Cambridge IGCSE Mathematics - Additional 0606 — 2025 Oct/Nov Paper 1 · Variant 3

This document has 16 pages. Any blank pages are indicated. [Turn over * 3 0 4 3 3 5 5 5 0 1 * Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/13 Paper 1 Non-calculator 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. ● Calculators must not be used in this paper. ● You must show all necessary working clearly. INFORMATION ● The total mark for this paper is 80. ● The number of marks for each question or part question is shown in brackets [ ]. DC (DE/SW) 347513/2 © UCLES 2025 ,










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2 0606/13/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 0606/13/O/N/25 © UCLES 2025 [Turn over Calculators must not be used in this paper. 1 Solve the following inequalities. (a) x x 6 0 2 H - - [3] (b) x x 3 4 2 1 - + [4] * 0000800000003 * ,










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4 0606/13/O/N/25 © UCLES 2025 2 Differentiate x e x 2 3 with respect to x. [3] 3 In this question you may use the values in the table below. i radians sini cosi n ta i r 6 2 1 2 3 3 3 r 3 2 3 2 1 3 Variables x and y are related by the equation sin y x 5 = where r x 0 10 G G . Use calculus to find the approximate change in x when y increases from 2 3 by the small amount 0.01. [4] * 0000800000004 * ,










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5 0606/13/O/N/25 © UCLES 2025 [Turn over 4 t s 0 v m s–1 T 4 8 The velocity–time graph represents the motion of a particle moving in a straight line. The acceleration during the first T seconds of the motion is ms 2 2 - . The total distance travelled is 27 m. (a) Calculate T. [4] (b) Calculate the acceleration during the last 4 seconds of the motion. [2] * 0000800000005 * ,










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6 0606/13/O/N/25 © UCLES 2025 5 The normal to the curve y x x x 2 3 2 1 3 2 = + - + at the point where x 0 = cuts the curve again at two other points. Find the x-coordinates of these two points. [8] * 0000800000006 * ,










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7 0606/13/O/N/25 © UCLES 2025 [Turn over 6 (a) B A C M a The diagram shows an equilateral triangle ABC with side a. M is the midpoint of AC and angle AMB = 90°. Use the diagram to find ° sec30 . [3] (b) Show that sec sec x x 1 1 1 1 - + + can be written as cosec cot x x 2 . [4] * 0000800000007 * ,










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8 0606/13/O/N/25 © UCLES 2025 7 The point O is the origin. Two points P and Q are such that PQ is in the same direction as i j 5 - + . (a) The point R is such that OR is in the same direction as PQ and the magnitude of OR is 3 26. Find OR. [3] (b) OP is in the same direction as i j 2 3 - and i j OQ 10 6 = + . Find OP. [4] * 0000800000008 * ,










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9 0606/13/O/N/25 © UCLES 2025 [Turn over 8 x O y y = x4 – 4x2 + 8 y = 12 – x2 The diagram shows part of each of the curves y x 12 2 = - and y x x 4 8 4 2 = - + . Find the area of the shaded region enclosed by the two curves. [7] * 0000800000009 * ,










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10 0606/13/O/N/25 © UCLES 2025 9 Solve the following equations. (a) ( ) log log x x 5 2 2 1 5 25 - - = [5] * 0000800000010 * ,










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11 0606/13/O/N/25 © UCLES 2025 [Turn over (b) 4 5 e e e y y 3 7 3 3 1 + = - - [6] * 0000800000011 * ,









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12 0606/13/O/N/25 © UCLES 2025 10 (a) An arithmetic progression has first term a and common difference d. Given that S S 3 20 10 # = , find a in terms of d. [3] * 0000800000012 * ,









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13 0606/13/O/N/25 © UCLES 2025 [Turn over (b) A geometric progression, A, has common ratio r, where r 1 1 . The terms of this progression are a1, a2, a3 … . Another geometric progression, B, has terms b1, b2, b3 … , where b a 1 2 = , b a 2 4 = , b a 3 6 = … . The sum to infinity of A is A S and the sum to infinity of B is SB . Find A S SB in terms of r. Give your answer in its simplest form. [5] * 0000800000013 * ,









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14 0606/13/O/N/25 © UCLES 2025 11 The lines x 0 = , x 4 = , y 3 = and y 1 = - are tangents to a circle. (a) Find the equation of the circle. [3] The line y x a 2 = + , where a is a constant, is also a tangent to the circle. (b) Show that ( ) ( ) x a x a 5 4 2 1 0 2 2 + - + - = , and hence find the possible values of a. Give your answers in exact form. [4] * 0000800000014 * ,









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15 0606/13/O/N/25 © UCLES 2025 12 (a) Write down the coefficient of xr in the binomial expansion of ( )x 2 59 + . [1] (b) For this expansion, find the value of r for which the coefficient of xr is equal to the coefficient of xr 1 + . [4] * 0000800000015 * ,









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16 0606/13/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 0606/13 Paper 1 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.

