Cambridge IGCSE Mathematics - Additional 0606 — 2024 May/June Paper 2 · Variant 2

This document has 16 pages. Any blank pages are indicated. [Turn over * 6 3 8 4 3 5 3 9 0 0 * Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/22 Paper 2 May/June 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/SG) 332420/2 © UCLES 2024 ,

  
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2 0606/22/M/J/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 = * 0019655485302 * ,

  
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3 0606/22/M/J/24 © UCLES 2024 [Turn over 1 (a) On the axes, sketch the graph of ( )( )( ) y x x x 2 5 3 1 = - + - , stating the intercepts with the coordinate axes. [3] O y x (b) Hence (i) solve the inequality ( )( )( ) x x x 2 5 3 1 0 G - + - [2] (ii) on the axes below, sketch the graph of ( )( )( ) y x x x 2 5 3 1 = - + - . [1] O y x * 0019655485303 * ,

  
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4 0606/22/M/J/24 © UCLES 2024 2 (a) Evaluate cos x x 4 d r r 3 2y . You must show all your working. [4] (b) Find x x x 4 3 1 1 d 3 - + e o y . [3] * 0019655485304 * ,

  
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5 0606/22/M/J/24 © UCLES 2024 [Turn over 3 (a) Determine whether the equation ( )( ) x x x x 5 3 4 1 3 2 1 - + + = + has two distinct real roots, two equal roots or no real roots. [4] (b) Solve the equation x x 12 4 3 3 - = . [4] * 0019655485305 * ,

  
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6 0606/22/M/J/24 © UCLES 2024 4 The polynomial p is such that ( )x x x x 6 12 5 p 3 2 = + - + . (a) Find the remainder when ( )x p is divided by x 2 - . [1] (b) (i) Show that x 2 1 - is a factor of ( )x p . [1] (ii) Hence write ( )x p as a product of linear factors. [3] (iii) Hence solve the equation sin sin sin 6 12 5 0 3 2i i i+ - + = for ° ° 0 90 G G i . [2] * 0019655485306 * ,

  
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7 0606/22/M/J/24 © UCLES 2024 [Turn over 5 A curve has equation y 5e e x 2 1 = + - . The tangent to the curve at the point where x 1 = cuts the x-axis at the point P. Find the equation of the tangent in the form y mx c = + , where m and c are exact values, and hence find the x-coordinate of P. [6] * 0019655485307 * ,

  
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8 0606/22/M/J/24 © UCLES 2024 6 (a) Show that sin cot cosec x x x 3 b l can be written as sin tan x x 2 . [3] (b) Solve the equation tan tan cos x x x 2 1 0 2 - = for r r x 1 1 - . [5] * 0019655485308 * ,

  
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9 0606/22/M/J/24 © UCLES 2024 [Turn over 7 Find the number of different ways the 9 letters of the word POLYMATHS can be arranged when (a) the O and A are not next to each other [2] (b) the letters MATHS are together in this order. [2] * 0019655485409 * ,

  
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10 0606/22/M/J/24 © UCLES 2024 8 An experiment was carried out and values of y for certain values of x were recorded. The table shows the values recorded. x 15 30 45 60 75 y 10 13 22 35 50 The relationship between y and x is modelled by y Aekx = , where A and k are constants. (a) Draw a straight line graph for lny against x. [2] ln y x 0 15 30 45 60 75 1 2 3 4 * 0019655485410 * ,

  
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11 0606/22/M/J/24 © UCLES 2024 [Turn over (b) Find the equation of the line in part (a) and hence find the values of A and k. Give each value correct to 1 significant figure. [5] (c) Find the value of x for which y 17 = . [2] * 0019655485411 * ,

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12 0606/22/M/J/24 © UCLES 2024 9 y x O y x x 32 4 48 2 = - - B(5, 12) D C A The diagram shows part of the curve y x x 32 4 48 2 = - - and the line AB. The curve and the line AB meet the x-axis at A and meet again at the point B(5, 12). The line CD extended is parallel to the y-axis and passes through the maximum point of the curve. Find the area of the shaded region. [9] * 0019655485412 * ,

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13 0606/22/M/J/24 © UCLES 2024 [Turn over Continuation of working space for Question 9. * 0019655485413 * ,

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14 0606/22/M/J/24 © UCLES 2024 10 The functions f and fg are defined by ( )x f ex 3 2 = + for x 0 1 ( )x fg e x 2 = for x 2 3 2 . (a) Explain why f 1 - exists. [1] (b) Find an expression for ( )x f 1 - and state the domain and range of f 1 - . [5] (c) Hence find and simplify an expression for ( )x g . [2] * 0019655485414 * ,

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15 0606/22/M/J/24 © UCLES 2024 11 In the binomial expansion of x 2 2 n + b l , the first three terms in increasing powers of x are b abx abx 8 9 2 + + . Find the values of the constants n, a and b. [8] * 0019655485415 * ,

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16 0606/22/M/J/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 * 0019655485416 * ,

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This document consists of 12 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/22 Paper 2 May/June 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 May/June 2024 series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 2 of 12 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.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 3 of 12 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.

