Cambridge IGCSE Mathematics - Additional 0606 — 2025 Feb/March Paper 1 · Variant 2

This document has 16 pages. [Turn over Cambridge IGCSE™ DC (CE/SG) 346699/1 © UCLES 2025 ADDITIONAL MATHEMATICS 0606/12 Paper 1 Non-calculator February/March 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 3 4 5 3 5 1 9 8 4 * ,










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2 0606/12/F/M/25 © UCLES 2025 List of formulas Equation of a circle with centre (a, b) and radius r. ( ) ( ) x a y b r 2 2 2 - + - = Curved surface area, A, of cone of radius r, sloping edge l. A rl r = Surface area, A, of sphere of radius r. A r 4 2 r = Volume, V, of pyramid or cone, base area A, height h. V Ah 3 1 = Volume, V, of sphere of radius r. V r 3 4 3 r = Quadratic equation For the equation , ax bx c 0 2 + + = 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 ( ) u a n d 1 n = + - ( ) { ( ) } S n a l n a n d 1 2 2 1 2 1 n = + = + - Geometric series u ar n n 1 = - ( ) ( ) S a r r r 1 1 1 n n ! = - - ( ) S r a r 1 1 1 = - 3 Identities sin cos A A 1 2 2 + = sec tan A A 1 2 2 = + cosec cot A A 1 2 2 = + Formulas for ABC T sin sin sin A a B b C c = = cos a b c bc A 2 2 2 2 = + - sin ab C 2 1 T = * 0000800000002 * ,










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3 0606/12/F/M/25 © UCLES 2025 [Turn over Calculators must not be used in this paper. 1 (a) On the axes, sketch the graphs of y x 4 1 = - and y x 3 2 = + , stating the intercepts with the axes. [3] O x y (b) Solve the inequality x x 4 1 3 2 G - + . [4] * 0000800000003 * ,










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4 0606/12/F/M/25 © UCLES 2025 2 0 x y – 7 3 – 180° 180° The diagram shows the curve cos y a bx c = + for ° ° x 180 180 G G - . It is given that a, b and c are integers. Find the values of a, b and c. [3] 3 The point A has coordinates ( , ) 3 6 - . The point B has coordinates ( , ) 7 8 - . Given that the line AB is the diameter of a circle, find the equation of the circle. [4] * 0000800000004 * ,










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5 0606/12/F/M/25 © UCLES 2025 [Turn over 4 The polynomial p is given by ( )x a x ax ax 2 2 p 2 3 2 = + + + , where a is a positive integer. It is given that x 2 1 + is a factor of ( )x p . (a) Find the value of a. [3] (b) Hence factorise ( )x p . [2] (c) Hence show that the equation ( )x 0 p = has only one real root. [1] * 0000800000005 * ,










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6 0606/12/F/M/25 © UCLES 2025 5 (a) Write x x 2 2 3 2 - + in the form x a b c 2 + + ` j , where a, b and c are constants. [3] It is given that ( )x x x 2 2 3 f 2 = - + , for x p G . (b) Write down the greatest value of p for which f has an inverse. [1] (c) Using this value of p, write down the range of f. [1] (d) Using this value of p, find an expression for f 1 - . [3] * 0000800000006 * ,










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7 0606/12/F/M/25 © UCLES 2025 [Turn over 6 It is given that tan 5 5 i = and ° ° 180 360 1 1 i . (a) Find the value of cosi. [2] (b) Find the value of sini. [1] (c) Find the value of sec cot i i + . Give your answer in the form a b c 5 + , where a, b and c are integers. [2] * 0000800000007 * ,










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8 0606/12/F/M/25 © UCLES 2025 7 O x y A y x 2 1 = + y x 1 5 2 = + + The diagram shows part of the curve y x 1 5 2 = + + and part of the line y x 2 1 = + intersecting at the point A. (a) Find the coordinates of A. [4] * 0000800000008 * ,










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9 0606/12/F/M/25 © UCLES 2025 [Turn over (b) Find the exact area of the shaded region. [6] * 0000800000009 * ,










