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

This document has 16 pages. Any blank pages are indicated. [Turn over * 8 3 2 6 3 1 2 2 2 4 * Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 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/FC) 336597/1 © UCLES 2024 ,










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2 0606/12/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 0606/12/O/N/24 © UCLES 2024 [Turn over 1 The curve cos y a bx c = + , where a, b and c are integers, passes through the points r, 6 2 - - b l and r, 9 2 1 e o. The curve has a period of r 3 2 . (a) Find the values of a, b and c. [4] (b) Find the least value of y on the curve for r x 0 2 G G , and state the value of x at which this occurs. [3] * 0000800000003 * ,










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4 0606/12/O/N/24 © UCLES 2024 2 It is given that ( ) f y x = , where ( ) ( )( ) f x x x 2 5 1 2 = - - . (a) Find the coordinates of the stationary points on the curve ( ) f y x = . [4] (b) On the axes, sketch the graph of ( ) f y x = , stating the intercepts with the axes. [3] O x y (c) Hence find the values of k for which ( ) f x k = has exactly one solution. [2] * 0000800000004 * ,










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5 0606/12/O/N/24 © UCLES 2024 [Turn over 3 In this question, all lengths are in centimetres and all angles are in radians. O A B C 5 12 The diagram shows a circle with centre O and radius 12, and a circle with centre C and radius 5. The circles intersect at the points A and B, such that OA and OB are tangents to the circle with centre C. (a) Show that the obtuse angle ACB is 2.35 radians, correct to 2 decimal places. [2] (b) Find the perimeter of the shaded region. [2] (c) Find the area of the shaded region. [3] * 0000800000005 * ,










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6 0606/12/O/N/24 © UCLES 2024 4 The function f is such that ( ) ( ) f ln x x 4 3 2 = - , for x a 2 , where a is as small as possible. (a) (i) Write down the value of a. [1] (ii) Write down the range of f. [1] (iii) Find f 1 - ( )x , stating its domain and range. [4] (iv) On the axes sketch the graphs of ( ) f y x = and f y 1 = - ( )x , stating the intercepts with the axes. [4] O x y * 0000800000006 * ,










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7 0606/12/O/N/24 © UCLES 2024 [Turn over (b) Given that ( ) ( ) g x x 2 1 4 2 1 = + + , for x 0 2 , solve the equation ( ) gg x 9 = . [3] * 0000800000007 * ,










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8 0606/12/O/N/24 © UCLES 2024 5 (a) Show that cot cot sec 1 2 2 2 i i i + = . [1] (b) Write down the derivative of tani with respect to i. [1] (c) Using part (a) and part (b), find the exact value of cot cot sin 1 d 2 2 r 3 i i i i + - 0 c e dd f p . [4] * 0000800000008 * ,










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9 0606/12/O/N/24 © UCLES 2024 [Turn over 6 (a) Find, in descending powers of x, the first 3 terms in the expansion of x x 2 2 10 + e o . Simplify each term as far as possible. [3] (b) Find the term independent of x in the expansion of x x 4 2 1 2 2 8 + e o . [2] * 0000800000009 * ,










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10 0606/12/O/N/24 © UCLES 2024 7 It is given that ( ) ln y x x 2 3 1 2 = + - , for x 3 1 2 . When x 1 = , y is increasing at the rate of h units per second. Find, in terms of h, the corresponding rate of change in x, giving your answer in exact form. [6] * 0000800000010 * ,










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11 0606/12/O/N/24 © UCLES 2024 [Turn over 8 The tangent to the curve ( ) e y x 2 5 x 2 1 = + at the point where x 2 = meets the x-axis at the point X and the y-axis at the point Y. Find the coordinates of the mid-point of XY, giving your answer in exact form. [8] * 0000800000011 * ,









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12 0606/12/O/N/24 © UCLES 2024 9 O x y y x 2 1 4 = + y x 2 6 1 = + The diagram shows part of the curve y x 2 1 4 = + and the straight line y x 2 6 1 = + . Find the area of the shaded region, giving your answer in the form lna b + , where a is an integer and b is a rational number. [8] * 0000800000012 * ,









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13 0606/12/O/N/24 © UCLES 2024 [Turn over Continuation of working space for question 9. * 0000800000013 * ,









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14 0606/12/O/N/24 © UCLES 2024 10 (a) The first 3 terms of an arithmetic progression are tan x 2 2 , tan x 2 5 , tan x 2 8 . Find the values of x, where ° x 180 180° G G - , for which the sum to 30 terms is 455 3. [5] * 0000800000014 * ,









