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

This document has 16 pages. [Turn over * 7 7 8 2 7 2 7 8 7 3 * Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 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/SG) 347114/4 © UCLES 2025 ,










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2 0606/11/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 0606/11/O/N/25 © UCLES 2025 [Turn over Calculators must not be used in this paper. 1 (a) Write down the amplitude and period of cos x 3 2 1 - . Amplitude = … Period = … [2] (b) Sketch the graph of cos y x 3 2 1 = - for ° ° x 0 360 G G . [3] 5 0 – 5 90° 180° 270° 360° x y * 0000800000003 * ,










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4 0606/11/O/N/25 © UCLES 2025 2 Find the set of values of k for which the line y k = x 3 - meets the curve y x x 2 2 = + at two distinct points. Give your answer in simplest surd form. [6] * 0000800000004 * ,










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5 0606/11/O/N/25 © UCLES 2025 [Turn over 3 (a) Solve the equation log log x 4 5 2 x 2 - = . [5] (b) Solve the equation 25 e e x x 3 7 2 2 = - - . [4] * 0000800000005 * ,










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6 0606/11/O/N/25 © UCLES 2025 4 (a) y O x Explain why this graph does not represent a function. [1] (b) The table shows the graphs of four different functions. is one-one is many-one is its own inverse y 0 x 4 4 y 0 x 4 4 y 0 x 4 4 y 0 x 4 4 Tick (✓) each correct box in the table. There may be more than one tick in a row or a column. [4] * 0000800000006 * ,










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7 0606/11/O/N/25 © UCLES 2025 [Turn over (c) Functions f and g are defined as follows. :f sin x x 7 for ° ° x a 30 G G :g x x 2 1 7 - for x 2 1 H It is given that the function gf exists. (i) Find the value of a so that the domain of gf is as large as possible. You may use the information that ° sin30 2 1 = . [2] (ii) For the domain found in part (i), find the range of the function gf. [2] (iii) Determine whether the function g2 exists. [1] * 0000800000007 * ,










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8 0606/11/O/N/25 © UCLES 2025 5 (a) In an arithmetic progression: • the first term is 3 • the sum of the first 10 terms is 4 times the sum of the first 5 terms. Find the common difference. [3] * 0000800000008 * ,










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9 0606/11/O/N/25 © UCLES 2025 [Turn over (b) The 1st, 2nd and 5th terms of another arithmetic progression are the 1st, 2nd and 3rd terms of a geometric progression. It is given that the 1st terms of the progressions are not 0. Find the common ratio, r, where r 1 ! , of the geometric progression. [4] * 0000800000009 * ,










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10 0606/11/O/N/25 © UCLES 2025 6 Solve the equation x x 5 6 2 - = . [5] 7 (a) Differentiate sin cos x x e x 1 3 + - with respect to x. [4] (b) Find tan x x 1 3 d 2 + ` j y . [3] * 0000800000010 * ,










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11 0606/11/O/N/25 © UCLES 2025 [Turn over 8 Solve the simultaneous equations y x x y 2 4 1 - =- and x y 1 6 = - . [5] * 0000800000011 * ,









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12 0606/11/O/N/25 © UCLES 2025 9 The diagram shows the velocity–time graph for a particle moving in a straight line. 0 1 2 3 Velocity Time (a) On the diagram below, sketch the speed–time graph for the motion of this particle. [1] 1 2 3 Speed 0 Time * 0000800000012 * ,









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13 0606/11/O/N/25 © UCLES 2025 [Turn over It is given that the velocity–time graph is part of a quadratic curve. The curve has gradient 0 when time is 0. (b) On the diagram below sketch a possible acceleration–time graph for the motion of the particle. [2] 1 2 3 Acceleration 0 Time * 0000800000013 * ,









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14 0606/11/O/N/25 © UCLES 2025 10 y x l y x x 9 18 2 = - + O The diagram shows the curve y x x 9 18 2 = - + and the line l, which is the normal to this curve at the point where x 5 = . Find the area of the shaded region. [11] * 0000800000014 * ,









