Cambridge IGCSE Mathematics - Additional 0606 — 2024 Oct/Nov Paper 2 · Variant 1

This document has 16 pages. [Turn over * 6 0 1 4 4 8 2 5 3 5 * Cambridge IGCSE™ DC (DE/CB) 337403/3 © UCLES 2024 ADDITIONAL MATHEMATICS 0606/21 Paper 2 October/November 2024 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 calculator where appropriate. ● You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator. ● 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. INFORMATION ● The total mark for this paper is 80. ● The number of marks for each question or part question is shown in brackets [ ]. ,










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2 0606/21/O/N/24 © UCLES 2024 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax bx c 0 2 + + = , x a b b ac 2 4 2 ! = - - Binomial Theorem ( ) a b a a b a b a b n n n r b 1 2 n n n n n r r n 1 2 2 f f + = + + + + + + - - - e e e o o o where n is a positive integer and ( )! ! ! n r n r r n = - e o Arithmetic series ( ) u a n d 1 n = + - ( ) { ( ) } S n a l n a n d 2 1 2 1 2 1 n = + = + - Geometric series u ar n n 1 = - ( ) ( ) S r a r r 1 1 1 n n ! = - - ( ) S r a r 1 1 1 = - 3 2. TRIGONOMETRY Identities sin cos A A 1 2 2 + = sec tan A A 1 2 2 = + ec cos cot A A 1 2 2 = + Formulae for ∆ABC sin sin sin A a B b C c = = cos a b c bc A 2 2 2 2 = + - sin bc A 2 1 T = * 0000800000002 * ,










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3 0606/21/O/N/24 © UCLES 2024 [Turn over 1 Show that tan cot i i + can be written as sec cosec i i. [3] * 0000800000003 * ,










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4 0606/21/O/N/24 © UCLES 2024 2 (a) Given that tan y x x = - , find x y d d . Write your answer in terms of tanx. [2] (b) Hence find tan x x d r 2 0 4y . Give your answer in exact form. [2] * 0000800000004 * ,










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5 0606/21/O/N/24 © UCLES 2024 [Turn over 3 (a) Solve the equation - 8 2 8 1 x x 1 1 # - = . [4] (b) It is given that a b b 3 3 3 2 - = + - ` ` j j , where a and b are integers. Find the possible values of a and b. [6] * 0000800000005 * ,










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6 0606/21/O/N/24 © UCLES 2024 4 A C B a b c O The diagram shows the triangle OAC. The point B lies on AC such that AB : BC = p : q, where p and q are constants ( ) p q !- . a OA = , O b B = and O c C = . Show that b a c q p q p = + + . [5] * 0000800000006 * ,










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7 0606/21/O/N/24 © UCLES 2024 [Turn over 5 Given that ( ) ( ) log log log log log p a p 1 1 2 5 12 a p a a a + + - + + = , find the value of p. [5] * 0000800000007 * ,










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8 0606/21/O/N/24 © UCLES 2024 6 In this question all lengths are in metres. F 5 5 6 E A C B D O 3 The diagram shows a shape ABCDEF. AB, BD and DE are three sides of a rectangle. O is the mid-point of BD. AFE is an arc of a circle whose centre is O. AB 3 = , BC CD 5 = = and BD 6 = . (a) Find the exact value of the perimeter of the shape, giving your answer in terms of r. [5] * 0000800000008 * ,










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9 0606/21/O/N/24 © UCLES 2024 [Turn over (b) Find the exact value of the area of the shape, giving your answer in terms of r. [3] * 0000800000009 * ,










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10 0606/21/O/N/24 © UCLES 2024 7 A curve has equation cos y x x 2 = . The normal to the curve at (r, r 2 - ) meets the x-axis and y-axis at points P and Q. Find the exact area of triangle POQ. [7] * 0000800000010 * ,










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11 0606/21/O/N/24 © UCLES 2024 [Turn over 8 A particle moves in a straight line so that its displacement from a fixed point O at time t seconds is x metres, where x t t t 8 3 2 = + - + and t 0 H . (a) Find the time when the particle changes direction. [3] (b) Show that the particle is moving towards O when t 0 = . [3] (c) Find the total distance travelled by the particle during the first 2 seconds of its motion. [4] * 0000800000011 * ,









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12 0606/21/O/N/24 © UCLES 2024 9 A curve has equation y x x c 8 2 = - + , where c is a constant. (a) Find the value of c in each of the following cases. (i) The curve crosses the x-axis at x 2 = . [1] (ii) The minimum value of y is 3. [3] (b) Find the range of values of c for which y is always greater than 0. [2] * 0000800000012 * ,









