Cambridge IGCSE Mathematics - Additional 0606 — 2025 May/June Paper 1 · Variant 3

This document has 16 pages. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/13 Paper 1 Non-calculator 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. ● 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 [ ]. * 0 4 1 6 2 0 7 4 2 2 * DC (SL/SG) 346682/2 © UCLES 2025 ,










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0606/13/M/J/25 © UCLES 2025 2 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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0606/13/M/J/25 © UCLES 2025 3 [Turn over Calculators must not be used in this paper. 1 (a) y x 10 8 6 4 2 12 0 – 8 – 6 – 4 – 2 – 1 1 2 – 2 The diagram shows the graph of ( )( )( ) y x a x b x c 2 = + + + where a, b and c are integers. Find values for a, b and c. [2] (b) Use the graph to find the values of x for which y 2 H . [3] * 0000800000003 * ,










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0606/13/M/J/25 © UCLES 2025 4 2 Find the values of k for which the equation x kx k 4 3 0 2 + + + = has two equal roots. [4] 3 The polynomial p is such that ( ) p x x Ax 30 3 = + + , where A is a constant. When ( ) p x is divided by x + 2 the remainder is 84. Write ( ) p x as a product of linear factors. [5] * 0000800000004 * ,










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0606/13/M/J/25 © UCLES 2025 5 [Turn over 4 Solve the equation ( ) log g lg l x 10 2 4 2 x - + = for x 0 2 . [5] * 0000800000005 * ,










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0606/13/M/J/25 © UCLES 2025 6 5 Solutions by accurate drawing will not be accepted. A circle, C, has equation ( ) ( ) x y 5 12 100 2 2 - + - = . (a) Find the equation of the tangent to C at the point (11, 4). Give your answer in the form ax by c + = , where a, b and c are integers. [4] (b) Show that C and the circle with equation x y 4 2 2 + = do not intersect. [2] * 0000800000006 * ,










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0606/13/M/J/25 © UCLES 2025 7 [Turn over 6 Find the x-coordinates of the points of intersection of the following curves. ln y x 4 = and ln y x 5 3 2 = - ` j Give your answers in exact form. [5] * 0000800000007 * ,










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0606/13/M/J/25 © UCLES 2025 8 7 In this question, all lengths are in centimetres and all angles are in radians. B O A i 6 The diagram shows a sector of a circle with centre O and radius 6. (a) It is given that the area of triangle AOB is cm 9 2. Find the value of sini. [2] (b) It is also given that the exact area of the shaded segment is ( )cm 15 9 2 r- . Find the exact length of the arc AB. [4] * 0000800000008 * ,










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0606/13/M/J/25 © UCLES 2025 9 [Turn over 8 0 1 2 y x y x x x 1 3 2 = + - - A B The diagram shows part of the curve y x x x 1 3 2 = + - - . The points A and B lie on the curve such that the x-coordinate of A is 1 and the x-coordinate of B is 2. (a) Find the y-coordinates of A and B. [1] (b) Show that the area of the shaded region enclosed by the line AB and the curve is ln a b 4 2 - , where a and b are integers to be found. [7] * 0000800000009 * ,










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0606/13/M/J/25 © UCLES 2025 10 9 The function f is defined by ( ) f x x x 2 9 10 – 2 = + - for x 0 3 G G . (a) (i) Write ( ) f x in the form ( ) a b x c 2 + + where a, b and c are constants. [3] (ii) Hence determine whether or not f 1 – exists. [2] * 0000800000010 * ,










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0606/13/M/J/25 © UCLES 2025 11 [Turn over (b) The function g is defined by ( ) ( ) g ln x x 3 5 2 = - for . x 0 2 5 1 G . (i) On the axes, sketch the graph of ( ) g y x = . State the exact values of the intercepts with the coordinate axes and the equation of any asymptote. [3] O y x (ii) Find an expression for ( ) g x 1 – . [2] (iii) Find the domain and range of g 1 – . Give each of your answers in exact form. [2] * 0000800000011 * ,









