Cambridge IGCSE Mathematics - Additional 0606 — 2024 Oct/Nov Paper 2 · Variant 3

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ * 3 7 9 8 1 0 3 5 0 2 * ADDITIONAL MATHEMATICS 0606/23 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 [ ]. DC (DE/CB) 336428/3 © UCLES 2024 ,










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2 0606/23/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 0606/23/O/N/24 © UCLES 2024 [Turn over 1 0 –1 –1 1 2 3 –2 –2 1 2 3 4 y x The diagram shows the graph of ( )( )( ) y x x x 1 1 2 = + - - . Use the graph to solve the inequality ( )( )( ) x x x 1 1 2 1 1 + - - . [3] * 0000800000003 * ,










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4 0606/23/O/N/24 © UCLES 2024 2 The function f is defined by ( ) f x x x 1 4 2 = - - for all real values of x. (a) Write ( ) f x in the form ( ) a x b 2 - + , where a and b are constants. [2] (b) Find the range of f. [1] The function g is defined by ( ) g x x x 1 4 2 = - - for x k H , where k is a constant. (c) State the least possible value of k such that g has an inverse. [1] (d) Using your value of k, find ( ) g x 1 - , stating its domain and range. [5] * 0000800000004 * ,










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5 0606/23/O/N/24 © UCLES 2024 [Turn over 3 (a) Show that ( )( ) tan sec tan sec tan 2 2 3 1 2 i i i i i+ - = - . [2] (b) Hence solve the equation ( )( ) tan sec tan sec 2 2 1 i i i i + - = for ° ° 0 180 G G i . [4] * 0000800000005 * ,










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6 0606/23/O/N/24 © UCLES 2024 4 The diagram shows a design for a logo. The logo is a sector of a circle, radius r cm, with angle a radians. r cm a The area of the logo is 9 cm2. (a) Show that the perimeter, P cm, of the logo is given by P r r 2 18 = + . [3] (b) Given that r can vary, find the stationary value of P and determine its nature. [5] * 0000800000006 * ,










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7 0606/23/O/N/24 © UCLES 2024 [Turn over 5 The tangent to the curve y x x 1 = + at the point where x 3 = meets the line y x 16 = - at the point A. Find the coordinates of A. [8] * 0000800000007 * ,










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8 0606/23/O/N/24 © UCLES 2024 6 (a) Find x x 3 2 1 d + y . [2] (b) Find, in terms of a, a e x d ( ) . x 1 2 0 5 - y . [3] * 0000800000008 * ,










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9 0606/23/O/N/24 © UCLES 2024 [Turn over 7 (a) In the expansion of x x2 8 + ` j in ascending powers of x, the 3rd and 6th terms are equal. Find the value of x. [3] (b) In the expansion of x x 2 n + b l in decreasing powers of x, the 6th term is a constant. (i) Find the value of the positive integer n. [2] (ii) Find the value of the 6th term. [2] * 0000800000009 * ,










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10 0606/23/O/N/24 © UCLES 2024 8 (a) Solve the equation sin x 4 2 1 = for r x 0 4 G G , giving your answers in terms of r. [2] (b) y x O y = sin 4x y 2 1 = The diagram shows parts of the graphs of sin y x 4 = and y 2 1 = . Find the exact area of the shaded region enclosed by the curve and the line. [5] * 0000800000010 * ,










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11 0606/23/O/N/24 © UCLES 2024 [Turn over 9 DO NOT USE A CALCULATOR IN THIS QUESTION. Write 2 10 16 11 10 1 + + + in the form p q 10 + , where p and q are integers. [4] * 0000800000011 * ,









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12 0606/23/O/N/24 © UCLES 2024 10 (a) Suzma is training for a marathon. In the first week she runs 10 km. Then each week she runs a distance that is 10% greater than the week before. The total distance that Suzma has run by the end of n whole weeks is more than 200 km. Find the smallest possible value of n. [4] (b) A geometric progression has 1st term a and common ratio r, where a 0 ! and r 1 ! . The 1st, 2nd and 3rd terms of the geometric progression are the 1st, 3rd and 7th terms of an arithmetic progression. Find the value of r. [4] * 0000800000012 * ,









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13 0606/23/O/N/24 © UCLES 2024 [Turn over 11 (a) There are 3 girls and 2 boys standing in a straight line. Find the number of possible orders in each of the following cases. (i) No girls are next to each other. [2] (ii) The 2 boys are not next to each other. [2] (b) 12 people, including Anjie and Bubay, are divided into 3 groups of 4 people. Anjie and Bubay must not be in the same group. Find the number of ways in which the 3 groups can be selected. [2] * 0000800000013 * ,









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14 0606/23/O/N/24 © UCLES 2024 12 A particle moves in a straight line. Its velocity, v ms 1 - , at time t seconds is given by cos sin v t t = - . (a) Find the acceleration, a ms 2 - , when r t 3 = . [2] The displacement of the particle from a fixed point O at time t is s metres. The particle passes through O when t 0 = . (b) Find the displacement at the time when the particle first changes direction after passing through O. [6] (c) Find an expression for a in terms of s. [1] * 0000800000014 * ,









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16 0606/23/O/N/24 © UCLES 2024 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. BLANK PAGE * 0000800000016 * ,









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This document consists of 9 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/23 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.

