Cambridge IGCSE Mathematics - Additional 0606 — 2023 May/June Paper 1 · Variant 1

This document has 16 pages. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 Paper 1 May/June 2023 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 [ ]. * 0 6 8 9 0 8 0 3 1 3 * DC (CJ/FC) 312487/2 © UCLES 2023

2 0606/11/M/J/23 © UCLES 2023 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/23 © UCLES 2023 [Turn over 1 (a) Write x x 5 14 8 2 - + in the form ( ) a x b c 2 + + , where a, b and c are constants to be found. [3] (b) Hence write down the coordinates of the stationary point on the curve y x x 5 14 8 2 = - + . [2] (c) On the axes below, sketch the graph of y x x 5 14 8 2 = - + , stating the coordinates of the points where the graph meets the coordinate axes. [3] O y x (d) Write down the range of values of k for which the equation x x k 5 14 8 2 - + = has 4 distinct roots. [2]

4 0606/11/M/J/23 © UCLES 2023 2 The polynomial p is such that ( ) p x ax x bx c 7 3 2 = + + + , where a, b and c are integers. (a) Given that p 2 1 32 = ll b l , show that a 6 = . [2] (b) Given that ( ) p x has a factor of x 3 4 - and a remainder of 7 when divided by x 1 + , find the values of b and c. [4]

5 0606/11/M/J/23 © UCLES 2023 [Turn over (c) Write ( ) p x in the form ( ) ( ) q x x 3 4 - , where ( ) q x is a quadratic factor. [2] (d) Hence write ( ) p x as a product of linear factors with integer coefficients. [1]

6 0606/11/M/J/23 © UCLES 2023 3 The points A and B have coordinates ( , ) 2 5 and ( , ) 10 15 - respectively. The point P lies on the perpendicular bisector of the line AB. The y-coordinate of P is 9 - . (a) Find the x-coordinate of P. [5] (b) The point R is the reflection of P in the line AB. Find the coordinates of R. [2]

7 0606/11/M/J/23 © UCLES 2023 [Turn over 4 0 10 10 v t V 20 30 45 The diagram shows the velocity–time graph for a particle travelling in a straight line with velocity, ms v 1 - , at time t seconds. When t 30 = the velocity of the particle is ms V 1 - . The particle travels 800 metres in 45 seconds. (a) Find the value of V. [2] (b) Find the acceleration of the particle when t 35 = . [2]

8 0606/11/M/J/23 © UCLES 2023 5 DO NOT USE A CALCULATOR IN THIS QUESTION. In this question, all lengths are in centimetres. (a) You are given that ° cos120 2 1 =- , ° sin120 2 3 = and ° tan120 3 =- . In the triangle ABC, , AB BC 5 3 6 5 3 6 = - = + and angle ° ABC 120 = . Find AC, giving your answer in the form a b where a and b are integers greater than 1. [4]

9 0606/11/M/J/23 © UCLES 2023 [Turn over (b) You are given that ° cos30 2 3 = , ° sin30 2 1 = and ° tan30 3 1 = . In the triangle PQR, PQ 3 2 5 = + and angle ° PQR 30 = . Given that the area of this triangle is 4 2 5 5 + , find QR, giving your answer in the form c d 5 + , where c and d are integers. [4]

10 0606/11/M/J/23 © UCLES 2023 6 (a) Show that sec cot tan cosec i i i i + = . [4]

11 0606/11/M/J/23 © UCLES 2023 [Turn over (b) Hence solve the equation , sec cot tan for 3 3 3 2 540 540 ° ° 2 1 1 z z z z + = - f p . [6]

12 0606/11/M/J/23 © UCLES 2023 7 (a) A team of 8 people is to be chosen from a group of 15 people. (i) Find the number of different teams that can be chosen. [1] (ii) Find the number of different teams that can be chosen if the group of 15 people contains a family of 4 people who must be kept together. [3] (b) Given that ( ) ( ) P P n n 9 243 n n 10 2 1 9 # # + = + - , find the value of n. [3]

13 0606/11/M/J/23 © UCLES 2023 [Turn over 8 A curve has the equation ( ) y x x 2 1 3 4 3 1 = + - . (a) Show that ( ) ( ) x y x x Ax B 2 1 3 4 d d 2 3 2 = + - + , where A and B are integers to be found. [5] (b) Find the coordinates of the stationary point on the curve. [2]

14 0606/11/M/J/23 © UCLES 2023 9 (a) The first three terms of an arithmetic progression are nl q, lnq4 and lnq7, where q is a positive constant. The sum to n terms of this progression is lnq 4845 . Find the value of n. [3] (b) The first three terms of a geometric progression are , p p x x 3 and p x - , where p is a positive integer. Find the nth term of this progression giving your answer in the form p( ) a bn x + . [3]

15 0606/11/M/J/23 © UCLES 2023 [Turn over (c) The first three terms of a different geometric progression are , , cos cos and cos 3 4 3 9 16 3 27 64 3 2 4 6 i i i for r 0 3 1 1 i . Find the set of values of i for which this progression has a sum to infinity. [5] Question 10 is printed on the next page.

