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

This document has 16 pages. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/21 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 1 3 3 6 7 0 5 5 3 * DC (CJ/CGW) 347692/1 © UCLES 2025 ,










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2 0606/21/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 0606/21/M/J/25 © UCLES 2025 [Turn over 1 A curve has equation y x x 2 3 2 = + - . (a) Use the method of completing the square to find the coordinates of the stationary point on the curve. [3] (b) On the axes, sketch the curve, stating the intercepts with the coordinate axes. [2] O y x * 0000800000003 * ,










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4 0606/21/M/J/25 © UCLES 2025 2 In this question, k is a constant. It is given that ( ) x k x k 2 3 2 0 2 + - + = has roots that are real and distinct. Find the set of possible values of k. [5] 3 A sports club has the following members. 5 runners, 4 swimmers, 3 gymnasts (a) These members stand in a straight line. Find the number of ways that this can be done when all the runners stand together, all the swimmers stand together and all the gymnasts stand together. [2] (b) Four of these members are selected for an event. Find the number of ways that this can be done when at least one runner, at least one swimmer and at least one gymnast must be selected. [3] * 0000800000004 * ,










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5 0606/21/M/J/25 © UCLES 2025 [Turn over 4 The coordinates of points A, B, C and D are as follows. ( , ) ( , ) ( , ) ( , ) A B C D 4 3 6 9 15 10 14 1 - - - The line L has equation y x 11 75 = - . The perpendicular bisector of the line AB meets L at the point E. Find the area of triangle CDE. [7] * 0000800000005 * ,










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6 0606/21/M/J/25 © UCLES 2025 5 Given that tan y x x 2 2 = , use calculus to find the approximate change in y as x increases from r 3 to r h 3 + , where h is small. [4] * 0000800000006 * ,










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7 0606/21/M/J/25 © UCLES 2025 [Turn over 6 (a) Using an appropriate quadratic factorisation, find the first three terms in the binomial expansion of ( ) x x 9 12 4 2 5 + + , in ascending powers of x. You must simplify your coefficients. [4] (b) Find the term independent of x in the expansion of x x 6 2 4 12 + 2 f p . [2] * 0000800000007 * ,










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8 0606/21/M/J/25 © UCLES 2025 7 (a) The function f is defined by ( ) f x 2 3 e x = + - for x R ! . On the axes, sketch the graph of ( ) f y x = and hence, on the same axes, sketch the graph of ( ) f y x 1 = - . Show clearly • the positions of any points where your graphs meet the coordinate axes • the positions of any asymptotes. [4] O y x * 0000800000008 * ,










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9 0606/21/M/J/25 © UCLES 2025 [Turn over (b) The function g is defined by ( ) g x 2 2 3 ex = - + for x 0 H . Given that g 1 - exists, find an expression for ( ) g x 1 - and state its domain. [4] * 0000800000009 * ,










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10 0606/21/M/J/25 © UCLES 2025 8 (a) Solve the equation ( ) cot cos x x 2 3 0 - = for r x 0 2 1 G . [3] * 0000800000010 * ,










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11 0606/21/M/J/25 © UCLES 2025 [Turn over (b) Solve the equation ( ) ( ) cosec sin 2 2 1 12 2 1 5 i i + - + = , where i is in radians and 1 2 G G i - . [6] * 0000800000011 * ,









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12 0606/21/M/J/25 © UCLES 2025 9 In this question, all lengths are in centimetres and all angles are in radians. 15 15 A and B B O O A 5 6r The diagram shows a sector AOB of a circle, centre O, radius 15. Angle AOB = r 5 6 . The sector is made into a cone with points A and B touching, as shown. (a) Find the curved surface area of the cone. [2] The top of the cone is a horizontal circle. (b) Find the circumference of the circular top. [2] (c) Hence find the radius of the circular top and the perpendicular height of the cone. [2] * 0000800000012 * ,









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13 0606/21/M/J/25 © UCLES 2025 [Turn over (d) Water is poured into the cone. When the depth of the water in the cone is h, the radius of the circular top of the water is r. (i) Find an expression for r in terms of h. [1] (ii) The water is poured into the cone at a constant rate of cm 27 3 per second. Find the rate at which the depth of the water is rising when the depth of the water is 4. [5] * 0000800000013 * ,









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14 0606/21/M/J/25 © UCLES 2025 10 A particle P moves in a straight line. t seconds after passing a fixed point, O, the acceleration of P, ms a 2 - , is given by . e for for a t t a t 2 0 4 19 5 4 10 t 2 8 2 G G G G = - = - - When t 3 = , the velocity of P is ms 3 1 1 - - and its displacement from O is m 4 1 - . (a) (i) Find the velocity of P when t 4 = . [3] (ii) Find the displacement of P from O when t 4 = . [3] * 0000800000014 * ,









