Cambridge IGCSE Mathematics - Additional 0606 — 2025 Oct/Nov Paper 2 · Variant 3

This document has 16 pages. [Turn over * 2 6 6 3 0 7 4 5 7 0 * Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/23 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 [ ]. DC (DE/SW) 347515/3 © UCLES 2025 ,










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2 0606/23/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 0606/23/O/N/25 © UCLES 2025 [Turn over 1 It is given that ( )x ax x bx 7 9 p 3 2 = - - + , where a and b are constants. x 3 - is a factor of ( )x p . When ( )x p is divided by x 2 + the remainder is 35 - . Find the values of a and b. [5] * 0000800000003 * ,










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4 0606/23/O/N/25 © UCLES 2025 2 (a) (i) x y – 4 – 4 4 8 – 8 – 8 0 4 8 y = f(x) The diagram shows the graph of ( ) y x f = . On the same diagram sketch the graph of ( ) y x f 1 = - . [1] (ii) Describe the relationship between the graph of ( )x f and the graph of ( )x f 1 - . [1] * 0000800000004 * ,










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5 0606/23/O/N/25 © UCLES 2025 [Turn over (b) A function g is defined by ( )x g e x 2 = - for x 2 H . (i) Find an expression for ( )x g 1 - . [3] (ii) Write down the range of g 1 - . [1] (iii) A function h is defined by ( )x x 1 2 h 2 = + for x 0 2 . Find an expression for ( )x gh in its simplest form. [2] * 0000800000005 * ,










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6 0606/23/O/N/25 © UCLES 2025 3 (a) (i) Write x x 6 2 - - in the form ( ) x a b 2 + + where a and b are constants. [2] (ii) Hence write down the coordinates of the stationary point on the curve y x x 6 2 = - - . [2] (b) On the axes, draw the graph of y x x 6 2 = - - for x 4 4 G G - . [3] y x – 4 – 2 0 2 4 – 8 – 6 – 4 – 2 2 4 6 8 10 12 14 (c) Use your graph to solve the inequality x x 6 4 2 1 - - . [2] * 0000800000006 * ,










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7 0606/23/O/N/25 © UCLES 2025 [Turn over 4 (a) Integrate the following with respect to x. (i) e x 5 2 - [2] (ii) x 4 3 1 - where x 3 4 1 [2] (b) Show that sec x x 2 1 2 1 3 3 d 2 r r 3 2 = - b e l o y . [3] * 0000800000007 * ,










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8 0606/23/O/N/25 © UCLES 2025 5 Six different digits are chosen from the nine digits 1, 2, 3, 4, 5, 6, 7, 8, 9. These digits are used to form a 6-digit number. Find how many 6-digit numbers can be formed in the following cases. (a) There are no restrictions. [1] (b) The number is greater than 700 000. [2] (c) The number is greater than 750 000. [3] * 0000800000008 * ,










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9 0606/23/O/N/25 © UCLES 2025 [Turn over 6 The line y x 3 4 = + meets the curve y x x 2 8 1 2 = + + at two points A and B. Find the equation of the perpendicular bisector of AB, giving your answer in the form ax by c 0 + + = , where a, b and c are integers. [9] * 0000800000009 * ,










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10 0606/23/O/N/25 © UCLES 2025 7 Solve the equation sec tan x x 3 3 3 0 2 + - = for ° ° x 0 120 G G . [6] * 0000800000010 * ,










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11 0606/23/O/N/25 © UCLES 2025 [Turn over 8 In this question the units are metres. 2 3 B P A Q O 2 The diagram shows a circle, centre O and radius 2. The chord AB has length 2 3. The point Q lies on the circle such that AQ = BQ. The arc APB is part of a circle, centre Q. (a) Find the exact value of angle AQB in radians. [2] (b) Hence find the area of the shaded region. Give your answer in terms of r. [6] * 0000800000011 * ,









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12 0606/23/O/N/25 © UCLES 2025 9 Two variables, x and y, are related by an equation of the form y Axb = , where A and b are constants. The following pairs of values of x and y are given. x 0.61 4.48 12.18 33.1 y 1.65 4.47 7.39 12.17 (a) On the axes below, use these values to draw the straight-line graph of lny against lnx. [2] ln y ln x – 1 0 2 1 4 3 – 1 1 2 3 * 0000800000012 * ,