16 0606/11/M/J/23 © UCLES 2023 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. 10 It is given that ( ) ( ) ln y x x 3 1 3 1 2 = + + . (a) Find x y d d . [3] (b) Hence find ( ) ( ) ln x x x 3 1 3 1 d + + y . [4]

This document consists of 9 printed pages. © UCLES 2023 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 Paper 1 May/June 2023 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 2023 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 2023 © UCLES 2023 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 descriptors 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 2023 © UCLES 2023 Page 3 of 9 Maths-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, 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 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 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

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 4 of 9 Question Answer Marks Guidance 1(a) 2 7 9 5 5 5         x 3 B1 for 2 5( )  x b B1 for 2 7 5        x B1 for 9 5  Alternative By comparing coefficients: 2 2 2 2 ) 5 14 8      a(x bx b c= x x . 2 14  abx x 2 8  ab c = (3) B1 for 5  a B1 for 14 10  b= oe B1 for 18 oe 10   c 1(b) 7 9 , 5 5        2 FTB1 for each, follow through on their a and b from (a) or SC1 if differentiation is used in their (a) or restarted in (b) d 10 -14 0 d   y x x then 7 9 , 5 5        1(c) 3 B1 for the correct shape. Must have the parabola part of the curve with maximum in the first quadrant and cusps on the x-axis. Ignore labelling of their maximum point if incorrect coordinates B1 for 4 , 0 5       and (2, 0), must have a correct shape in the first quadrant. B1 for (0, 8) must have a correct shape. 1(d) 0 < k < 9 5 2 B1FT follow through from 0 and their b in part (a) 2(a) p(x) = 3ax2 + 14x + b p(x) = 6ax + 14 leading to 3a + 14 = 32 a = 6 2 M1 for attempt to differentiate twice and substitute 1 2  x

