Cambridge IGCSE Mathematics - Additional 0606 — 2025 Oct/Nov Paper 2 · Variant 2

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/22 Paper 2 October/November 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 [ ]. * 8 3 7 7 5 4 4 0 9 9 * DC (CJ/SW) 347168/3 © UCLES 2025 ,










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2 0606/22/O/N/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/O/N/25 © UCLES 2025 [Turn over 1 The line y x 4 3 = - meets the curve y x x 3 5 2 2 = + - at the points A and B. (a) Find the coordinates of A and B. [4] (b) The perpendicular bisector of the line AB cuts the coordinate axes at the points P and Q. Given that O is the origin, find the area of the triangle POQ. [5] * 0000800000003 * ,










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4 0606/22/O/N/25 © UCLES 2025 2 A D B C O The diagram shows the shaded region ABCD. The lines AC and BD each have a length of 12 cm. The lines AC and BD bisect each other at the point O. The lines AD and BC are parallel and each have a length of 4 cm. The arcs AB and DC are part of a circle centre O. (a) Find the obtuse angle AOB. Give your answer in radians. [3] Use your answer to part (a) to find (b) (i) the perimeter of the shaded region [2] (ii) the area of the shaded region. [3] * 0000800000004 * ,










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5 0606/22/O/N/25 © UCLES 2025 [Turn over 3 Find the exact value of the term independent of x in the expansion of x x 2 3 1 4 2 10 2 2 + - e ` o j . [6] * 0000800000005 * ,










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6 0606/22/O/N/25 © UCLES 2025 4 Variables x and y are such that when ey is plotted against x3, a straight-line graph is obtained. This line passes through the points (1, 13.5) and (7.5, 0.5). (a) Find y in terms of x. [4] (b) Find the values of x for which your equation is valid. [2] * 0000800000006 * ,










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7 0606/22/O/N/25 © UCLES 2025 [Turn over 5 A 6-character password is to be formed from the following characters. Letters b f g k m Numbers 3 5 7 9 Symbols * ! @ Each character can be used at most once in any 6-character password. (a) Find the number of 6-character passwords that can be formed if there are no further restrictions. [1] (b) Find the number of 6-character passwords that can be formed if the password starts and ends with a symbol. [2] (c) Find the number of 6-character passwords that can be formed if the password: • starts with either a symbol and then a number, or a number and then a symbol and • ends with 2 letters. [2] * 0000800000007 * ,










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8 0606/22/O/N/25 © UCLES 2025 6 In this question lengths are in centimetres and time, t, is in seconds. A particle P is moving in a straight line with a speed of 26 in the direction of the vector 5 12 - e o. (a) Find the velocity vector of P. [2] When t 0 = , P passes through a point A which has position vector 3 6 e o. (b) Write down the position vector of P at time t. [2] At the same time that P passes through A, a particle Q passes through a point B. The position vector of Q at time t is given by t t 8 5 2 25 - - e o. The distance between P and Q at time t is d. (c) Show that d mt nt r 2 2 = + + , where m, n and r are integers to be found. [3] (d) Hence show that P and Q do not collide. [1] * 0000800000008 * ,










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9 0606/22/O/N/25 © UCLES 2025 [Turn over 7 (a) Given that cos y x x 2 = , find x y d d . [2] (b) Hence find sin x x x 2 d y . [4] * 0000800000009 * ,










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10 0606/22/O/N/25 © UCLES 2025 8 An arithmetic progression has first term t and common difference 1.5. The 4th, 8th and 20th terms of this arithmetic progression form the 1st, 2nd and 3rd terms of a geometric progression. (a) Find the value of t. [5] (b) Find the common ratio of the geometric progression. [2] * 0000800000010 * ,










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11 0606/22/O/N/25 © UCLES 2025 [Turn over 9 It is given that ( ) ( ) f ln x x 2 5 = + for x a 2 , where a is a constant. (a) Write down the least possible value of a. [1] (b) Using your value of a, write down the range of f. [1] It is also given that ( )x x 1 g 2 = + for x R ! . (c) Using your value of a, solve the equation ( )x 4 fg = . Give your answers in exact form. [3] * 0000800000011 * ,









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12 0606/22/O/N/25 © UCLES 2025 10 y x O y = 2 + 5e x y = 4 – 3e 2x The diagram shows parts of the graphs of y 2 5ex = + and y 4 3e x 2 = - . Find the area of the shaded region. Give your answer in the form ln a b 3 + , where a and b are exact constants. [10] * 0000800000012 * ,









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13 0606/22/O/N/25 © UCLES 2025 [Turn over Additional working space for Question 10. * 0000800000013 * ,









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14 0606/22/O/N/25 © UCLES 2025 11 (a) Solve the equation tan tan x x 2 4 2 0 2 - = for ° ° x 0 180 G G . [4] * 0000800000014 * ,









