Cambridge IGCSE Mathematics - Additional 0606 — 2025 Oct/Nov Paper 1 · Variant 2

This document has 16 pages. [Turn over * 0 1 2 2 4 2 6 3 6 7 * Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 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) 347167/2 © UCLES 2025 ,










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2 0606/12/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 0606/12/O/N/25 © UCLES 2025 [Turn over Calculators must not be used in this paper. 1 (a) On the axes, sketch the graph of ( )( )( ) y x x x 1 5 2 5 = - - - , stating the intercepts with the axes. [3] O x y (b) Hence solve the inequality ( )( )( ) x x x 1 5 2 5 0 G - - - . [2] * 0000800000003 * ,










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4 0606/12/O/N/25 © UCLES 2025 2 The polynomial p is such that ( )x x ax x b 2 13 p 3 2 = + + + , where a and b are integers. It is given that x 2 + is a factor of ( )x p . When ( )x p is divided by x 1 + there is a remainder of 6. (a) Find the values of a and b. [4] (b) Show that the equation ( )x 0 p = has only one real root. [3] * 0000800000004 * ,










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5 0606/12/O/N/25 © UCLES 2025 [Turn over 3 The sum of the first two terms of a geometric progression is 9. The sum to infinity of this geometric progression is 25. (a) Find the possible values of the common ratio of this geometric progression. [4] (b) Find the first term of this geometric progression for each possible value of the common ratio. [2] * 0000800000005 * ,










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6 0606/12/O/N/25 © UCLES 2025 4 (a) Show that x x 2 5 3 2 + + can be written in the form ( ) x a b 2 2 + + , where a and b are constants to be found. [2] (b) Hence write down the coordinates of the stationary point on the curve y x x 2 5 3 2 = + + . [2] A function f is such that ( )x x x 2 5 3 f 2 = + + , for x p H , where p is a constant. It is given that f 1 - exists. (c) (i) Write down the least possible value of p. [1] * 0000800000006 * ,










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7 0606/12/O/N/25 © UCLES 2025 [Turn over (ii) Using your value of p, sketch the graphs of ( ) y x f = and ( ) y x f 1 = - . Label each graph. State the intercepts of each of the graphs with the axes. [5] O x y * 0000800000007 * ,










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8 0606/12/O/N/25 © UCLES 2025 5 (a) Write lg lg a b 5 4 3 - - as a single base 10 logarithm. [3] (b) Solve the equation ( ) log log x 1 5 0 ( ) x 5 1 + - = + . [4] * 0000800000008 * ,










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9 0606/12/O/N/25 © UCLES 2025 [Turn over 6 Solve the equation x x 2 10 5 2 + - = . [5] * 0000800000009 * ,










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10 0606/12/O/N/25 © UCLES 2025 7 The point A has coordinates (-2, 4). The point B has coordinates (6, 10). The point C has coordinates (12, 2). (a) Find the gradients of the lines AB, AC and BC. [2] (b) Hence find the equation of the circle which passes through the points A, B and C. [4] * 0000800000010 * ,










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11 0606/12/O/N/25 © UCLES 2025 [Turn over 8 On the axes, sketch the graph of ( ) ln y x 5 4 3 = + . State the intercepts with the axes. State the equation of any asymptote. [4] O x y * 0000800000011 * ,









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12 0606/12/O/N/25 © UCLES 2025 9 A curve ( ) y x f = is such that ( ) ( ) x x 2 5 f 2 3 = + - ll . The curve has gradient 3 2 at the point (2, 2). (a) Find the coordinates of the stationary point on the curve. [11] * 0000800000012 * ,









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13 0606/12/O/N/25 © UCLES 2025 [Turn over Continuation of working space for Question 9(a). (b) Determine the nature of this stationary point. [1] * 0000800000013 * ,









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14 0606/12/O/N/25 © UCLES 2025 10 Show that r ( ) ( ) sin cos sin cos k d 2 2 r r 2 2 3 i i i i i + + - = ` j y , where k is an integer to be found. [5] 11 Solve the equation ( ) n C C 4 n n 1 5 2 7 - = + + . [3] * 0000800000014 * ,









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15 0606/12/O/N/25 © UCLES 2025 [Turn over 12 A curve has equation y x 2 e x2 = - ` j for x 2 1 . (a) Find the value of x y d d when x 0 = . [4] When x 0 = , y is increasing at the rate of 0.5 units per second. (b) Find the corresponding rate of change of x. [2] Question 13 is printed on the next page. * 0000800000015 * ,









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16 0606/12/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. 13 For x 1 1 G G - it is given that sin x 2 3 i = - and cot y 3 2i = . Find y in terms of x. [4] * 0000800000016 * ,









