Cambridge IGCSE Mathematics - Additional 0606 — 2024 May/June Paper 1 · Variant 3

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/13 Paper 1 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 [ ]. * 9 1 7 9 2 4 2 7 6 4 * DC (PQ/SW) 332389/2 © UCLES 2024

2 0606/13/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 =

3 0606/13/M/J/24 © UCLES 2024 [Turn over 1 0 y x 80° – 80° – 2 2 4 6 – 160° – 240° – 320° 160° 240° 320° The diagram shows the graph of sin y a bx c = + , for ° ° x 320 320 G G - , where a, b and c are constants. Find the values of a, b and c. [3] 2 Solve the equation 3 2 11 2 3 0 x x 2 1 - + = + ` ` j j , giving your answers correct to 2 decimal places. [4]

4 0606/13/M/J/24 © UCLES 2024 3 (a) Find the coordinates of the stationary points on the curve ( ) ( ) y x x 2 1 3 2 = + - . [4] (b) On the axes, sketch the graph of ( ) ( ) y x x 2 1 3 2 = + - , stating the intercepts with the axes. [3] O y x

5 0606/13/M/J/24 © UCLES 2024 [Turn over (c) Write down the values of k for which the equation ( ) ( ) x x k 2 1 3 2 + - = has exactly one solution. [2] 4 Find x 1 e d x 2 0 2 2 + ` j y , giving your answer in exact form. [5]

6 0606/13/M/J/24 © UCLES 2024 5 When e y 2 is plotted against x3, a straight line graph that passes through the points (2, 5) and (6.4, 7.2) is obtained. (a) Find y in terms of x. [4] (b) Find the values of x for which y exists. [2]

7 0606/13/M/J/24 © UCLES 2024 [Turn over 6 It is given that ln y x x 2 2 1 2 = + + ` j , x 2 ! - . (a) Find x y d d . [3] (b) Given that x increases from 1 to h 1+ , where h is small, find the approximate corresponding change in y. [2] (c) When x 1 = , the rate of change in y is 3 units per second. Find the corresponding rate of change in x. [2]

8 0606/13/M/J/24 © UCLES 2024 7 (a) A 6-digit number is to be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The 6-digit number cannot start with 0. Each digit can be used at most once in any 6-digit number. Find how many of these 6-digit numbers are divisible by 5. [3] (b) The number of combinations of ( ) n 1 + objects taken 13 at a time is equal to 16 times the number of combinations of n objects taken 12 at a time. Find the value of n. [3]

9 0606/13/M/J/24 © UCLES 2024 [Turn over 8 The line L is the normal to the curve ( ) y x 3 5 6 2 1 = + at the point where x 2 = . The point ( , )k 2 - , where k is a constant, lies on L. Find the exact value of k. [7]

10 0606/13/M/J/24 © UCLES 2024 9 In this question, all lengths are in metres, and time, t, is in seconds. A particle P moves in a straight line such that, t seconds after leaving a fixed point O, its displacement, s, is given by cos s t t 4 4 2 4 = - + . (a) Find the velocity, v, of P at time t. [2] (b) On the axes, sketch the velocity–time graph for P for t 0 G G r, stating the intercepts with the axes in exact form. [5] O v t r

11 0606/13/M/J/24 © UCLES 2024 [Turn over (c) Find the acceleration of P at time t. [1] (d) Find the times when the acceleration of P is zero for t 0 G G r. Give your answers in terms of r. [2]

12 0606/13/M/J/24 © UCLES 2024 10 (a) In an arithmetic progression, the first term is a and the common difference is d. The sum of the first three terms of this arithmetic progression is 42. The product of the first three terms of this arithmetic progression is 6720 - . (i) Show that ( ) a a d 2 480 + = - . [3] (ii) Hence, given that a is positive, find the values of a and d. [4]

13 0606/13/M/J/24 © UCLES 2024 [Turn over (b) In a geometric progression, the 3rd term is 4 e x 4 and the 10th term is 512 e x 11 . Find the first term and the common ratio. [5]

14 0606/13/M/J/24 © UCLES 2024 11 Solve the following simultaneous equations, giving your answers in exact form. log log x y 8 12 5 3 81 + = log log y x 4 3 2 9 3 + = [6]

15 0606/13/M/J/24 © UCLES 2024 12 Solve the equation r sec 3 2 2 i- = J L KK N P OO for r r 2 2 G G i - . Give your answers in exact form. [5]

16 0606/13/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

This document consists of 9 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/13 Paper 1 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.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 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.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 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.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 4 of 9 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 Guidance 1 4 a  B1 3 8 b  or 0.375 oe B1 2 c  B1 2     2 6 2 11 2 3 0 x x    soi B1 1 3 2 , 2 3 2 x x   2 M1 for solution of their quadratic equation in 2x or appropriate substitution to obtain 2 ... x  1.58, 0.58  A1 For both

