Cambridge IGCSE Mathematics - Additional 0606 — 2020 Oct/Nov Paper 1 · Variant 1

Cambridge IGCSE™ * 3 0 8 1 1 3 5 6 0 0 * DC (PQ/FC) 187997/3 © UCLES 2020 [Turn over This document has 16 pages. Blank pages are indicated. ADDITIONAL MATHEMATICS 0606/11 Paper 1 October/November 2020 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/O/N/20 © UCLES 2020 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/O/N/20 © UCLES 2020 [Turn over 1 - 2 - 1 0 4 24 y x The diagram shows the graph of ( ) y x p = , where ( )x p is a cubic function. Find the two possible expressions for ( )x p . [3] 2 (a) Write down the amplitude of cos x 1 4 3 + b l. [1] (b) Write down the period of cos x 1 4 3 + b l. [1] (c) On the axes below, sketch the graph of cos y x 1 4 3 = + b l for x 180 180 ° ° ° G G - . - 60 60 120 180 x y 0 - 120 - 180 1 2 3 4 5 6 [3]

4 0606/11/O/N/20 © UCLES 2020 3 (a) Write p q p r qr 3 1 3 2 3 1 - ` ` j j in the form p q r a b c, where a, b and c are constants. [3] (b) Solve x x 6 5 1 0 3 2 3 1 - + = . [3]

5 0606/11/O/N/20 © UCLES 2020 [Turn over 4 It is given that sin tan y x x 3 = . (a) Find the exact value of x y d d when x 3 r = . [4] (b) Hence find the approximate change in y as x increases from 3 r to h 3 r + , where h is small. [1] (c) Given that x is increasing at the rate of 3 units per second, find the corresponding rate of change in y when x 3 r = , giving your answer in its simplest surd form. [2]

6 0606/11/O/N/20 © UCLES 2020 5 (a) (i) Find how many different 4-digit numbers can be formed using the digits 1, 3, 4, 6, 7 and 9. Each digit may be used once only in any 4-digit number. [1] (ii) How many of these 4-digit numbers are even and greater than 6000? [3]

7 0606/11/O/N/20 © UCLES 2020 [Turn over (b) A committee of 5 people is to be formed from 6 doctors, 4 dentists and 3 nurses. Find the number of different committees that could be formed if (i) there are no restrictions, [1] (ii) the committee contains at least one doctor, [2] (iii) the committee contains all the nurses. [1]

8 0606/11/O/N/20 © UCLES 2020 6 A particle P is initially at the point with position vector 30 10 e o and moves with a constant speed of 10ms 1 - in the same direction as 4 3 - e o. (a) Find the position vector of P after t s. [3] As P starts moving, a particle Q starts to move such that its position vector after t s is given by t 80 90 5 12 - + e e o o. (b) Write down the speed of Q. [1] (c) Find the exact distance between P and Q when t 10 = , giving your answer in its simplest surd form. [3]

9 0606/11/O/N/20 © UCLES 2020 [Turn over 7 It is given that ( ) ( ) ln x x 5 2 3 f = + for x 2 3 2- . (a) Write down the range of f. [1] (b) Find f 1 - and state its domain. [3] (c) On the axes below, sketch the graph of ( ) y x f = and the graph of ( ) y x f 1 = - . Label each curve and state the intercepts on the coordinate axes. O y x [5]

10 0606/11/O/N/20 © UCLES 2020 8 (a) (i) Show that ( )( ) cosec sin sin sec 1 1 2 2 i i i i + - = . [4] (ii) Hence solve ( )( ) cosec sin sin 1 4 3 2 i i i + - = for ° ° 180 180 G G i - . [4]

11 0606/11/O/N/20 © UCLES 2020 [Turn over (b) Solve sin cos 3 3 2 3 3 2 r r z z + = + e e o o for 0 3 2 G G r z radians, giving your answers in terms of r. [4]