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10 0606/12/F/M/25 © UCLES 2025 8 The first term of a geometric progression is 10. This geometric progression has a positive common ratio r. The first term of an arithmetic progression is also 10. This arithmetic progression has a negative common difference d. The second term of the geometric progression is the same as the fourth term of the arithmetic progression. The third term of the geometric progression is the same as the sixth term of the arithmetic progression. (a) Find the values of r and d. [6] (b) Determine whether the geometric progression has a sum to infinity. [1] * 0000800000010 * ,










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11 0606/12/F/M/25 © UCLES 2025 [Turn over 9 Solve the equation ( ) . log log x 1 4 2 3 ( ) x 2 1 + - = + [5] * 0000800000011 * ,









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12 0606/12/F/M/25 © UCLES 2025 10 The point P lies on the curve y x 5 2 3 2 = + ` j . The x-coordinate of P is 5. The normal to the curve at P intersects the line x y 11 + = at the point Q. The point R is the reflection of Q in the tangent to the curve at P. Find the coordinates of R. [9] * 0000800000012 * ,









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13 0606/12/F/M/25 © UCLES 2025 [Turn over Continuation of working space for Question 10. * 0000800000013 * ,









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14 0606/12/F/M/25 © UCLES 2025 11 O N A X B M In the diagram, OA a = and OB b = . The point M is the midpoint of OB. The point N is such that . ON NA 3 = The lines BN and AM intersect at the point X. , BX BN m = where m is a constant. , MX MA n = where n is a constant. (a) Find OX in terms of a, b and m. [3] * 0000800000014 * ,









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15 0606/12/F/M/25 © UCLES 2025 [Turn over (b) Find OX in terms of a, b and n. [2] (c) Hence find the values of m and n. [4] Question 12 is printed on the next page. * 0000800000015 * ,









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16 0606/12/F/M/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 It is given that y xe x 3 2 = + . (a) Find x y d d . [3] (b) Hence find . x x e d x 3 2 + y [4] * 0000800000016 * ,









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This document consists of 12 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 February/March 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 February/March 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 February/March 2025 © Cambridge University Press & Assessment 2025 Page 2 of 12 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 February/March 2025 © Cambridge University Press & Assessment 2025 Page 3 of 12 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 February/March 2025 © Cambridge University Press & Assessment 2025 Page 4 of 12 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 Benefit of the doubt Communication mark Incorrect point 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 Misread Omission