0606/13 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.

0606/13 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.

0606/13 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

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 5 of 16 Annotation Meaning Method mark awarded three 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

0606/13 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) (x − 3)(x + 2) M1 x * 3, x * −2 A1 * can be any inequality sign or = x ⩾ 3, x ⩽ −2 A1 Do not allow and between inequalities Do not accept 2 3 − x 

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 16 Question Answer Marks Guidance 1(b) x * 3 B1 * can be any inequality sign or = −3x + 4 < x + 2 or 3x − 4 > − x − 2 M1 x * 1 2 A1 * can be any inequality sign or = 1 3 2 x   A1 Do not accept 1 , 3 2 x x  unless and is between the inequalities Alternative method: 9x2 − 24x + 16 * x2 + 4x + 4 (M1) Square both sides * can be any inequality sign or = (2x − 1)(x − 3) * 0 (M1) * can be any inequality sign or = Critical values 1 , 3 2 (A1) 1 2 < x < 3 (A1) Do not accept 1 , 3 2 x x   unless ‘and’ is between the inequalities 2 3 2 3 2 e 3 e x x x x + isw 3 B1 for 3e3x M1 for use of product rule A1 FT their 3e3x 3 π 15 x = B1 Allow for π 5 3 x = d d y x = 5cos 5x 5 2   =     B1 For differentiation 0.01 5 2 y their x x    = = oe DM1 Dep on previous B1 FT their 5 2 Allow if 0.01 5cos5 x x  = δx = 0.004 or 1 250 A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 16 Question Answer Marks Guidance 4(a) Velocity at time T is 2T soi B1 ( ) 1 8 4 2 27 2 T T + −  = M1 For an attempt trapezium area or three areas and = 27 using their 2T (T − 3)(T − 9) = 0 M1 Dep on previous M1 for solving their 3-term quadratic by factorising, completing the square or quadratic formula 3 A1 Alternative method Velocity at time T is 2T leading to 2 v T = soi (B1) 1 8 4 27 2 2 v v   + −  =     (M1) For an attempt trapezium area = 27 using their 2 v (18 )( 6) 0 v v − − = (M1) Dep on previous M1 for solving their 3-term quadratic by factorising, completing the square or quadratic formula 6 then 3 v T = = (A1) 4(b) 0 2 oe 8 4 their T a − = − M1 For an attempt on gradient using their T −1.5 A1 Must be negative

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 16 Question Answer Marks Guidance 5 d d y x = 3x2 + 3x – 2 M1 Allow one error in the coefficient of x2or x Gradient of tangent at A = −2 nfww A1 Gradient of normal at A 1 2 = M1 Dep for an attempt −1 ÷ their gradient – must come from differentiation Equation of normal is y − 1 = 1 2 x oe M1 Dep for the normal equation using their normal gradient and ( ) 0, 1 3 2 1 3 1 2 1 2 2 x x x x + = + − + M1 Dep For solving simultaneously curve with normal FT their normal equation x(2x2 + 3x − 5) = 0 oe nfww A1 [x](x − 1)(2x + 5) = 0 M1 For an attempt solve their 3-term quadratic or trying to solve their cubic using long division x = 1 and 5 2 x = − nfww A1 6(a) 2 3 or 2 3 3 3 B2 for   3 2 a BM = or 2 2 1 sec 30 1 3   = +     or 2 2 2 cos 30 1 a a     = −      oe B1 for 2 2 2 4 a BM a   = −   soi