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13 0606/21/O/N/24 © UCLES 2024 [Turn over 10 (a) A class contains 7 girls and 8 boys. A group of 6 is selected from the class. The group must contain at least 3 girls and at least 2 boys. Find the number of different groups that can be selected. [3] (b) A 5-character code is to be formed from the following characters. Letters A B C D E F Numbers 1 2 3 No character may be used more than once in any code. The characters may be arranged in any order. Find the number of different codes that can be formed using 4 letters and 1 number. [3] * 0000800000013 * ,









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14 0606/21/O/N/24 © UCLES 2024 11 (a) ( )x x x 1 f = - for , x x 10 10 1 ! G G - . The diagram shows the graph of ( ) y x f = . y x 10 – 10 0 – 5 – 10 – 5 10 5 5 (i) Use the diagram to explain why f is a function. [1] (ii) Find ( )x ff , giving your answer in its simplest form. [2] * 0000800000014 * ,









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15 0606/21/O/N/24 © UCLES 2024 [Turn over (iii) Using your answer to part (ii) state the relationship between the functions f and f 1 - . [1] (iv) Explain how the diagram shows the relationship between f and f 1 - . [1] (b) A function g is defined by ( )x x x 1 g = - for x 2 H . Find the range of g. [1] (c) A function h is defined by ( )x x x 1 3 2 h = + for the largest possible domain. State the domain of h. [1] Question 12 is printed on the next page. * 0000800000015 * ,









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16 0606/21/O/N/24 © UCLES 2024 12 Two arithmetic progressions, A and B, each have 100 terms. Their terms are denoted by a1, a2, a3, a4, … a100 and b1, b2, b3, b4, … b100 respectively. It is given that a b 1 1 100 = = and a b 298 100 1 = = . (a) Find n such that a b 45 n n - = . [6] (b) Find the smallest m such that a b 2 m m 2 . [3] 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 9 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/21 Paper 2 October/November 2024 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 2024 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 2024 © Cambridge University Press & Assessment 2024 Page 2 of 9 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 2024 © Cambridge University Press & Assessment 2024 Page 3 of 9 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. 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 October/November 2024 © Cambridge University Press & Assessment 2024 Page 4 of 9 Question Answer Marks Partial Marks 1 sin cos cos sin     + M1 2 2 sin cos cos sin     + A1 1 cos sin  oe and completion to given answer sec cosec   A1 Alternative 1 tan tan     + =   = 2 tan 1 tan + (M1) 2 sec tan   (A1) 2 sec cos sin    oe and completion to given answer sec cosec   (A1) 2(a) d d y x   =     tan2x nfww B2 B1 for sec2x − 1 2(b)   π 4 0 tan x x − M1 π 4 1− or exact equivalent A1 3(a)   2 1 1 8 8 2 0 x x  − − =     oe or   2 3 3 2 2 2 0 x x  − − =     oe B1 ( 1 8 x − 2)( 1 8 x + 1) = 0 oe or ( 3 2 x − 2)( 3 2 x + 1) = 0 oe M1 FT their 3-term quadratic in 1 8 x oe 1 8 x = 2 1 8 1 x   = −   or 3 2 x = 2 3 2 1 x    = − oe A1 x = 3 nfww A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 5 of 9 Question Answer Marks Partial Marks 3(b) 2 2 3 3 a a − + [= b + (3 − b) 3 ] B1 Equates coefficients to form two equations M1 FT their( 2 2 3 3 a a − + ) = b + (3 − b) 3 providing of equivalent difficulty a2 + 3 = b and −2a = 3 – b oe DM1 FT their( 2 2 3 3 a a − + ) = b + (3 − b) 3 a2 − 2a [= 0] or b2 −10b + 21 [= 0] DM1 a = 2, 0 or b = 3 or 7 A1 a = 2, b = 7 and a = 0, b = 3 A1 4 b − a = p p q + (c − a) oe AND Correct completion to given answer q p q p + = + a c b 5 B4 for b − a = p p q + (c − a) OR B1 for p AB AC p q = + soi B1 for AB = b − a soi B1 for AC = c – a soi Alternative 1 q(b − a) = p(c – b ) oe AND Correct completion to given answer q p q p + = + a c b (5) B4 for q(b − a) = p(c – b ) OR B1 for q AB = p BC soi B1 for AB = b − a soi B1 for BC = c – b soi Alternative 2 b = a + p p q + (c – a) oe or b = c + q p q + (a – c) oe AND Correct completion to given answer q p q p + = + a c b (5) B4 for b = a + p p q + (c – a) oe or b = c + q p q + (a – c) oe OR B1 for OB OA AB = + or OB OC CB = + soi B1 for p AB AC p q = + or q CB CA p q = + soi B1 for AC = c – a or CA = a – c soi