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 5 of 12 Annotation Meaning 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 point Not from wrong working 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 February/March 2025 © Cambridge University Press & Assessment 2025 Page 6 of 12 Question Answer Marks Partial Marks 1(a) Fully correct, ruled graphs with all intercepts indicated y 1 4 2 3 − 2 3 M2 for a graph of correct shape with vertex on the x-axis and 1 and 4 oe appropriately indicated or a graph of correct shape with vertex on the x-axis and 2 3 − and 2 oe appropriately indicated or M1 for a graph of correct shape with vertex on the x-axis and 1 or 4 oe appropriately indicated or a graph of correct shape with vertex on the x-axis and 2 3 − or 2 oe appropriately indicated 1(b) 4(x – 1) * 3x + 2 oe and 4(x – 1) * –3x – 2 oe OR  2 7 44 12 *0 − + x x where * is = or any inequality sign M2 M1 for 4(x – 1) * –3x – 2 oe OR M1 for (4(x – 1))2 = (3x + 2)2 oe Critical values 2 7 and 6 A1 2 6 7 x mark final answer A1 2 5 = a B1 3 = b B1 2 = − c B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 7 of 12 Question Answer Marks Partial Marks 3 ( ) ( ) 2 2 2 1 74 − + + = x y oe nfww or 2 2 4 2 69 0 + − + − = x y x y oe nfww 4 M1 for centre: 3 7 6 8 , 2 2 −+ −       or ( ) 2, 1 − soi M2 for [r = ] 74 or [r2 = ] 74 or [diameter =] 2 74 or c = –69 or M1 for ( ) ( ) 2 2 2 3 7 6 8   = −− + −−   AB oe or ( ) ( ) 2 2 2 3 2 6 ( 1)   = −− + − −   r their their oe or ( ) ( ) 2 2 2 7 2 8 ( 1)   = − + −− −   r their their oe Alternative method M2 for 6 ( 8) 1 ( 3) 7 − −−  = − −− − y y x x or M1 for e.g. P(x, y) on the circumference means AP ⊥ BP → 1  = − AP BP m m M1 for 2 2 2 48 1 4 21 + − = − − − y y x x oe FT 6 ( 8) ( 3) 7   − −−    −− −   y y their x x with at most one sign error 4(a) 3 2 2 1 1 1 2 2 0 2 2 2       − + − + − + =             a a a oe, soi M1 Simplifies and solves for a M1 FT their quadratic in a providing 1 p 0 2   − =     oe attempted 4 = a A1 4(b) ( )( ) 2 2 2 1 4 1 + + x x mark final answer 2 M1 for ( ) 2 2 2 1 2 2 their a x x   + +     or ( ) 2 8 2 + x found as quadratic factor 4(c) Discriminant of 2 4 1 + x is 0 16 0 −  oe and no real roots oe or 2 2 4 1 0 4 1 x x + = → = − oe and therefore no solution oe [statement that only solution is 1 2 = − x ] 1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 8 of 12 Question Answer Marks Partial Marks 5(a) 2 1 5 2 2 2   − +     x 3 B2 for 2 1 2 2   −     x or B1 for 2 1 2   −     x or a = 2 and b = –0.5 oe B1 for 5 2 or c = 2.5 oe 5(b) 1 2 B1 FT –their b 5(c) 5 f 2 ≥ B1 FT their c 5(d) ( ) 1 1 1 5 f 2 2 2 −     = − −       x x oe or ( ) 1 2 8 20 f 4 − − −   =   x x oe 3 M1 FT for a complete method to find the inverse with a correct order of operations FT an expression of the form f(x) = ( ) 2 + + a x b c A1 FT for ( ) 1 1 1 5 f 2 2 2 −     =  −       x x oe FT their 1 2 and their 5 2 6(a) 5 cos 6 = − oe, isw B2 B1 for 5 cos 6 = oe or 5 cos 6 =  oe 6(b) 1 sin 6 = − oe, isw B1 6(c) 5 [1] 6 5 − 2 M1 for [sec + cot =] 1 cos+ 1 tan or 1 cos+ cos sin  soi or for 1 5 cos 5 + their oe or 1 cos cos sin    + their their their

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 9 of 12 Question Answer Marks Partial Marks 7(a) ( )( ) ( ) 2 1 1 5 2 1 + + = + + x x x or ( )( ) 2 1 1 5 − + = x x oe M1   2 2 6 0 + − = x x A1 Factorises or solves their 3-term quadratic e.g. (x + 2)(2x – 3) M1 3 , 4 2       A1 7(b) Correct plan, difference of integrals oe: 0 5 2 d area correct trapezium 1   + −   +    k x x soi where k > 0 M1 or ( ) 0 0 5 2 d 2 1 d 1   + − +   +     k k x x x x soi Area of trapezium: ( ) 1 3 4 1 2 2   +     their their oe B1 or   2 3 3 0 2 2  + −     their their or 15 4 oe Integral: ( ) 1.5 0 5ln 1 2  + +    their x x B2 B1 for 5ln x + 1 or for nln(x + 1) where n is a constant and n > 0 Correct substitution of upper and lower limits into ( ) 5ln 1 2 + + x x M1 FT their integral providing at least B1 awarded and their 1.5 Area of shaded region: 5 3 5ln 2 4  −     isw A1 Alternative method Correct plan, integral of difference of functions: 0 5 2 1 d 1   − +   +    k x x x soi where k > 0 (M1) Integral: ( ) 1.5 2 0 5ln 1   + + −   their x x x (B3) B2 for 5ln(x + 1) and one other correct term or nln(x + 1) + x – x2 where n is a constant and n > 0 or B1for 5ln x + 1 or for nln(x + 1) where n is a constant and n > 0 Area of shaded region: 5 3 5ln 2 4  −     isw (2) M1FT for correct substitution of upper and lower limits; FT their integral providing at least B1 awarded and their 1.5

