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

This document has 16 pages. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/21 Paper 2 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. ● 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 [ ]. * 8 9 5 8 2 8 0 0 3 8 * DC (CJ/CT) 347115/2 © UCLES 2025 ,










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2 0606/21/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 0606/21/O/N/25 © UCLES 2025 [Turn over 1 The polynomial p is such that ( )x x ax bx 2 p 3 2 = + + - , where a and b are constants. It is given that: • x 2 + is a factor of ( )x p • when ( )x p is divided by x 3 - the remainder is 40. Find the values of a and b. [5] * 0000800000003 * ,










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4 0606/21/O/N/25 © UCLES 2025 2 y x 4 – 4 0 – 2 – 4 – 6 – 2 4 6 2 2 The diagram shows the graph of y x 3 3 = + . Use a graphical method to solve the inequality . x x 3 3 2 H + - [3] * 0000800000004 * ,










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5 0606/21/O/N/25 © UCLES 2025 [Turn over 3 (a) 3 men and 3 women are standing in a line. The 3 men are standing next to each other. Find how many different arrangements are possible. [2] (b) In the 13-letter word M U L T I B R A N C H E D, there are 4 vowels, U, I, A and E. 7 different letters are selected from these 13 letters. Find how many different selections are possible if the selection includes at least 2 vowels. [3] * 0000800000005 * ,










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6 0606/21/O/N/25 © UCLES 2025 4 (a) Show that cos sin sin cos sec x x x x x 1 1 2 - + - = . [3] (b) Hence solve the equation cos sin sin cos for 2 1 2 1 2 2 3 0 720 ° ° G G i i i i i - + - = . [3] * 0000800000006 * ,










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7 0606/21/O/N/25 © UCLES 2025 [Turn over 5 Solutions to this question by accurate drawing will not be accepted. A circle has centre (4, 2) and meets the x-axis at (-2, 0). (a) Find the equation of the circle. [2] (b) Find, in exact form, the coordinates of the points where the circle meets the y-axis. [3] * 0000800000007 * ,










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8 0606/21/O/N/25 © UCLES 2025 6 Variables x and y are such that when lny is plotted against x a straight-line graph is obtained. The line passes through the points ( , ) ln 1 15 and ( , ) ln 2 75 . Show that y Abx = where A and b are integers to be found. [4] * 0000800000008 * ,










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9 0606/21/O/N/25 © UCLES 2025 [Turn over 7 Show that the curve ( ) ln y x x x 2 2 = - + has exactly one stationary point. Find the x-coordinate of this point. [6] * 0000800000009 * ,










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10 0606/21/O/N/25 © UCLES 2025 8 In this question, the units are metres and seconds. A particle P is travelling in a straight line through a fixed point O. At time t its acceleration, a, is given by ( ) , where a t t 2 3 0 2 H = - . When t 3 = , P has a velocity of 6. (a) (i) Find an expression for the velocity, v, of P at time t. [3] (ii) Find the time when P is at rest. [3] * 0000800000010 * ,










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11 0606/21/O/N/25 © UCLES 2025 [Turn over When t 2 5 = , the displacement of P from O is 4. (iii) Find the displacement of P from O when t 3 = . [4] (b) Use calculus to find the approximate change in v when t increases from 2 5 by the small amount 0.02. [3] * 0000800000011 * ,









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12 0606/21/O/N/25 © UCLES 2025 9 (a) 8 cm 5 cm 5 cm rad θ The diagram shows an isosceles triangle. Find the value of i in radians. [2] (b) 8 cm This diagram shows a shape made of two arcs. Each arc is part of a circle with radius 5 cm. The height of the shape is 8 cm. Use your answer to part (a) to find (i) the perimeter of the shape [2] * 0000800000012 * ,









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13 0606/21/O/N/25 © UCLES 2025 [Turn over (ii) the area of the shape. [3] 10 In this question a, b and n are constants. When ( ) ax 5 2 n + is written in ascending powers of x, the first three terms are b x x 640 30240 2 2 + + . Find the value of a and the possible values of b. [5] * 0000800000013 * ,









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14 0606/21/O/N/25 © UCLES 2025 11 A cylinder has radius r cm and height h cm. The total surface area, including the two ends, is cm A 2. The volume of the cylinder is cm 330 3. (a) Show that r A r r 2 660 2 = + . [3] (b) Given that r can vary, find the value of r that gives a stationary value for A and show that this value is a minimum. [5] * 0000800000014 * ,









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15 0606/21/O/N/25 © UCLES 2025 [Turn over 12 In this question, the x- and y-directions are east and north respectively. The units are metres and seconds. Boat A starts from the origin O and moves with constant speed ms 3 5 1 - on a bearing of 030°. After 100 seconds boat B starts from point P, which has position vector 0 1000 e o . Boat B moves with constant speed ms 10 1 - on a bearing of 060°. (a) Find the velocity of each boat in vector form. [2] (b) Show that the two boats will collide. [5] Question 13 is printed on the next page. * 0000800000015 * ,









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16 0606/21/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 Solve the equation sin sin cos 2 3 3i i i = for r r 2 2 G G i - . [6] * 0000800000016 * ,









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This document consists of 13 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/21 Paper 2 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.