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 6 of 9 Question Answer Marks Partial Marks 5 loga 5( 1) 2 p p p + +       = loga 12 oe, nfww or 5( 1) 2 p p p + + = 12 oe, nfww M2 M1 for correct use of one log law in a correct equation e.g. log ( 1) log log ( 2) log 5 log 12 a a a a a p p p + + − + + = or log ( 1) log ( 2) log 12 log 5 a a a a p p p + − + = − 1 1 or log 2 log log 5 log 12 a p a a p p a + + + + = 5p2 − 7p − 24 = 0 A1 (5p + 8)(p − 3) = 0 or formula DM1 FT their 3-term quadratic p = 3 and no other solution nfww A1 6(a) OA: 2 3 or 12 soi B1 Correct method to find angle AOE e.g. 2tan(…) = 3 3 oe or 36 12 12 2(12)cos AOE = + − oe or π − 2tan−1 3 3 oe M1 Angle AOE: 2π 3 soi, isw A1 Arc AFE: their 2 3 × their 2π 3 soi or 4 3 3 π M1 FT their 2π 3 and their 2 3 Perimeter: 10 + 2 3 + 4 3 3 π or exact equivalent, cao A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 7 of 9 Question Answer Marks Partial Marks 6(b) Sum of three correct areas 4π + 12 + 3 3 cao 3 M2 FT their 2π 3 and their 2 3 for sector AOE: 1 2 × (their 2 3 )2 × their 2π 3 oe or 4π and 1 6 4 2  or 2× 1 2 ×3× 3 oe or M1 FT their 2π 3 and their 2 3 for sector AOE: 1 2 × (their 2 3 )2 × their 2π 3 oe or 4π 7 Attempts product rule M1 d d y x = 2cosx − 2xsinx oe A1 When x = π, d d y x = −2 DM1 FT their π d d x y x = Gradient of normal: 1 2 M1 FT π 1 d d x y t r x hei = − y + 2π = 1 2 (x − π) oe A1 FT their normal gradient (5π, 0) or (0, −5 2 π ) oe, soi A1 Area of triangle POQ: 2 25 π 4 nfww A1 8(a) 3t2 + 2t − 1 B1 (3t − 1)(t + 1) M1 FT their 3-term quadratic in t = 0 soi t = 1 3 and no other solutions A1 8(b) At t = 0, x = 8 and v = −1 B2 B1 for t = 0, x = 8 or t = 0, v = −1 Conclusion: Since x is positive and v is negative [the particle is moving towards O.] B1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 8 of 9 Question Answer Marks Partial Marks 8(c) t = 1 3 x = 211 27 or 7.814[814…] rot to 4 or more sf B1 Distance t = 0 to t = 1 3 : 8 – 211 27 or 5 27 or 0.1851[85…] rot to 4 or more sf and Distance t = 1 3 to t = 2: 18 – 211 27 or 275 27 or 10.18[51…] rot to 4 or more sf M2 M1 for distance t = 0 to t = 1 3 : 8 – 211 27 or 5 27 or 0.1851[85…] rot to 4 or more sf or t = 1 3 to t = 2: 18 – 211 27 or 275 27 or 10.18[51…] rot to 4 or more sf Total: 10.4 or 10.37[037…] rot to 4 or more sf A1 9(a)(i) c = 12 B1 9(a)(ii) d d y x = 2x − 8 = 0 or x2 − 8x + c = (x − 4)2 − 16 + c M1 3 = −16 + c or 3 = 42 − 8×4 + c DM1 c = 19 A1 9(b) c > 16 B2 B1 for c * 16 where * is = or an incorrect inequality sign 10(a) 7 8 7 8 3 3 4 2 C C C C  +  M2 M1 for 7 8 7 8 3 3 4 2 or C C C C   2940 A1 10(b) 6 4 3 5 P  oe M2 M1 for   6 4 1 P  or 6 4 3 P  or 6 4 5 P  oe 5400 A1 11(a)(i) Valid explanation e.g. The line x = k, where –10 ⩽ k ⩽ 10 cuts the curve in one point only B1 11(a)(ii) 1 1 1 x x x x − −− M1 Correct simplification to x A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 9 of 9 Question Answer Marks Partial Marks 11(a)(iii) Valid explanation e.g. f and f−1 are the same or f is self-inverse oe B1 11(a)(iv) Valid explanation e.g. The curve is symmetrical about line y = x B1 11(b) 1 < g ⩽ 2 B1 11(c) x ≠ −1 3 B1 12(a) dA = 3 and dB = −3 B2 B1 for dA = 3 or dB = −3 an = 1 + (n −1)×3 oe, isw or an = 3n − 2 B1 bn = 298 + (n −1)×(−3) oe, isw or bn = −3n + 301 B1 Solves 3n − 2 − (−3n + 301) = 45 oe to find a value of n M1 FT their an − (−3n + 301) or 3n – 2 –their bn n = 58 A1 12(b) 3m − 2 * 2(−3m + 301) oe where * is = or any inequality sign M1 FT their an and their bn from part (a) m * 67.1[11…] where * is = or any inequality sign A1 68 A1