Cambridge IGCSE Mathematics - Additional 0606 — 2025 May/June Paper 1 · Variant 2

This document has 16 pages. [Turn over * 1 3 0 5 7 2 4 1 8 4 * Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 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 [ ]. DC (DE/SW) 346668/1 © UCLES 2025 ,










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2 0606/12/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 0606/12/M/J/25 © UCLES 2025 [Turn over Calculators must not be used in this paper. 1 y x – 0.5 0 2 3 – 2 The diagram shows the graph of ( ) y x f = , where f is a cubic polynomial. Find expressions for the two possible functions ( )x f . Write each expression in fully factorised form. [3] * 0000800000003 * ,










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4 0606/12/M/J/25 © UCLES 2025 2 Solve the equation x x 1 6 3 1 3 1 + = . [4] * 0000800000004 * ,










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5 0606/12/M/J/25 © UCLES 2025 [Turn over 3 A circle with centre C has the equation x y x y 10 4 24 0 2 2 + - - + = . (a) Show that the line y x 2 3 = - is a tangent to this circle. [3] (b) Given that this tangent touches the circle at the point P, find the coordinates of P. [2] (c) Find the equation of the circle which has its centre at P and passes through the origin. [3] * 0000800000005 * ,










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6 0606/12/M/J/25 © UCLES 2025 4 (a) Find sin d r 0 i i y . [2] (b) Given that r 0 2 1 1 a , show that cot tan sec a a a + can be written as sina. [3] * 0000800000006 * ,










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7 0606/12/M/J/25 © UCLES 2025 [Turn over 5 The polynomial p is such that ( )x x x ax b 3 7 = - + + p 3 2 , where a and b are integers. It is given that ( ) 1 21 p - = l and that x 2 - is a factor of ( )x p . (a) Find the values of a and b. [4] (b) Hence write ( )x p as a product of linear factors with integer coefficients. [3] (c) Using your values of a and b, solve the equation a b 3 7 0 e e e y y y 6 4 2 - + + = . [3] * 0000800000007 * ,










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8 0606/12/M/J/25 © UCLES 2025 6 When lny is plotted against x3, a straight line passing through the points (2, 5) and (-8, 25) is obtained. (a) Find y in terms of x. [4] (b) Find the value of x when y e25 = . [2] * 0000800000008 * ,










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9 0606/12/M/J/25 © UCLES 2025 [Turn over 7 A geometric progression has a 4th term of k 27 8 6 and a 6th term of k 243 32 10 , where k is a constant. The common ratio of this geometric progression is positive. (a) Find the common ratio in terms of k and the value of the first term of this geometric progression. [4] (b) Given that this geometric progression has a sum to infinity of 3, find the possible values of k. [3] * 0000800000009 * ,










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10 0606/12/M/J/25 © UCLES 2025 8 It is given that ( ) ln y x x 3 16 2 2 = + + . (a) Find x y d d when x 0 = . Give your answer in the form lnp, where p is a constant. [5] (b) Given that x increases from 0 to h, where h is small, write down the approximate change in y. [1] * 0000800000010 * ,










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11 0606/12/M/J/25 © UCLES 2025 [Turn over 9 It is given that ( ) ( ) ln x x 2 3 4 f = - , for x a 2 , and that f 1 - exists. (a) Find the least possible value of a. [1] (b) For your value of a, find the range of f. [1] (c) For your value of a, find an expression for ( )x f 1 - . [2] (d) It is given that the equation ( ) ( ) x x f f 1 = - has two roots. For your value of a, sketch the graphs of ( ) y x f = and ( )x y f 1 = - on the axes. Label each graph. State the intercepts of each graph with the axes. State the equations of any asymptotes. [4] y x O * 0000800000011 * ,









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12 0606/12/M/J/25 © UCLES 2025 10 rad i A B C E D F O r r The diagram shows the shape OABCDEF. AOF is a straight line. OAB and OEF are sectors of a circle with centre O and radius r. Angle BOA = angle EOF. OCD is a sector of a circle with centre O and radius r 3 4 . Angle COD is i radians. The point B lies on the line OC and the point E lies on the line OD. The line BE is parallel to the line AOF. (a) Find, in terms of r and i, the area of the shaded region BCDE. [3] * 0000800000012 * ,









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13 0606/12/M/J/25 © UCLES 2025 [Turn over (b) rad i A B C E D F O r r The diagram shows the shape from part (a) with region OABEF shaded. Find, in terms of r and i, the perimeter of the shaded region. [5] * 0000800000013 * ,









