Cambridge IGCSE Mathematics - Additional 0606 — 2021 Oct/Nov Paper 1 · Variant 1

This document has 16 pages. Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 Paper 1 October/November 2021 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 [ ]. * 7 8 7 6 4 3 4 7 5 0 * DC (CJ/FC) 212434/2 © UCLES 2021 [Turn over

2 0606/11/O/N/21 © UCLES 2021 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/21 © UCLES 2021 [Turn over 1 - 90° - 6 90° 180° 270° 360° x y - 180° - 270° - 360° - 4 - 2 0 2 The diagram shows the graph of sin y a b x c = + for , x 360 360 ° ° G G - where a, b and c are integers. (a) Write down the period of . sin a b x c + [1] (b) Find the value of a, of b and of c. [3]

4 0606/11/O/N/21 © UCLES 2021 2 Points A and C have coordinates ( , ) 4 6 - and ( , ) 2 18 respectively. The point B lies on the line AC such that . AB AC 3 2 = (a) Find the coordinates of B. [2] (b) Find the equation of the line l, which is perpendicular to AC and passes through B. [2] (c) Find the area enclosed by the line l and the coordinate axes. [2]

5 0606/11/O/N/21 © UCLES 2021 [Turn over 3 (a) Find the vector which has magnitude 39 and is in the same direction as 5 12 - b l. [2] (b) Given that a 1 2 = - b l and b 5 4 = - b l, find the constants m and n such that a b 5 6 4 m n + = e o . [4]

6 0606/11/O/N/21 © UCLES 2021 4 (a) Given that r q pr pq p q r a b c 3 2 3 = - - ` j , find the value of each of the constants a, b and c. [3] (b) Solve the equation x x 3 8 5 0 5 4 5 2 - + = . [4]

7 0606/11/O/N/21 © UCLES 2021 [Turn over 5 The polynomial ( ) p ax bx x x 6 4 3 2 = + + + , where a and b are integers, is divisible by x 2 - . When ( ) p x l is divided by x 1 + the remainder is 7 - . (a) Find the value of a and of b. [5] (b) Using your answers to part (a), find the remainder when ( ) p x ll is divided by x. [2]

8 0606/11/O/N/21 © UCLES 2021 6 A curve with equation ( ) f y x = is such that d d e x y x 4 6 x 2 2 3 = + . The curve has a gradient of 5 at the point ,0 3 5 b l. Find ( ) f x . [7]

9 0606/11/O/N/21 © UCLES 2021 [Turn over 7 The first three terms, in ascending powers of x, in the expansion of ax 2 n + ` j can be written as bx cx 64 2 + + , where n, a, b and c are constants. (a) Find the value of n. [1] (b) Show that b c 5 768 2 = . [4] (c) Given that b 12 = , find the exact value of a and of c. [2]

10 0606/11/O/N/21 © UCLES 2021 8 O D B C A 6 cm 5 cm The diagram shows a circle, centre O, radius 5 cm. The lines AOB and COD are diameters of this circle. The line AC has length 6 cm. (a) Show that angle AOC = 1.287 radians, correct to 3 decimal places. [2] (b) Find the perimeter of the shaded region. [2]

11 0606/11/O/N/21 © UCLES 2021 [Turn over (c) Find the area of the shaded region. [3]

12 0606/11/O/N/21 © UCLES 2021 9 (a) Find the coordinates of the stationary points on the curve y x x 2 1 3 2 = + - ` `j j . Give your answers in exact form. [4]

13 0606/11/O/N/21 © UCLES 2021 [Turn over (b) On the axes below, sketch the graph of y x x 2 1 3 2 = + - ` `j j , stating the coordinates of the points where the curve meets the axes. [4] O y x (c) Hence write down the value of the constant k such that x x k 2 1 3 2 + - = ` `j j has exactly 3 distinct solutions. [1]

14 0606/11/O/N/21 © UCLES 2021 10 (a) Jess runs on 5 days each week to prepare for a race. In week 1, every run is 2 km. In week 2, every run is 2.5 km. In week 3, every run is 3 km. Jess increases the distance of the run by 0.5 km every week. (i) Find the week in which Jess runs 16 km on each of the 5 days. [2] (ii) Find the total distance Jess will have run by the end of week 8. [3]

15 0606/11/O/N/21 © UCLES 2021 [Turn over (b) Kyle also runs on 5 days each week to prepare for a race. In week 1, every run is 2 km. In week 2, every run is 2.5 km. In week 3, every run is 3.125 km. The distances he runs each week form a geometric progression. (i) Find the common ratio of the geometric progression. [1] (ii) Find the first week in which Kyle will run more than 16 km on each of the 5 days. [3] (iii) Find the total distance Kyle will have run by the end of week 8. [3] Question 11 is printed on the next page.

