Cambridge IGCSE Mathematics - Additional 0606 — 2021 May/June Paper 1 · Variant 1

This document has 16 pages. Any blank pages are indicated. Cambridge IGCSE™ * 2 1 7 2 6 1 8 6 7 8 * ADDITIONAL MATHEMATICS 0606/11 Paper 1 May/June 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 [ ]. DC (LK/FC) 202078/4 © UCLES 2021 [Turn over

2 0606/11/M/J/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/M/J/21 © UCLES 2021 [Turn over 1 (a) On the axes, sketch the graph of ( )( )( ) y x x x 5 1 3 2 2 = + - - , stating the intercepts with the coordinate axes. [3] y O x (b) Hence find the values of x for which ( )( )( ) x x x 5 1 3 2 2 0 2 + - - . [2] 2 Find ( ) d x x x 1 1 1 1 2 3 5 - - - e o y , giving your answer in the form ln a b + , where a and b are rational numbers. [5]

4 0606/11/M/J/21 © UCLES 2021 3 The polynomial ( ) p x ax x bx 9 6 3 2 = - + - , where a and b are constants, has a factor of x 2 - . The polynomial has a remainder of 66 when divided by x 3 - . (a) Find the value of a and of b. [4] (b) Using your values of a and b, show that ( ) ( ) ( ) p q x x x 2 = - , where ( ) q x is a quadratic factor to be found. [2] (c) Hence show that the equation ( ) p x 0 = has only one real solution. [2]

5 0606/11/M/J/21 © UCLES 2021 [Turn over 4 The first 3 terms in the expansion of a x x 1 3 3 5 + - ` b j l , in ascending powers of x, can be written in the form bx cx 27 2 + + , where a, b and c are integers. Find the values of a, b and c. [8]

6 0606/11/M/J/21 © UCLES 2021 5 The functions f and g are defined as follows. ) (f x x x 4 2 = + for x R ! ( ) g e x 1 x 2 = + for x R ! (a) Find the range of f. [2] (b) Write down the range of g. [1] (c) Find the exact solution of the equation ) ( fg x 21 = , giving your answer as a single logarithm. [4]

7 0606/11/M/J/21 © UCLES 2021 [Turn over 6 (a) (i) Find how many different 5-digit numbers can be formed using the digits 1, 3, 5, 6, 8 and 9. No digit may be used more than once in any 5-digit number. [1] (ii) How many of these 5-digit numbers are odd? [1] (iii) How many of these 5-digit numbers are odd and greater than 60 000? [3] (b) Given that ( ) C C n 45 1 n n 4 1 5 # # = + + , find the value of n. [4]

8 0606/11/M/J/21 © UCLES 2021 7 (a) In this question, all lengths are in metres and time, t, is in seconds. –10 0 10 20 30 40 50 displacement time 10 20 30 40 The diagram shows the displacement–time graph for a runner, for t 0 40 G G . (i) Find the distance the runner has travelled when t 40 = . [1] (ii) On the axes, draw the corresponding velocity–time graph for the runner, for t 0 40 G G . [2] –1 0 2 4 6 velocity time 1 –2 –3 – 4 –5 –6 3 5 10 20 30 40

9 0606/11/M/J/21 © UCLES 2021 [Turn over (b) A particle, P, moves in a straight line such that its displacement from a fixed point at time t is s. The acceleration of P is given by t2 4 2 1 + - ` j , for t 0 2 . (i) Given that P has a velocity of 9 when t 6 = , find the velocity of P at time t. [3] (ii) Given that s 3 1 = when t 6 = , find the displacement of P at time t. [3]

10 0606/11/M/J/21 © UCLES 2021 8 DO NOT USE A CALCULATOR IN THIS QUESTION. A curve has equation y x x 2 3 1 2 = - + - ` j . The x-coordinate of a point A on the curve is 3 2 3 1 - + . (a) Show that the coordinates of A can be written in the form , p q r s 3 3 + + ` j, where p, q, r and s are integers. [5]

11 0606/11/M/J/21 © UCLES 2021 [Turn over (b) Find the x-coordinate of the stationary point on the curve, giving your answer in the form a b 3 + , where a and b are rational numbers. [3]

12 0606/11/M/J/21 © UCLES 2021 9 (a) (i) Write xy y x 6 3 4 2 + + + in the form ( )( ) ax b cy d + + , where a, b, c and d are positive integers. [1] (ii) Hence solve the equation sin cos cos sin 6 3 4 2 0 i i i i + + + = for ° ° 0 360 1 1 i . [4]