0606/23 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.

0606/23 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

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 4 of 9 Question Answer Marks Partial marks 1 −0.8, 0.55, 2.25 B1 x < −0.8 0.55 < x < 2.25 B2 B1 for either inequality correct 2(a) 5 − (x + 2)2 B2 B1 for −(x + 2)2 or (x + 2)2 or a = 5 and b = 2 2(b) f ⩽ 5 or f(x) ⩽ 5 B1 FT their 5 2(c) −2 B1 FT − their 2 2(d) Complete method to find inverse function: Swaps the variables and rearranges or rearranges and swaps the variables at some point in their solution M1 For M1 FT their part (a) provided it is in the form ( ) 2 a x b − + [g−1(x) =] – 2 + 5 x − or ( ) ( ) 2 1 4 4 4 1 1 g 2 x x − −+ − −   =   or ( ) ( ) ( )( ) 2 1 4 ( 4) 4 1 1 g 2 x x − −− − − − − −   =   − oe, isw A2 A1 for [g−1(x) =] – 2  5 x − or [g−1(x) =] ( ) ( ) 2 4 4 4 1 1 2 1 x − − − = or [g−1(x) =] ( ) ( )( ) ( ) 2 4 4 4 1 1 2 1 x −−  − − − = − [Domain:] x ⩽ 5 B1 FT their part (b) provided it is in form f(x) ⩽ a where a is a constant [Range:] g–1 ⩾ their −2 B1 FT their value of k in part (c) 3(a) 4tan2 θ − sec2 θ M1 e.g. 2 2 2 sin 1 4 cos cos    − Justified completion to given answer e.g. 4tan2 θ − (1 + tan2 θ) = 3tan2 θ − 1 or 3tan2 θ – (sec2 θ − tan2 θ) = 3tan2 θ – 1 A1 e.g. 2 2 2 2 2 sin sin cos 4 cos cos      + − = 4tan2 θ − tan2 θ – 1 = 3tan2 θ − 1 3(b) tan θ = [±] 2 3 oe M2 M1 for tan2 θ = 2 3 oe [θ =] 39.2 or 39.23 to 39.232 [θ =] 140.8 or 140.76 to 140.77 and no extras in range A2 A1 for either one correct, ignoring extras

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 5 of 9 Question Answer Marks Partial marks 4(a) Correct statement for area e.g. 2 1 2 r  = 9 or 1 2 rs = 9 B1 Finds an equation that can be used to eliminate  or s e.g. 2 18 r = or 18 r r = or 18 s r = M1 Correct substitution and completion to given answer e.g. 2 18 18 2 2 P r r r r r = +  = + A1 4(b) Correct derivative: 2 18 2 r − oe isw B1 2 18 2 r − = 0 and solves for r M1 FT their d d P r = 0 providing of the form 2 + k n r where n and k are non-zero integers r = 3 only soi A1 P = 12 only soi A1 [2nd derivative =] 3 36 r oe and which is positive for r = 3 →minimum or as r > 0 [2nd derivative =] > 0 →minimum or [2nd derivative =] 3 36 3 oe →minimum OR correctly finds the values of the first derivative at 3  h where h is small → minimum A1 dep on r = 3 and no other value

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 6 of 9 Question Answer Marks Partial marks 5 [When x = 3] y = 2 3 soi B1 Correct derivative: ( ) 1 2 2 1 1 1 2 x x x x −  + − + or ( ) ( ) 1 1 2 1 2 2 1 1 1 2 x x x x − − − − + +  + M2 M1 for an attempt to differentiate using the quotient rule oe Gradient of tangent: ( ) 1 2 2 1 3 4 4 2 3 −  − M1 dep on attempt at a derivative which includes ( ) 1 2 1 x − + −5 36 isw A1 dep on having a correct derivative 5 13 36 12 y x = − + oe or y – 2 3 = 5 36   −     (x − 3) oe soi A1 FT their non-zero 2 3 and their non- zero −5 36 A(15, –1) B2 B2 dep on all previous marks awarded B1 dep for x = 15 or y = −1 6(a) ( ) ( ) 1 2 2 3 2 3 x c + + oe 2 B1 for ( ) 1 2 3 2 k x + oe, soi where k is a constant and k ≠ 0 6(b) 1 2 e 1 2 2 a − − + oe, isw 3 B2 for ( ) 1 2 1 e 2 x − − oe or B1 for ke(1 − 2x) or ke(1 – 2a) where k is a constant, k ≠ 0 7(a) x = 3 28 56 oe, nfww, isw 3 B2 for 28x10 and 56x13 OR [x8](28x2) and [x8](56x5) or B1 for 28x10 or 56x13 OR [x8](28x2) or [x8](56x5) 7(b)(i) n = 10 2 B1 for the correct term in any form e.g. 5 5 5 2 n n C x x −      or for n − 5 = 5 oe, soi 7(b)(ii) 8064 2 B1 for 10C5 × 25 oe e.g. 252  32