12 0606/11/O/N/20 © UCLES 2020 9 (a) Given that ln x x x 1 2 3 1 3 d a 1 - + = e o y , where a 0 2 , find the exact value of a, giving your answer in simplest surd form. [6]

13 0606/11/O/N/20 © UCLES 2020 [Turn over (b) Find the exact value of sin cos x x x 2 3 1 2 d 0 r 3 r + - + e b l o y . [5]

14 0606/11/O/N/20 © UCLES 2020 10 (a) An arithmetic progression has a second term of 8 and a fourth term of 18. Find the least number of terms for which the sum of this progression is greater than 1560. [6]

15 0606/11/O/N/20 © UCLES 2020 (b) A geometric progression has a sum to infinity of 72. The sum of the first 3 terms of this progression is 8 333. (i) Find the value of the common ratio. [5] (ii) Hence find the value of the first term. [1]

16 0606/11/O/N/20 © UCLES 2020 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 the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge. BLANK PAGE

This document consists of 9 printed pages. © UCLES 2020 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 Paper 1 October/November 2020 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 October/November 2020 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 October/November 2020 © UCLES 2020 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 October/November 2020 © UCLES 2020 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 October/November 2020 © UCLES 2020 Page 4 of 9 Question Answer Marks Guidance 1 ( )( )( ) 3 2 1 4 = ± + + − y x x x 3 B1 for 3 B1 for ( )( )( ) 2 1 4 + + − x x x B1 for ± 2(a) 4 B1 2(b) o 1080 or 6π B1 2(c) 3 B1 for shape, it must be symmetrical about the y-axis. B1for y-intercept of 5 B1 for ( ) o 180 ,3 ± 3(a) 3 2 = a or 3 2 p B1 10 3 = b or 10 3 q B1 7 3 = − c or 7 3 − r B1 3(b) 1 1 3 3 3 1 2 1 0    − − =          x x M1 For recognising as a quadratic in 1 3 x and attempt to solve to obtain 1 3 = x k 1 1 3 3 1 1 , 3 2 x x = = leading to 1 27 = x or 0.0370 1 8 = x or 0.125 2 Dep M1 for a valid method of solving 1 3 = x k where 0 > k A1 for both 4(a) 2 2 d sin 3sec 3 tan3 cos d sin × − = y x x x x x x 3 B1 for 2 3sec 3x M1 for differentiation of a quotient or equivalent product A1 for all other terms apart from 2 3sec 3x correct When π 3 = x d 2 3 d = y x A1 4(b) 2 3h B1 FT on their answer to (a)

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 5 of 9 Question Answer Marks Guidance 4(c) d d d d d d × = y x y x t t d 2 3 3 d × = y t M1 For correct use of rates of change using their answer to (a) d 6 3 d = y t A1 5(a)(i) 360 B1 5(a)(ii) Starts with 6: 1 4 3 1 × × × 12 = B1 Starts with 7 or 9 : 2 4 3 2 = × × × 48 = B1 Total = 60 B1 Alternative Ending in 4: 1 3 360 36 6 5 × × = (B1) Allow unsimplified Ending in 6: 1 2 360 24 6 5 × × = (B1) Allow unsimplified Total = 60 (B1) 5(b)(i) 1287 B1 5(b)(ii) 7 5 1287 −C or 1 doctor: 210 2 doctors: 525 3 doctors: 420 4 doctors: 105 5 doctors: 1 M1 For their (b)(i) 7 5 −C or listing all the other separate cases which must be evaluated, allow 1 error 1266 A1 5(b)(iii) 45 B1 6(a) Velocity vector = 8 6 −       2 M1 for obtaining 5 30 8 10 6 t −     +         B1 FT for ( ) 30 velocity vector 10  +     their t 6(b) 13 B1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 6 of 9 Question Answer Marks Guidance 6(c) 50 : 70 P −       30 : 210 −       Q M1 Using 10 = t to find position vector of each particle 2 2 20 140 + M1 Dep on previous M mark, for use of Pythagoras on difference of the 2 position vectors 100 2 A1 7(a) f ∈ B1 Allow y but not x 7(b) ( ) 5ln 2 3 = + x y 5 e 2 3 = + x y M1 For a complete attempt to obtain inverse ( ) 5 1 e 3 f 2 − − = x x A1 Must be using correct notation Domain ∈ x B1 FT on their (a). Must be using correct notation 7(c) 5 B1 for shape of ( ) f = y x B1 for shape of ( ) 1 f − = y x B1 for 5ln3 or 5.5 and 1 − on both axes for ( ) f = y x B1 for 5ln3 or 5.5 and 1 − on both axes for ( ) 1 f − = y x B1 All correct, with apparent symmetry which may be implied be previous 2 B marks or by inclusion of , = y x and implied asymptotes, may have one or two points of intersection 8(a)(i) ( ) 2 1 1 1 sin sin sin θ θ θ   + −     B1 For use of 1 cosec sin θ θ = , may be implied 2 1 sin 1 sin sin θ θ θ + − − M1 For expansion of brackets 2 1 cos θ M1 For simplification and use of identity 2 sec θ A1 For final result, must see 2 1 cos θ

