Cambridge IGCSE Mathematics - Additional 0606 — 2025 May/June Paper 2 · Variant 3

This document has 16 pages. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/23 Paper 2 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. ● You should use a scientific calculator where appropriate. ● You must show all necessary working clearly. ● Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. ● For r, use either your calculator value or 3.142. INFORMATION ● The total mark for this paper is 80. ● The number of marks for each question or part question is shown in brackets [ ]. * 5 3 1 6 3 2 5 7 9 2 * DC (PQ/CT) 347695/1 © UCLES 2025 ,










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2 0606/23/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 0606/23/M/J/25 © UCLES 2025 [Turn over 1 Solve the inequality ( )( ) x x 2 4 5 0 G + - . [2] 2 Given that y x x 4 5 2 3 = - , show that x y d d can be written as ( ) x x 2 2 5 3 3 + . [3] 3 Variables x and y are such that when y is plotted against x3 a straight line graph passing through the points (2, 5) and (10, 21) is obtained. Find y in terms of x. [4] * 0000800000003 * ,










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4 0606/23/M/J/25 © UCLES 2025 4 The function f is defined by ( )x x 2 1 f = - for x R ! . (a) Explain why the function f 2 can be formed. [1] (b) On the axes, sketch the graph of ( ) y x f 2 = . State any intercepts with the coordinate axes. [4] O y x (c) It is given that ( )x ax b f 2 G + for x 1 3 G G - and for no other values of x. Find the values of a and b. [3] * 0000800000004 * ,










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5 0606/23/M/J/25 © UCLES 2025 [Turn over 5 In this question, all angles are in radians. (a) Write down the period of tan x 5 4 1 + b l . [1] (b) On the axes, sketch the graph of tan y x 5 4 1 = + b l for x 2 4 G G r r - . State the intercept with the y-axis. Show clearly the positions of any asymptotes. [3] 0 y x 2r – 2r 4r * 0000800000005 * ,










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6 0606/23/M/J/25 © UCLES 2025 6 (a) Five of the digits 1 2 3 4 5 6 7 8 9 are used to make a 5-digit number. Find how many ways this can be done when the 5-digit number is greater than 50 000. [2] (b) In a group of 13 people, 7 have red hair and 6 have brown hair. (i) The 13 people stand in line for a photograph. No person with red hair is standing next to another person with red hair. Find the number of different ways this can be done. [2] (ii) Chris chooses 5 people from this group. Find the number of ways Chris can do this if at least 1 person chosen has brown hair. [2] * 0000800000006 * ,










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7 0606/23/M/J/25 © UCLES 2025 [Turn over 7 (a) Find the term independent of x in the expansion of x x 3 2 4 15 - e o . [2] (b) In the expansion of ( ) ax 1 9 + the coefficient of x3 is 7 times the coefficient of x2. Given that a is a positive constant, find the value of a. [3] * 0000800000007 * ,










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8 0606/23/M/J/25 © UCLES 2025 8 (a) It is given that e tan y x x 3 2 = + . Use calculus to find the approximate change in y as x increases from 0.1 to . h 0 1+ , where h is small. [5] * 0000800000008 * ,










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9 0606/23/M/J/25 © UCLES 2025 [Turn over (b) A curve is such that ( ) sin x y x 3 d d r = + . The curve passes through the point , 9 3 4 r e o. Find the exact y-coordinate of the point on the curve where x 12 5r = . [4] * 0000800000009 * ,










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10 0606/23/M/J/25 © UCLES 2025 9 (a) The 1st term of an arithmetic progression is 9. The last term of this progression is 159. The sum of all the terms is 2604. The 12th term of this arithmetic progression is the 1st term of a geometric progression. The 8th term of this arithmetic progression is the 2nd term of the geometric progression. Find the sum of the first 6 terms of the geometric progression. [7] * 0000800000010 * ,










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11 0606/23/M/J/25 © UCLES 2025 [Turn over (b) A different geometric progression has 1st term sini. The common ratio of this progression is cosi where ° ° 45 135 G G i . (i) Show that this progression has a sum to infinity. [1] (ii) Show that this sum to infinity can be written as cosec cot i i + . [4] * 0000800000011 * ,









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12 0606/23/M/J/25 © UCLES 2025 10 In this question, time is in seconds. (a) At time t 0 = , particle P starts from the point with position vector j 30 - . P travels with speed ms 58 1 - in the direction i j 20 21 + . Find the position vector of P at time t. [3] (b) Also at time t 0 = , particle Q starts from the point with position vector i j 10 18 - + . Q travels with speed ms 75 1 - at an angle a above the positive x-axis, where tan 24 7 a = . Find the position vector of Q at time t. [4] * 0000800000012 * ,









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13 0606/23/M/J/25 © UCLES 2025 [Turn over (c) Determine whether P and Q collide. [2] 11 Given that 437 P P n n 1 5 3 = + , use an algebraic method to find the value of n. [4] * 0000800000013 * ,









