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

This document has 16 pages. [Turn over * 8 8 8 7 5 1 2 3 8 9 * Cambridge IGCSE™ DC (DE/JG) 332317/3 © UCLES 2024 ADDITIONAL MATHEMATICS 0606/11 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 [ ].

2 0606/11/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 0606/11/M/J/24 © UCLES 2024 [Turn over 1 (a) On the axes, sketch the graph of ( )( )( ) y x x x 5 1 2 2 1 5 =- + + - , stating the intercepts with the axes. [3] O y x (b) Hence solve the inequality ( )( )( ) . x x x 5 1 2 2 1 5 0 H - + - + [2]

4 0606/11/M/J/24 © UCLES 2024 2 DO NOT USE A CALCULATOR IN THIS QUESTION. The polynomial p is such that ( ) . x x x x 6 35 34 45 p 3 2 = - + + (a) Find ( )x p in the form ( ) ( ) , x x r 2 5 q - + where ( )x q is a polynomial and r is a constant. [3] (b) Hence write the expression ( )x 5 p - as a product of linear factors. [2] (c) Hence write down the solutions of the equation ( ) . x 5 p = [1]

5 0606/11/M/J/24 © UCLES 2024 [Turn over 3 (a) Write ( ) ( ), lg lg x x 1 1 2 1 2 + - - - where , x 1 2 as a single logarithm to base 10. Give your answer in its simplest form. [4] (b) Solve the equation ( ) , log log x 4 1 9 5 ( ) x 5 1 + = + giving your answers in the form , a b c + where a, b and c are constants. [5]

6 0606/11/M/J/24 © UCLES 2024 4 (a) The first three terms, in ascending powers of x, in the expansion of ( ) px 3 n + are , x qx 243 810 2 + + where n, p and q are constants. Find the values of n, p and q. [5] (b) Find the term independent of y in the expansion of . y y 2 3 1 2 6 - e o Give your answer in exact form. [2]

7 0606/11/M/J/24 © UCLES 2024 [Turn over 5 (a) The diagram shows the graph of , cos y a bx c = + for , x 360 360 ° ° G G - where a, b and c are constants. Find the values of a, b and c. [3] y x 0 – 60° 60° 120° 180° 240° 300° 360° 2 1 – 6 – 5 – 4 – 3 – 2 – 1 – 7 – 8 – 9 – 10 – 120° – 180° – 240° – 300° – 360° (b) The line y p = is a tangent to the curve sin y 3 2 6 . i = - Write down the possible values of p. [2]

8 0606/11/M/J/24 © UCLES 2024 6 Find ( ) , x x x 2 3 2 3 5 3 d 2 2 4 - - - c e dd e o giving your answer in exact form. [4] 7 Given that cot x 2 3 i + = and , sin y i = find y in terms of x. [3]

9 0606/11/M/J/24 © UCLES 2024 [Turn over 8 Solve the equation r sin 4 2 3 1 2 a- = b l for r r . 2 2 G G a - Give your answers in terms of r. [5]

10 0606/11/M/J/24 © UCLES 2024 9 (a) Solve the following simultaneous equations. 1 e e x y x y 3 2 # = + - x y 256 2 = [5]

11 0606/11/M/J/24 © UCLES 2024 [Turn over (b) Solve the equation , 10 11 6 e e ( ) ( ) x x 2 1 1 2 - = - - giving your answer in exact form. [4]

12 0606/11/M/J/24 © UCLES 2024 10 In this question, all distances are in metres and time, t, is in seconds. A particle P is at a fixed point O at time . t 0 = The velocity, v, of P is given by sin v t 3 2 = for . t 0 H (a) Find the exact value of t for which the velocity is zero for the first time after P leaves O. [2] (b) Find an expression, in terms of t, for the displacement of P from O at time t. [4]

13 0606/11/M/J/24 © UCLES 2024 [Turn over (c) Find the distance travelled by P for r. t 0 G G [3]

14 0606/11/M/J/24 © UCLES 2024 11 The tangent to the curve ( ) y x 3 1 3 1 = - at the point where x 3 = meets the coordinate axes at the points A and B. The point with coordinates ( , ) a a lies on the perpendicular bisector of the line AB. Find the exact value of a. [10]

15 0606/11/M/J/24 © UCLES 2024 [Turn over Continuation of working space for Question 11. Question 12 is printed on the next page.