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13 0606/23/O/N/25 © UCLES 2025 [Turn over (b) Use your graph to find the values of A and b. [4] * 0000800000013 * ,









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14 0606/23/O/N/25 © UCLES 2025 10 In this question the units are metres and seconds. A particle moves along a straight line through a point A. Its displacement, s, from A at time t is given by s t t 2 100 10 100 2 = + + . The diagram shows the displacement–time graph for the first 30 seconds of the motion. s t 30 O (a) Find the value of t when s is a maximum. [6] * 0000800000014 * ,









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15 0606/23/O/N/25 © UCLES 2025 [Turn over (b) The particle passes through its starting point again at time t T = . (i) Find the total distance travelled by the particle during the first T seconds of its motion. [2] (ii) Use algebra to find T. [3] Question 11 is printed on the next page. * 0000800000015 * ,









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16 0606/23/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. 11 A circle has equation x y 25 0 2 2 + - = . A second circle has the same radius as the first circle, and the coordinates of its centre are both positive. The two circles intersect at the points A and B. The line AB has length 6 and is parallel to the line y x = - . Find the equation of the second circle in the form x y ax by c 0 2 2 + + + + = , where a, b and c are constants. [5] * 0000800000016 * ,









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This document consists of 15 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/23 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.

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 2 of 15 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 October/November 2025 © Cambridge University Press & Assessment 2025 Page 3 of 15 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 October/November 2025 © Cambridge University Press & Assessment 2025 Page 4 of 15 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 October/November 2025 © Cambridge University Press & Assessment 2025 Page 5 of 15 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 October/November 2025 © Cambridge University Press & Assessment 2025 Page 6 of 15 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 27a − 63 − 3b + 9 = 0 oe B1 −8a − 28 + 2b + 9 = −35 oe B1 Eliminates one unknown and solves for a or b M1 FT their equations provided both equations are linear with terms in a and b a = 2, b = 0 nfww A2 A1 for each

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 15 Question Answer Marks Guidance 2(a)(i) B1 • Correct inverse function drawn with correct shape over correct domain and range • 2nd intersection at roughly (4, 4) 2(a)(ii) Reflection in y = x oe B1 2(b)(i) Complete method including change of subject and swop of variables at some point M2 M1 for ln 2 = − y x or ln 2 = − x y soi ( ) ( ) 2 1 g ln 2 −     = + x x A1 Final answer 2(b)(ii) g−1(x) ⩾2 B1 2(b)(iii) 1 e x or e x or 1 e − x B2 Final answer B1 for 2 1 2 2 e + − x soi oe 3(a)(i) 2 1 25 2 4   − −     x oe isw B2 B1 for 2 1 2   −     x or for ( ) 2 25 4 + − x a OR B1 for 1 25 , 2 4 = − = − a b stated 3(a)(ii) ( ) , −their a theirb oe B2 STRICT FT from part (i) B1 STRICT FT for x-coordinate = –their a or for y-coordinate = their b

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 15 Question Answer Marks Guidance 3(b) Correct graph -4 -2 2 4 2 4 6 8 10 12 14 x y 0 B3 B2 for correct shape with cusps and intercepts at (–2, 0), (3, 0) and (0, 6) and maximum in 1st quadrant or B1 for correct shape with cusps on x-axis and incorrect intercepts or B1 for correct shape with no cusps (e.g. having minima) and intercepts at (–2, 0), (3, 0) and at (0, 6) 3(c) their −2.7 < x < −1, 2 < x < their 3.7 B2 STRICT FTB2 dep on a graph with correct intercepts earning at least B1 in part (b) and their curve crossing y = 4 in four places STRICT FTB1 for either ‘their −2.7 < x < −1’ or ‘ 2 < x < their 3.7’ OR B1 FT for their 4 critical values providing at least B1 awarded for part (b) soi 4(a)(i) 5 2 1e 5 − x (+ c) 2 B1 for 5 2 − x ke where k is a non-zero constant 4(a)(ii) 1ln(4 3 ) 3 − −x (+ c) 2 B1 for 1 ln 4 3 3 − −x or ( ) 4 3 − kln x where k is a non-zero constant 4(b) 1 2tan 2       x M2 M1 for 1 tan 2       k x where k is a non-zero constant Fully justified completion to given answer e.g. π π 2tan 2tan 4 6 − = 3 2 1 3   −       A1 Or 2(tan tan ) 4 6   − = 3 2 1 3   −       5(a) 60 480 Final answer B1