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14 0606/23/M/J/25 © UCLES 2025 12 The function f is defined by ( ) ( . ) ( . ) cos sin x x x 15 3 1 5 7 3 1 5 13 f 2 = + + + - for . . x 0 3 0 5 G G - , where x is in radians. (a) Solve the equation ( )x 0 f = . [5] * 0000800000014 * ,









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15 0606/23/M/J/25 © UCLES 2025 [Turn over (b) Find the x-coordinates of the two stationary points on the curve ( ) y x f = . [5] Question 13 is printed on the next page. * 0000800000015 * ,









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16 0606/23/M/J/25 © UCLES 2025 13 Solve the equation 64 2 3 0 x x 3 3 1 + - = + . [4] 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 13 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/23 Paper 2 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.

0606/23 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.

0606/23 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.

0606/23 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

0606/23 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

0606/23 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

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 13 Question Answer Marks Partial Marks 1 –2 ⩽ x ⩽ 5 4 mark final answer B2 B1 for the correct critical values 2 Attempts to find y = 4x – 5x–2 M1 at least one term correct 3 d 4 10 d − = + y x x A1 Correct completion to given answer 3 3 2(2 5) + x x A1 Alternative version 4 4 4 d 12 8 10 d − + = y x x x x x or equivalent unsimplified form (2) M1 for an attempt at the quotient rule with correct structure ( )( ) ( ) 2 2 3 2 2 ( 12 ) 4 5 2 d d − − = x their x x their x y x x Correct completion to given answer 3 3 2(2 5) + x x (A1) 3 3 = + y mx c soi M1 m = 21 5 10 2 − − oe or 2 M1 c = 5 – 2(2) oe or 1 M1 FT their m y = (2x3 + 1)2 oe, isw A1 Alternative method 3 = + y mx csoi (M1) 21 10 = + m c and 5 2 = + m c and solves to find m = 2 or c = 1 (M1) Finds the other unknown (M1) FT their m or c y = (2x3 + 1)2 oe, isw (A1) 4(a) Range of f is equal to the domain of f oe 1

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 13 Question Answer Marks Partial Marks 4(b) 2 f ( ) 4 3 = − x x soi B2 B1 for correct order of composition Correct sketch with both intercepts indicated 3 4 3 O y x B2 FT their (4x – 3) providing B1 has been awarded and the expression is of the form mx + c B1 FT for correct sketch with one intercept correct 4(c) [When x = –1] y = 7 [When x = 3] y = 9 B1 FT their (4x – 3) providing it has been formed using a correct order of composition and the expression is of the form mx + c a = 9 7 3 1 − + oe or 7 = –a + b and 9 = 3a + b and solves correctly for one unknown M1 FT their y-coordinates a = 1 2 , b = 15 2 A1 5(a) 4 B1 5(b) Correct sketch with y-intercept and asymptotes marked -2π 2π 4π 1 x y 3 B1 for correct shape for both sections B2 for asymptotes drawn at –2, 2 and the y-intercept marked as 1; must have attempted correct shape or B1 for any two of these; must have attempted correct shape or If 0 scored then SC1 for correct graph between –2π and 2π with asymptotes marked at –2, 2 and the y-intercept marked as 1

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 13 Question Answer Marks Partial Marks 6(a) 8400 2 M1 for 5  8  7  6  5 oe 6(b)(i) 3 628 800 2 M1 for 7!  6! 6(b)(ii) 1266 2 M1 for 13 5 C – 7 5 C 7(a) –729 729 2 B1 for ( ) 5 10 15 2 5 4 3   −     C x x oe 7(b) 3 2 9(8)(7) 9(8) 7 6 2 =  a a oe M2 M1 for 3 3 9(8)(7) [ ] 6 a x oe or 2 2 9(8) 2     a x oe a = 3 nfww A1 8(a) 3 2 3 2 2 d 3e tan e sec d + + = + x x y x x x 3 B1 for 3 2 3 2 d(e ) 3e d + + = x x x M1 for correct structure of product rule A1 FT their 3 2 3 2 d(e ) 3e d + + = x x x for all other components of product rule correct Correct small changes relationship e.g. 0.1 δ d d = = x y y their h x M1 13.1h or 13.07[68…]h A1 dep on first three marks awarded 8(b) 1 : cos(3 π) 3 − + y x + c B2 B1 for 1cos(3 π) 3 − + x or for cos(3 π) + k x +c where k is1 3 or k < 0 4 1 3π cos π 3 3 9   = − + +     c M1 FT their k providing B1 has been awarded 1 5π 7 7 2 cos π 3 4 6 6   −   = − + + =         y or exact equivalent A1 9(a) n = 31 OR n = 150 + d d 2 M1 for   9 159 2604 2 + = n oe OR M1 for 159 9 ( 1) = + − d n oe

