Cambridge IGCSE Mathematics - Additional 0606 — 2021 Feb/March Paper 1 · Variant 2

This document has 16 pages. Cambridge IGCSE™ DC (CE/SW) 202024/2 © UCLES 2021 [Turn over ADDITIONAL MATHEMATICS 0606/12 Paper 1 February/March 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 [ ]. * 3 2 5 1 6 3 3 3 7 5 *

2 0606/12/F/M/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 0606/12/F/M/21 © UCLES 2021 [Turn over 1 Find the exact solutions of the equation . ln ln x x 3 5 2 5 1 0 2 + - = ` j [4]

4 0606/12/F/M/21 © UCLES 2021 2 y x 2r - r - r 2r 0 1 2 3 4 5 6 7 y = a sin bx + c The diagram shows the graph of sin y a bx c = + where x is in radians and x 2 2 G G r r - , where a, b and c are positive constants. Find the value of each of a, b and c. [3]

5 0606/12/F/M/21 © UCLES 2021 [Turn over 3 The line AB is such that the points A and B have coordinates ( , ) 4 6 - and (2, 14) respectively. (a) The point C, with coordinates (7, a) lies on the perpendicular bisector of AB. Find the value of a. [4] (b) Given that the point D also lies on the perpendicular bisector of AB, find the coordinates of D such that the line AB bisects the line CD. [2]

6 0606/12/F/M/21 © UCLES 2021 4 (a) Show that x x 2 5 3 2 + - can be written in the form a x b c 2 + + ` j , where a, b and c are constants. [3] (b) Hence write down the coordinates of the stationary point on the curve with equation y x x 2 5 3 2 = + - . [2] (c) On the axes below, sketch the graph of y x x 2 5 3 2 = + - , stating the coordinates of the intercepts with the axes. [3] y x O (d) Write down the value of k for which the equation x x k 2 5 3 2 + - = has exactly 3 distinct solutions. [1]

7 0606/12/F/M/21 © UCLES 2021 [Turn over 5 In this question all lengths are in kilometres and time is in hours. Boat A sails, with constant velocity, from a point O with position vector 0 0 e o. After 3 hours A is at the point with position vector 12 9 - e o. (a) Find the position vector, OP, of A at time t. [1] At the same time as A sails from O, boat B sails from a point with position vector 12 6 e o, with constant velocity 5 8 - e o. (b) Find the position vector, OQ, of B at time t. [1] (c) Show that at time t PQ t t 26 36 180 2 = + + 2 . [3] (d) Hence show that A and B do not collide. [2]

8 0606/12/F/M/21 © UCLES 2021 6 (a) A geometric progression has first term 10 and sum to infinity 6. (i) Find the common ratio of this progression. [2] (ii) Hence find the sum of the first 7 terms, giving your answer correct to 2 decimal places. [2]

9 0606/12/F/M/21 © UCLES 2021 [Turn over (b) The first three terms of an arithmetic progression are log 3 x , ( ) log 3 x 2 , ( ) log 3 x 3 . (i) Find the common difference of this progression. [1] (ii) Find, in terms of n and log 3 x , the sum to n terms of this progression. Simplify your answer. [2] (iii) Given that the sum to n terms is log 3081 3 x , find the value of n. [2] (iv) Hence, given that the sum to n terms is also equal to 1027, find the value of x. [2]

10 0606/12/F/M/21 © UCLES 2021 7 DO NOT USE A CALCULATOR IN THIS QUESTION In this question all lengths are in centimetres. 17 1 - 17 4 + 17 1 + B C A D E The diagram shows a trapezium ABCDE such that AB is parallel to EC and ABCD is a rectangle. It is given that BC 17 1 = + , ED 17 1 = - and . DC 17 4 = + (a) Find the perimeter of the trapezium, giving your answer in the form a b 17 + , where a and b are integers. [3] (b) Find the area of the trapezium, giving your answer in the form c d 17 + , where c and d are integers. [2]

11 0606/12/F/M/21 © UCLES 2021 [Turn over (c) Find tanAED, giving your answer in the form e f 17 8 + , where e and f are integers. [2] (d) Hence show that sec AED 81 9 17 32 2 = + . [2]

12 0606/12/F/M/21 © UCLES 2021 8 (a) (i) Show that . sin tan cos sec x x x x + = [3] (ii) Hence solve the equation sin tan cos 2 2 2 4 i i i + = for r 0 4 G G i , where i is in radians. [4]

13 0606/12/F/M/21 © UCLES 2021 [Turn over (b) Solve the equation ( °) cot y 38 3 + = for ° ° y 0 360 G G . [3]

14 0606/12/F/M/21 © UCLES 2021 9 The polynomial ( )x x x x 2 3 1 p 3 2 = - - + has a factor x 2 1 - . (a) Find ( )x p in the form ( ) ( ) x x 2 1 q - , where ( )x q is a quadratic factor. [2] y x O B A y x 1 = y x x 2 3 1 2 =- + + The diagram shows the graph of y x 1 = for x 0 2 , and the graph of y x x 2 3 1 2 =- + + . The curves intersect at the points A and B. (b) Using your answer to part (a), find the exact x-coordinate of A and of B. [4]

15 0606/12/F/M/21 © UCLES 2021 [Turn over (c) Find the exact area of the shaded region. [6] Question 10 is printed on the next page.