13 0606/11/M/J/21 © UCLES 2021 [Turn over (b) Solve the equation r sec 2 1 4 2 3 1 z+ = b l for r r 1 1 z - , where z is in radians. Give your answers in terms of r. [5]

14 0606/11/M/J/21 © UCLES 2021 10 In this question all lengths are in centimetres. O A B C 25 15 10 The diagram shows a shaded shape. The arc AB is the major arc of a circle, centre O, radius 10. The line AB is of length 15, the line OC is of length 25 and the lengths of AC and BC are equal. (a) Show that the angle AOB is 1.70 radians correct to 2 decimal places. [2] (b) Find the perimeter of the shaded shape. [4]

15 0606/11/M/J/21 © UCLES 2021 (c) Find the area of the shaded shape. [5]

16 0606/11/M/J/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. BLANK PAGE

This document consists of 9 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 Paper 1 May/June 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 May/June 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 May/June 2021 © UCLES 2021 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 2021 © UCLES 2021 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 2021 © UCLES 2021 Page 4 of 9 Question Answer Marks Guidance 1(a) 3 B1 for a well-drawn cubic graph in the correct orientation with both arms extending beyond x-axis B1 for 1, 2 = − = x x and 2 3 = x either on the graph or stated with a cubic graph B1 for 20 = y either on the graph or stated with a cubic graph 1(b) 2 1 3 −< < x B1 Must be found from a cubic graph 2 > x B1 2 ( ) 5 3 1 ln 1 1   − +   −   x x 2 B1 for ( ) ln 1 − x B1 for 1 1 + − x 1 1 ln 4 ln 2 4 2     + − +         M1 Dep on at least one B mark, for correct use of limits 1 ln2 4 − 2 A1 for ln2 A1 for 1 4 − oe 3(a) ( ) p 2 : 8 36 2 6 0 − + − = a b B1 ( ) p 3 : 27 81 3 6 66 − + − = a b B1 M1 Dep on at least one of the previous B marks, for attempt to solve their equations and obtain a solution for both a and b 6, 3 = = − a b A1 For both 3(b) ( )( ) 2 2 6 3 3 − + + x x x 2 M1 for attempt at quadratic factor either by observation to obtain 2 6 3 + + x px or by algebraic long division to obtain at least 2 6 3 ... + x x A1 all correct