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 15 Question Answer Marks Guidance 5(b) 20 160 Final answer 2 M1 for 8 5 3 or 3 8 7 6 5 4   P or 9 6 1 3P oe 5(c) 16 800 Final answer 3 M1 for starts 7 then 5 or 6 or 8 or 9: 3360 M1 for starts 8, 9: 13 440 OR M2 their(b) – 3360 or M1 for starts 7 then 1 or 2 or 3 or 4: 3360 6 2 3 4 2 8 1 + = + + x x x oe M1 Eliminates one unknown   2 2 5 3 0 + − = x x or   2 – 2 –5 3 0 + = x x A1 Writes in solvable form (2x − 1)(x + 3) or (−2x + 1)(x + 3) oe M1 Uses quadratic formula or factorises their 3-term quadratic 1 11 , 2 2       (−3, −5) oe A1 Equation of their perpendicular bisector: 1 1 5 4 3 4   − = − +     y x or 1 1 3 6 = − − y x oe M3 FT their midpoint and perpendicular gradient from correct use of their(0.5, 5.5) and their (−3, −5) M1 FT for midpoint: (−1.25, 0.25) or ( ) 0.5 ( 3) 5.5 ( 5) , 2 2 + − + − their their their their M1 FT for perpendicular gradient: ( ) ( ) 1 1 5 5.5 3 3 0.5 oe or ⊥ − = − − − − − m their their their their OR M1 FT for equating lengths ( ) ( ) ( ) ( ) 2 2 2 2 3 5 – 0.5 – 5.5 + + + = + x y x y M1 FT for expanding soi 2 2 2 2 6 10 34 11 30.5 + + + + = + − − + x y x y x y x y

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 15 Question Answer Marks Guidance 6 2x + 6y + 1 = 0 final answer A2 A1 for correct equation in a different form isw e.g. 1 1 5 4 3 4   − = − +     y x or 1 1 3 6 = − − y x or 0.25 7 1.25 21 − = − + y x or 1 3 0 2 + + = x y or 2x + 6y = –1 oe OR A1 FT their perpendicular bisector for a final answer in the form 𝑎𝑥+ 𝑏𝑦+ 𝑐= 0 Dep on previous M1 M1 and an attempt to find the perpendicular bisector 7   2 0 ( 3 ) 1 tan 3 tan3 + + − = x x M1 Uses 2 2 sec 3 1 tan 3 = + x x 2 tan 3 tan3 2 + − x x [= 0] A1 Writes in solvable form (tan 3x + 2)(tan 3x − 1) [= 0] tan 3x = −2, tan 3x = 1 M1 Factorises or solves their 3-term quadratic in tan 3x Correct triple angle: 3x = 116.6 or 296.6 or 45 or 225 A1 [x =] 15, 75, 38.9 or 38.85[5…] 98.9 or 98.85[5…] and no extras in range 0 120 x  A2 A1 for any two correct angles, ignoring extras in range

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 15 Question Answer Marks Guidance 8(a) π 3 2 M1 for angle AOB: 2π 3 or angle AQO: π 6 or angle OAB: π 6 or e.g. cos (AOB) = 1 2 − or 1 sin 2  =     AOB 3 2 or AQ or BQ = 2 3 If 0 scored, SC1 for answer of 1.047 rads

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 15 Question Answer Marks Guidance 8(b) 2π 2 3 3 − or 6 3 2 3 − oe nfww 6 B5 correct calculation that simplifies to give shaded area e.g. ( ) 4 3 2 3 3 3    − − −     OR B1 for sector AOB: 2 1 2π 2 2 3   oe B1 for AOB or AOQ or BOQ: 2 1 2π 2 sin 2 3   or 1 2 2 3sin 2 6   oe B1 for sector AQB: ( ) 2 1 π 2 3 2 3   oe B1 for AQB: ( ) 2 1 π 2 3 sin 2 3 oe OR B2 FT their exact angles for a correct calculation for any two of these areas OR B2 for sector AOB: 4.188; AOB or AOQ or BOQ: 1.732; sector AQB: 6.282; AQB: 5.195[5] from using 1.047 rads OR B1FT their exact angles for a correct calculation for any one of these areas OR B1 for one of these areas using 1.047 rads 9(a) Points plotted at lnx –0.5 1.5 2.5 3.5 soi lny 0.5 1.5 2 2.5 and single, ruled, straight line of best fit drawn B2 B1 for at least 3 correctly plotted points