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 13 Question Answer Marks Partial Marks d = 5 2 M1FT for   31 159 9 30 = = + u d or   31 18 30 2604 2 + = d FT their derived value of n OR M1 FT for 1 150 150 18 1 2604 2   + +     + − =             d d d d d FT their derived expression for n in terms of d r = 11 16 oe, isw or 0.6875 2 M1FT   12 9 11( 5) or 64 = + u their and   8 9 7( 5) or 44 = + u their soi FT their value of d 6 6 64(1 0.6875 ) 1 0.6875   − = =   −   S 183 or 183.2 or 183.17[4…] A1 9(b)(i) cos 45 = 0.707… cos 135  = –0.707… therefore –1< cos < 1 oe OR –1< cos < 1 oe → 0 <  < 180 oe B1 9(b)(ii) sin 1 1 cos     =   − −   a r M1 sin 1 cos 1 cos 1 cos     +  − + A1 2 sin (1 cos ) 1 cos    + − = 2 sin (1 cos ) sin    + A1 1 cos cosec cot sin sin      + = + A1 10(a) [magnitude direction =] 2 2 20 21 + oe B1 [velocity vector =] ( ) 58 20 21 29 + i j M1 Correctly written column vectors are acceptable all through Q10 [position vector =] –30j + t(40i + 42j) or (40t)i + (–30 + 42t)j oe A1

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 13 Question Answer Marks Partial Marks 10(b) [direction vector =] 24i + 7j or x-component: cos = 24 25 and y-component: sin = 7 25 B1 [velocity vector =] ( ) 2 2 75 24 7 24 7 + + i j oe M1 FT their direction vector 72 21 + i j oe A1 [position vector =] –10i + 18j +t(72i + 21j) or (–10 + 72t)i + (18 + 21t)j A1 FT their velocity vector 10(c) Solves 40t = –10 + 72t and –30 + 42t = 18 + 21t or solves one equation and substitutes the value of t into the other equation M1 FT their position vectors of P and Q at time t Shows correctly that the values of t are not consistent e.g. 10 32 = t and 48 21 = t oe so do not collide oe A1 11 A correct and simplified equation e.g. 437 1 3 + = − n n B2 B1 for an equation with ( 1)! ! +   =     n n n + 1 and 437 or ( 3)! oe ( 4)!   − =   −   n n n – 3 and 437 OR for ( 1)! ! 437( 4)! ( 3)! + = − − n n n n n2 – 2n – 440 [= 0] B1 dep on B2 n = 22 as only solution B1 dep on all previous marks awarded

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 12 of 13 Question Answer Marks Partial Marks 12(a) 2 15(1 sin (3 1.5)) 7sin(3 1.5) 13[ 0] − + + + − = x x M1 2 15sin (3 1.5) 7sin(3 1.5) 2[ 0] + − + − = x x A1 (5sin(3 1.5) 1)(3sin(3 1.5) 2)[ 0] + + + − = x x M1 Factorises or solves their 3-term quadratic in sin(3x + 1.5) dep on previous M1 3x + 1.5 = – 0.201[3…] or 3x + 1.5 = 0.729[7…] or 3x + 1.5 = 2.41[1…] or 1 1 3 1.5 sin 5 −  + = −     x their or 1 2 3 1.5 sin 3 −  + =     x their M1 dep on previous M1 FT providing –1 ⩽ their 1 5 − ⩽ 1 and/or –1 ⩽ their 2 3 ⩽ 1 x = 0.304 or 0.3039[54…] and x = –0.257 or –0.2567[57…] and no extras A1

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 13 of 13 Question Answer Marks Partial Marks 12(b) 30(cos(3 1.5))( 3sin(3 1.5)) 21cos(3 1.5) [ 0] x x x + − + + + = M2 M1 for correctly differentiating cos(3x + 1.5) or sin(3x + 1.5) ( ) (3cos(3 1.5)) 7 30sin(3 1.5) [ 0] + − + = x x oe M1 FT a derivative of the form cos(3 1.5)sin(3 1.5) cos(3 1.5)[ 0] + + + + = a x x b x where a < 0 and b > 0 3x + 1.5 = π 2 or 1.57[0…] or ( ) 1 cos 0 − or 3x + 1.5 = 0.235[5…] or 1 7 3 1.5 sin 30 −  + =     their x their M1 dep on previous M1 FT (their 3)cos(3x + 1.5)  ( ) ( )   7 30 sin(3 1.5) − + their their x providing their 3  0 and their 30 > their 7 x = 0.0236 or 0.02359[87…] and x = 0.469 or 0.4686[96…] and no extras A1 dep on all previous marks awarded 13   2 4 8 8 3 0  + − = x x or ( )   2 3 3 4 2 2 3 0  + − = x x oe B1 ( )( ) 4(8 ) 3 8 1 − + x x or ( )( ) 3 3 4(2 ) 3 2 1 − + x x oe M1 FT their 3-term quadratic in 8x or 23x 3 8 4 = x or 3 3 2 4 = x oe, nfww A1 8 log (0.75) = x or 2 1log (0.75) 3 or –0.138[34…] oe A1 and no other solutions