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15 0606/22/O/N/25 © UCLES 2025 (b) Solve the equation ( . ) cosec y 1 2 4 + = , where y is in radians and y 5 2 1 1 - . [6] * 0000800000015 * ,









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16 0606/22/O/N/25 © UCLES 2025 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. BLANK PAGE * 0000800000016 * ,









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This document consists of 16 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/22 Paper 2 October/November 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 October/November 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 October/November 2025 © Cambridge University Press & Assessment 2025 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 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 2025 © Cambridge University Press & Assessment 2025 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, 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 October/November 2025 © Cambridge University Press & Assessment 2025 Page 4 of 16 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 October/November 2025 © Cambridge University Press & Assessment 2025 Page 5 of 16 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 October/November 2025 © Cambridge University Press & Assessment 2025 Page 6 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 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 Question Answer Marks Guidance 1(a) Eliminates one unknown: e.g. 2 4 3 3 5 2 − = + − x x x M1 Writes in solvable form: 2x2 – x – 6 [ = 0] or – 2x2 + x + 6 [ = 0] A1 Solves or factorises their 3-term quadratic e.g. (2x + 3)(x – 2) M1 dep previous M1 ( ) 1.5, 9 − − ( ) 2, 5 A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 16 Question Answer Marks Guidance 1(b) Equation of their perpendicular bisector soi: 1 1 2 4 4   + = − −     y x or y = 1 4 − x 31 16 − oe, isw M3 FT their midpoint and perpendicular gradient from correct use of their ( ) 1.5, 9 − − and their ( ) 2, 5 M1 FT for midpoint: ( 1.5) 2 ( 9) 5 , 2 2 − + − +       their their their their or ( ) 0.25, 2 − M1 FT for perpendicular gradient: 1 1 oe or ( 9) 5 4 ( 1.5) 2 ⊥ − = − − − − − m their their their their OR M1 FT their( ) 1.5, 9 − − and their ( ) 2, 5 for (x –2)2 + (y – 5)2 = (x +1.5)2 + (y + 9)2 soi M1 FT their( ) 1.5, 9 − − and their ( ) 2, 5 for expanding 2 2 2 2 4 10 29 3 18 83.25 + − − + = + + + + x y x y x y x y soi 961 128 isw or 7.51 or 7.507 to 7.508 oe cao A2 not from wrong coordinates A1 for sight or use of e.g. 31,0 4   −     and 31 0, 16   −     nfww or x = 31 4 − and y = 31 16 − nfww or x = 31 4 and y = 31 16 nfww

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 16 Question Answer Marks Guidance 2(a) Complete and correct method e.g.: 1 1 π 2sin 3 −  −     or 1 7 π cos 9 −  −     or 3.142 0.6796... − oe or 1 2 2cos 6 −      or 2  1.2309…oe or 1 36 36 128 cos 2(6)(6) −  + −     or 1 7 cos 9 −  −     oe M2 M1 for a correct useful angle or implicit calculation for angle AOB e.g. angle = BOC 1 1 2sin 3 −      or 0.6796… or 1 36 36 16 cos 2(6)(6) −  + −     or 1 7 cos 9 −      or half angle BOC = 1 1 sin 3 −      or 0.339… or half angle = AOB 1 2 cos 6 −      or 1 4 2 sin 6 −        or 1.230… or128 36 36 2(6)(6)cos = + − AOB oe 2.46 or 2.461 to 2.462[…] A1 2(b)(i) 6 2.46 their soi M1 FT their 2.46 providing it is less than 2 37.5 or 37.52 to 37.55 A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 16 Question Answer Marks Guidance 2(b)(ii) 2 1 6 2.46 2  their or ( ) 2 1 2 6 2.46 2          their OR 2 1 6 0.68[0] 2  their or ( ) 2 1 2 6 0.68[0] 2          their M1 FT their 2.46 providing it is less than 2 Correct plan e.g. : 2 sectors + 2 triangles ( ) ( ) 2 2 1 1 2 6 2.46 2 6 sin 2 2        +            their theirBOC or ( ) ( ) 2 2 1 1 2 6 2.46 2 6 sin 2.46 2 2        +            their their or ( ) 2 1 1 2 6 2.46 4 2 2       +          their theirAB oe or 2 segments + 1 rectangle ( ) 2 1 2 6 2.46 sin 2.46 4 2     − +      their their theirAB oe OR whole circle radius 6 – 2 segments cut off by AD and BC   62 – ( ) 2 1 2 6 sin 2     −     theirBOC theirBOC M1 dep previous M1 111 or 111.2 to 111.3 A1 3 2 10 9 8 2 2 3 3 2 10 2 45 2   +   +      x x oe, soi and 2 4 1 8 16 − + x x soi B3 B2 for 2 10 9 8 2 2 3 3 2 10 2 45 2   +   +      x x oe, soi or B1 for two correct terms in 2 10 9 8 2 2 3 3 2 10 2 45 2   +   +      x x oe, soi B1 for 2 4 1 8 16 − + x x ( ) 10 2 15360 8 103680 16 their their their their +  − +  M2 FT their expansions for M2 or M1 M1 FT for any two correct terms in this sum 1 537 024 A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 16 Question Answer Marks Guidance 4(a) 3 e = + y mx c soi B1 Correct method to find m e.g. 13.5 0.5 1 7.5 − − or 13 = – 6.5m OR m = –2 soi M1 Correct method to find c e.g. 13.5 = their(–2) [1] + c oe or 0.5 = their(–2) (7.5) + c oe or 0.5 = 7.5(13.5 – c ) + c oe OR c = 15.5 soi M1 ( ) 3 ln 15.5 2 = − y x mark final answer A1 4(b) 3 7.75  x oe, nfww 2 M1 FT for 3 15.5 2 0 −  x oe, soi; FT 3 + a bx > 0 from ( ) 3 ln = + y a bx 5(a) 665 280 B1 5(b) 30 240 2 M1 for 3  10  9  8  7  2 oe 5(c) 26 880 2 B1 for a final answer of 13 440 or M1 for 2  3  4  8  7  5  4 oe 6 All through this question allow correctly written vectors in i j form but, incorrectly written column vectors e.g. 10 24     −   i j do not earn full marks unless the correct form is recovered 6(a) 10 24     −   oe, isw 2 B1 for the magnitude of the direction vector is 13 soi or M1 for 2 2 5 ( 12) + − oe or M1 for 26cos1.18 26sin1.18     −   oe 6(b) 3 10 6 24 t    +    −    oe, isw 2 M1 FT for 3 10 6 24 t their      +      −      FT their velocity vector