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 4 of 12 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 Partial Marks 1(a) Correct graph and intercepts O x y -3 1 2.5 -15 B3 B1 for correct shape; the ends must extend above and below the x-axis B1 for correct roots indicated; must have attempted a cubic shape B1 for correct y-intercept indicated; must have attempted a cubic shape

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 5 of 12 Question Answer Marks Partial Marks 1(b)(i) 3 ⩽ x ⩽ 1 , x ⩾ 2.5 mark final answer B2 FT their (a) providing it is an equivalent cubic shape and has 3 stated or indicated roots for B2, B1 or SC1 B1 for one correct inequality out of two If 0 scored then SC1 for 3 < x < 1 , x > 2.5 or 3 < x ⩽ 1 , x > 2.5 or 3 ⩽ x < 1 , x > 2.5 1(b)(ii) Graph of correct shape, with cusps, positive y-intercept and x-intercepts which match (a) B1 FT their (a) providing it is an equivalent cubic shape 2(a) 2 3 4sin 4 x         B2 B1 for sin 4 x k where k > 0 or k = 4 4sin 4sin 8 12    M1 FT provided at least B1 awarded 0.495 or 0.4954[57…] rot to 4 or more sf A1 dep on all previous marks awarded 2(b) 2 1 ln(4 3) ( ) 4 2 x x c     oe, isw or 2 1 ln( 0.75) ( ) 4 2 x x c     oe, isw B3 B2 for 1 ln(4 3) 4 x  or 1 ln( 0.75) 4 x  or B1 for 1 ln4 3 4 x  or 1 ln 0.75 4 x  or ln(4 3) k x  or ln( 0.75) k x  where k ≠ 1 4 and B1 for 2 2 x  oe

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 6 of 12 Question Answer Marks Partial Marks 3(a)   2 7 9 5 0 x x    or   2 7 9 5 0 x x     B2 B1 for two terms correct in 2 7 9 5 0 x x    or at most one term incorrect in 2 2 12 11 2 5 2 3 x x x x      oe 29 4(7)(5)  or 2 ( 9) 4( 7)( 5)     oe M1 FT their 3-term quadratic 59 and no real roots or 81 – 140 < 0 and no real roots oe A1 3(b)   2 3 3 4 12 0 x x    oe soi or 3 and y x  2 4 12 0 y y   oe soi B1 Factorises or solves their 3-term quadratic in 3 x M1 FT their 3-term quadratic in 3 x or a stated substituted unknown 8 216 x x   A2 A1 for 3 3 2 6 x x   4(a) 33 B1 4(b)(i) 3 2 1 1 1 6 12 5 0 2 2 2                       oe or 1 1 12 6 5 0 8 4 2          oe or 3 1 6 5 0 4 4     oe B1 4(b)(ii) Finds the quadratic factor 3x2 + 2x – 5 M2 M1 for any two terms correct in 3x2 + 2x – 5 (2x – 1)(3x + 5)(x – 1) oe A1 If 0 scored then SC2 for justifying x – 1 as a factor and writing down (2x – 1)(3x + 5)(x – 1) without any incorrect work seen 4(b)(iii) [sin 0.5  ] 30  nfww   sin 1  90  nfww and no value of  from 3sin + 5 = 0 B2 B1for sin = 0.5 or 30  or B1 for sin = 1 nfww or  = 90 nfww