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 16 Question Answer Marks Guidance 6(b) ( )( ) 2sec sec 1 sec 1 x x x − + soi B1 For adding the two fractions ( ) 2 2sec sec 1 x x − B1 Dep on previous B1 for expanding and simplifying the denominator 2 2sec tan x x or 2 2 tan cos x x B1 Dep on previous B1 for use of 2 2 sec 1 tan x x = + 2sec cot tan x x x  or 2 2cos sin x x oe or 2 tan sin x x leading to 2cosec x cot x AG B1 Dep on previous B1 with at least one more correct step to get to the given answer Alternative method 1 1 1 1 1 1 cos cos x x + − + leading to cos cos 1 cos 1 cos x x x x + − + (B1) = 2 2 2 cos cos cos cos 1 cos x x x x x + + − − (B1) Dep on previous B1 for adding the two fractions = 2 2cos sin x x (B1) Dep on previous B1 For use of 2 2 sin 1 cos x x = − = 2cosec x × cos sin x x oe leading to = 2cosec x cot x AG (B1) Dep on previous B1 with at least one more correct step to get to the given answer 7(a) OR   =    k(−i + 5j) or 1 5 k −       B1 soi 2 2 1 5 26 + = B1 For finding the magnitude of PQ  allow unsimplified −3i + 15j or 3 15 −       B1 Allow 3(−i + 5j) or 1 3 5 −       Do notisw

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 16 Question Answer Marks Guidance 7(b) a(2i − 3j) + b(−i + 5j) = 10i +6j oe B1 2a − b = 10 oe, −3a +5b = 6 oe or for 10 2 1 6 3 5 a a − − = + or 10 2 6 5 3 b b + − = − M1 Dep on previous B1 for equating like vectors a = 8 or b = 6 A1 16i − 24j oe A1 Allow 8(2i − 3j) or 2 8 3     −   8 −x4 + 3x2 + 4 (= 0) M1 For attempt to solve simultaneously Allow one term error (x2 − 4)(x2 + 1) = 0 oe M1 Dep for an attempt to solve their 3-term quadratic equation in terms of x2 x = 2 and −2 only A1 4 2 ( 3 4)d x x x + + −  M1 For an attempt to integrate their quadratic−quartic or the other way around with one term correctly integrated Allow without limits 2 5 3 2 4 5 x x x −   − + +     A1 All correct with correct limits of 2 and -2 or with limits of 0 and 2 or -2 and 0 then double the area of the integral −32 5 + 8 + 8 − ( 32 5 − 8 − 8) M1 Dep for correct use of their limits into their integral substituted correctly If their integration – must see substitution of limits to award the mark = 96 5 or 19.2 A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 16 Question Answer Marks Guidance 9(a) ( ) 2 5 5 log 5 2 log 1 oe x x − − = 2 B1 for 5 5 25 5 log log log log 25 2 x x x = = B1 for ( ) ( ) 2 5 5 2log 5 2 log 5 2 x x − = − oe ( ) 2 5 5 2 log 1 x x − = M1 Dep on one of the previous B1 For correct use of rules of logarithms e.g. dividing their logarithms to the same base Accept 5 log 5 for 1 25x2 − 25x + 4 = 0 oe M1 Dep M1 for forming a 3-term quadratic equation, with an attempt to solve must have at least 2 B marks x = 4 5 only A1 Alternative method 1 ( ) 2 25 25 1 log 5 2 log 2 x x − − = (2) B1 for 25 25 5 25 log (5 2) log (5 2) log (5 2) 1 log 5 2 x x x − − − = = oe B1 for 25 1 log 5 2 = or for 25 1 1 log 25 2 2 = soi ( ) 2 25 5 2 1 log 2 x x − = or ( ) 25 25 1 log 5 2 log 5 2 x x − = (M1) Dep on one of the previous B1 For correct use of rules of logarithms e.g. dividing their logarithms to the same base or multiplying Accept 25 log 5 for 1 2 25x2 − 25x + 4 = 0 oe (M1) Dep M1 for forming a 3-term quadratic equation, with an attempt to solve must have at least 2 B marks x = 4 5 only (A1)