16 0606/12/F/M/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. 10 A curve has equation y x x 2 10 1 2 2 3 = + - ` j for x 1 2 . (a) Show that x y d d can be written in the form ( x x Ax Bx C 2 10 1 2 2 2 1 2 + - + + ) ` ` j j , where A, B and C are integers. [5] (b) Show that, for x 1 2 , the curve has exactly one stationary point. Find the value of x at this stationary point. [4]

This document consists of 10 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 March 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 March 2021 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 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 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 3 of 10 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 4 of 10 Question Answer Marks Guidance 1 ( )( ) 3ln5 1 ln5 1 0 − + = x x 1 ln5 3 = x , ln5 1 = − x M1 For recognition of a quadratic in ln5x and attempt to solve to obtain ln5 = x k 1 3 1 e , 5 = x 1 3 ln5 3 e , e 5 − oe 1 1 ln5 1 e , , e 5e 5 − −− = x oe 3 Dep M1 for dealing with their ln5 = x k correctly once A1 for 1 3 1 e 5 = x oe isw A1 for 1 5e = x oe isw 2 3 = a B1 1 2 = b B1 4 = c B1 3(a) Gradient of line perp to AB 3 4 = − B1 Mid-point of AB ( ) 1, 10 − soi B1 ( ) 3 10 1 4 − = − + y x soi M1 For attempt at straight line using their perp gradient and their mid-point ( ) 3 10 7 1 4 − = − + a 4 = a A1 Allow 4 = y 3(b) ( ) 9, 16 − 2 B1 for 9 = − x B1 FT on their a, dep on M1 from (a) for 16 = y or 20 −their a B1 for 9,16 − 4(a) 2 5 49 2 4 8   + −     x 3 B1 for 2 5 4   = +     b x or ( ) 2 1.25 + x B1 for 49 8 = − c or 6.125 −

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 5 of 10 Question Answer Marks Guidance 4(b) 5 49 , 4 8   − −     oe 2 B1 for 5 4 − as part of a set of coordinates or 5 4 = − x , FT on − their b B1 for 49 8 − as part of a set of coordinates or 49 8 = − y FT on their c Need to be using their answer to (a) and not using differentiation as ‘Hence’. B1 for 5 49 , 4 8 − − 4(c) 3 B1 for correct shape, with maximum in the second quadrant and cusps on the x-axes and reasonable curvature for 3 < − x and 0.5. > x B1 for ( ) 3, 0 − and ( ) 0.5, 0 either seen on the graph or stated, must have attempted a correct shape B1 for ( ) 0, 3 either seen on the graph or stated, must have attempted a correct shape 4(d) 49 8 oe B1 FT on their c from (a) Allow 49 8 from other methods 5(a) 4 3 −       t or 0 4 0 3 −    +       t oe B1 5(b) 12 5 6 8 −     +         t or 12 5 6 8 −     +   t t B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 6 of 10 Question Answer Marks Guidance 5(c) 12 5 4 6 8 3 − −       = + −              PQ t t M1 For ( ) ( ) their b their a − , or ( ) ( ) their a their b − Allow unsimplified. Both vectors must be in terms of t 12 6 5 −     +   t t soi B1 ( ) ( ) ( ) 2 2 2 12 6 5 = − + +  PQ t t ( ) 2 2 26 36 180 = + +  PQ t t A1 Allow FT for use of modulus with 12 6 5 −     −−   t t and simplification to obtain the given result. 5(d) Attempt to solve or consider the discriminant of 2 26 36 180 0 + + = t t M1 Must be using the equation from part (c) as ‘Hence’. Conclusion from either ( )( ) 2 36 4 26 180 0 − < or 0 > t A1 Must have stated somewhere that ( ) 2 0 =  PQ oe has been considered not just ( ) 2 .  PQ 6(a)(i) 10, 6 1 = = − a a r 10 6 6 = −r M1 For use of first term and sum to infinity to obtain an equation in r only 2 3 = − r A1 6(a)(ii) ( ) ( ) 7 7 1 10 1 − = − their r S their r M1 For sum formula with . 1 < their r 7 6.35 = S A1 6(b)(i) log 3 x B1 6(b)(ii) ( ) ( ) 2log 3 1 log 3 2 = + − n x x n S n M1 For use of sum formula with their (i) ( ) 1 1 1 log 3, log 3 , log 3 2 2 2 + + + n n x x x n n n n A1 Allow other similar equivalents 6(b)(iii) ( )1 3081 2 + = n n M1 For a correct attempt to solve their (ii) 3081log 3 = x to obtain an answer for n. Must be a 3 term quadratic in n only. 78 = n A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 7 of 10 Question Answer Marks Guidance 6(b)(iv) ( ) 78 1027 79 log 3 2 = x or 3081 log 3 x M1 For using their 78 in a sum equation or using 3081 to obtain x 27 = x A1 7(a) ( ) ( ) 2 2 2 17 1 17 1 = − + + AE 18 2 17 18 2 17 = + + − M1 For attempt to find AE. Must see at least 3 terms in an expansion that is not the difference of two squares to be convinced a calculator is not being used. 6 = AE A1 Perimeter = 4 17 8 + + their AE 4 17 14 = + B1 FT on their AE 7(b) Area = ( )( ) 1 3 17 7 17 1 2 + + oe ( ) 1 51 3 17 7 17 7 2 = + + + oe M1 For attempt at a trapezium or triangle and rectangle. Must see at least 3 terms in an expansion that is not the difference of two squares to be convinced a calculator is not being used. Allow one arithmetic slip. Area 29 5 17 = + A1 7(c) 17 1 17 1 tan 17 1 17 1 + + = × − + AED M1 For attempt at rationalisation. 9 17 8 + A1 Must come from 18 2 17 16 + to be convinced a calculator is not being used. 7(d) 2 2 sec tan 1 = + AED AED ( ) 2 9 17 1 64 + = + 81 17 18 17 64 64 + + + oe if ( ) 2 9 17 64 + and 1 are considered separately. M1 For use of their (c) in the correct identity and attempt to simplify to obtain a single fraction. Must see at least 3 terms in an expansion that is not the difference of two squares to be convinced a calculator is not being used. Allow one arithmetic slip 81 9 17 32 + oe A1 cao