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 7 of 12 Question Answer Marks Partial Marks 5   2 1 d 10e 0 d x y x    oe M2 M1 for 2 1 d e d x y k x   , k ≠ 10 or SCM1 for 2 1 d 10e d x y c x    where c is algebraic or numerical 1 d 10e d x y x   and y = 6e A1 FT their d d y x provided M1 or SCM1 awarded and a value is found 6e 10e( 1) y x    oe or y = 10ex + c and 6e = 10e + c M1 FT their 1 d d x y x  and their y y = 10ex  4e isw A1 [x-coordinate of P = ] 0.4 oe, isw A1 dep on correct equation of tangent with exact values 6(a) Convincing correct statement from which the answer can be easily determined e.g. 3 1 sin sin sin cos x x x x        oe or 2 1 1 sin sin sin cot x x x x         oe or 3 3 1 cos 1 sin sin sin sin cos x x x x x x               oe or 3 3 1 1 1 sin sin tan sin tan sin x x x x x x                oe 2 M1 for either cosecx correctly written as 1 sin x oe seen in a correct expression or for cotx correctly written as cos sin x x or 1 tan x oe seen in a correct expression e.g.   3 sin cosec tan x x x  or 3 cos sin cosec sin x x x x        or 3 1 cos sin sin sin x x x x        or 3 1 1 sin sin tan x x x        Correct completion to given answer: 2 sin tan x x A1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 8 of 12 Question Answer Marks Partial Marks 6(b) Factorises: 2 1 tan cos 0 2         x x oe or 2 1 tan sin 0 2 x x         oe or correctly rewrites and then factorises: 2 1 sin cos 0 2 x x         oe or 2 1 sin sin 0 2 x x         oe or   2 tan 1 tan 0 x x   oe M1 Note: division by tanx is M0 Note: division by sinx is M0 [tanx = 0 or sinx = 0] [x =] 0 A1 cosx =  1 2  oe or sinx =  1 2  oe or tan x = []1 M1 nfww [x = ] π 4  or  0.785 or  0.7853 to  0.7854, 3π 4  or  2.36 or  2.356 to  2.3562 A2 with no extras in range; nfww A1 for any two out of four correct, ignoring extras 7(a) 282 240 2 M1 for 9! – 2!  8! oe 7(b) 120 2 M1 for 5! or 5P5 oe 8(a) Points plotted at x 15 30 45 60 75 lny 2.3 2.5 or 2.6 3.1 3.5 or 3.6 3.9 soi and single, ruled, straight line of best fit drawn B2 B1 for at least 4 correctly plotted points

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 9 of 12 Question Answer Marks Partial Marks 8(b) lny = 0.03x + 1.8 2 M1 for m = awrt 0.02 to awrt 0.03 or c = awrt 1.7 to awrt 2.0 or for the straight line form in terms of lnA and k: lny = lnA + kx[lne] A = 6 or A = 7 and k = 0.03 or k = 0.02 B3 Must have been found using linear points or linear equation B2 for A = 6 or A = 7 or A in range: awrt 6 or awrt 7 or B1 FT for lnA = their 1.8 or A = etheir1.8 and B1 FT for k = 0.03 or 0.02 or k in range: awrt 0.02 or awrt 0.03 Maximum of 2 marks if one or both values not rounded to 1 sf If B0 scored, award SC1 for A = 6 or A = 7 and SC1 for k = 0.03 or k = 0.02 found not using transformed data

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 10 of 12 Question Answer Marks Partial Marks 8(c) A value of x in range 33 ⩽ x ⩽ 37.5 nfww, isw 2 M1 for lny = 2.8 or 2.83[32...] OR A value of x in range 29.5 ⩽ x <33 or 37.5 < x ⩽ 45 nfww OR M1 STRICT FT for ln17 ln their A x their k   STRICT FT their stated lnA and their stated k or their stated linear equation in (b) OR 1 17 ln x their k theirA        STRICT FT their stated A and their stated k or their stated exponential equation in (b)

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 11 of 12 Question Answer Marks Partial Marks 9 A(2, 0) or x = 2 [x = 6] 2 M1 for factorising or solving 2 32 4 48 0 x x    Finds equation CD/ x-coordinate of C or D or maximum point : x = 4 2 M1 for 2 6 2  or 32 – 8x = 0 D(4, 8) or [equation AB is y =] 4x – 8 2 M1 for y = 2 12 3  or mAB = 12 0 5 2   soi 4 2 3 2 32 4 48 2 3 x x x         or 4 2 3 2 28 4 40 2 3 x x x         B1 must be seen Correct plan including correct substitution of upper and lower limits at some point e.g. 4 2 3 2 4 1 16 48 ( 4 2) 8 3 2 their x x x their their             or   4 4 2 3 2 2 2 4 16 48 2 8 3 their their x x x x x           or 4 2 3 2 4 14 40 3 their x x x         M1 dep on attempt to integrate FT their 4 and their 8 if needed or FT their 4 and their 4x – 8 of the form mx + c if needed or FT their 4 and     2 32 4 4 48 ( 8) their x x their      providing clear evidence of the derivation of this has been seen 40 3 isw or 13.3[33….] nfww A1 dep on all previous marks awarded 10(a) Valid explanation using f: f is one-one oe B1 10(b) Complete method to find inverse function: Swaps the variables and rearranges or rearranges and swaps the variables M1 Condone one sign or arithmetic error but must have the correct order of operations 1 f ( ) ln 3 x x         isw or 1 3 f ( ) ln e x x        oe isw A2 A1 for  1 f ( ) ln 3 x x         or  1 3 f ( ) ln e x x        oe Domain f1: x > e3 B1 Range f1: f1 < 0 B1

0606/22 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 12 of 12 Question Answer Marks Partial Marks 10(c) g(x) = f1(e2x) soi or 2 g( ) lne 3 x x   M1 FT their expression for f1 2 3 x   A1 If 0 scored, allow SCB1 for 2 3 x   found from solving (g(x))2 + 3 = 2x and using existence of composite functions to deduce that the square root must be negative