Cambridge IGCSE Mathematics - Additional 0606 — 2023 May/June Paper 2 · Variant 1

This document has 16 pages. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/21 Paper 2 May/June 2023 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/CGW) 313043/3 © UCLES 2023 * 2 6 3 0 0 7 0 6 6 4 *

2 0606/21/M/J/23 © UCLES 2023 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 0606/21/M/J/23 © UCLES 2023 [Turn over 1 Variables x and y are such that when lg y is plotted against x a straight line passing through the points (1, 5) and (2.5, 8) is obtained. Show that y A b x # = where A and b are constants to be found. [4]

4 0606/21/M/J/23 © UCLES 2023 2 The function g is defined for ° ° x 0 120 G G by ( ) cos x x 2 4 6 g = + . (a) On the axes, sketch the graph of ( ) y x g = . [3] 30° 30° 0 60° 60° 90° 90° 120° 120° y x – 10 – 10 10 10 (b) State the amplitude of g. [1] (c) State the period of g. [1]

5 0606/21/M/J/23 © UCLES 2023 [Turn over 3 y x – 2.5 – 2 – 3.5 – 4 – 3 – 1.5 – 1 – 0.5 0.5 0 1 1.5 2 7 8 6 5 4 3 13 12 11 10 9 2 1 – 1 The diagram shows the graph of ( ) y x h = where ( ) ( ) ( ) x x a b cx h 2 = + + and a, b and c are integers. The curve meets the x-axis at the points (-2, 0) and (1.5, 0) and the y-axis at the point (0, 12). (a) Find the values of a, b and c. [2] (b) Use the graph to solve the inequality ( )x 9 h G . [3]

6 0606/21/M/J/23 © UCLES 2023 4 (a) Solve the equation 5 6 3 y y 2 1 # = - , giving your answer correct to 3 decimal places. [3] (b) Solve the equation e 4 3 0 e x x 2 2 - + = - , giving your answers in exact form. [4]

7 0606/21/M/J/23 © UCLES 2023 [Turn over 5 The volume, V, of a sphere of radius r is given by V r 3 4 3 r = . The volume of a sphere is increasing at a constant rate of cm s 24 3 1 - . Find the rate of increase of the radius when the radius is 6 cm. [4]

8 0606/21/M/J/23 © UCLES 2023 6 (a) The position vectors of the points P, Q and R relative to an origin O are 4 7 e o, 8 5 e o and x y e o respectively. The point R lies on PQ extended such that QR PR 3 2 = . Use a vector method to find the values of x and y. [3] (b) You are given that i is a unit vector due east and j is a unit vector due north. Three vectors, a, b and c are in the same horizontal plane as i and j and are such that a b c + = . The magnitude and bearing of a are 5 and 210°. The magnitude and bearing of c are 10 and 330°. (i) Find a and c in terms of i and j. [2]

9 0606/21/M/J/23 © UCLES 2023 [Turn over (ii) Find the magnitude and bearing of b. [5]

10 0606/21/M/J/23 © UCLES 2023 7 (a) y x O y x x 6 2 = - y x = 5 The diagram shows the curve y x x 6 2 = - for x 0 5 G G and the line y x = . Find the area of the shaded region. [4]

11 0606/21/M/J/23 © UCLES 2023 [Turn over (b) (i) Find ( ) cos d x x x 2 6 1 3 - + e o y . [3] (ii) Find ( ) d x x x 1 2 4 2 + y . [3]

12 0606/21/M/J/23 © UCLES 2023 8 (a) y x 0 3 The diagram shows the graph of ( ) f y x = where f is defined by ( ) f x x x 5 1 3 = + for x 0 3 G G . (i) Given that f is a one-one function, find the domain and range of f 1 - . [3] (ii) Solve the equation ( ) f x x = . [2] (iii) On the diagram above, sketch the graph of ( ) f y x 1 = - . [2]

13 0606/21/M/J/23 © UCLES 2023 [Turn over (b) The functions g and h are defined by ( ) g x x 8 3 3 3 = + for x 1 H , ( ) h e x x 4 = for x k H . (i) Find an expression for ( ) g x 1 - . [2] (ii) State the least value of the constant k such that gh(x) can be formed. [1] (iii) Find and simplify an expression for gh(x). [1]

14 0606/21/M/J/23 © UCLES 2023 9 In this question all lengths are in centimetres and all angles are in radians. (a) The area of a sector of a circle of radius 24 is cm 432 2. Find the length of the arc of the sector. [4] (b) O α D B A C y The diagram shows an isosceles triangle, OAB, with AO AB y = = and height AD. OCD is a sector of the circle with centre O. Angle AOB is a. (i) Find an expression for OB in terms of y and a. [1] (ii) Hence show that the area of the shaded region can be written as ( ) cos sin cos y 2 2 2 a a a a - . [3]

15 0606/21/M/J/23 © UCLES 2023 [Turn over 10 In the expansion of ax x b 2 9 + e o , where a and b are constants with a 0 2 , the term independent of x is -145 152 and the coefficient of x6 is - 6912. Show that a b 12 2 =- and find the value of a and the value of b. [7] Question 11 is printed on the next page.