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15 0606/11/O/N/25 © UCLES 2025 [Turn over Continuation of working space for Question 10. 11 The number of permutations of n items taken 4 at a time is equal to 6 1 # the number of permutations of 2n items taken 3 at a time. (a) Show that n satisfies the equation n n 3 19 20 0 2 - + = . [4] (b) Hence find the value of n. [2] Question 12 is printed on the next page. * 0000800000015 * ,









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16 0606/11/O/N/25 © UCLES 2025 12 Solutions to this question by accurate drawing will not be accepted. Relative to an origin O, the position vectors of the points A, B and C are OA 1 7 = e o, OB 7 4 = e o and OC k 1 2 = e o, where k is a scalar constant. Given that C lies on the line AB, use a vector method to find the ratio AC : AB. [6] 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. * 0000800000016 * ,









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This document consists of 12 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 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 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

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 5 of 12 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

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 6 of 12 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 180° or π B1 Allow without degrees sign 1(b) Correct curve 3 B1 for amplitude 3 B1 for period 180° B1 for correct end points 5 -5 90 180 270 360 0 x y

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 12 Question Answer Marks Guidance 2 2 2 3 x x kx + = − ( ) 2 2 3 0 x k x    + − + =   M1 ( ) 2 2 12*0 k − − M1 Dep on previous M mark *Allow use of any inequality or equality 2 4 8*0 k k − − A1 k = 2 4 4 4 8 2  +  oe M1 Solve their 3-term quadratic equation in k k = 2 ± 12 or 4 48 2  oe A1 k < 2 − 2 3 , k > 2 + 2 3 oe A1 Allow 12 for 2 3 3(a) 2 2 2 log 2 log log 4 5 x x − = or = 2 1 log x M1 Correct use of change of base ( ) 2 2 2 log 4log 5 0 x x − −= A1 Arrange as 3-term quadratic equation in 2 log x ( )( ) 2 2 log 5 log 1 0 x x − + = or 2 log x = 2 4 4 4 ( 5) 2  −− M1 Dep on previous M mark 2 log x = 5 or −1 A1 For either x = 32 or x = 1 2 A1 Both Alternative method log log 2 4 5log 2 x x x x − = (M1) ( ) 2 5log 2 4log 2 1 0 x x + −= (A1) ( )( ) 5log 2 1 log 2 1 0 x x − + = or 2 4 4 4 ( 5) 10 log 2 x − −− = (M1) Dep on previous M mark 1 log 2 5 x = or −1 (A1) For either x = 32 or x = 1 2 (A1) Both

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 12 Question Answer Marks Guidance 3(b) 2 2 3 (7 ) ex x −− − = 25 oe M1 Arrange as equation with one power of e 2 2 10 ln25 x − = A1 2 ln25 10 2 x + = A1 x = ln25 10 2 +  oe or ln5 5  + A1 isw 4(a) Many-many (mapping) oe or A vertical line meets graph more than once (or twice) oe B1 4(b) ✓ ✓ ✓ ✓ ✓ 4 B1 for each correct row 4(c)(i) sin x ≥ 1 2 M1 May be implied. Allow > a = 150 seen A1 4(c)(ii) 0 ≤ gf(x) ≤ 1 2 2 B1 for each correct value with correct sign associated with it 4(c)(iii) No. Range of g  domain of g oe B1 5(a) 10 2 (6 + 9d) = 4 × 5 2 (6 + 4d) M2 M1 for either sum correct d = 6 A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 12 Question Answer Marks Guidance 5(b) ar2 = a + 4(ar − a) M1 For ar = a + d, 2 4 ar a d = + or for ( ) 2 ar a k ar a = + − (k > 2) with a correct attempt to eliminate d r2 − 4r + 3 = 0 A1 (r − 3)(r − 1) = 0 or formula M1 Dep for solving their 3-term quadratic r = 3 only A1 Alternative method 4 a d a d a a d + + + = (M1) d = 2a (A1) r = 2 a a a + (M1) Dep r = 3 (A1) 6 2 2 5 6 5 6 0 x x x x − =  − − = (x − 6)(x + 1) = 0 M1 or x = 5 25 4 ( 6) 2  −− x = 6 or −1 A1 2 2 5 6 5 6 0 x x x x − = −  − + = M1 (x − 3)(x − 2) = 0 M1 or x = 5 25 4 6 2  − x = 3 or 2 A1 Alternative method ( ) 2 2 5 36 x x − = (M1) 4 3 2 10 25 36 0 x x x − + − = (A1) (x − 3)(x − 2)(x − 6)(x + 1) = 0 (M1) Attempt to factorise into 4 brackets x = −1 or 2 or 3 or 6 (A1A1) A1 for any two