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0606/13/M/J/25 © UCLES 2025 12 10 O A B C b a The diagram shows four points, O, A, B and C. A, B and C lie in a straight line and are such that AC AB 3 1 = . OA a = and OB b = . (a) Find OC in terms of a and b. Simplify your answer. [3] * 0000800000012 * ,









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0606/13/M/J/25 © UCLES 2025 13 [Turn over (b) The line OA is extended to the point D such that OA : AD = 2 : 7. Point E lies on CD such that OE b m = . Find the value of m. [5] * 0000800000013 * ,









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0606/13/M/J/25 © UCLES 2025 14 11 A particle P moves in a straight line and passes through a fixed point O. At time t seconds, its displacement from O, s metres, is given by s t t t 6 2 3 = + - for t 0 3 G G s t t 12 3 1 3 2 = - - for t k 3 G G where k is a constant. It is given that, for t k 3 G G , the velocity of P is positive and its acceleration is negative. (a) The maximum velocity of P occurs when t 2 = . On the axes below, sketch a velocity–time graph for the first k seconds of the motion of P. [4] 0 v ms–1 t s k * 0000800000014 * ,









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0606/13/M/J/25 © UCLES 2025 15 [Turn over (b) The total distance travelled by P for t k 0 G G is 57 metres. Given that when t 3 = the distance and displacement of P from O are equal, find the value of k. [6] Question 12 is printed on the next page. * 0000800000015 * ,









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0606/13/M/J/25 © UCLES 2025 16 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. 12 The normal to the curve y x ax 4 7 2 = + + at the point where x 2 = has equation x y b 4 + = . Find the values of a and b. [6] * 0000800000016 * ,









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This document consists of 11 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/13 Paper 1 Non-calculator 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.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 2 of 11 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 May/June 2025 © Cambridge University Press & Assessment 2025 Page 3 of 11 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 May/June 2025 © Cambridge University Press & Assessment 2025 Page 4 of 11 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

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 11 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

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 6 of 11 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

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 11 Question Answer Marks Partial Marks 1(a) 1 a = B1 2, 1 b c = = − or 2, 1 c b = = − B1 1(b) Critical values: awrt –1.7, –1, awrt 1.2 B1 –1.7 ⩽ x ⩽ –1 B1 FT their reasonable attempts at –1.7 x ⩾ 1.2 B1 FT their reasonable attempts at 1.2 2 ( ) ( ) 2 4 4 3 k k − + soi M1 For correct use of discriminant 2 16 4 12 k k − − oe A1 ( )( ) 4 3 1 0 k k + − = oe M1 Dep on previous M mark for solving a 3 term quadratic equated to zero to obtain 2 values for k k = 3 ,1 4 − A1 Mark final answer 3 A = –31 B1 x – 1 or x – 5 or x + 6 B1 For finding a factor  ( ) 2 1 30 x x x − + −  ( ) 2 5 5 6 x x x − + −  ( ) 2 6 6 5 x x x + − + 2 M1 for a quadratic factor with 2 correct terms or for one further linear factor (x – 1)(x – 5)(x + 6) A1 4 2 2lg log 10 x x = B1 2 lg lg2 4 x x  =   +   soi M1 For use of power and subtraction rules 2 2 8 0 x x − −= A1 ( )( ) 4 2 0 x x − + = M1 Dep on previous M mark 4 x = only A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 11 Question Answer Marks Partial Marks 5(a) Centre ( ) 5, 12 soi B1 Gradient of radius = 12 4 4 5 11 3 −   = −   −   M1 Equation of normal: ( ) 3 4 11 4 y x − = − M1 Must be using their perpendicular gradient and the correct point 3 4 17 x y − = A1 5(b) Distance between centres = 13 Sum of radii = 12 M1 For both 13 12  so no intersection A1 6 ( ) ( ) 2 8 ln 10ln 3 0 x x − + = 2 M1 for elimination of y and re-arrangement to a 3 term quadratic in terms of ln x ln 0.75, ln 0.5 x x = = 2 Dep on previous M mark for correct method of solution 0.75 0.5 e , e x x = = A1 For both 7(a) 1 6 6 sin 9 2   = oe M1 sin = 1 2 A1 7(b) 1 (36)( sin ) 15π 9 2 their   − = − oe M1 5π 6 = A1 5π 2 M1 for 5π 6 6 their , but not π 6 8(a) 5 2 , 1 B1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 11 Question Answer Marks Partial Marks 8(b) Area of trapezium = 7 4 B1 FT on their y coordinates Area under curve = 2 1 3 2 1 d 1   −+   +    x x x B1 Condone omission of limits here ( ) 3ln 1 2ln x x x + − + 2 M1 for at least one correct log term ( ) ( ) 3ln3 2 2ln 2 3ln 2 1 − − − − M1 Dep on previous M mark 27 ln 1 2 − M1 Dep on previous M mark for simplification using log rules correctly 11 27 ln 4 2   −     A1 9(a)(i) 2 1 9 2 8 4 x   − −     3 B2 for 2 9 2 4 a x   − −     or B1 for 2 9 4 a b x   + −     B1 for a = 1 8 oe 9(a)(ii) Turning point 9 1 , 4 8       B1 FT on their a and c 1 f = does not exist as given domain includes the turning point. oe B1 Dep on previous B mark, turning point must be correct 9(b)(i) 3 B1 for correct shape ; may not be over correct domain but must have positive y-intercept and appear to tend to a vertical asymptote in the 4th quadrant B1 for (0, 3 ln 5) and (2, 0) stated or marked on graph; must have attempted correct shape B1 for asymptote at x = 2.5 O y x 3ln5 2 x = 2.5