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This document consists of 14 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 2 of 14 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 October/November 2025 © Cambridge University Press & Assessment 2025 Page 3 of 14 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 October/November 2025 © Cambridge University Press & Assessment 2025 Page 4 of 14 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 5 of 14 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 6 of 14 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) 3 B1 for the correct shape with a distinct minimum in 4th quadrant and a distinct maximum in the first quadrant. No turning over at end points, suggesting further stationary points. B1 for 1, 2.5 and 5 on the x-axis or stated. Independent mark for critical values seen correctly marked on the x-axis or stated with no contradiction and no other critical x-values B1 for 25 marked correctly on the y-axis or stated with no contradiction. Independent mark.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 14 Question Answer Marks Guidance 1(b) 1 2.5 x  B1 Mark final answer 5 x B1 Mark final answer 2(a) ( ) ( ) p 2 16 4 26 0 − = − + − + = a b B1 ( ) ( ) p 1 2 13 6 − = −+ − + = a b B1 7, 14 = = a b 2 M1 to solve their equations to obtain at least one value, must be using ( ) p 2 − or ( ) p 1 − 2(b) ( )( ) 2 2 2 3 7 0 + + + = x x x soi 2 M1 for attempt to factorise or use of long division on their ( ) p x Correct use of discriminant on their quadratic factor or correct attempt to solve their quadratic factor equated to zero with valid conclusion B1 FT on their quadratic factor. Quadratic factor must have a negative discriminant and have a correct value if evaluated 3(a) 9 + = a ar or ( ) 2 1 9 1 − = − a r r or ( ) 2 1 9 1 − = − a r r B1 25 1 = − a r B1 2 16 25 = r B1 Dep B mark on both previous B marks 4 5 =  r or 0.8  B1 Dep B1 mark on all previous B marks 3(b) 5, 45 = a 2 B1 for each 4(a) 2 5 1 2 4 8   + −     x 2 B1 for 2 5 2 4   +     x B1 for 1 8 − 4(b) 5 1 , 4 8   − −     2 B1FT on their – a B1FT on their b B0 if calculus used as question says ‘Hence’ 4(c)(i) 5 4 − or 5 4 = − p or 5 4 − x B1 FT on their – a from part (a). Allow if from calculus

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 14 Question Answer Marks Guidance 4(c)(ii) 5 B1 for ( ) f = y x drawn as a one-one function in the first, second and third quadrants or first and second quadrants. Must have correct curvature Dep B1 for ( ) 1 f − = y x as a reflection of their ( ) f = y x in the line = y x . Maybe implied by intercepts, but must have the correct curvature B1 for ( ) f = y x clearly passing through 1 = − x correctly soi. Allow if domain is unrestricted but must have correct quadratic shape or curvature B1 for ( ) f = y x passing through 3 = y only on the y-axis soi. Allow if domain is unrestricted but must have correct quadratic shape or curvature B1 for – 1 and 3 marked correctly for ( ) 1 f − = y x allow for unrestricted range. Must have correct curvature. 5(a) 5 4 lg 1000       a b or 5 3 4 lg 10       a b or ( ) 3 5 4 lg 10− − a b or ( ) 5 4 lg 0.001 − a b or 5 4 3 lg 10 −       a b or equivalent correct single logarithm to base 10. 3 B2 for ( ) 5 4 lg 1000   a b or B2 for 5 4 lg 1000        a b B1 for 3 lg1000 = or 3 lg10 soi B1 for use of the power rule at least once

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 14 Question Answer Marks Guidance 5(b) ( ) ( ) 5 5 1 log 1 0 log 1 + − = + x x oe B1 For change of base seen in the equation, allow for ( ) ( ) 5 5 5 log 5 log 1 0 log 1 + − = + x x ( ) ( ) 2 5 log 1 1 + = x leading to ( )  5 log 1 1 + =  x M1 For correct simplification to a 2- term quadratic and at least one solution, may be implied by a substitution 4 = x A1 SC B1 if ‘spotted’ from a correct equation 4 5 = − x A1 A0 if rejected or crossed out Alternative method 1 1 1 1 log 5 0 log 5 + + − = x x oe (B1) For change of base seen in the equation, allow for ( ) 1 1 1 log 1 log 5 0 log 5 + + + + − = x x x x ( ) 2 1 log 5 1 + = x  1 log 5 1 + =  x (M1) For correct simplification to a 2- term quadratic and at least one solution, may be implied by a substitution. 4 = x (A1) 4 5 = − x (A1) A0 if rejected or crossed out Alternative method 2 ( ) ( ) lg 1 lg5 0 lg5 lg 1 + − = + x x oe (B1) For change of base seen in the equation, allow for ( ) ( ) lg10 lg 1 0 lg 1 + − = + x x ( ) ( ) ( ) 2 2 lg 1 lg5 + = x leading to ( )  lg 1 lg5 + =  x (M1) For correct simplification to a 2- term quadratic and at least one solution, may be implied by a substitution 4 = x (A1) SC B1 if ‘spotted’ from a correct equation 4 5 = − x (A1) A0 if rejected or crossed out

