Cambridge IGCSE Mathematics - Additional 0606 — 2022 Feb/March Paper 1 · Variant 2

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ * 7 0 1 1 6 7 4 5 8 2 * DC (CJ/CGW) 303740/2 © UCLES 2022 ADDITIONAL MATHEMATICS 0606/12 Paper 1 February/March 2022 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 0606/12/F/M/22 © UCLES 2022 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/22 © UCLES 2022 [Turn over 1 Find the values of k such that the line y kx 9 1 = + does not meet the curve ( ) . y kx x k 3 2 1 4 2 = + + + [5]

4 0606/12/F/M/22 © UCLES 2022 2 DO NOT USE A CALCULATOR IN THIS QUESTION. Solve the equation ( ) ( ) x x 3 5 3 2 3 5 1 0 2 - + + - = , giving your solutions in the form a b 3 + , where a and b are rational numbers. [6]

5 0606/12/F/M/22 © UCLES 2022 [Turn over 3 The curve with equation sin y a bx c = + , where a, b and c are constants, passes through the points r ( , ) 4 11 and r, 3 4 5 - e o. It is given that sin a bx c + has period r 16 . (a) Find the exact values of a, b and c. [4] (b) Using your answer to part (a), find the coordinates of the minimum point on the curve for r x 0 16 G G . [4]

6 0606/12/F/M/22 © UCLES 2022 4 (a) Show that ( ) x x 2 1 1 2 1 4 2 - + - can be written as ( ) x x 2 1 2 3 2 - + . [1] (b) Find ( ) d x x x 2 1 2 3 2 2 5 - + y , giving your answer in the form n a b 1 + , where a and b are constants. [5]

7 0606/12/F/M/22 © UCLES 2022 [Turn over 5 Variables x and y are such that ( ) n y x x 1 3 2 3 2 = - . (a) Find d d x y. [3] (b) Hence find the approximate change in y when x increases from 2 to h 2+ , where h is small. [2] (c) At the instant when x 2 = , y is increasing at the rate of 4 units per second. Find the corresponding rate of increase in x. [2]

8 0606/12/F/M/22 © UCLES 2022 6 The normal to the curve tan y x 1 3 = + at the point P with x-coordinate r 12, meets the x-axis at the point Q. The line r x 12 = meets the x-axis at the point R. Find the area of the triangle PQR. [8]

9 0606/12/F/M/22 © UCLES 2022 [Turn over 7 A curve ( ) f y x = is such that d d ( ) x y x 2 3 2 2 3 1 = - - . The curve passes through the point ( , . ) 2 10 2 - . The gradient of the tangent to the curve at ( , . ) 2 10 2 - is –6. Find ( ) f x . [8]

10 0606/12/F/M/22 © UCLES 2022 8 In this question, all lengths are in metres and all times are in seconds. A particle A is moving in the direction 20 21 - e o with a speed of 58. (a) Find the velocity vector of A. [1] (b) Given that A is initially at the point with position vector 5 3 - e o, write down the position vector of A at time t. [1] A particle B starts to move such that its position vector at time t is t t 44 2 35 4 - - + e o. (c) Find the displacement vector AB at time t. [2]

11 0606/12/F/M/22 © UCLES 2022 [Turn over (d) Hence find the distance AB, at time t, in the form pt qt r 2 + + , where p, q and r are constants. [2] (e) Find the value of t when the distance AB is 6, giving your answer correct to 2 decimal places. [2]

12 0606/12/F/M/22 © UCLES 2022 9 (a) The function f is such that ( ) n( ) f x x 1 5 2 = + , for x a 2 , where a is as small as possible. (i) Write down the value of a. [1] (ii) Hence find the range of f. [1] (iii) Find f 1 - ( )x , stating its domain. [3] (iv) On the axes, sketch the graphs of ( ) f y x = and f y 1 = - ( )x , stating the exact values of the intercepts of the curves with the coordinate axes. [4] y O x

13 0606/12/F/M/22 © UCLES 2022 [Turn over (b) The function g is such that g x x 4 2 1 7 | - , for x 0 2 . Solve the equation g ( )x 2 2 =- . [3]

14 0606/12/F/M/22 © UCLES 2022 10 (a) The first three terms of an arithmetic progression are , , sin sin sin x x x 3 5 3 9 3 . Find the exact values of x, where r x 0 2 G G , for which the sum to twenty terms is equal to 390. [6]