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 13 of 16 Question Answer Marks Guidance 9(a) Alternative method 2 ( ) 5 5 1 1 log 5 2 log 2 2 x x − − = Leading to ( ) 5 5 5 2 log log 5 x x − = (2) B1 for 25 5 1 log log 2 x x = soi B1 for 5 1 log 5 2 = soi ( ) 2 5 2 5 x x − = or ( ) 5 2 5 x x − = (M1) Dep on one of the previous B1 For correct use of rules of logarithms e.g. dividing their logarithms to the same base Accept 25 log 5 for 1 2 25x2 − 25x + 4 = 0 or 5 5 2 0 x x − − = (M1) Dep M1 for forming a 3-term quadratic equation, with an attempt to solve must have at least 2 B marks x = 4 5 only (A1) Alternative method 3 log(5 2) log 1 log5 2log5 2 x x − − = Leading to ( ) 2 5 2 log log5 x x − = (2) B1 for 5 log(5 2) log (5 2) log5 x x − − = soi B1 for 25 log log 2log5 x x = soi ( ) 2 5 2 5 x x − = (M1) Dep on one of the previous B1 For correct use of rules of logarithms e.g. dividing their logarithms to the same base 25x2 − 25x + 4 = 0 (M1) Dep M1 for forming a 3-term quadratic equation, with an attempt to solve must have at least 2 B marks x = 4 5 only (A1)

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 14 of 16 Question Answer Marks Guidance 9(b) 3 7 3 1 3 3 1 e 4e 5 y y y − + − − + − + = oe or 3 4 3 1 3 1 3 e 4e 5e y y y − + − − + = or 3 7 1 3 3 4 e 5 e e y y − − + = or 6 4 3 8 e 4e 5 e y y + + = or 3 7 3 3 e 4 5e e e e y y + = or 3 1 6 3 3 1 e 4 5 e e e y y − − + = B1   6 8 3 4 e 4e 5 0 y y − − + − = or   3 4 4 3 e 4 5 e 0 y y − − + − = or ( )   2 7 3 3 3 e e 4e e 5 e 0 y y − − − + = or ( )   2 3 1 3 3 1 6 e 4e e 5 e 0 y y − −− + = B1 Dep for arranging as 3-term quadratic ( )( ) 3 4 3 4 e 1 e 5 0 y y − − − + = or ( ) ( ) ( ) ( ) 3 3 4 3 e e e e e 5 0 y y − − − + = or (e3y −1 + 5e3)( e3y −1 - e3) = 0 M1 Dep for an attempt at solution 3 4 e 1 y− = or 3 1 3 e e y −= A1 3y − 4 = ln 1 oe soi or 3y – 1 = 3 A1 4 3 y = A1 10(a) ( ) ( ) 20 10 2 19 3 2 9 2 2 a d a d + =  + oe M1 20a + 190d = 30a + 135d oe A1 Correct removal of brackets a = 11 2 d oe A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 15 of 16 Question Answer Marks Guidance 10(b) SA = 1 1 a r − soi B1 Allow a for 1a rb = r2 soi B1 SB = 2 2 1 a r     −   = 1 2 1 a r r − B1 Allow a for 1a with r2 k = B A S S = 1 2 1 a r their r − × 1 1 r a − 1 2 1 1 1 B A a r S their r a S r − = − M1 Dep on the first and second B1 for their B A S S Allow a for 1a 1 r r + cao A1 11(a) Centre (2, 1) B1 Accept x = 2 and y = 1 Radius 2 soi B1 ( ) ( ) 2 2 2 1 4 x y − + − = oe B1 FT their centre and radius ISW Do not accept 22 for 4 11(b) ( ) ( ) 2 2 2 2 1 4 x x a − + + − = M1 Substitute y = 2x + a into their circle equation ( ) 2 2 4 4 4 4 1 x x x a x − + + + − + ( ) 2 1 4 a − = A1 Equivalent must be seen as AG leading to ( ) ( ) 2 2 5 4 2 1 0 x a x a + − + − = AG ( ) ( ) 2 2 1 6 2 20 1 0 a a − − − = M1 for finding the discriminant of the given equation and equating to zero 2 6 11 0 a a + − = A1 isw from a correct answer a = −3 ± 20 or 6 80 2 − oe isw 12(a) 59 59 2 r r r C x −   oe B1 Allow for 59 59 2 r r −       or ( ) 59! 59 ! ! r r − 259 − r

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 16 of 16 Question Answer Marks Guidance 12(b) ( ) 59 59! 2 59 ! ! r r r − − = ( ) ( ) 58 59! 2 58 ! 1 ! r r r − − + oe B1 Allow if in terms of n and r e.g.: ( ) ! 2 ! ! n r n n r r − − = ( ) ( ) 1 ! 2 ( 1) !( 1)! n r n n r r − + − + + 2 59 r − = 1 1 r + B2 Dep on previous B1 for correctly obtaining either 2 59 r − or 1 1 r + r = 19 B1 Dep on previous B1 marks