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15 0606/12/O/N/24 © UCLES 2024 (b) The first 3 terms of a geometric progression are r cos 5 2 2 i- b l, r cos 2 20 4 i- b l, r cos 2 80 6 i- b l, where r r 6 6 7 G G i - . Find the values of i for which this geometric progression has a sum to infinity. [6] * 0000800000015 * ,









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16 0606/12/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 10 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 2 of 10 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 3 of 10 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 4 of 10 Question Answer Marks Guidance 1(a) b = 3 B1 Use of y = a cosbx + c, with their b and either set of given coordinates M1 their 2 3   b c = –2 A1 a = 5 A1 1(b) Minimum when cos bx = −1 soi B1 Allow for their b, 2 3   b 3  = x B1 Allow if b is correct y = −7 B1 FT on their c – their a Alternative d 15sin3 d y x x = − (B1) FT on their a, b and c, 2 3   b When d d y x = 0, 3  = x (B1) Allow if b is correct y = −7 (B1) FT on their c – their a 2(a) ( ) 2 f ( ) 2( 1)(2 5) 2( 1)  = − − + − x x x x oe or 2 6 18 12 − + x x M1 For use of product rule or expansion and differentiation 2 2( 1)(2 5) 2( 1) 0 + − + − = x x x oe or 2 6 18 12 0 − + = x x oe M1 Dep for equating their quadratic f ( ) x to zero and attempt to solve to obtain x = … x =1, y = 0 x = 2, y = −1 2 A1 for any correct pair, must be from correct working only 2(b) 3 B1 for a correct cubic shape B1 for a correct cubic shape in the correct position, touching the x-axis once in the 4th quadrant and intersecting once with the positive x-axis B1 for all intercepts and no extras 2(c) 1 − k B1 0  k B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 5 of 10 Question Answer Marks Guidance 3(a) 1 12 2tan 5 −  =     ACB oe M1 (2 1.176 ) =  ACB = 2.35 to 2 dp A1 Must see justification to 2 dp 3(b) Arc length = 5 ACB B1 Perimeter = 35.8 B1 Allow awrt 35.8 3(c) Area = 2 1 (12 5) 5 2.35 2    −       M2 M1 for area of kite or area of sector M1 dep for kite area – sector area 30.6 A1 Allow greater accuracy Any use of fractions gets A0 4(a)(i) 2 3 B1 Allow 2 3  x , 2 3 = a , but not 2 3 = x unless it is replaced with a correct answer 4(a)(ii) oe B1 Must be using correct notation 4(a)(iii) 4 3 2 e − = x y or 4 3 2 e − = y x M1 For valid attempt to reach this stage 1 4 1 f ( ) e 2 3 −   = +       x x A1 Must be using correct notation Domain  x Range 1 2 f 3 − B2 B1 for each, must be using the correct notation. 4(a)(iv) 4 B1 for the shape of y = f (x) in the first and fourth quadrants only B1 dep on previous B1 for (1, 0) B1 for a correct shape for f −1 ( x), or FT on their y = f (x) with correct shape in first quadrant for symmetry about y = x soi B1 dep on previous B1, for (0, 1) and at least one point of intersection with y = f (x) correct in the first quadrant

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 6 of 10 Question Answer Marks Guidance 4(b) 1 1 2 2 2 (2 1) 4 1 4      + + + +           x B1 1 2 (2 1) 8 + = x B1 Dep x = 31.5 oe B1 Dep on both previous B marks Alternative g(x) = 9, x = 12 (B1) g(x) = 12 (B1) Dep x = 31.5 oe (B1) Dep on both previous B marks 5(a) 2 2 2 2 2 cosec 1 sin cot sin cos      =  2 2 1 sec cos   = = B1 Sufficient detail is needed Do not award if  is consistently omitted Alternative 1 2 2 2 2 2 sin cos sin cos sin      + 2 2 1 sec cos   = = (B1) Sufficient detail is needed Do not award if  is consistently omitted Alternative 2 2 1 1 cot  + 2 tan 1 + (B1) Sufficient detail is needed Do not award if  is consistently omitted 5(b) 2 sec  B1 5(c) ( ) 2 sec sin d    −  soi B1 tan cos   + B2 B1 for each 1 3 2 − or exact equivalent B1 Dep on 3 previous B marks 6(a) 10 7 4 20 180 + + x x x 3 B1 for each correct term

