Cambridge IGCSE Mathematics - Additional 0606 — 2024 Feb/March Paper 1 · Variant 2

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 February/March 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 [ ]. * 1 7 7 4 6 3 4 2 5 2 * DC (PQ/CT) 327629/3 © UCLES 2024

2 0606/12/F/M/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 =

3 0606/12/F/M/24 © UCLES 2024 [Turn over 1 Given that cos y 2 4 3i = + , for ° ° 120 120 G G i - , (a) write down the amplitude of y [1] (b) write down the period of y. [1] (c) On the axes, sketch the graph of y. [3] y i 0 100° 120° 80° 60° 40° 20° – 20° – 40° – 60° – 80° – 100° – 120° 2 4 6 8 – 4 – 6 – 8 – 2

4 0606/12/F/M/24 © UCLES 2024 2 (a) Given that log log log log a 12 6 3 4 p p p p + - = , find the value of a. [3] (b) Find the exact solutions of the equation log log x 4 9 3 x 3 = . [4]

5 0606/12/F/M/24 © UCLES 2024 [Turn over 3 The curve C has equation ln y x 3 3 = + ` j. The normal to C at the point where x 1 = meets the line y x = at the point P. Find the exact coordinates of P. [7]

6 0606/12/F/M/24 © UCLES 2024 4 A function f is such that ( )x 2 f e x 3 = + - , x R ! . (a) Write down the range of f. [1] (b) Find an expression for f 1 - . [2] (c) On the axes, sketch the graphs of ( ) y x f = and ( ) y x f 1 = - , stating the coordinates of the points where the curves meet the coordinate axes. State the equations of any asymptotes. Label your curves. [4] y x O

7 0606/12/F/M/24 © UCLES 2024 [Turn over A function g is such that ( )x x 4 g 2 3 = + , x 0 H . (d) Find the exact solution of the equation ( ) . x 12 gf = [4]

8 0606/12/F/M/24 © UCLES 2024 5 The polynomial p is such that ( )x x ax x b 5 39 p 3 2 = + + + , where a and b are constants. (a) Given that x 3 + is a factor of both ( )x p and ( )x pl , find the values of a and b. [5] (b) Hence solve the equation ( )x 0 p = . You must show your working. [3]

9 0606/12/F/M/24 © UCLES 2024 [Turn over (c) Hence, using your values for a and b, solve the equation cosec cosec cosec a b 5 2 2 39 2 0 3 2 i i i + + + = for ° ° 0 360 G G i . [5]

10 0606/12/F/M/24 © UCLES 2024 6 In this question all distances are in metres and all times are in seconds. (a) v ms–1 t s 0 10 30 70 120 V (i) The diagram shows the velocity–time (v–t) graph of a particle travelling in a straight line. The particle travels a distance of 2750 m in 120 s. Find the velocity, V, of the particle when t 70 = . [2] (ii) Find the acceleration of the particle for t 70 120 1 1 . [2]

11 0606/12/F/M/24 © UCLES 2024 [Turn over (b) A different particle moves in a straight line such that its velocity, v ms 1 - , t seconds after leaving a fixed point O, is given by v t t 5 2 2 1 = + ` j . (i) Find the exact acceleration of the particle when t 2 = . [4] (ii) Explain why the particle does not change direction for t 0 2 . [1]

12 0606/12/F/M/24 © UCLES 2024 7 (a) Find ( ) x x 5 2 d 2 4 3 2 - - y , giving your answer in exact form. [4] (b) Find ( ) x x x 2 1 4 2 1 8 d 2 0 2 1 + + + J L KK N P OO y , giving your answer in the form ln a b + , where a and b are integers. [5]