16 0606/11/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. 12 (a) It is given that ln y x x 3 2 = for . x 0 2 Find . x y d d Give your answer in the form ln x A B x 3 3 + , where A and B are integers. [4] (b) Hence find . ln x x x 3 d 3 < [4]

This document consists of 10 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 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.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 4 of 10 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(a) 3 B1 for correct shape with max and min in the correct quadrant. Ignore labelling of their maximum point if incorrect coordinates. B1 for 1 5, 2, 2   marked on x-axis. Must have a cubic shape B1 for 2 marked on the y-axis. Must have a cubic shape 1(b) 1 5, 2 2   x x    2 B1 for each If B0 then SC1 for 1 5, 2 2     x x

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 5 of 10 Question Answer Marks Guidance 2(a)  q x = 2 3 10 8   x x r = 5 3 M1 for a valid attempt to obtain the quotient by algebraic long division, synthetic division, factorising (2x – 5) or by forming an identity A1 for each 2(b)     2 p 5 3 10 8 2 5 x x x x         M1 For a correct attempt to factorise     3 2 4 2 5    x x x A1 Must see p(x) 5 as a product of linear factors 2(c) 2 5 , 4, 3 2  x B1 FT on their 3 terms quadratic q(x) 3(a) 1 lg10  soi B1     2 2 10 1 lg 1   x x oe M1 For correct use of logs power rule or multiplication or division rule. i.e.: Award M1 for example: 2 2 2 2 2 2 2 2 2log( 1) log( 1) log( 1) log( 1) log( 1) log10 log( 1) log10( 1) ( 1) log( 1) log( 1) log ( 1)                   x x x x x x x x x x x     2 10 1 ( 1) lg 1    x x x DM1 Dep on previous M1 for a correct attempt to factorise and an attempt to simplify   10 1 lg 1   x x A1 3(b)     5 5 9 4log 1 log 1       x x or     1 1 4 9log 5 log 5        x x B1 For change of base     2 5 9 log 1 4   x or     2 1 4 log 5 9   x 2 M1 for a correct method in forming a quadratic equation Dep on previous B1 3 2 1 5   x M1 Dep on dealing with logarithms correctly Allow if ± is missing 1 1 5 5, 1 5 25   x A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 6 of 10 Question Answer Marks Guidance 4(a) 5  n B1 4 5 3 810   p M1 For considering the second term. Allow for use of their n 2  p A1 3 2 10 3    p q M1 For considering the third term. Allow for use of their n and 2 p 1080  q A1 4(b)    2 4 6 2 2 1 2 3        C y y M1 For identifying the correct term. Condone errors with brackets and coefficients. Could be implied by a correct answer 80 3 oe A1 Must be exact. Allow 240 2 or 26 9 3 5(a) 5  a , 3 4  b (oe), 4  c 3 B1 for each 5(b)     1, 5 p p   2 B1 for each Do not allow if written as inequalities 6   ln 2 3  x B1 Allow unsimplified 1 3 5  x oe B1 1 ln5 1 7   M1 For correct application of limits in their integral, Must be in the correct form ln(2 3) 3 5    b a x x 6 ln5 7  cao A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 7 of 10 Question Answer Marks Guidance 7 Use of 2 2 cot 1 cosec    and 1 cosec sin    B1 2 2 4 cot 9   x gets B0 M0 A0 unless recovered   2 1 3 2 1        x y oe M1   2 1 3 2 1    y x oe A1 Alternative Method: Use of cos cot sin     leading to 2 2 cos (3 2)    y x oe M1 For use of the identity to substitute cot to get cos 2 3   x y and rearranging to get 2 2 cos (3 2)    y x Use 2 2 sin cos 1     to get 2 1 (3 2)    y y x oe M1 For use of the identity to substitute 2 cos to get 2 2 1 sin (3 2)     y x and use of 2 sin y   2 1 3 2 1    y x oe A1 8  π 1 sin 2 3 2          B1 Condone if ± is missing for this mark 5π π π π , , , 12 4 12 4   oe in terms of π 4 M1 for correct attempt to obtain one solution using correct order of operations A3 for 4 correct solutions and no extras in the range, or A2 for 3 correct solutions or A1 for 2 correct solutions 9(a) For use of 0e 1 or ln1 = 0 B1   4 0   x y B1 For simplifying powers of e leading to a linear equation in x and y 3 4 256  x or 3 256 16  y M1 For attempt to obtain a cubic equation in one variable using their linear 4 0   x y with an attempt to solve 4, 16   x y 2 A1 for each A0 for y = ±16