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 7 of 9 Question Answer Marks Guidance 8(a)(ii) 2 3 cos 4 θ = B1 For relating to and making use of (a) 3 cos 2 θ = ± M1 For attempt to solve, may be implied by one correct solution o o o o 150 , 30 , 30 , 150 θ = − − 2 A1 for any correct pair A1 for a second correct pair and no extra solutions within the range 8(b) 2π tan 3 1 3 φ   + =     B1 2π π 5π 9π 3 , , 3 4 4 4 φ + = 7π 19π 3 , 12 12 φ = M1 For correct order of operations 7π 36 φ = A1 19π 36 φ = A1 9(a) ( ) 1 1 ln ln 2 3 2   − +     a x x 2 B1 for ln x B1 for ( ) 1 ln 2 3 2 + x ( ) 1 1 ln ln 2 3 ln5 2 2 − + + a a M1 For correct application of limits, must have at least one B1 5 ln 2 3 + a a M1 Dep on previous M mark, for application of log laws 2 5 18 27 0 − − = a a M1 Dep on previous M mark for equating toln3 and simplification to a 3 term quadratic = 0 9 6 6 5 + = a A1 Must have one solution only

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 8 of 9 Question Answer Marks Guidance 9(b) 1 π 1 cos 2 sin2 2 3 2   − + + −     x x x 3 B1 for 1 π cos 2 2 3   − +     x B1 for 1 sin2 2 + x B1 for −x 1 1 2π π cosπ sin 2 2 3 3 1 π cos 2 3   − + −       −−     M1 For correct use of limits in their integral, must have at least one B1 term 3 3 π 4 4 3 + − A1 10(a) 8 + = a d 3 18 + = a d 2 B1 for both equations M1 for attempt to solve their equations 3, 5 = = a d A1 For both ( ) ( ) 6 1 5 1560 2 + − > n n M1 For correct use of sum formula with their a and d, allow equality 2 5 3120 0 + − > n n M1 For attempt to solve, allow equality, to obtain at least one critical value Positive critical value 24.9 25terms A1 10(b)(i) 72 1 = − a r and either 2 333 8 + + = a ar ar or ( ) 3 1 333 1 8 − = − a r r B1 For both ( ) 72 1 = − a r and ( ) 2 333 1 8 + + = a r r oe ( )( ) 2 333 72 1 1 8 − + + = r r r or ( ) 3 333 72 1 8 − = r M1 For attempt to obtain an equation in terms of r only 3 333 1 576 − = r A1 0.75 = r 2 M1 for attempt to solve their equation in r

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 9 of 9 Question Answer Marks Guidance 10(b)(ii) 18 = a B1 FT on their r provided 1 < r