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 7 of 9 Question Answer Marks Partial marks 8(a) [x =] 24 , 5π 24 and no extras in range 2 B1 for [x =] 24  or 5π 24 ignoring extras or for 4x = 6  and 5π 6 soi 8(b) Correct and complete plan soi e.g. 5 24 24 1 sin(4 ) d 2 their their x x     −      or 8 24 1 2 sin(4 ) d 2 their x x     −      M1 FT their limits in radians from part (a) or 5π 24 π 24 sin4 d their their x x  – 1 5π π 2 24 24 their their   −     Integrates sin 4x: 1 cos4 4 x − B2 B1 for cos4 k x where k < 0 or k = 1 4 Substitutes exact limits in correct order: 1 5π 1 π cos4 cos4 4 24 4 24       − −−             oe soi M1 FT their exact limits and their 1 cos4 4 x − providing at least B1 awarded 3 4 12  − or exact equivalent A1 9 18 12 10 2 10 + + B1 ( ) 18 12 10 2 10 2 10 2 10 their their + −  + − M1 FT 10 2 10 a b + + where a and b are integers 36 18 10 24 10 120 6 − + − − A1 14 −10 nfww A1 dep on all previous marks awarded Alternative 16 11 10 2 10 [ 1] 2 10 2 10 + −  + + − (M1) 32 16 10 22 20 110 [ 1] 6 − + − + − (A1)   78 6 10 6 6 − + − − oe (A1) e.g. 13 10 [ 1] − + 14 −10 nfww (A1) dep on all previous marks awarded

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 8 of 9 Question Answer Marks Partial marks 10(a) 1.1 *3 n B2 where * is any inequality sign or = B1 for ( ) 10 1.1 1 *200 1.1 1 n − − or ( ) 10 1 1.1 *200 1 1.1 n − − or for r = 1.1 soi nlog1.1 * log3 oe or log1.1 3 [* n] M1 FT 1.1 * 3 n their providing B1 has been awarded for a correct sum to n terms and (their 3) > 0 [n =]12 A1 dep on all previous marks awarded 10(b) r = 2 only nfww B4 B3 for a correct equation or equations which can be solved directly for r e.g.  ( ) ( ) 1 1 2 4 a r a r r d − − = = or  ( )  2 1 ( 1) 2 6 a r a r d − − = = or [a](r2 − 3r + 2) [= 0] oe or ( )( ) ( ) 2 1 1 6 1 r r r + − = − or r = 4 2 d d or   2 a d r a + = or   6 2 a d r a d + = + and a = 2d or B2 for a correct equation or equations which need to be rearranged to find r e.g.   2 2 4 ar a ar ar d − − = = or   2 2 6 ar a ar a d − − = = or    2 6 2 1 1 d d a r r = = − − or ( ) 1 2 a r d − = and ( ) 1 4 ar r d − = or either   2 a d r a + = or   6 2 a d r a d + = + and 4d 2 – 2ad [= 0] or B1 for ar = a + 2d oe and B1 for ar2 = a + 6d oe

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 9 of 9 Question Answer Marks Partial marks 11(a)(i) 12 2 B1 for 3! × 2! or 2 3 2 3 P P  oe 11(a)(ii) 72 2 B1 for 5! − 4! × 2! oe or 6 × 2! × 3! oe 11(b) 4200 2 B1 for 10 7 4 3 3 4 C C C       oe or 10 6 3 4 3 3 C C C       oe 12(a) Correct derivative: −sin t − cos t M1 1 3 2 + − oe or −1.37 or –1.366[02…] rot to 3 or more dp A1 Mark final answer 12(b) [v = 0 ] cos t − sin t = 0 t = 4  B2 B1 for cos t − sin t = 0 Correct integral: sin t + cos t (+ c) M1 0 = sin0 + cos0 + c M1 FT their attempt to integrate s = sin t + cos t −1 A1 2 − 1 oe, isw or 0.414[21…] A1 dep on all previous marks awarded Alternative [v = 0 ] cos t − sin t = 0 4 t  →= (B2) B1 for cos t − sin t = 0 Correct integral: sin t + cos t (M1) with limits t = 4  and t = 0 (A1) Must be in radians Substitutes limits into correct integral (M1) FT their t = 4  from attempt at solving v = 0 2 − 1 oe, isw or 0.414[21…] (A1) dep on all previous marks awarded 12(c) 1 s −− oe isw B1