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 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.

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 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.

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 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

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 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

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 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 Question Answer Marks Guidance 1 –8 + 4a –2b – 2 = 0 oe B1 27 + 9a + 3b – 2 = 40 oe B1 Solves their linear equations in a and b to find the value of one unknown M1 a = 2 and b = −1 nfww A2 A1 for a = 2 or b = −1 2 Correct graph of y = | x – 2 | AND x ⩽ −2.5, x ⩾ − 0.25 3 B1 for correct graph of y = | x – 2 | and B1 for x ⩽ −2.5 or x ⩾ − 0.25 or B1 STRICT FT for their critical values from the two intersections of their straight-line section of graph providing it has negative gradient

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 13 Question Answer Marks Guidance 3(a) 4! × 3! oe M1 144 A1 3(b) 4C2 × 9C5 + 4C3 × 9C4 + 4C4 × 9C3 oe M2 M1 for one correct product 1344 A1 Alternative method 13C7 − (4C1 × 9C6 + 9C7) oe (M2) M1 for 13C7 − 4C1×9C6 or 13C7 − 9C7 oe 1344 (A1) 4(a) 2 2 (1 sin ) cos cos (1 sin ) − + − x x x x or 2 2 1 2sin sin cos cos (1 sin ) − + + − x x x x x oe M1 Common denominator 2 2sin cos (1 sin ) − − x x x oe A1 Use of cos2 x + sin2 x = 1 2(1 sin ) cos (1 sin ) − − x x x and correct completion to given answer, 2sec x A1 Fully justified with sufficient detail 4(b) 2 cos 2 3 = oe M1 At least one correct half angle: 48.19[89…] or 311.8[10…] A1 [ = ] 96.4 or 96.37[9…], 623.6 or 623.62[06…] A1 and no other angles in range 5(a) (x − 4)2 + (y − 2)2 = 40 oe, isw or 2 2 8 4 20 0 + − − − = x y x y oe, isw B2 B1 for r2 = 40 or r = 40 oe, soi 5(b) 16 + (y − 2)2 = 40 oe M1 FT their equation of the circle providing it is the equation of a circle (y − 2)2 = 24 or y2 − 4y − 20 = 0 A1 Obtain correct quadratic equation in any form (0, 2 ± 24 ) isw or equivalent exact form A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 13 Question Answer Marks Guidance 6 ln y = mx+ c soi B1 Correct method to find m e.g. m = ln75 ln15 2 1 − − oe or –m = ln15 – ln 75 oe OR m = ln5 oe M1 FT their c if necessary Correct method to find c e.g. ln15 = their(ln5) [1] + c oe or ln75 = their(ln5) (2) + c oe or ln75 = 2(ln15 – c ) + c oe OR c = ln 3 oe, soi M1 FT their m if necessary y = 3 × 5x nfww, mark final answer A1 7 d d y x = 1 − 2 2 2 2 + + x x x 2 B1 for 2 2 d( ln( 2 )) 1 f ( ) d 2 − + =  + x x x x x x soi Equates their first derivative to 0 and simplifies as far as 2(x + 1) = x (x+ 2) oe M1 FT their d d y x x2 − 2 = 0 or x2 = 2 A1 x = 2 [x = − 2 ] A1 Justification that the negative solution should be rejected e.g. x = − 2 results in y = 2 − ln(2 − 2 2 ) which is impossible B1 8(a)(i) [Velocity:] 3 (2 3) 3 2 −  t (+ c) M1 Allow for k(2t − 3)3 3 (2 3 3) 6 − + c = 6 [c = 3 2 ] M1 Correct calculation which can be used to find c [Velocity at time t] = 3 (2 3) 6 − t + 3 2 oe A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 13 Question Answer Marks Guidance 8(a)(ii) 3 (2 3) 6 − t + 3 2 = 0 M1 FT their expression for v in terms of t providing of the form 3 (2 3) − + t b a where a and b are non-zero constants 2t − 3 = 3 9 − or 2t − 3 = –2.08[0…]oe A1 t = 3 3 9 2 + −oe isw or t = 0.46[0] or 0.4599[…] A1 8(a)(iii) [Displacement:] ( ) 4 (2 3) 3 4 2 6 2 − + +  t t d M1 FT their expression for v in terms of t providing of the form 3 (2 3) − + t b a where a and b are non-zero constants 4 (2 2.5 3) 3 (2.5) 4 48 2  − + + = d M1 Correct calculation which can be used to find d [Displacement:] 4 (2 3) 3 1 48 2 12 − + − t t A1 [Displacement at t = 3] 293 48 oe isw or 6.1[0] oe A1 8(b) d d v t = (2t − 3)2 oe, soi or ( ) 2 d (2 3) d = − their v k t t where k is a non- zero constant B1 FT their expression for v in terms of t providing of the form 3 (2 3) − + t b a where a and b are non-zero constants, if necessary Correct small changes relationship: δ 0.02 = v their(2 × 2.5 − 3)2 oe M1 FT 2.5 d d = t v their t δv = 0.08 A1 9(a) 2sin−1 4 5       or 1 64 25 25 cos 2(5)(5) −  − −   −   oe M1 1.85 or 1.854 to 1.855 A1 9(b)(i) 5 × their 1.85 [×2] M1 FT their 1.85 providing it is less than 2 18.5 or 18.54 to 18.55 A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 13 Question Answer Marks Guidance 9(b)(ii) 1 2 × 52 × their 1.85 M1 FT their 1.85 providing it is less than 2 Correct plan e.g. 1 2 2         52 (their 1.85– sin(their 1.85) M1 dep previous M1 22.4 or 22.21 to 22.38 nfww A1 10 n = 7 soi B1 5 2 7 6 5 2 30240 2     = a oe B1 5 × 7 × 26 × a = b2 B1 a = 3 only, nfww B1 b = ± 6720 oe, isw, nfww B1 If first B1 only scored then SC1 for a =3 5 or 45 or 6.71 or 6.708[2…] only and SC1 for b = 1344 5  or ±54.8[2…] 11(a) Correct use of πr2h = 330 to find an expression that can be used to eliminate h e.g. 2 330 π = h r or 330 π = rh r M2 M1 for πr2h = 330 soi Correct substitution and completion to given answer e.g. 2 2 2 330 660 2π 2π 2π π +  = + r r r r r A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 13 Question Answer Marks Guidance 11(b) Correct derivative: 4πr − 660r−2, isw B2 B1 for one correct term 4πr − 2 660 r = 0 and solves for r M1 FT their derivative providing that at least one term is correct 3 660 4π = r oe, isw or 3.74 or 3.744[94…] rot to 3 or more dp A1 Second derivative 2 2 d d S r = 4π + 1320r−3 and When r = 3.74, > 0 oe [hence minimum] or 2 2 d d S r = 12 or 37 to 38 [hence minimum] or as r > 0, > 0 [hence minimum] OR correctly finds the values of the first derivative at 3.74 oe  h, where h is small [hence minimum] A1 dep on previous mark 12(a) 5 3cos60 5 3sin60         oe or 5 3 2 15 2         oe 10cos30 10sin30       oe or 5 3 5         oe 2 B1 for each or for correct components not stated in vector form