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 13 of 15 Question Answer Marks Guidance 9(b) Uses linear points to find b = awrt 0.50 soi 2 M1 for b = 2.5 0.5 3.5 0.5 − + or 2.5 2 3.5 2.5 − = − b oe OR M1 for 0.5 = –0.5b + c oe and 2.5 = 3.5b + c oe and correctly eliminates c. If 0 scored, SC1 for b = awrt 0.50 from use of given exponential equation Uses linear points to find 0.7 to 0.8 e = A or 2.01 to 2.23 isw soi 2 M1 for ln A = 0.75 If 0 scored, SC1 for A = 2.12 or 2.117[00…] or e0.75 from use of given exponential equation 10(a) Differentiates 2 2 100 : + t ( ) 1 2 2 1 2 100 4 oe 2 − +  t t or differentiates 1 2 2 (2 100) − + t : ( ) 3 2 2 1 2 100 4 2 −   − +      t t B2 B1 for ( ) ( ) 1 2 2 1 2 100 f 2 − +  t t OR for ( ) ( ) 3 2 2 1 2 100 f 2 −   − +      t t where f(t) is a function of t Correct structure of the quotient rule: ( )( ) ( ) ( ) ( ) 1 2 2 2 2 2 1 2 100 10 10 100 2 100 4 2 2 100 −   + − + +      + t t t t t OR Correct structure of product rule applied to ( ) 1 2 2 10 100 (2 100) − + + t t : (10t + 100) ( ) 3 2 2 1 2 100 4 2 −   − +        t t + 10 1 2 2 (2 100) − + t M1 FT their ( ) 2 d 2 100 d + t x or their ( ) 1 2 2 d 2 100 d −   +       t x as appropriate

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 14 of 15 Question Answer Marks Guidance 10(a) Correct d d s t : ( )( ) ( ) ( ) ( ) 1 2 2 2 2 2 1 2 100 10 10 100 2 100 4 2 2 100 −   + − + +        + t t t t t oe , isw OR (10t + 100) ( ) 3 2 2 1 2 100 4 2 −   − +        t t + 10 2 0.5 (2 100)− + t oe , isw A1 Equates their d d s t to 0 M1 Dep on previous M1 t = 5 nfww A1 Dep on all previous marks and sight of e.g. ( ) ( )  2 10 2 100 2 10 100 0 + − + = t t t or 1000 – 200t [= 0] or ( ) 2 1 2 100 10 100 5 + = + t t t 10(b)(i) ( ) 2 150 10 − oe, isw or 4.49 or 4.494 to 4.495 B2 Final answer B1 for t = 0, s = 10 and t = 5, s = 150 or 5 6 or 12.24[74] soi 10(b)(ii) 2 10 100 2 100 + + t t = their 10 M1 FT their 10 providing it is positive and greater than their 5 ( ) 2 2 2 (10 100) (10) 2 100 + = + t t M1 FT their10 Dep on previous M1 T = 20 A1

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 15 of 15 Question Answer Marks Guidance 11 Radius: 5 soi B1 Implied by 2 25 = r or 25 used as 2 r in their 2nd circle equation Centre: ( ) 4 2,4 2 soi oe B2 B1 for (5.66, 5.66) or better OR B1 for distance between centres is 8 soi 2 2 8 8 2 2     − + −         x y = 25 oe or ( ) ( ) ( ) 2 2 2 2 4 2 2 4 2 2 4 2 25 0 + − − + − = x y x y oe M1 FT their( ) 4 2,4 2 in form (k, k) where k is a non-zero constant e.g. 2 2 8 8 2 2     − + −         x their y their = 25 or 2 2 2 – 2 – 2 – 25 0 oe ( ) + + = x y theirx theiry their 2 2 + x y − 8 2 x − 8 2 y + 39 = 0 A1 Must be exact