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 11 Question Answer Marks Partial Marks 9(b)(ii) ( ) 1 3 1 g 5 e 2 x x −   = −     2 M1 for an attempt at the full method to find the inverse function with at most one sign or arithmetic slip 9(b)(iii) Domain: x ⩽ 3ln5 B1 Range: 0 ⩽ g–1 < 2.5 B1 10(a) AB = − b a B1 OC = a + 3(b – a) or b + 2(b – a) B1 FT on their − b a OC = 3b – 2a B1 10(b) 4.5 OD = a or 3.5 AD = a B1 Allow unsimplified ( ) 3 2 4.5 DC their their = − − b a a or ( ) 4.5 3 2 CD their their = − − a b a B1 FT Allow unsimplified ( ) 4.5 3 6.5 OE their their  = +  − a b a oe or ( ) ( ) 3 2 6.5 3 OE their their  = − +  − b a a b M1 9 13 = oe or 4 13 = oe A1 27 13 = A1 11(a) 2 1 12 3 v t t = + − B1 2 12 3 v t = − B1 2 B1 for correct shape for 0 ⩽ t ⩽ 3, with v-intercept of 1 and end point (3, 10) soi and maximum point B1 for a line with negative gradient for 3 ⩽ t ⩽ k, start point (3, 10) soi and end point above the t-axis

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 11 Question Answer Marks Partial Marks 11(b) When t = 3, s = 30 B1 1 2 (10 12 )( 3) 57 30 2 3 k k + − − = − oe or     2 12 3 30 57 30 3 k k − − − = − oe 2 M1 for attempt at area using t k = k = 6 or k = 30 2 A1 for   2 1 12 60 0 3k k − + − = or   2 36 180 0 k k − + = oe k = 6 A1 as v is negative when k = 30 12 3 d 8 d y a x x = − + oe B1 When d 2, 1 d y x a x = = − soi B1 FT on their d d y x 1 1 ( 1) 4 their a − = − − or their (a – 1) = 4 M1 FT on their ( )1 a − 5 a = A1 When x = 2, 2 4( 18) b their = + oe M1 Dep b = 74 A1