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 13 Question Answer Marks Guidance 12(b) 5 3 2 15 2         t oe and 0 5 3 ( 100) 1000 5    + −           t oe 2 B1 for each 5 3 2 t = 5 3 (t − 100) and 15 2 t = 1000 + 5(t − 100) oe OR finds the magnitude d, or d 2, for 5 3 2 15 2         t t – 5 3 500 3 5 500   −     +   t t and equates to 0 2 M1 FT for one correct equation from their displacement vectors or for their 5 3 2 15 2         t t – 5 3 500 3 5 500   −     +   t t t = 200 obtained from both equations OR as the only solution from d = 0 [hence collide] A1 Alternative method ( ) 5 3 2 15 2 100   +       t oe and 0 5 3 1000 5    +           t oe (2) B1 for each ( ) 5 3 2 100 + t = 5 3 t and ( ) 15 2 100 + t = 1000 + 5t oe OR finds the magnitude d, or d 2, for 5 3 2 250 3 7.5 750   +     +   t t – 5 3 5 1000       +   t t and equates to 0 (2) M1 FT for one correct equation from their displacement vectors or for their 5 3 2 250 3 7.5 750   +     +   t t – 5 3 5 1000       +   t t t = 100 obtained from both equations OR as the only solution from d = 0 [hence collide] (A1)

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 13 of 13 Question Answer Marks Guidance 13 sinθ(2sin2θ − 3cosθ) [= 0] B1 [sinθ = 0] θ = 0 and no other solutions in range from this factor B1 2(1– cos2θ) = 3cosθ oe M1 Write in solvable form: − 2cos2θ − 3cosθ + 2 [= 0] oe A1 Factorises or solves their 3-term quadratic in cosθ e.g. (2cosθ − 1)(cosθ + 2) [= 0] M1 θ = ± π 3 or ± 1.05 or ± 1.047[19…] and no other solutions in range from this factor A1