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14 0606/12/M/J/25 © UCLES 2025 11 A P Q B R O The diagram shows the triangle OAB, where a OA = and b OB = . The point P lies on OA such that O O P A 4 3 = . The point Q lies on AB such that AQ AB 3 1 = . The straight line through P and Q meets the straight line through O and B at the point R. It is given that b OR m = and PR PQ n = , where m and n are constants. (a) Find OR in terms of a, b and n. [6] * 0000800000014 * ,









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15 0606/12/M/J/25 © UCLES 2025 [Turn over (b) Hence find the values of m and n. [3] Question 12 is printed on the next page. * 0000800000015 * ,









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16 0606/12/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. 12 A curve is such that its gradient at the point (x, y) is given by ( ) x 5 2 3 1 - . The curve passes through the point ,2 5 32 b l. Find the coordinates of the stationary point on the curve. [6] * 0000800000016 * ,









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This document consists of 13 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 2 of 13 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 3 of 13 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 4 of 13 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 13 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 6 of 13 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 13 Question Answer Marks Guidance 1 ( )( )( ) 1 2 2 1 3 3 x x x  + + − isw 3 B1 for  with 3 factors or expanded out cubic B1 for 1 3 in a product with 3 factors B1 for ( )( )( ) 2 2 1 3 x x x + + − and no extra terms or ( )( )( ) 2 2 0.5 3 x x x + + − and no extra terms 2 ( ) 2 1 3 3 6 0 x x + − = oe soi B1 For multiplying through by 1 3 x Allow if substitution is used ( ) 1 1 3 3 3 2 0 x x    + − =          oe M1 Allow if substitution is used 27 x = − A1 A0 if rejects 27 x = − 8 x = A1 3(a) ( ) 2 5 30 45 0 x x − + = or equivalent 3 term quadratic 2 M1 for attempt to eliminate one variable Use of discriminant or solution to show one repeated root or a single solution A1 Must have conclusion e.g. Discriminant = 0 (so tangent) oe One solution only (so tangent) Alternative 1 Centre of the circle (5, 2) Equation of radius perpendicular to given line ( ) 1 2 5 2 y x − = − − Intersection at ( ) 3,3 (2) M1 for a complete method to obtain the point of intersection Show that P lies on the circle or that 5 CP = (A1) Alternative 2 The line y kx n = + touches the circle ( ) ( ) 2 2 2 x p y q r − + − = if ( ) ( ) 2 2 2 1 kp q n r k − + − = + so ( ) ( ) ( ) 2 2 2 5 2 3 5 1 2 −  + − = + (2) M1 for a complete method showing substitution, allow one slip 25 = 25, so a tangent (A1)

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 13 Question Answer Marks Guidance 3(b) (3, 3) 2 B1 for each 3(c) Radius = 3 2 oe soi B1 ( ) ( ) 2 2 3 3 x y − + − soi B1 FT on their answer to (b) only ( ) ( ) 2 2 3 3 18 x y − + − = oe B1 ISW any expansion 4(a)   π 0 cos − B1 May be implied by ( ) ( ) cosπ cos0 − −− 2 B1 4(b) 1 cos cos sin sin cos      + B1 For ratios in terms of sine and cosine 2 2 1 cos cos sin sin cos      + B1 For simplifying the denominator to one term 1 cos 1 cos sin    = sin B1 Must show sufficient correct detail Alternative 1 cos 1 tan tan   + or sec 1 tan tan   + (B1) For ratios in terms of tangent and cosine or secant and tangent 2 1 cos sec tan    or 2 sec sec tan    oe (B1) For dealing with the denominator tan sin cos sec cos      =  oe sin = (B1) Must show correct sufficient detail