15 0606/12/F/M/22 © UCLES 2022 (b) The first three terms of a geometric progression are , , cos cos cos y y y 20 10 5 2 3 . (i) Explain why this progression has a sum to infinity. [2] (ii) Find the value of y, where y is in radians and y 0 2 1 1 , for which the sum to infinity is 9. Give your answer correct to 2 decimal places. [4]

16 0606/12/F/M/22 © UCLES 2022 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. BLANK PAGE

This document consists of 9 printed pages. © UCLES 2022 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 February/March 2022 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 February/March 2022 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 February/March 2022 © UCLES 2022 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2022 © UCLES 2022 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2022 © UCLES 2022 Page 4 of 9 Question Answer Marks Guidance 1 ( ) 2 9 1 3 2 1 4 kx kx k x + = + + + , leading to ( ) [ ] 2 3 3 3 0 kx x k + − + = M1 For equating the two equations and attempt to obtain a 3 term quadratic equation equated to zero. ( ) ( ) 2 3 3 4 3 k k − − × oe M1 Dep on previous M mark for attempt to use the discriminant in any form 2 3 10 3 k k − + oe M1 Dep on previous M mark for simplification to a 3 term quadratic expression in terms of k Critical values 3 and 1 3 A1 For both 1 3 3 k < < A1 Mark the final answer 2 x = ( ) ( ) ( )( ) ( ) 2 2 3 5 2 3 5 4 3 5 3 1 2 3 5 3 − + ± + − − − − M1 For the use of the quadratic formula x = ( ) ( ) 2 3 5 12 20 3 25 12 20 3 2 3 5 3 − + ± + + + − − M1 For expansion of the square root, must see at least 4 terms ( ) 12 2 3 2 3 5 3 x − − = − oe, ( ) 2 2 3 2 3 5 3 x − = − oe A1 For both ( ) 12 2 3 3 5 3 3 5 3 2 3 5 3 x − − + = × + − oe or ( ) 2 2 3 3 5 3 3 5 3 2 3 5 3 x − + = × + − oe with an attempt to simplify M1 For attempt to rationalise at least one of their solutions (must be similar structure) Sufficient detail must be seen, at least 3 terms in the numerator 1 3 2 2 + A1 Must have sufficient detail shown 2 3 11 33 − A1 Must have sufficient detail shown

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2022 © UCLES 2022 Page 5 of 9 Question Answer Marks Guidance 3(a) 1 8 b = B1 4π 11 sin 8 a c = + 4π 5 sin 3 8 a c −   = +   ×   M1 For attempt to form two simultaneous equations using given points, together with an attempt to obtain at least one unknown. Allow use of their b. 4 a = A1 7 c = A1 3(b) Using symmetry M1 For e.g. period is 16π , symmetrical about the line 8π x = For obtaining max at 4π and min at 12π M1 12π x = A1 3 y = A1 Alternative method 1 Minimum value when 3 y = (B2) FT on their a c −+ When 3, 12π. y x = = (2) M1 for attempt to solve their 3 sin a bx c = + using their values of a, b and c to get x = … Alternative method 2 Min occurs 3 4 through sine cycle so 12π x = (B2) When 12π, 3 x y = = (2) M1 for attempt to solve ( ) sin 12π y a b c = + using their values of a, b and c Alternative method 3 d cos d y ab bx x = ( )cos 0 ab bx = (M1) 4π, 12π x = (M1) Dep for attempt to solve to obtain x = 12π x = (A1) 3 y = (A1) cao

