  1 oe or 48 or Ends with 0: 5  4  3  1 oe or 60 OR Starts 5, ends 0: 1  4  3  1 oe or 12 or Does not start 5: 4  4  3  2 oe or 96 5(b) Correct simplified equation: ( ) 2 1 33 12 11 + =   n oe, nfww B2 B1 for an equation with simplified algebraic component ( ) 2 1 + n oe or simplified numerical component33 12 11   oe, nfww 65 = n B1 dep on B2; must be the only solution

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 13 of 18 Question Answer Marks Partial marks 6 3 = r nfww B1 stated or very clearly implied 2 d 2π 4π d = r r t soi or 2 d 1 d 2 = r t r oe and d 8π d = S r r M1 Uses relevant chain rule e.g. d d d d d d =  V V r t r t to find the derivatives and values of d d r t and d d S r d 1 d 18 = r t and d 24π d = S r soi A1 Correct chain rule including d d S t B1 e.g. d d d d d d =  S S r t r t or d d d d d d =  r S r t t S oe d 4π d 3 = S t isw or 4.19 or 4.188 to 4.1888 2 M1 for d 1 24π d 18 =  S t Alternative method 1 3 = r nfww (B1) stated or very clearly implied 2 d 4π d = V r r and d 8π d = S r r (M1) Finds correct expressions for and values of d d V r and d d S r d 36π d = V r and d 24π d = S r soi (A1) Correct chain rule including d d S t (B1) e.g. d d d d d d d d =   S S r V t r V t or d d d d d d =  S S V t V t d 4π d 3 = S t isw or 4.19 or 4.188 to 4.1888 (2) M1 for d 1 24π 2π d 36π =   S t or 2 2π 3 

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 14 of 18 Question Answer Marks Partial marks 6 Alternative method 2 3 3 4π = V r nfww (B1) stated or very clearly implied 2 3 3 4π 4π   =       V S → 2 1 3 3 d 3 2 4π d 4π 3 −   =       S V V (M1) d 2 d 3 = S V oe soi (A1) Correct chain rule including d d S t (B1) e.g. d d d d d d =  S S V t V t d 4π d 3 = S t isw or 4.19 or 4.188 to 4.1888 (2) M1for d 2 2π d 3 =  S t Alternative method 3 3 = r nfww (B1) stated or very clearly implied 3 4 , π , 4π 3 4π   = =       S S r V ( ) 1 2 3 2 d 4 1 3 π d 3 2 4π =   V S S (M1) S = 36 and d 3 d 2 = V S oe soi (A1) Correct chain rule including d d S t (B1) e.g. d d d d d d =  S S V t V t d 4π d 3 = S t isw or 4.19 or 4.188 to 4.1888 (2) M1 for d 1 2π d 1.5 =  S t

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 15 of 18 Question Answer Marks Partial marks 7(a) Forms a correct equation from which the logarithms can be eliminated : has consistent powers of lnx in all terms ( ) ( ) 12ln 1 4ln *43 24ln 2 + −   n x n x x oe, soi OR uses log laws to combine terms ( ) 6 2 ( 1) 24 43 ln *ln −   n n n x x x or better 3 where * is = or any inequality sign M2 for a = 6 and d = 4 or a = 6lnx soi and d = 4lnx soi or a = lnx6 soi and d = 6 2 ln 3 x soi or a = 4 3 ln 2 x soi and d = lnx4 soi or for 12 4( 1) 24 43 2 2 ln ln *ln  −  + n n n x x x oe or M1 for any correct expression for d or for a correct expression for the sum to n terms using their a and their d providing their d is not one of the given terms  2 2 4 1032 *0 + − n n oe A1 Solves their 3-term quadratic in n M1 dep on attempt at the sum of an AP n = 22 cao, nfww A1 dep on all previous marks awarded

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 16 of 18 Question Answer Marks Partial marks 7(b) Correct term e.g. [u25 = ] 6ln 24 4ln +  x x or 6 6 2 ln 24 ln 3 +  x x or 4 4 3 ln 24ln 2 + x x or ( ) 3 7 2 2ln 24 2ln 5ln +  − x x x oe or ( ) 3 2 3 2ln 24 5ln 2ln +  − x x x oe B1 Forms and correctly simplifies an equation e.g. 102ln 408 = x or 6 17ln 408 = x or 4 51ln 408 2 = x or 6 336 240 ln 408  = x x x or 6 240 144 ln 408  = x x x M1 FT their u25 providing it is an expression for the term of an AP which requires the simplification/collection of two natural logarithms 4 e = x A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 17 of 18 Question Answer Marks Partial marks 8 For A: 1 3 = x B1 2 3 d 15 10 d = − + y x x x oe. isw OR 2 4 d 15 10 d − + = y x x x x oe, isw B2 B1 for each correct term OR for 2 4 d 15 ... d − + = y x x x oe or 4 ... 10 + x x Solves d 0 d = y their x as far as x = … M1 FT their d d y x providing at least one term is correct 2 45 : , 3 4       B oe, nfww A2 A1 for each correct coordinate Correct plan e.g. 2 3 1 2 3 15 5 1 45 2 1 d 2 4 3 3     − −   −          their their x their their their x x or 2 3 1 2 3 15 5 135 45 d 4 4     − − −          their their their x their x x x M1 2 15 5 5 d 15ln   − = +      x x x x x oe B2 B1 for 15lnx or klnx + 5 x oe 2 15 1 15ln 15ln 15 3 2 3     + − +         or exact equivalent M1 FT their 1 3 and their 2 3 providing at least B1 awarded for integration Area of shaded region: 15ln2 – 9.375 or 75 15ln2 8 − or exact equivalent A1 dep on all previous marks awarded 9(a) 1 tan3 3 = x or 3 3 or 0.57735… B2 B1 for 3 3 cos3 sin3 = x x oe 3x = 30 or 210 or –150 or –330 M1 Finds one correct and valid triple angle May be in radians for this mark implied by one correct value of x x: 10, 70, –50, –110 and no extras in range A2 A1 for any two correct angles ignoring extras

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 18 of 18 Question Answer Marks Partial marks 9(b) π sin 0 3   + =     y B1 π 1 cos 3 2   + =     y B1 π 3 + y = 0 or  or 2 or π 3 or 5π 3 M1 Finds one correct and valid compound angle May be in degrees for this mark y: 2π 5π , , 3 3 4π 0, 3 oe, nfww A2 A1 for any two correct angles ignoring extras 10 n = 6 stated; nfww B2 B1 for first term: ( ) 2 3 n x or ( ) 6 2 3x soi ( ) ( ) 5 2 10 10 6 3 2 729 972 − + = x a x x oe and ( ) ( ) ( ) ( ) 4 2 2 5 2 8 8 2 6 5 3 2 6 3 729 2  − +  − + = x a x a x bx x oe B3 B2 for one correct equation OR B1 for second term: ( )( ) 1 2 3 − − n n a x or ( )( ) 5 2 6 3 −a x or 10 1458 − ax soi B1 for third term: ( )( ) ( ) 2 2 2 1 3 2 − − − n n n a x or ( )( ) ( ) 4 2 2 6 5 3 2 −a x or 1215a2 x8 soi B1 for   2 4 2 1 1 + + x x soi 1 3 = a B2 B1 for 1458 1458 972 − + = a soi b = –108 B2 B1 for 1215a2 –2916a + 729 soi