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 8 of 10 Question Answer Marks Guidance 8(a)(i) sin sin cos cos + x x x x B1 2 2 sin cos cos + x x x oe B1 1 sec cos = x x B1 Poor notation is B0 8(a)(ii) sec 4 2 θ = 1 cos 2 4 θ = M1 For use of (i) and dealing with sec to obtain 1 cos 2 4 θ = 1.3181, 4.9651 2 θ = 2.64 θ = or 0.839π 9.93 θ = or 3.16π 3 Dep M1 for a correct attempt to solve to obtain at least one solution for θ A1 for one correct solution A1 for a second correct solution and no extra solutions 8(b) ( ) o 1 tan 38 3 + = y o 172 = y o 352 = y 3 M1 for dealing with cot and a correct attempt to obtain at least one correct solution, allow for o 8 − A1 for one correct solution A1 for a second correct solution and no extra solutions 9(a) ( )( ) 2 2 1 1 − − − x x x M1 For attempt at factorisation by observation or by algebraic long division ( )( ) 2 2 1 1 − − − x x x A1 cao 9(b) At A 1 2 = x B1 2 1 0 − −= x x M1 For a valid attempt to solve their quadratic equation, allow for decimal solutions 1 5 2 ± = x soi A1 At B 1 5 2 + = x A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 9 of 10 Question Answer Marks Guidance 9(c) 1 d ln =  x x x B1 [ ] ( ) 1 5 2 1 2 ln ln 1 5 + = + x B1 Allow 1 5 1 ln ln 2 2   + −       2 2 3 2 3 2 3 1 d 3 2 x x x x x x   − + + = − + +        M1 M1 for attempt at 2 3 2 3 3 2 − + + x x x , must have 2 correct terms. 1 2 2 3 0 2 3 3 2 x x x   − + +     2 1 3 1 1 3 8 2 4 2     = − × + × +         oe M1 Dep for correct application of the limits 0 and 1 2 and attempt to evaluate – may be implied by 0.792 or 19 . 24 19 24 A1 ( ) 19 ln 1 5 24 + + A1 isw 10(a) ( )( )( ) ( ) ( ) 1 3 2 2 2 2 2 1 6 2 10 2 10 1 − + − + − x x x x x 3 B1 for ( ) 1 2 2 3 4 2 10 2 × × + x x oe M1 for correct attempt at differentiation of a quotient A1 for all the other terms correct ( ) ( ) ( ) 1 2 2 2 2 2 10 4 6 10 1   +   − −   −       x x x x 2 A2 for all 3 terms correct in the quadratic A1 for 2 terms correct and 1 incorrect term in the quadratic A0 for 1 term correct or no terms correct in the quadratic

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 10 of 10 Question Answer Marks Guidance 10(b) 2 4 6 10 0 − − = x x ( )( ) 2 5 1 0 − + = x x M1 For attempt to solve their quadratic = 0 and obtain at least one solution or state that their quadratic equation has no real roots. 5 2 = x A1 Rejecting 1 = − x correctly A1 May be implied by the statement 1. > x Discounting ( ) 1 2 2 2 10 0 + = x B1