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 16 Question Answer Marks Guidance 6(c) 3 10 8 5 6 24 2 25 + −     −     − −     t t t t oe or 8 2 4 +     +   t t oe, soi B1 or 8 5 3 10 2 25 6 24 − +     −     − −     t t t t or 8 2 4 −−     −−   t t oe soi Finds d 2 or d for 3 10 8 5 6 24 2 25   + −     −       − −       t t their t t or 8 5 3 10 2 25 6 24   − +     −      − −       t t their t t oe, soi M1 FT 3 10 6 24 t their      +      −      2 2 5 40 80 = + + d t t mark final answer A1 6(d) Correct argument including d e.g. When 0 4 = → = − d t , but t ⩾ 0 [so no collision] or When 4 0 = −→ = t d , but t ⩾ 0 [so no collision] or d > 0 when t ⩾ 0 so d  0 [so no collision] B1 dep on part (c) being correct 7(a) [1] ( ) cos2 2sin 2 + − x x x oe, isw, nfww 2 M1 for ( ) d cos2 d = x x 2sin2 − x soi or M1 FT for correct structure of product rule FT their( ) 2sin 2 − x

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 16 Question Answer Marks Guidance 7(b) 1 1 cos2 + sin2 + 2 4 − x x x c oe, isw, nfww 4 M3 for 1 1 cos2 sin2 2 4 − + x x x or for 1 2 sin2 d cos2 sin2 2 − = − x x x x x x +c or for 1 cos2 sin2 2 − + + x x k x c , where k > 0 oe, or M2 for 1 2 sin2 d cos2 sin2 2 − = − x x x x x x or for 1 cos2 sin2 2 − + x x k x , where k > 0, or for 1 1 sin2 d cos2 cos2 d 2 2 = − +   x x x x x x x oe, or M1 for 2 sin2 d cos2 cos2 d − = −   x x x x x x x or 2 sin2 d cos2 cos2 d = − +   x x x x x x x or cos2 cos2 d 2 sin2 d = + −   x x x x x x x oe, soi