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 10 of 12 Question Answer Marks Partial Marks 8(a) 10 10 3 = + r d soi B1 2 10 10 5 = + r d soi B1 2 10 3 10 10 5 10 +  = +     d d oe or ( ) 2 5 10 10 10 10 3 − = + r r oe M1 2 9 10 0 + = d d or 2 3 5 2 0 − + = r r oe A1 Correct method to find d or r A1 FT their quadratic in d or r 10 9 = − d and 2 3 = r A1 8(b) Appropriate determination for their r e.g. 1  r so the geometric progression has a sum to infinity oe B1 STRICT FT their r 9 Correct change of base e.g. 2 2 2 4log 2 log ( 1) 3 log ( 1) + − = + x x or ( ) ( ) ( ) 1 1 1 log ( 1) 4log 2 3 log 2 + + + + − = x x x x B1 ( ) ( ) ( )   2 2 2 log 1 3log 1 4 0 + − + − = x x oe or ( ) ( ) ( )   2 1 1 4 log 2 3log 2 1 0 + + + − = x x oe B1 Factorises or solves e.g. ( ) ( ) ( ) ( ) 2 2 log 1 4 log 1 1 0 + − + + = x x or ( ) ( ) ( ) ( ) 1 1 4log 2 1 log 2 1 0 + + − + = x x M1 FT their 3-term quadratic in a suitable logarithm providing correct change of base seen ( ) 2 log 1 4 + = x their and ( ) ( ) 2 log 1 1 x their + = − or ( )1 1 log 2 4 + = x their and ( ) ( ) 1 log 2 1 x their + = − and correctly solves as far as x = … at least once M1 dep on previous M1 FT ( ) 2 log 1 + = x a or ( )1 log 2 + = x b where a and b are constants 1 15, 2 = = − x x nfww mark final answer A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 11 of 12 Question Answer Marks Partial Marks 10 [When x = 5 ] 9 = y B1 ( ) 1 3 d 2 5 2 5 d 3 − = +  y x x oe B2 B1 for ( ) 1 3 d 10 5 2 , d 3 − = +  y k x k x 5 d 10 d 9 = = x y x B1 FT their 5 d d = x y x providing previous B1 awarded Equation of normal e.g. ( ) 1 9 5 10 9 − = − − y their x their oe, soi M1 FT 5 1 d d = − x y their x and their y-coordinate Eliminates one variable e.g. ( ) 1 11 9 5 10 9 − − = − − x their x their M1 dep on previous M mark For Q: 25, 36 = − = x y A2 A1 for each Coordinates of R: (35, –18) or ( ) ( ) ( ) ( ) 10 25 , 18 36 − − − their their B1 FT their coordinates of Q 11(a) 3 4     = −       OX b + a b oe or ( ) 3 3 1 4 4     = − −       OX a + b a oe 3 B2 for 3 4 = − BN a b oe, soi or B1 for 3 4 = ON a oe, soi 11(b) 1 1 2 2     = −       OX b + a b oe or ( ) 1 1 2     = − −       OX a + b a oe 2 B1 for 1 2 = − MA a b 11(c) 3 1 1 4 2 2       − = −         b + a b b + a b soi M1 FT their answer to (a) in terms of a, b and  and (b) in terms of a, b and  3 4   = or 1 1 1 2 2   − = − M1 FT their answer to (a) in terms of a, b and  and (b) in terms of a, b and  3 4   = and 1 1 1 2 2   − = − A1 4 3 , 5 5   = = A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 12 of 12 Question Answer Marks Partial Marks 12(a) ( ) 3 2 3 2 d e 3e d + + = x x x soi B1 ( ) 3 2 3 2 d 3e [1]e d + + =  + x x y x their x oe, isw 2 FT their 3 2 3e + x M1 for correct structure of product rule 12(b) 3 2 3 2 1 1 e e 3 9 + + − + x x x c oe, nfww 4 B3 for 3 2 3 2 3 2 1 3 e d e e ( ) 3 + + + = − +  x x x x x x A or better or B2 for 3 2 3 2 3 2 3 e d e e d + + + = −   x x x x x x x or better or 3 2 3 2 3 2 1e 3 e d e 3 + + + + =  x x x x x x or 3 2 3 2 3 2 e d e + e 3 + + + =  x x x x x x k where k = 1 9 or 1 3 − or B1 for ( ) 3 2 3 2 3 2 3 2 3 e d 3 e e d e d x x x x x x x x x + + + + = + −    or 3 2 3 2 d d 3 e d e d d + + = +    x x y x x x x x or better