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 5 of 9 Question Answer Marks Guidance 2(b) p 4 3       : 80 + 4b + 3c = 0 oe M1 Must have 3 terms. For use of 4 3  x , at least once and equating to 0 with an attempt at simplification leading to an equation in b and c only Allow one sign error. p(–1): –b + c = 6 oe M1 Must have 3 terms. For use of x = –1 and equating to 7 with an attempt at simplification leading to an equation in b and c only 14, 8 b c   2 M1 dep on both previous M marks and attempt to solve simultaneously to obtain both b and c A1 for both 2(c) (3x – 4)(2x2 + 5x + 2) 2 B1 for two terms correct in the quadratic factor. Allow if seen as a quotient in long division. For both marks, need to see both factors together. 2 4 (6 15 6) 3          x x x from synthetic method gets 0 marks unless recovered. 2(d) (3x – 4)(2x + 1)(x + 2) B1 Must be all integers 3(a) Mid-point (6, –5) B1 Gradient of AB = 5 2  B1 Perpendicular gradient 2 5 M1 For their perp gradient –9 + 5 = 2 5 (x – 6) oe x = –4 2 Dep M1 for attempt at the equation of the perpendicular bisector with their mid-point and their perpendicular gradient and use of y = –9 3(b) (16, –1) 2 B1 for each, FT on 12 – their a for the x-coordinate. 4(a) Area under graph = 800 1 2 (10  10) + (10  10) + 1 2 (10(10 + V)) + 15 2 V = 800 M1 For attempt to find the area, allow one error and one omission. V = 48 A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 6 of 9 Question Answer Marks Guidance 4(b) ( ) 15 their V M1 Allow omission of negative sign. 16 5  ms–2 oe A1 FT on their V but must be negative. 5(a) (5 3 – 6)2 + (5 3 + 6)2 – 2(5 3 – 6) (5 3 + 6)cos 120° soi M1 For the correct use of the cosine rule Condone missing brackets if intention is clear 75 + 36 – 60 3 + 75 + 36 + 60 3 + 75 – 36 M1 M1 Dep must see sufficient detail to be sure that a calculator is not being used. This is the minimum acceptable 75 + 36 – 60 3 + 75 + 36 + 60 3 + 39 261 A1 Maybe implied by 261 3 29 A1 5(b) 2 5 5 1 (3 2 5) sin30 4 2       QR soi M1 For the correct use of the area of the triangle. Condone missing brackets if intention is clear 2 5 5 3 2 5 2 5 5 3 2 5 or 3 2 5 3 2 5 3 2 5 3 2 5           M1 M1 dep for a correct attempt to rationalise their QR .must be the same two terms in the numerator and denominator to rationalise 6 15 5 4 5 50 9 20     11 5 44 11   M1 M1 dep, must see sufficient detail to be sure that a calculator is not being used. This is the minimum acceptable 6 15 5 4 5 50 11     4 – 5 A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 7 of 9 Question Answer Marks Guidance 6(a) cos sin sin cos 1 cos       soi 2 B1 for tan  and cot  in terms of sin and cos. B1 for 1 sec cos    2 2 cos sin cos sin cos        soi oe M1 For dealing with the fractions in the numerator. 1 sin cosec  cso A1 For correct use of cos2  + sin2  = 1 to obtain the given answer. 2 1 tan 1 tan tan cos 1 tan cos           soi oe (2) B1 for sec  = 1 cos M1 for dealing with the fractions in the numerator. 2 sec cos tan    (B1) For correct use of tan2  + 1 = sec2  1 sin cosec  cso (A1) For correct use of tan  and sec2  to obtain the given answer. 6(b) 2 1 2 sin 3             or 1 sin 3 2  soi or tan 1 3  soi B2 B1 for ± missing –405°, –135°, 135°, 405° 4 M1 for one correct positive or negative solution of their sin 3 k A1 for another correct solution M1Dep for one negative or positive solution A1 for another correct solution and no extras in the range. 7(a)(i) 6435 B1 Must be evaluated not just 15 8 C 7(a)(ii) With family of 4: 330 B1 Must be evaluated not just 11 4 C or implied by a correct answer Without family of 4: 165 B1 Must be evaluated not just 11 8 C or implied by a correct answer Total: 495 B1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 8 of 9 Question Answer Marks Guidance 7(b) 2 ( 9) ! ( 243)( 1)! ( 10)! ( 1 9)!        n n n n n n n(n + 9) = n2 + 243 oe 2 B1 for either 10 10 ! ( 1)! ( -10)! ( 1 9)!     n n n n P or P n n B1 dep for n(n + 9) = n2 + 243 n = 27 B1 8(a) 2 1 3 3 2 1 (2 1) 3(3 4) 2(3 4) 3 (2 1)         x x x x oe or by using the product rule 2 1 3 1 2 3 1 3 (3 4) (2 1) 2 3 (2 1) (3 4)           x x x x 3 B1 for 2 3 1 3 (3 4) 3    x oe M1 for an attempt to differentiate a quotient. A1 for all terms other than 2 3 1 3 (3 4) 3    x correct. Allow unsimplified. 2 3 2 (3 4) (2 1)    x x ((2x + 1) – 2(3x – 4)) M1 M1 dep for attempt to factorise, must be in the form 2 3 2 (3 4) (2 1)    x x [(ax + 1) – b(3x – 4)] 2 2 3 9 4 (2 1) (3 4)    x x x A1 8(b) (2.25, 0.255) 2 B1 FT for their x-coordinate only. Do not allow FT if they score M0 in part (a) 9(a) n = 57 cso 3 B1 for 2 n (2 ln q + 3(n – 1)ln q) oe soi Allow if in indices form i.e.: 2 n (ln q2 + (n – 1)ln q3) B1 for 3n2 – n – 9690 = 0 oe soi 9(b) Common ratio = p–2x B1 Allow unsimplified nth term = p3x(p–2x)n–1 soi B1 Allow unsimplified p(5–2n)x B1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 9 of 9 Question Answer Marks Guidance 9(c) Common ratio = 2 4 cos 3 3  B1 Allow unsimplified. Must be convinced it is the common ratio not just writing the first term e.g. r = or seeing 4 2 16 cos 3 9 4 cos 3 3   2 2 2 4 cos 3 (*) 1 or 3 4 cos 3 (*) 1 or 3 4 cos 3 (*) 0 3     oe soi B1 3 cos3 (*) or 2 3 cos3 (*) or 2 cos3 (*) 0     soi B1 5π π 3 (*) 3 (*) 6 6   and soi B1 Seeing 5π π or 18 18 implied the first 3 marks 18 <  ⩽ 18  B1 10(a) 2 d 3(3 1) 2 3(3 1)ln(3 1) d 3 1        y x x x x x Simplified to: d 3(3 1)(1 2ln(3 1)) d     y x x x 3 B1 for 3 3 1  x M1 for attempt to differentiate a product A1 for all terms other than 3 3 1  x correct. Allow unsimplified. 10(b) 2 (3 1) ln(3 1)   k x x soi B1 2 3 (3 1)d ( ) 2      x x x x c B1 May be a multiple May be seen as 2 1 1 (3 1) 3 2    x 2 2 1 3 (3 1) ln(3 1) 6 4 2      x x x x c B2 B1 for two correct algebraic terms For B2 must have (+ c)