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 14 Question Answer Marks Guidance 6 2 2 10 5 + − = x x leading to 2 2 15 0 + − = x x with a valid attempt to solve to obtain 2 values for x M1 5 , 3 2 = − x A1 For both Mark final answer for this equation, A0 if – 3 is rejected 2 2 10 5 + − = − x x leading to 2 2 5 0 + − = x x M1 Must be a correct 3-term quadratic equated to zero 1 41 4 − = x oe 2 Dep M1 for a valid method of solution to obtain 2 values for x. Mark final answer for this equation, A0 if negative root is rejected Denominator of final answer must be positive 7(a) Grad 6 8 = AB oe Grad 2 14 = − AC oe Grad 8 6 = − BC oe 2 B1 for 2 correct unsimplified 7(b) Centre ( ) 5, 3 B1 FT on their diameter, allow for ( ) 2, 7 or ( ) 9, 6 Radius2 = 50 oe soi 2 M1 for use of their centre and coordinates of either A, B or C to find the square of the radius or the radius. Centre must not be A, B or C. or Finding the length between any 2 points A, B or C and halving it. ( ) ( ) 2 2 5 3 50 − + − = x y oe A1 Allow ( ) 2 5 2 for radius2

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 14 Question Answer Marks Guidance 8 4 B1 for correct shape in first three quadrants tending towards a vertical asymptote in the second and third quadrants. Dep B1 for curve passing through x = 0.5 − oe on the x-axis , or stated and no other points on the x-axis Dep B1 for curve passing through y = 5ln3 oe on the y-axis , or stated and no other points on the y-axis B1 for 3 4 = − x oe shown on graph or stated as the equation of the asymptote. May be implied by a dotted or straight vertical line through 3 4 = − x . This is not a dependent mark.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 14 Question Answer Marks Guidance 9(a) ( ) ( ) ( ) 1 2 f 2 5 −  = − + + x x c 2 M1 for ( ) 1 2 2 5 − + k x Allow unsimplified for A1 ( ) ( ) 1 2 f 2 5 1 −  = − + + x x soi 2 Dep M1 for use of ( ) 2 f 3  = x and 2 = x in their ( ) f x in the form ( ) 1 2 2 5 − + + k x c . Allow term in ( ) 1 2 2 5 − + x unsimplified. Need to see attempt at substitution. ( ) ( ) ( ) 1 2 f 2 5 = − + + x x x d 2 M1 for ( ) 1 2 2 5 + p x ( ) ( ) 1 2 f 2 5 +3 = − + x x x soi 2 Dep M1 for use of 2 = x and 2 = y in their ( ) f x in the form ( ) 1 2 2 5 + + + rx p x d . Allow term in ( ) 1 2 2 5 + x unsimplified. Need to see an attempt at substitution. When ( ) f 0  = x , 2 = − x 2 M1 for solution of their ( ) f 0  = x , must be in the form ( ) 1 2 2 5 − + + p x q to obtain a value for x. 0 = y A1 9(b) ( ) f 2 1 − = [so positive] a min point or correct use of first derivative test considering x values greater than 5 2 − B1 Need to see evidence of substitution for either test. B0 if 2 = − x obtained incorrectly in part (a) 10 ( ) ( ) 2 2 sin cos sin cos 2     + + − = 2 M1 for expansion of both terms with an attempt to simplify either using the correct identity or eliminating terms in 2sin cos   2 d 2   =  M1 Dep M1 3π 2 π 2 2 d 3π π = −  = 2π 2 Dep M1 for correct use of limits

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 13 of 14 Question Answer Marks Guidance 11 ( )( ) ( ) ( ) ( ) 4 1 ! 2 ! 4 !5! 5 !7! − + + = − − n n n n n leading to 7 6 2 = + n 2 B1 for 7 6  or 42 seen as a result of simplification of the numerical factorials B1 for 2 + n seen as a result of simplification of the algebraic factorials 40 = n B1 12(a) ( ) ( ) 2 2 2 2 2 e e d d 2 − − = − x x x x y x x or ( ) ( ) 2 2 1 2 d 2 e 2 e 2 d − − = − − − x x y x x x x 3 B1 for 2 2 ex x M1 for differentiation of a quotient or equivalent correct product A1 all other terms correct 1 4 − A1 Must be from correct working 12(b) For use of d d d d d d  = y x y x t t M1 FT on 1 0.5 1 4    −     their 2 − A1 Must be from correct working in part (a)

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 14 of 14 Question Answer Marks Guidance 13 2 sin 3  − = x B1 2 2 cos 3 sin     =     y soi B1 Use of 2 2 sin cos 1   + = to obtain 2 2 2 1 3 3 2 3   −   −         =   −             x y x B1 Dep on both previous B marks ( ) 2 3 3 1 2       = −       −     y x oe B1 Dep on all previous B marks No fractions within fractions Alternative method 1 3 cosec 2 = −x (B1) 2 cot 3 = y (B1) Use of 2 2 cot 1 cosec   + = to obtain ( ) 2 9 1 3 2 + = − y x (B1) Dep on both previous B marks ( ) 2 9 3 1 2     = −   −   y x oe (B1) Dep on all previous B marks No fractions within fractions Alternative method 2 ( ) 2 3 cosec 1  = − y or 2 cosec 1 3  = − y soi (B1) 2 2 1 cosec sin   = soi (B1) 2 sin 3  − = x (B1) ( ) 2 9 3 1 2     = −   −   y x oe (B1) Dep on all previous B marks No fractions within fractions