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 12 Question Answer Marks Guidance 7(a) 1 3 1 3 1 3 2 e (cos sin ) (sin cos ) ( 3)e (e ) x x x x x x x − − − − − + − 4 B1 for −3e1 − 3x B1 for cos sin x x − M1 for differentiation of quotient or equivalent correct product A1 FT on their −3e1 − 3x and their cos sin x x − 7(b) ( ) 2 sec 3 d x x  M1 Use of 2 2 sec 3 1 tan 3 x x = + 1 3 tan3x + c oe 2 A1 if + c omitted 8 ( ) 2 2 2 1 6 8 2 12 y y y y − − = − + oe M1 Equation in one variable only, in any form. Allow one arithmetical error   2 16 10 1 0 y y − + = A1 (2y − 1)(8y − 1) [= 0] M1 Attempt to factorise or use formula for their 3-term quadratic y = 1 2 , x = −2 y = 1 8 , x = 1 4 2 Correct pairs seen or implied A1 for any two correct values 9(a) As v–t graph but negative portion reflected in x-axis B1 9(b) Straight line with positive gradient B1 Straight line starting at O B1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 12 Question Answer Marks Guidance 10 Equation of normal y = −x + 3 soi 2 M1 for a correct attempt to obtain an equation of the normal using calculus and perpendicular gradient 2 3 9 18 x x x −+ = − + or 2 9 18 0 x x − + = M1 x = 3 [or 5] A1 5 2 3 ( 9 1 ) 3 8 ) d (x x x x   −+ −   +  −   3 5 2 2 3 3 9 3 18 2 2 x x x x x       −        + − − +             or 3 5 2 3 3 4 15 x x x   − + −     4 M1 for attempt to subtract and integrate Allow if order is reversed FT on their normal equation for M1 A1 for 3 2 3 9 18 2 x x x − + soi A1 for 2 3 2 x x − + soi Ignore limits at this stage ( ) 125 3 100 75 ( 9 36 45) − + − −−+ − oe 2 M1 dep for attempt to use their limits Allow one arithmetic error for M1 May include area of triangle rather than area using the normal 4 3 A1 Curve – line gives 4 3 − , final answer must be 4 3 Alternative method 2 9 18 0 x x − + = (M1) x = 3 [or 5] (A1) 5 2 3 ( 9 18) d x x x     − +  = 3 5 2 3 3 9 18 2 x x x   − +       (3) M1 for attempt to integrate A1 for one term correct soi Ignore limits at this stage 125 3 225 81 90 (9 54) 2 2    − + − − +     soi (2) M1 dep for attempt to use their limits Allow one arithmetic error for M1 Area of triangle = 1 2 2 2  (2) M1 for a complete method Shaded area = 10 3 − 2 = 4 3 (2) M1 for a difference between their areas Must be positive for A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 12 Question Answer Marks Guidance 11(a) (2 )! ! ( 4)! 6(2 3)! n n n n − − = B1 n(n − 1)(n − 2)(n − 3) = 2 (2 1)(2 2) 6 n n n − − 2 Dep on previous B1 B1 for simplification of one side e.g. 3(n − 2)(n − 3) = 2(2n− 1) 3n2 − 19n + 20 = 0 AG B1 Dep on all previous B marks With sufficient detail shown 11(b) (3n − 4)(n − 5) = 0 n = 5 only 2 M1 for attempt to solve the given equation 12 AB  = 7 4    − 1 7    6 3     =     −     M1 AC  = k 1 2    − 1 7    1 2 7 k k  −    =     −     M1 Use of AB  // AC  or AC mAB =  to obtain either 1 2 7 6 3 k k − − − = or 1 6 k m −= and 2 7 3 k m − = − oe M1 Form equation without vectors FT their AB  and AC  k = 3 A1 AC  = 3 1 2 3 7 −     −   = 2 1     −   or 1 3 m = M1 Substitute their k into their AC  AC : AB = 1 : 3 or 1 3 A1