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 13 of 16 Question Answer Marks Guidance 8(a) At least two correct terms from: 4.5 10.5 28.5 + + + t t t soi B1 condone a for t r = 10.5 4.5 + + t t or r = 28.5 10.5 + + t t or r2 = 28.5 4.5 + + t t oe M1 FT their distinct terms of the form t + 1.5k where k is a positive integer 10.5 28.5 4.5 10.5 + + = + + t t t t oe A1 or( ) 28.5 4.5 10.5 10.5 + +  = + + t t t t oe Clears algebraic fractions and expands e.g. 2 2 21 110.25 33 128.25 + + = + + t t t t A1 must be seen –1.5 oe nfww A1 and no other values dep on all previous marks awarded Alternative method At least two correct terms from: 4.5 10.5 28.5 + + + t t t soi (B1) One correct equation in a and r e.g. a –4.5 = ar – 10.5 or ar – 10.5 = ar2 – 28.5 or a – 4.5 = ar2 – 28.5 oe (M1) FT their distinct terms of the form t + 1.5k where k is a positive integer Correct equation in r e.g. 6 18 1 ( 1) = − − r r r OR 2 2 18 18 4.5 10.5 − = − − − r r r r r (A1) OR correct equation in a e.g. 2 6 24 +   + =     a a a a   6 18 = r r = 3 OR 2 2 18 4.5( ) 18 10.5( ) − − = − − r r r r r and e.g. 2 18 10.5 = − − r t r r or a = 3 (A1) must be seen OR 2 2 24 12 36 + = + + a a a a and a = 3 –1.5 oe nfww (A1) and no other values dep on all previous marks awarded 8(b) 3 nfww 2 and no other values M1 for ( 1.5) 10.5 ( 1.5) 4.5 − + − + their their their their oe or for 10.5 4.5 + − their a their their their a oe or for 28.5 10.5 1 ( 1.5) 10.5 − + − + their their their their oe FT their t providing their t  0 or their a providing their a  0 or 1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 14 of 16 Question Answer Marks Guidance 9(a) 5 2 − oe B1 9(b) f  ℝ oe B1 9(c) ( ) ( )   2 ln 2 1 5 4 + + = x oe, soi B1 ( ) 2 4 2 1 5 e + + = x or their (2x2 + 7) = e4 M1 4e 7 2 − =  x or exact equivalent A1 Alternative method Solves ln(2 5) 4 + = x to find x = 4e 5 2 − and states ( ) 4e 5 g 2 − = x OR finds 1 e 5 f ( ) 2 − − = x x and states ( ) 4e 5 g 2 − = x (2) M1 for solving ( ) f 4 = x to find x = k and states ( ) g = x k , where k is a constant OR for stating ( ) 1 g f (4) − = x 4 2 e 5 1 2 − + = x → 4e 5 1 2 − = −  x or exact equivalent (A1)

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 15 of 16 Question Answer Marks Guidance 10 Writes as a 3-term quadratic in ex in solvable form:   2 3e 5e 2 0 + − = x x soi M1 condone one sign or arithmetic error Solve or factorises their 3-term quadratic in ex M1 dep previous M1 1 e 3 = x or ex = 0.333[3...] e 2   = −   x A1 1 ln 3 = x or ln3 − only solution, nfww A1 must be exact value, but ignore decimal if also stated Correct plan: ( ) 0 2 1 ln3 3e 5e 2 d + −  x x their x OR ( ) ( ) 0 0 2 1 1 ln ln 3 3 2 5e d 4 3e d + − −   x x their their x x soi M1 dep on use of 2 + 5ex = 4 – 3e2x to find the lower limit 2 3 e 5e 2 2 + − x x x OR ( ) 2 3 2 5e and 4 e 2   + −     x x x x B2 B1 for 2 2 3 3e d e 2   =    x x x soi B1 for all other terms correct 1 1 2ln ln 3 3 3 3 1 5 e 5e 2ln 2 2 3   + − + −       OR 1 ln3 1 5 2ln 5e 3   − +       OR 1 2ln3 3 1 3 4ln e 2 3 2   − − −       oe M1 FT correct substitution of limits 0 and k where k is exact into ae2x + bex + cx or aex + bx or ae2x + bx at least once, providing limits are consistent 14 2ln3 3 − dep on all previous marks awarded 2 M1 dep on correct plan for one correct term out of two or for 1 2ln 3 + a with constant a 11(a)   tan 2 0 = → x Any two of 2 0, 180, 360 = x M1 [x = ] 0, 90, 180 nfww and no extras from tan 2x = 0 in range 0 ⩽ x ⩽180 A1   tan 2 4 = → x 2 = 75.96... or 255.96... x M1 38 [.0] or 37.98[18…] isw 128 [.0] or 127.98[18…] isw and no extras from tan 2x = 4 in range 0 ⩽ x ⩽180 A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 16 of 16 Question Answer Marks Guidance 11(b) ( ) 1 sin 1.2 4 + = y B1 Two correct values for y + 1.2 soi y + 1.2 = 0.2526[80…] y + 1.2 =  – 0.2526[80…] or 2.888[91…] y + 1.2 = – – 0.2526[80…] or –3.394[27…] soi M2 M1 for any one correct value for y + 1.2 –0.947 or –0.9473[19…] isw 1.69 or 1.688[91…] isw –4.59 or –4.594[27…] isw and no others in range –5 < y < 2 A3 A2 for two correct angles, ignoring extras in range or A1 for one correct angle, ignoring extras in range