13 0606/12/F/M/24 © UCLES 2024 [Turn over 8 (a) A 5-digit number is to be formed using 5 different numbers selected from 1, 2, 3, 4, 5, 6, 7, 8 and 9. No digit may be used more than once in any 5-digit number. (i) Find how many 5-digit numbers can be formed. [1] (ii) Find how many of these 5-digit numbers are greater than 50 000 and even. [3] (b) A team of 9 people is to be chosen from 6 doctors, 4 dentists and 2 nurses. Find how many possible teams include at least 2 doctors, at least 2 dentists and at least 2 nurses. [3]

14 0606/12/F/M/24 © UCLES 2024 9 (a) The first three terms of an arithmetic progression are lg 2 i , lg 5 i and lg 8 i . (i) Given that the sum to n terms of this progression is lg 4732 i, find the value of n. [5] (ii) This sum is equal to 14196 - . Find the exact value of i. [1]

15 0606/12/F/M/24 © UCLES 2024 (b) The first three terms of a geometric progression are lg 3 z , lgz and lg 3 1 z . (i) Determine whether this geometric progression has a sum to infinity. [2] (ii) Find the nth term of this geometric progression, giving your answer in the form lg 3A z, where A is a function of n. [3] (iii) Find the value of z, given that the 20th term is 3 18 - . [1]

16 0606/12/F/M/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

This document consists of 8 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 February/March 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 February/March 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 February/March 2024 © Cambridge University Press & Assessment 2024 Page 2 of 8 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 February/March 2024 © Cambridge University Press & Assessment 2024 Page 3 of 8 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 February/March 2024 © Cambridge University Press & Assessment 2024 Page 4 of 8 Question Answer Marks Guidance 1(a) 4 B1 1(b) o 120 B1 1(c) 3 To score marks, must have minimum points in the correct quadrants and symmetry about the y-axis. B1 for correct  intercepts o 40 ,  o 80  and no others B1 for y-intercept of 6 B1 for a completely correct shape with no errors. 2(a) 3 12 log log 4 6 = p p a soi 2 B1 for correct use of addition and subtraction rule B1 for correct use of power rule 32 = a B1 2(b) 3 3 9 4log log = x x or 4 9log 3 log 3 = x x soi B1 For change of base ( ) 2 3 9 log 4 = x or ( ) 2 4 log 3 9 = x soi B1 1.5 3 = x or exact equivalents 2 B1 for each solution 3 2 3 3 3 + x x B1 When 1 , = x d 3 d 4 = y x oe M1 For finding the value of their d d y x ln4 = y B1 ( ) 4 ln4 1 3 − = − − y x 2 M1 for attempt at normal equation using their d d y x and their y Allow A1 if 4 ln4 3 = + c seen 4 3ln4 4 3ln4 , 7 7 + +       2 M1 dep for attempt to use = y x and obtain at least one solution 4(a) f 2  B1 4(b) ( ) ( ) 1 1 f ln 2 3 − = − − x x or 1 1 ln 3 2     −   x isw 2 M1 for a complete attempt at inverse, allow sign slip but brackets must be used correctly.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2024 © Cambridge University Press & Assessment 2024 Page 5 of 8 Question Answer Marks Guidance 4(c) 4 B1 for correct ( ) f = y x with y-intercept of 3. Must have correct asymptotic behaviour and be in the first and second quadrant. B1dep for correct reflection of ( ) f = y x to obtain ( ) 1 f − = y x with x-intercept of 3. Must have correct asymptotic behaviour and be in the first and fourth quadrant. B1 for asymptote of 2 = y stated or drawn through 2 = y , must have a correctly shaped ( ) f = y x B1 for asymptote of 2 = x stated or drawn or drawn through 2 = x , must have a correctly shaped ( ) 1 f − = y x 4(d) ( ) 3 3 2 2 e 4 − + + x soi B1 For correct order 3 2 e 4 − + = x M1 For forming an equation, must be correct order 1ln2 3 = − x 2 M1 dep for correct attempt to solve for x. 5(a) ( ) 2 p 15 2 39  = + + x x ax soi B1 ( ) p 3 : 135 6 39 0 − − + = a oe B1 ( ) p 3 : 135 9 117 0 − − + − + = a b oe B1 29 = a B1 9 = − b B1 5(b) ( ) ( ) 2 3 5 14 3  +  + −   x x x 2 M1 for attempt by any valid method, to obtain a quadratic with 2 correct terms or correct follow through on their a and b. 1 3, 5 = − x A1 For both