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 13 Question Answer Marks Guidance 5(a) ( ) 2 p 9 14 x x x a  = − + soi B1 ( ) p 1 9 14 21 a − = + + = 2 a = − B1 ( ) p 2 24 28 2 0 a b = − + + = soi B1 Each term must be simplified, allow using their a 8 b = B1 5(b)  ( ) 2 2 3 4 x x x − − − soi 2 M1 for quadratic factor with two terms correct A1 must be from correct a and b ( )( )( ) 2 1 3 4 x x x − + − A1 A1 must be from correct a and b 5(c) ( ) 2 e 2 0 y − = soi B1 1 4 ln 2 3 y = oe, 1 ln2 2 y = oe 2 B1 for one correct solution B2 for both solutions and no extra solution. 6(a) 3 ln y mx c = + soi B1 c may be shown as a log term 25 8m c = − + 5 2m c = + M1 For at least one correct equation, may be used with gradient of 2 − 2, 9 m c = − = A1 2 9 y x = − + seen implies M1A1 3 9 2 e x y − = oe A1 6(b) 3 25 9 2x = − M1 For equating exponential indices and obtaining x = … or use of 3 25 9 2 their x their − = − 2 x = − only A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 13 Question Answer Marks Guidance 7(a) 6 10 3 5 8 32 , 27 243 k k ar ar = = soi B1 May be implied by a correct 5th term 8 16 81 k 2 2 3 k r = 2 M1 for solution of their equations to obtain either r 2 mk = or a c = , where m is an unsimplified numeric constant and c is a non-zero constant e.g. Use of 2 6th term 4th term r = 1 a = A1 7(b) 2 1 3 2 1 3 k = − soi B1 For use of sum to infinity formula with their values of a where a is numeric and r which is in terms of k2 1 k =  2 B1 for each must be from correct work 8(a) ( ) ( ) ( ) 2 2 2 6 2 ln 3 16 d 3 16 d 2 x x x y x x x + − +   + =     + 3 B1 for 2 6 3 16 x x + M1 for differentiation of a quotient or equivalent product A1 for all other terms apart from 2 6 3 16 x x + correct When d ln16 0, d 4 y x x = = − 1 ln 2 = from correct work 2 M1 dep for attempt to substitute in 0 x = and obtain a single log term For A1, all previous marks must have been awarded. 8(b) Change = 1 ln 2 h oe B1 FT on their single log answer to part (a) B0 for 1 ln 2 h or their 1 ln 2 h 9(a) 4 3 or 4 3 a = B1 Award B1 when a correct answer is seen 9(b) oe B1 May be in terms of f or y but not x 9(c) ( ) 2 1 4 e f 3 x x − +  =   oe 2 M1 for a complete valid method, allow a sign error A1 must be using the correct notation

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 13 Question Answer Marks Guidance 9(d) 4 B1 for ( ) f y x = in the 1st and 4th quadrants and appropriate asymptotic behaviour B1 dep on previous B mark for ( ) 1 f y x − = intersecting twice with ( ) f y x = , in the 1st and 2nd quadrants and appropriate asymptotic behaviour B1 dep on first B1 for 5 3 marked correctly on each axis or stated, and no other intercepts B1 for 4 3 y = and 4 3 x = (independent) either on the graph or at the side. 10(a) 2 2 1 4 1 sin 2 3 2 r r     −     oe 3 B1 for 2 1 4 2 3 r        B1 for 2 1 sin 2 r  or 2 sin cos 2 2 r   or 2 2 sin 2 tan 2 r   oe Allow final answer unsimplified 10(b) Arc length = ( ) π 2 r  − B1 ( ) 2 2 1 cos BE r  = − or 2 sin 2 r  or sin π sin 2 r   −       oe 2 M1 for complete method to find BE or BE2 , using either cosine rule, sine rule or basic trig using r and a correct angle. A0 if an error in rearranging to obtain BE is made. Allow unsimplified Perimeter = ( ) ( ) 2 π 2 2 1 cos r r r   − + + − oe 2 M1 for a correct plan using their lengths 2 arc lengths + 2r + their BE Allow A1 for an unsimplified answer

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 12 of 13 Question Answer Marks Guidance 11(a) ( ) 1 3 OQ = + − a b a or ( ) 2 3 OQ = + − b a b 2 1 3 3 OQ   = +     a b 2 M1 for attempt to use ( ) OQ k = + − a b a or ( ) OQ k = + − b a b , where k can be positive or negative Allow unsimplified 3 4 PQ their OQ = − + a M1 3 2 1 4 3 3 PQ = − + + a a b 1 1 3 12 PQ   = −     b a A1 Allow unsimplified 3 4 OP = a B1 Allow anywhere 3 1 1 4 3 12 OR   = + −     a b a oe B1 Must be simplified Alternative ( ) 1 1 4 3 PQ = + − a b a 1 1 3 12 PQ   = −     b a (4) B1 for 1 4 PA = a M1 for ( ) 1 3 AQ = − b a , allow a sign error i.e. use of ( ) − a b M1 for attempt to use 1 4 PQ their theirAQ = + a Allow unsimplified 3 4 OP = a (B1) 3 1 1 4 3 12 OR   = + −     a b a or 3 1 4 12 3 OR    = − +     a a b or 3 4 12 3 OR     = − +     a b (B1) Must be simplified

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 13 of 13 Question Answer Marks Guidance 11(b) 1 3   = B1 For equating b vectors 3 1 0 4 12  − = oe B1 For equating a vectors to zero 9, 3   = = B1 12 When d 2 0, d 5 y x x = = B1 Must come from first derivative and not the second derivative ( ) ( ) ( ) 4 3 3 5 2 20 y x c = − + 2 M1 for ( ) 4 3 5 2 k x − ( ) ( ) 4 3 32 3 5 2 2 5 20 c =  − + oe M1 Dep for attempt to find c using their ( ) 4 3 5 2 k x − 4 c = A1 Allow an unsimplified fraction Stationary point 2 , 4 5       B1 Dep on first B mark