16 0606/11/O/N/21 © UCLES 2021 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. 11 (a) Solve the equation cosec cot 3 5 5 2i i - = for ° ° 0 180 G G i . [4] (b) Solve the equation r sin 3 2 1 z+ =- b l , where z is in radians and r r G G z - . Give your answers in terms of r. [4]

This document consists of 8 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 Paper 1 October/November 2021 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 2021 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 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 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 2021 © UCLES 2021 Page 3 of 8 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 2021 © UCLES 2021 Page 4 of 8 Question Answer Marks Guidance 1(a) o 1080 B1 1(b) 4 = a B1 3 = b B1 2 = − c B1 2(a) ( ) 0, 14 2 B1 for x-coordinate B1 for y-coordinate 2(b) 1 14 2 − = − y x 2 M1 for finding the gradient of a perpendicular line and attempt at the straight line equation using their B A1 Allow unsimplified 2(c) Area = 1 14 28 2 × × M1 Must be a complete method making use of their answer to (b) 196 A1 3(a) 13 soi B1 For finding the magnitude of 12 5     −   36 15     −   B1 3(b) 10 4 4 λ μ + = − or 5 6 5 λ μ −+ = 2 M1 for equating like vectors Dep M1 for attempt to solve their simultaneous equations to obtain 2 solutions 20 11 μ = − A1 15 22 λ = − A1 4(a) 7 2 = a B1 1 = b B1 1 6 = c B1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 5 of 8 Question Answer Marks Guidance 4(b) 2 2 5 5 3 5 1 0    − − =          x x 2 M1 for recognition of a quadratic in 2 5 x Dep M1 for solution and a correct attempt to get at least one solution for x 3.59 A1 1 A1 5(a) 0 8 4 12 4 = + + + a b B1 For ( ) p 2 ( ) 2 p 3 2 6 ′ = + + x ax bx M1 For an attempt to obtain ( ) p′ x 3 2 6 7 − + = − a b M1 Dep for ( ) p 1 ′ − 0 2 4 = + + a b 13 3 2 − = − a b M1 Dep on both previous M marks for solution of equations to obtain both a and b 3 2 = − = a b A1 5(b) ( ) p 18 4 ′′ = − + x x M1 For differentiation of their ( ) p′ x to obtain ( ) p′′ x 4 A1 FT on twice their b. 6 ( ) 3 2 d e 2 d = + + x y m x c x M1 ( ) 3 2 d 2e 2 d = + + x y x c x A1 5 2 = + c 3 = c M1 Dep on previous M mark ( ) 3 3 f e ... x x p qx = + M1 3 3 2 2 e ... 3 3 x y x = + A1 5 2 3 3 = + d 1 = d M1 Dep on previous M mark ( ) ( ) 3 3 2 2 f e 3 1 3 3 = + + + x x x x A1 7(a) 6 B1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 6 of 8 Question Answer Marks Guidance 7(b) 192 = b a B1 May be implied by the term in x 2 240 = c a B1 May be implied by the term in 2 x 2 2 240 192 = c b M1 For elimination of a 2 5 768 = b c A1 For correct manipulation to verify the given answer 7(c) 1 16 = a B1 15 16 = c B1 8(a) 3 sin 2 5 = AOC or ( ) 2 2 2 6 5 5 2 5 5 cos = + − × × AOC M1 For a complete method to find AOC 1.2870 = AOC AOC = 1.287 A1 AG Must see 1.2870 = AOC or better before rounding for A1 8(b) Arc length = 1.287 5 × B1 Perimeter = 32.4 B1 8(c) Sector area = 2 1 5 1.287 2 × × B1 Area of triangle = 2 1 5 sin1.287 2 × × B1 Total area = 28.1 B1 9(a) ( )( ) ( ) 2 d 2 2 1 3 2 3 d = + − + − y x x x x or 2 d 6 22 12 d = − + y x x x M1 For differentiation of a quotient, or expansion and subsequent differentiation ( )( ) 0 2 3 3 2 = − − x x M1 Dep for simplification, equating to zero and attempt to solve ( ) 3, 0 A1 2 343 , 3 27       A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 7 of 8 Question Answer Marks Guidance 9(b) 4 B1 for correct shape with maximum in the first quadrant B1 for 1 , 0 2   −     and ( ) 3, 0 with a cubic curve with one max only B1 for ( ) 0, 9 with a cubic curve with one max only B1 All correct with a cusp at 1 2 = − x and a minimum at 3 = x 9(c) 343 27 B1 FT on their answer from (a) 10(a)(i) ( ) 2 1 0.5 16 + − = n oe M1 For use of ( ) 1 + − a n d 29 = n A1 10(a)(ii) ( ) ( ) ( ) 8 2 2 7 0.5 2 + M1 For use of sum formula, may be implied if distances have been multiplied by 5 first. ( ) ( ) ( ) 8 2 2 7 0.5 5 2 + × M1 For multiplication by 5 150 (km) A1 10(b)(i) 1.25 = r oe B1 10(b)(ii) ( ) 1 2 1.25 16 −> n or ( ) 1 2 1.25 16 −= n M1 For use of 1 − n ar ln8 1 ln1.25 −> n or ln8 1 ln1.25 −= n M1 Dep for correct method of solution to obtain 1 − n 11 A1 10(b)(iii) ( ) 8 2 1.25 1 1.25 1 − − M1 For use of sum formula may be implied by multiplication by 5 ( ) 8 2 1.25 1 5 1.25 1 − × − M1 For multiplication by 5 198 (km) A1 Allow greater accuracy

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 8 of 8 Question Answer Marks Guidance 11(a) 2 3cot 5cot 2 0 θ θ − − = M1 For use of correct identity and simplification to a 3 term quadratic equated to zero. 1 tan 3, tan 2 θ θ = − = M1 Dep for solution of quadratic and dealing with cot o 108.4 A1 o 26.6 A1 11(b) π π 3 6 φ + = − M1 For a correct order of operations π 2 φ = − A1 π 7π 3 6 φ + = M1 For a correct order of operations 5π 6 φ = A1