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15 0606/21/M/J/25 © UCLES 2025 [Turn over (b) Find the displacement of P from O when t 10 = . [5] Question 11 is printed on the next page. * 0000800000015 * ,









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16 0606/21/M/J/25 © UCLES 2025 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. 11 A geometric progression has first term a and common ratio r, where r 0 2 . The sum of the 2nd and 3rd terms of the progression is 168. The sum of the 4th and 5th terms of the progression is 94.5. (a) Find the 6th term of the progression. [7] (b) Find the sum to infinity of the progression. [1] * 0000800000016 * ,









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This document consists of 12 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/21 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.

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 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.

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 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.

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 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 Method mark awarded three

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 12 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

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 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

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 12 Question Answer Marks Partial Marks 1(a) ( ) 2 1 4 x + − oe (−1, −4) B3 B2 ( ) 2 1 4 y x = + − oe or B1 ( ) 2 1 y x k = + + , where k is a constant 1(b) Correct curve x y −3 −3 1 O B2 B1 for correct shape in 4 quadrants with minimum in the 3rd quadrant B1 for an attempt at a quadratic graph with intercepts marked at –3 on the y-axis and –3 and 1 on the x-axis 2 ( )( ) 2 (3 2) 4 2 [*0 k k − − ] M1 Uses 2 4 b ac − * could be any inequality sign or =   2 9 20 4 *0 k k − + A1 Correctly simplifies to 3-term quadratic ( )( ) 9 2 2 k k − − M1 FT Factorises or solves their 3-term quadratic Critical values 2 9 , 2 A1 k < 2 9 k > 2 A1 Mark final answer 3(a) 103 680 2 M1 for k  5!  4!  3! where k = 1, 2, 3, 6 3(b) 270 3 M2 for 5 4 3 5 4 3 1 2 1 1 1 2 5 4 3 2 1 1 C C C C C C C C C   +   +   oe OR M1 for at least one correct product If 0 scored then SC1 for a final answer of 540 or 264

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 12 Question Answer Marks Partial Marks 4 mAB = 9 3 6 4 −− −− oe isw or 6 5 − M1 [m⊥ =] 5 6 or 1 AB their m − M1 FT their mAB providing that difference in difference in y x attempted [Midpoint] 4 6 3 9 , 2 2 −+ −       oe isw or (1, 3) − M1 Finds equation of perpendicular bisector: ( ) 1 ( 3) 1 6 5 − − = − −   −     y their x their their oe M1 FT their perpendicular gradient providing gradient is 1 AB their m − and using their midpoint Their midpoint must not be any of A, B, C or D 5 23 11 75 6 6 x x − = − M1 Eliminates one variable dep on previous M mark 11x – 75 = their perp bisector equation E(7, 2) A1 [Area CDE =] 40 A1 Dependent on all previous marks 5 2 2 d 2 tan sec d 2 2 2 y x x x x x = + oe 2 B1 for 2 d 1 tan sec d 2 2 2 x x x  =     soi or M1 FT for a product rule of correct structure with their d tan d 2 x x       3 x y dy their h dx   = = or 3 x dy their h dx  =  M1 Correct small changes relationship 1.94h A1 Dependent on B2 6(a) (3x + 2)10 oe M2 M1 for (3x + 2)2 soi 1024 + 15 360x + 103 680x2 A2 A1 for any two terms correct or 10 9 8 2 2 10(2 )(3 ) 45(2 )(3 ) x x + + or for 3 correct terms seen but not summed e.g. a list After M2 A0, allow SC1 for an answer of 59 049x10 + 393 660x9 + 1 180 980x8

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 12 Question Answer Marks Partial Marks 6(b) 51 963 120 2 M1 for 4 8 4 12 4 2 6 2 x C x           soi 7(a) Correct pair of graphs with both asymptotes and both intercepts correctly shown 4 B1 for correct shape for f; must have positive y-intercept and appear to tend to an asymptote in the 1st quadrant B1 for (0, 5) indicated; must have attempted correct shape for f B1 for asymptote at y = 3; must have attempted correct shape for f B1 for a correct reflection of their f in the line y = x y = x does not have to be drawn Maximum of B3 if not fully correct 7(b) 3 e 2 2 y x = − − oe OR 3 e 2 2 x y = − − oe and variables swapped at some point M1 Rearranges to ey = … e.g. 2 1 e 2 y x x − = − or 1 2 e 2 − = − y x x or 3 e 2 2 − = − − y x 3 ln 2 2 x   −   −   oe, isw A1 1 ⩽ x < 2 B2 B1 x ⩾ 1 oe OR x < 2 oe OR 1< x ⩽ 2 8(a) 3 cos 0 or tan 2 x x = = oe M1 [x =] π or 1.57[07...] 2 0.983 or 0.9827[93…] A2 and no extra solutions in range A1 for either ignoring extras 8(b) 2 12sin(2 1) 5 sin(2 1)   − + = + or 12 2cosec(2 1) 5 cosec(2 1)   + − = + M1 • Writes in terms of cosec(2 +1) or sin(2 +1) only. 5 3 5 3 O y x f ( ) y x = 1 f ( ) y x − = y = x