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 5 of 9 Question Answer Marks Guidance 3(c) Discriminant of ( ) q x = 2 3 4 6 3 − × × 63 = − which is < 0 M1 For calculation of discriminant and confirmation that it is < 0 ( ) q 0 = x has no real solutions hence ( ) p 0 = x has only one real solution A1 For a correct conclusion from correct work. 4 ( ) 3 3 2 2 3 3 3   + = + + +   a x a a x ax x B1 5 2 5 10 1 1 ... 3 3 9   − = − +     x x x 2 M1 allow one sign error or one arithmetic slip 3 27, 3 = = a a B1 Term in x: 2 3 5 3 3 − = a a b M1 For multiplying their terms, must have sum of 2 relevant products = b 18 = − b A1 Term in 2 : x ( ) 2 3 5 10 3 3 3 9 − + = a a a c M1 For multiplying their terms, must have sum of 3 relevant products = c 6 = − c A1 5(a) f ⩾ –4 2 M1for a valid method to find the least value of 2 4 + x x A1 for f ⩾ –4, y ⩾ –4 or f(x) ⩾ –4 5(b) g 1 > B1 Allow 1 > y or ( ) g 1 > x 5(c) ( ) ( )[ ] 2 2 2 1 e 4 1 e 21 + + + = x x M1 4 2 e 6e 16 0 + − = x x ( )( ) 2 2 e 8 e 2 0 + − = x x M1 Dep for quadratic in terms of 2e x and attempt to solve to obtain 2e = x k 2e 2 = x 1 ln 2 = x k M1 Dep on both previous M marks, for attempt to solve 2e = x k ln 2 = x or 1 2 ln2 A1 6(a)(i) 720 B1 6(a)(ii) 480 B1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 6 of 9 Question Answer Marks Guidance 6(a)(iii) [Starts with 6 or 8]: 192 B1 [Starts with 9]: 72 B1 Total = 264 B1 Alternative [Ends with 9]:48 (B1) [Ends with 1,3 or 5]:216 (B1) Total = 264 (B1) 6(b) ( ) ( )( ) ( ) ( ) 1 1 ! 45 ! 4 !4! 1 5 !5! + + = − + − n n n n n B1 ( ) 2 1 45 5 + = n leading to 15 1 = + n or 2 2 224 0 + − = n n M2 M1 for 15 M1 for n + 1 OR M1 for 2 2 224 0 + − = n n oe M1 for ( )( ) 14 16 0 − + = n n 14 = n only A1 7(a)(i) 110 (m) B1 7(a)(ii) B2 B1 for a line joining ( ) 0,5 and ( ) 10,5 B1 for a line joining ( ) 10, 2 − and ( ) 40, 2 − 7(b)(i) ( ) ( ) 1 2 2 4 = + + v t c M1 For ( ) 1 2 2 4 + k t 9 4 = + c M1 Dep for attempt to find c using 9 = v and 6 = t in their v ( ) 1 2 2 4 5 + + t A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 7 of 9 Question Answer Marks Guidance 7(b)(ii) ( ) ( ) 3 2 1 2 4 5 3 = + + + s t t d M1 For ( ) 3 2 2 4 + k t 1 64 30 3 3 = + + d M1 Dep for attempt to find d using 1 3 = s and 6 = t in their s ( ) 3 2 1 2 4 5 51 3 + + − t t A1 8(a) 3 1 2 3 2 3 2 3 + + = × − + x leading to 1 5 3 3 = + x M1 For attempt to rationalise and simplify showing all working 5 3 3 = + x A1 Either: Using 5 3 3 = + x ( )( ) 2 3 52 30 3 5 3 3 1 = − + + + − y 14 8 3 4 3 3 = + + + Or: Using x = 3 1 2 3 + − ( ) ( ) ( ) 2 2 3 1 3 1 2 3 1 2 3 2 3 + + = − + − − − y 4 2 3 3 1 2 3 2 3 + + + − + = − ( ) ( ) ( ) 4 3 3 2 3 2 3 2 3 + + = × − + 8 3 6 12 3 3 1 + + + = M1 For complete method, showing all steps. Allow one slip in arithmetic 11 3 18 + 2 A1 for 18 A1 for 11 3

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 8 of 9 Question Answer Marks Guidance 8(b) ( ) d 2 2 3 1 d = − + y x x M1 For attempt at differentiation to obtain form of d 1 d = + y kx x ( ) 0 2 2 3 1 = − + x ( ) ( ) ( ) 2 3 1 2 2 3 2 3 + = − × − + x leading to ( ) 2 3 1 2 1 − + = x M1 Dep on previous M for equating to zero, rationalisation and attempt to simplify 3 1 2 = −− x A1 9(a)(i) ( )( ) 3 2 2 1 + + y x B1 9(a)(ii) ( )( ) 3cos 2 2sin 1 0 θ θ + + = 2 cos 3 θ = − , 1 sin 2 θ = − M1 For relating to part (i) and a correct attempt to obtain cos ... θ = or sin ... θ = o o 131.8 , 228.2 θ = o o 210 , 330 θ = 3 M1 for solving one of the equations to obtain one correct solution A1 for any two correct solutions A1 for a further two correct solutions with no extra solutions within the range 9(b) π 3 cos 2 4 2 φ   + =     oe B1 5π π 19π 23π , , , 24 24 24 24 φ = − − 4 M1 for solving to obtain one correct positive solution M1 for solving to obtain one correct negative solution A1 for any two correct solutions A1 for a further two correct solutions with no extra solutions within the range 10(a) 7.5 sin 2 10 = AOB M1 For a valid method 1.696 = AOB = 1.70 to 2 dp A1 Must see greater accuracy to justify given answer

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 9 of 9 Question Answer Marks Guidance 10(b) 2 2 2 10 25 2 10 25cos 2     = + − × ×         AOB AC M1 For a complete and valid method to find AC awrt 19.9 = AC A1 Major arc AB = awrt 45.9 or awrt 45.8 B1 Perimeter = awrt 85.5 or awrt 85.6 A1 10(c) Area of major sector AOB = ( ) 2 1 10 2π 2 × −AOB M1 awrt 229 A1 Area of kite OACB = 1 15 25 2 × × B1 Allow working with 2 separate triangles Area of their major sector plus area of their kite M1 Total area = awrt 417 A1