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 7 of 10 Question Answer Marks Guidance 6(b) ( ) 4 4 8 2 4 2 1 4 2       C x x M1 May be implied by working to obtain r = 4 1120 A1 From correct working 7 ( ) 2 2 2 6 ( 2) ln 3 1 d 3 1 d ( 2)   + − −     −   =   +   x x x y x x x or ( ) ( ) 1 2 2 2 6 ( 2) ( 2) ln 3 1 3 1 − − + − + − − x x x x x oe 3 B1 for 2 6 3 1 − x x M1 for correct attempt at differentiation of a quotient or a correct product A1 for all terms apart from 2 6 3 1 − x x correct. When x = 1, d 9 ln2 d 9 − = y x M1 For use of x = 1 in their d d y x , must see a substitution if in decimal form unless 0.923 obtained from a correct derivative d 9 d 9 ln2 = − x h t or exact equivalent 2 M1 for 9 ln2 9 −       h their , with x = 1 substituted in A0 if using small changes 8 1 1 2 2 d e (2 5) e (2 5) d − = + + + x x y k x x x M1 1 1 2 2 d e (2 5) e (2 5) d − = + + + x x y x x x A1 When x = 2, 2 d 10e d 3 = y x M1 Dep allow unsimplified Allow for using their d d y x When x = 2, y = 3e2 B1 Tangent: 2 2 10e 3e ( 2) 3 − = − y x M1 Allow for using their d d y x and their y When y = 0, x = 11 10 A1 Must be simplified Must be from correct work When x = 0, y = 2 11e 3 − A1 2 11 11e , 20 6   −     A1 FT on their coordinates for x and y, but must be exact and simplified

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 8 of 10 Question Answer Marks Guidance 9 8 6 1 2 1 = + + x x 2 12 8 7 0 + − = x x M1 Attempt to obtain a 3-term quadratic in one variable equated to zero. 1 2 = x 2 M1 dep for solution, see guidance ln(2 1) + k x M1 Area under curve =   1 2 0 ln(2 1) + their k x = ln(2( ) 1) ( 0) + − k their x M1 Dep on previous M1 for correct application of limits using their x, k and zero Allow unsimplified Area under curve = 2ln 2 A1 Not from incorrect work Area under straight line = 5 8 or 0.625 oe B1 Shaded area = ln4 – 5 8 A1 Not from incorrect work

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 9 of 10 Question Answer Marks Guidance 9 Alternative Either 8 6 1 2 1 = + + x x 2 12 8 7 0 + − = x x (M1) Attempt to obtain a 3-term quadratic in one variable equated to zero. 1 2 = x (2) M1 dep for solution, see guidance y = 2 (A1) Award only if attempt at integration with respect to y is subsequently seen Or 2 1 2 = − x y and 2 1 6 − = y x oe (M1) For rearranging both equations to obtain x or 2x in terms of y 2y2 + 2y – 12 = 0 (M1) Dep for attempt to obtain a 3-term quadratic in one variable equated to zero. y = 2 (2) M1 dep for solution, see guidance Then area enclosed between curve, y-axis and the line y = 2 = 4 2 1 ln 2   −    their k y y ln4 2 ln2 1 = −− + k k (M1) For correct application of limits using their y = 2, k and 4 Allow unsimplified 2ln2 – 1 (A1) Not from incorrect work Area enclosed by straight line, the y axis and the line y = 2 , 3 8 = (B1) Shaded area = ln4 – 5 8 (A1) Not from incorrect work 10(a) 30(4tan2 (29 3tan2 )) 455 3 2 +  = x x M1 For attempt to use sum formula with correct a and d 3 455 3 tan2 , 3 1365 = x A1 x = –165o, – 75o, 15o, 105o 3 M1 for 1 correct solution (allow if in radians or from use of tan 2x = 0.577 or tan 2x = 0.58 e.g. 14.99o, 15.06o.) A1 for a second correct solution A1 for 2 further correct solutions and no extras in the range

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 10 of 10 Question Answer Marks Guidance 10(b) 2 π 4cos 2    = −     r B1 2 π 4cos 1 2    −      or 2 π 1 4cos 1 2    − −      or 2 π 0 4cos 1 2    −      M1 For use of sum to infinity condition π 6 = − , π 6 , 5π 6 , 7π 6 2 M1 dep for one correct critical value A1 for all critical values and no extras in the range π 7π 6 6  − π π 6 6  −   (excluding 0) 5π 7π 6 6     (excluding ) 2 A1 for each correct set of values