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2022 © UCLES 2022 Page 6 of 9 Question Answer Marks Guidance 4(a) ( ) ( ) ( ) 2 2 2 1 4 2 3 2 1 2 1 x x x x − + + = − − B1 Alternative method ( ) ( ) ( ) ( ) 2 2 3 3 2 1 4 2 1 4 4 3 2 1 2 1 x x x x x x − + − + − = − − ( )( ) ( ) ( ) 3 2 2 1 2 3 2 3 2 1 2 1 x x x x x − + + = = − − (B1) 4(b) Use of ( ) 2 1 4 d 2 1 2 1 x x x + − −  to obtain ( ) ( ) 1 2 ln 2 1 2 2 1 x x − − − 2 B1 for ( ) 1 ln 2 1 2 x − or equivalent B1 for ( ) 2 2 1 x − − , allow unsimplified ( ) ( ) 5 2 1 2 ln 2 1 2 2 1 x x   − −   −     1 2 1 2 ln9 ln3 2 9 2 3     − − −         M1 For application of limits, must be in the form ( ) ( ) ln 2 1 2 1 b a x x − + − 4 ln 3 9 = + 2 A1 for ln 3 A1 for 4 9 5(a) ( ) ( ) 2 2 2 4 3 3ln 2 3 2 3 d d 9 x x x x y x x     × − −   −   = oe 3 B1 for ( ) 2 4 2 3 x x − M1 for attempt to differentiate a quotient or product A1 for all terms other than ( ) 2 4 2 3 x x − correct. 5(b) When d 2, 0.133 d y x x = = M1 For substitution of 2 x = into their d d y x and use of h 0.133h A1 5(c) d 4 d 0.133 x t = M1 For 4 d value of from (b) d y their x 30.2 A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2022 © UCLES 2022 Page 7 of 9 Question Answer Marks Guidance 6 2 d 3sec 3 d y x x = 2 M1 for 2 sec 3 a x When π , 2 12 x y = = B1 When π d , 6 12 d y x x = = M1 Gradient of perpendicular is 1 6 − M1 For 1 d d y their x − , must be numeric Equation of normal 1 π 2 6 12 y x   − = − −     M1 For attempt at a normal equation using their 1 6 − and 2 Area of triangle = 12 2 M1 dep for attempt at correct area using their 2 and their π 12 12 + 7 ( ) 2 3 1 2 3 2 x − − 2 M1 for ( ) 2 3 1 2 3 , 2 a x a − ≠− Allow unsimplified When d 2, 6 d y x x = − = − leading to 4 c = − 2 M1 Dep for correct attempt to find the value of the arbitrary constant ( ) 5 3 1 2 3 10 x − nfww 2 M1 for ( ) 5 3 1 2 3 , 10 b x b − ≠ Allow unsimplified When 2, 10.2 x y = − = leading to 1 d = − M1 Dep on previous M mark for attempt to find the value of a second arbitrary constant ( ) 5 3 1 2 3 4 1 10 y x x = − − − A1 8(a) 40 42 −       B1 Allow 20 2 21 −      

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2022 © UCLES 2022 Page 8 of 9 Question Answer Marks Guidance 8(b) 5 40 3 42 t −     +     −     B1 FT on their answer to (a), must be numeric but not 20 21 −       8(c) 35 4 5 40 44 2 3 42 t t t t − + −     −     − −+     M1 Allow if in the incorrect order, FT on their (b), must have correct structure 5 1 2 1 t t −     +   A1 8(d) ( ) ( ) 2 2 5 1 2 1 AB t t = − + + M1 For attempt at modulus and square root using their answer to (c) 2 29 6 2 t t − + A1 8(e) 2 29 6 4 0 t t − − = M1 For attempt to solve the square of their answer to (d) 6 0 −= 0.49 only A1 9(a)(i) 0.4 − B1 9(a)(ii) ( ) f x ∈ oe B1 9(a)(iii) ( ) ln 5 2 x y = + oe e 5 2 x y = + oe M1 For a correct attempt to find the inverse ( ) 1 e 2 f 5 x x − − = A1 Must be in the correct form x∈ B1 9(a)(iv) O x y -0.2 -0.2 ln2 ln2 y = f(x) y = f-1(x) 4 B1 for two correctly shaped graphs in the correct quadrants B1 for a correct graph for ( ) f y x = with correct intercepts B1 for a correct graph for ( ) 1 f y x − = with correct intercepts B1 all correct with symmetry implied, exact intercepts and two points of intersection

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2022 © UCLES 2022 Page 9 of 9 Question Answer Marks Guidance 9(b) ( ) 1 1 2 2 2 4 4 g x x       = − −             M1 For a correct order of operations 1 1 2 2 4 4 2 x       − − = −             leading to 1 2 8, x = M1 Dep on previous M mark for a correct attempt at a solution. Must deal with 1 2 x correctly to obtain the final solution x = 64 A1 10(a) Common difference = 4sin3x soi B1 ( ) ( ) 20 390 2sin3 19 4sin3 2 x x = + M1 M1 for attempt at sum to 20 terms using their common difference, equating to 390 and attempt to solve to obtain sin3 ... x = sin 3 0.5 x = A1 π 5π , 18 18 x = 3 M1 for a correct attempt to solve, may be implied by one correct solution, allow if not exact A1 for 1 correct solution A1 for a second correct solution and no others in the range 10(b)(i) Common ratio = 0.5cos y B1 0.5 0.5cos 0.5 y − < < B1 Correct use of common ratio 1 < 10(b)(ii) 20cos 9 1 0.5cos y y = − B1 For attempt to use sum to infinity equation correctly and solve 18 cos or 0.367... 49 y = 2 M1 for solution of their equation, must have r as a multiple of cos ,y to obtain cos ... y = 1.19 A1