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 5 of 9 Question Answer Marks Guidance 3(a)      2 d 2 1 4 2 1 3 d y x x x x      or 2 d 12 16 11 d y x x x    M1 For correct attempt to differentiate a product or expansion and then differentiate. 1 11 , 2 6 x  2 M1 for setting their quadratic d d y x to zero and attempting to solve to obtain two x values. 0, y  686 27  or 25.4  A1 For both Must be separate inequalities 3(b) 3 B1 for correct shape with max on the negative x-axis and a min in the 4th quadrant B1 for correct x intercepts 1 2  and 3. Must have a cubic curve B1 for correct y intercept 3 . Must have a cubic curve 3(c) 0 k  B1 686 27 k  or 25.4 k  B1 FT on their 686 27  or 25.4  from part (a) 4 2 4 1 2e e x x   B1 For correct integrand 4 2 e e 4 x x x   B2 B1 for 2 correct terms in integral 8 4 e 1 2 e 1 4 4                 M1 For correct application of limits. Must be in the form 4 2 e e x x m n x   8 4 3 e e 4 4   A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 6 of 9 Question Answer Marks Guidance 5(a) 2 3 e y mx c   soi B1 May be seen later Either 7.2 6.4m c   or 5 2m c   Or gradient = m = 1 2 M1 For correct use of coordinates at least once or for finding the gradient and equating it to m 1 , 4 2 m c   A1 For both 3 1 ln 4 2 2 x y         oe A1 5(b) 3 4 0 2 x   M1 Allow their 3 4 2 x  > 0 2 x  A1 6(a)        2 2 2 4 2 ln 2 1 2 1 2 x x x x x      3 B1 for   2 4 2 1 x x  M1 for correct attempt at a quotient or equivalent product A1 for all other terms correct 6(b) When d 4 ln3 1, d 9 y x x    or 0.322 B1 4 ln3 9 h  or 0.322h B1 FT on their 4 ln3 9  or 0.322 6(c) d 3 d 0.322 x t their  M1 9.31or 9.32 A1 7(a) Ending in 0: 15120 or 9 5P or 9 8 7 6 5   oe B1 Ending in 5: 13440 or 8 8 1 4 P P  or 8 8 7 6 5   oe B1 Total: 28560 B1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 7 of 9 Question Answer Marks Guidance 7(b)       1 ! 16 ! 12 !13! 12 !12! n n n n     soi B1 1 13 16 n   oe B1 Dep on previous B mark for either 1 n  or 13 16  207 n  B1 Dep on previous 2 B marks 8 When 2 x  , 12 y  B1   1 2 d 15 5 6 d 2 y x x    2 M1 for   1 2 15 5 6 , 2 k x k    When d 15 2, d 8 y x x   , A1 perp gradient = 8 15  M1 Dep on previous M mark. Allow for use of their gradient to obtain perp gradient Normal equation:   8 12 2 15 y x    M1 Allow for using their perp gradient and their y When 2 x  212 15 k  or 2 1415 A1 9(a) 4 8sin 2 v t   2 M1 for 4 sin2 , 8 v k t k    9(b) 5 B1 for a complete correct cycle between 0 t  and π t  B1 for 4 v  , must have a sine curve M1 for correct attempt to find one t-intercept using their v, must be in the form 4 sin 2 k t  and 0 π t  Allow M1 for o 105 or o 165 seen A1 for 7π 12 t  A1 for 11π 12 t  9(c) 16cos2 a t  B1 FT on their v which must be in the form 4 sin 2 k t  9(d) π 3π , 4 4 t  2 B1 for each Must be from cos2 0 t 

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 8 of 9 Question Answer Marks Guidance 10(a)(i) 14 a d   soi B1    2 6720 a a d a d    B1   14 2 6720 a a d     , leading to   2 480 a a d   B1 Dep on both previous B marks AG, needs sufficient detail 10(a)(ii)    14 14 480 d d    or   28 480 a a   M1 For attempt to obtain a quadratic equation in one variable 2 676 d  or 2 28 480 0 a a    A1 40, 26 a d   A2 A1 for each 10(b) 4 2 e 4 x ar  B1 11 9 e 512 x ar  B1 7 7 e 128 x r  oe M1 For a correct attempt to solve their 2 equations. Must be the same structure to obtain 7 ... r  e 2 x r  A1 2 e x a  A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 9 of 9 Question Answer Marks Guidance 11 81 3 12log 3log y y  oe or 9 3 4log 2log x x  oe B1 For either 3 3 8log 3log 5 x y   oe B1 3 3 2log 3log 2 x y   oe B1 1 3 3 y  , 3 x  3 M1 for attempt to solve their equations. Must have at least 2 B marks and obtain 2 solutions A1 for each Alternative 12 3 81 log 12log 4 y y  oe or 4 3 9 log 4log 2 x x  oe B1 32 12 20 3 x y  or 8 3 53 x y  oe B1 4 6 4 3 x y  or 2 3 2 3 x y  oe B1 1 3 3 y  , 3 x  3 M1 for attempt to solve their equations, which must both be in the correct form. Must have at least 2 B marks and obtain 2 solutions A1 for each 12 π 1 cos 3 2 2          oe soi B1 7π π 5π , , 18 18 18  4 M1 dep on previous B mark, for one of the following π 5π π π 5π 3 , , , 2 3 3 3 3    A1 for one correct solution A1 for a second correct solution A1 for a third correct solution and no extra solutions in the range.