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2024 © Cambridge University Press & Assessment 2024 Page 6 of 8 Question Answer Marks Guidance 5(c) cosec2 3 = − soi B1 1 sin2 3 = − o o o o o 2 19.47 , 199.47 , 340.53 , 559.47 , 700.53 = − o o o o 99.7 , 170.3 , 279.7 , 350.3 = 4 M1 a correct double angle M1 for correct order of operations to obtain one correct solution. May be implied by e.g. a correct solution or o 9.7 = − or a correct angle in radians A1 for 2 correct solutions A1 for a further 2 correct solutions and no extra solutions within the range 6(a)(i) ( ) 1 1 300 10 40 50 2750 2 2 + + + = V V or ( ) ( ) 1 1 700 40 10 50 2750 2 2 +  − + = V V oe M1 Allow one slip, but must be considering complete area 50 = V A1 6(a)(ii) 1 − nfww 2 M1 FT their V for a correct gradient calculation 6(b)(i) ( ) ( ) 1 1 2 2 2 2 d 1 2 5 5 d 2 −     =   + + +           v t t t t t soi 3 B1 for ( ) 1 2 2 1 2 5 2 −   + t t M1 for a correct attempt at a product A1 for all correct apart from ( ) 1 2 2 1 2 5 2 −   + t t 13 3 A1 6(b)(ii) There is no change of sign for v as v is always positive, so no change in direction. oe B1 7(a) ( ) 1 3 5 2 − a x M1 ( ) 1 3 3 5 2 5 − x oe A1 1 3 3 18 2 5   −       or exact equivalent 2 Dependent M1 for correct use of limits

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2024 © Cambridge University Press & Assessment 2024 Page 7 of 8 Question Answer Marks Guidance 7(b) ( ) 2ln 2 1 + x oe B1 4 2 1 − + x oe B1 ( ) 4 2ln 2 4 2   − −−     M1 For correct substitution of limits, must be using the form ( ) ln 2 1 2 1 + + + b a x x ln4 2 + 2 A1 for each term 8(a)(i) 15 120 B1 8(a)(ii) Total: 3780 3 B1: Starts with 5, 7 or 9: 2520 soi B1: Starts with 6 or 8: 1260 soi Alternative Total: 3780 (3) B1: Ends with 2 or 4: 2100 soi B1: Ends with 6 or 8: 1680 soi 8(b) 2 nurses, 2 dentists, 5 doctors = 36 2 nurses, 3 dentists, 4 doctors = 60 2 nurses, 4 dentists, 3 doctors = 20 2 M1 for two correct cases Total = 116 A1 Alternative 1 dentist only = 4 No nurses = 10 1 nurse only = 90 (M1) Total = 116 (2) M1 for attempt to subtract at least 2 correct cases from 220 9(a)(i) 3lg = d B1 ( ) ( ) ( ) 2 2lg 1 3lg 4732lg 2    + − = n n M1 For use of the sum formula to obtain an equation in lg only, using their a and d and 4732lg 2 3 9464 0 + − = n n A1 56 = n only 2 M1 for attempt to solve their quadratic equation in n 9(a)(ii) 0.001 oe B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2024 © Cambridge University Press & Assessment 2024 Page 8 of 8 Question Answer Marks Guidance 9(b)(i) 1 3 = r soi B1 1  r oe, so has a sum to infinity B1 Dep on previous B1 9(b)(ii) nth term ( ) 1 3 1 lg 3  −    n B1 2 3 lg −n 2 B1 for ( ) 1 3lg 3  −n or 1 3lg 3  − n 9(b)(iii) 10 B1