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 12 Question Answer Marks Partial Marks   2 12sin (2 1) 5sin(2 1) 2 0   + + + − = or   2 2cosec (2 1) 5cosec(2 1) 12 0   + − + − = oe A1 4 sin(2θ + 1) −1)(3 sin(2θ + 1) + 2) [= 0] or (2 cosec(2θ + 1) + 3)(cosec(2θ + 1) − 4) [= 0] M1 Factorises or solves their 3-term quadratic in sin(2 +1) or cosec(2 +1) dep on previous M1 2 +1 = 0.253 or 0.2526[80…] OR 2 +1 = – 0.73[0] or – 0.7297[27…] OR 2 +1 = 1 1 sin 4 their −      OR 2 +1 = 1 2 sin 3 their −  −     M1 dep on previous M1 FT providing 1 1 1 4 their − and/or 2 1 1 3 their − − – 0.374 or – 0.3736[59…] 0.944 or 0.9444[56…] – 0.865 or – 0.8648[63…] 1.44 or 1.435[66…] A2 and no extra solutions in range A1 for two correct solutions ignoring extras in range 9(a) 135 isw or 424 or 424.1 to 424.2 2 M1 for 2 1 6π 15 2 5   oe 9(b) 18 or 56.5 to 56.6 2 M1 for 6π 15 5  oe 9(c) radius = 9, height = 12 2 B1 for radius = 9 OR M1 for a correct Pythagoras’ statement e.g. their r2 + height2 = 152 9(d)(i) 3 4 r h = oe B1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 12 Question Answer Marks Partial Marks 9(d)(ii) 2 1 3 π 3 4 V h h   =      oe B1 FT their r = kh where k is a positive constant e.g. ( ) 2 1 3 V their kh h  =  2 d 9π d 16 V h h = oe B1 FT their k where k is a positive constant e.g. ( ) 2 2 d π d V their k h h =  Relevant, correct chain rule including d d h t soi d d d d d d h V h t t V =  oe B1 e.g. d d d d d d h V V t t h =  16 27 9π 16   oe M1 0.955 or 0.9549[29…] A1 10(a)(i) ( ) 3 [ ] 2 3 t v t c = − + M1 Or   4 3 3 2 3 t t   −     3 10 [ ] 2 3 3 t v t = − − A1 Or substituting in limits 3 3 4 3 1 2 4 2 3 3 3 3     − − − −         v = 10 A1 10(a)(ii) 2 correct terms in ( ) 4 2 10 [ ]12 3 t s t t d = − − + M1 FT an expression in form 3 at bt + Or 2 correct terms in   4 4 2 3 10 12 3 t t t   − −     1 4 2 10 12 12 3 t s t t = − − + A1 A1 substituting in limits 4 4 2 2 4 10 3 10 1 4 4 3 3 12 3 12 3 4     − −  − − −  −         s = 4 A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 12 of 12 Question Answer Marks Partial Marks 10(b) ( ) 8 2 5 [ ]19 e 2 t v t c − = + + oe M1 8 2 5 137 [ ]19 e 2 2 − = + − t v t oe A1 FT (their 10) – 78.5   ( ) 2 8 2 19 5 137 e 2 4 2 t t s t d − = − − + oe M1 dep on previous M1 FT their 137 2 − provided not 0 or algebraic   2 8 2 19 5 137 509 e 2 4 2 4 t t s t − = − − + oe, cao A1 s = 392 or 392.2[49…] A1 11(a) 2 168 ar ar + = or 3 4 94.5 ar ar + = oe soi B1 ( ) ( ) 3 1 94.5 1 168 ar r ar r + = + oe M1 Correct equation where a can be directly eliminated 168r2 = 94.5 oe M1 r = 3 4 only nfww A1 3 9 168 4 16 a a + = oe M1 FT their r a = 128 A1 6 243 8 u = only oe, isw B1 11(b) 512 B1