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 8 of 10 Question Answer Marks Guidance 9(b) 1 2 2 1 1 e e    x x B1 Correct use of the reciprocal   2 2 1 2 1 10 e 11e 6 0      x x or   2 1 2 1 2 6 e 11e 10 0      x x M1 For a correct attempt to form a quadratic equation in 2 1 e  x or in 1 2 e x 2 1 3 e 2  x 2 1 2 e 5         x or 1 2 2 e 3   x 1 2 5 e 2         x M1 Dep on attempt to solve their quadratic equation 1 1 3 ln 2 2 2   x or exact equivalent only A1 Allow 3 1 ln 2 2   x and 1 3 ln( ) 2 2  x e for A1 A0 if negative root not discounted Alternative Method: 2 2 10 6 11   x x e e e e B1 Correct use of the reciprocal 2 2 2 2 10( ) 11 ( ) 6 0    x x e e e e M1 For correct attempt to form a quadratic equation 2 11 19 20   x e e e M1 Dep on attempt to solve their quadratic equation 1 3 1 3 1 ln( ) ln 2 2 2 2 2          x e A1 Allow 3 1 ln 2 2   x and 1 3 ln( ) 2 2  x e for A1 A0 if negative root not discounted 10(a) π 2  t 2 M1 for attempt at solution of 3sin 2 0  t implied by 90 or π 2 A0 for t = 90 10(b) 3 cos2 2  t 2 M1 for cos2 , 6  k t k When t = 0, s = 0 so 3 2  c M1 Dep on attempt to find c using their s 3 3 cos2 2 2   s t A1 Must be an expression for s

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 9 of 10 Question Answer Marks Guidance 10(c) Distance travelled = π 2 0 3 3 2 cos2 2 2         t using symmetry or 3 3 π 2 cos 2 2 2 2               using symmetry π π 2 π 0 2 3 3 3 3 cos2 cos2 2 2 2 2                t t oe 2 M1 dep on their s from (b) unless restarted. Must be in the form of cos2 k t (+c) Condone the use of 0 to π as limits. Limits must be the correct way round and subtracted. M1dep for correct substitution of limits at least once and correct use of symmetry 6 A1 11 When 3, 2   x y B1   2 3 d 3 1 d    y x x M1 Allow for   2 3 3 1   k x When d 1 3, d 4   y x x A1 Tangent equation:   1 2 3 4    y x M1 Allow using their y and their d d y x Intercepts on the axes: 5 5, 4   x y A1 For both Midpoint of AB: 5 5 , 2 8        M1 Dep on M mark for equation of tangent FT on their intercepts Grad of perp bisector = 4  M1 FT on their d d y x or from their x and y intercept Perp bisector equation: 5 5 4 8 2          y x M1 Allow using their – 4 and their midpoint 5 5 4 8 2          a a M1 Dep on previous M mark for use of   , a a in their perp bisector equation 15 8  a , 1.875  A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 10 of 10 Question Answer Marks Guidance 12(a) 2 4 1 2 ln3 d d        x x x y x x x oe 2 3 d 3 2 ln3 d 3       y x x x x x 3 B1 for 1 x or 3 3x M1 for a correct attempt at a quotient or a product A1 for all terms apart from log derivative correct. 3 1 2ln3  x x A1 12(b) 2 3 3 3 2 ln3 1 2ln3 d ln3 1 ln3 2 d d oe x x x x x x x x x x x x        M1 For using integration as a reverse of differentiation (reverse of part (a)) Allow using their A and B but do not allow any extra terms added or subtracted from their part (a) 3 1 d         x x 2 1 2  x oe nfww B1 FT on their A 2 2 1 ln3 4 2    x c x x oe 2 M1 dep for rearranging simplification and an attempt to integrate 3  ax . Allow if +c is missing A1 must include +c