16 0606/21/M/J/23 © UCLES 2023 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. 11 The line with equation x y k 3 + = , where k is a positive constant, is a tangent to the curve with equation x y y 2 9 0 2 2 + + - = . Find the value of k and hence find the coordinates of the point where the line touches the curve. [9]

This document consists of 11 printed pages. © UCLES 2023 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/21 Paper 2 May/June 2023 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 2023 series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 2 of 11 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.

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 3 of 11 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.

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 4 of 11 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 Question Answer Marks Partial Marks 1 lg 2 3 y x   OR lgb = 2 and lgA = 3 B2 B1 for 8 5 lg 2.5 1 y x c           soi or lg 3 y m x   soi OR lgb = 8 5 2.5 1   or lgA = 3 soi 2 3 10 x y   or 3 lg 2 10 y x  OR b = 100 and A = 1000 M1 FT their m and c 3 10 100 x y   oe mark final answer A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 5 of 11 Question Answer Marks Partial Marks 2(a) Correct curve 30 60 90 120 -10 10 0 y x 3 B2 for correct cosine shape over 2 cycles with midline at y = 2 and consistent amplitude or B1 for attempt at cosine shape over 2 cycles with consistent amplitude B1 for a consistent amplitude of 2; must have attempted correct shape Maximum of 2 marks if not fully correct 2(b) 4 1 2(c) 60 1 3(a) a = 2, b = 3, c = 2 2 B1 for any two correct 3(b) 3 ⩽ x ⩽ 0.5 or x ⩾ 1 3 B1 for the critical values 3, 0.5, 1 B1 for 3 ⩽ x ⩽ 0.5 B1 for x ⩾ 1 4(a) (2 1)log5 log6 log3 y y    oe or 2 log5 log30 log3 y y   oe OR [rearranges 2 5 6 3 5 y y   and collects powers to a single power in y] 2 5 30 3 y       oe M1 Collects terms and factorises: (2log5 log3) log6 log5 y    oe or (2log5 log3) log30 y   oe OR takes logs 2 25 3 5 log 30 oe or log log30 3 y y         oe M1 FT if of equivalent difficulty 1.604 A1 cao

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 6 of 11 Question Answer Marks Partial Marks 4(b) 4 2 e 4e 3 0 x x    or 2 2 2 (e ) 4e 3 0 x x    oe M1 condone one error Factorises: 2 2 (e 1)(e 3) x x   oe or solves 4 2 e 4e 3 0 x x    oe M1 FT   2 2 2 (e ) + e 0 x x a b c   2 2 e 1, e 3 x x   A1 x= 0, 1 ln3 2 x  or exact equivalent A1 5 2 d 4π d V r r  oe and 6 d 144π d r V r   B1 d d d d d d r V r t t V   soi B1 Not if chain rule for d d t r unless answer is inverted 24 their 144π M1 their 144π must come from an attempt at differentiation 0.0531 or 0.05305[16….] rot to 4 or more sig figs A1 6(a) 8 4 3 2 5 7 x x y y                  oe OR 8 5 x QR y          and 4 7 x PR y          and 3x – 24 = 2x – 8 and 3y – 15 = 2y – 14 M1 x = 16 A1 dep on vector method y = 1 A1 dep on vector method

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 7 of 11 Question Answer Marks Partial Marks 6(b)(i) a = 2.5i 5 3 2  j isw B1 c = 5i + 5 3 j isw B1 6(b)(ii) b = (5 + 2.5)i +   5 3 2.5 3  j oe soi B1   2 2 ( 2.5) 7.5 3 r    M1 FT their b of the form xi + yj providing neither component is zero [r = ] 13.2 or 13.22875… rot to 4 or more sf A1 dep on B1 tan = 7.5 3 2.5       oe or awrt 79.1 or tan  = 2.5 7.5 3       oe or awrt 10.9 M1 FT their b of the form xi + yj 349[.106…] rot to 3 or more sf A1 dep on B1 Alternative method 2 2 2 [ ]10 5 2 10 5 cos120 r       (M1) [r = ] 13.2 or 13.22875… rot to 4 or more sf (A1) sin sin 120 10 5 7 their their  or sin sin 120 5 5 7 their their  (M1) FT consistent use of their 120 and their r [ =] awrt 40.9 or    awrt 19.1 (A1) 349[.106…] rot to 3 or more sf (A1)

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 8 of 11 Question Answer Marks Partial Marks 7(a) 5 2 3 0 6 2 3 x x        B1 Area under line: 0.5 5 5 oe B1 Fully actioned correct plan: 3(25)  3 5 3  3 0 3(0) 3         0.5 5 5 oe M1 125 6 oe isw A1 dep on all previous marks Alternative method 5 2 0 (5 )d x x x   (B1) 5 2 3 0 5 2 3 x x        (B1) Correct use of correct limits 2.5(25)  3 5 3  3 0 2.5(0) 3        (M1) 125 6 oe isw (A1) dep on all previous marks 7(b)(i) 2 (2 6) sin 2 2 x x     (+ c) oe, isw 3 B1 for sinx B2 for 2 (2 6) 2 2 x    or B1 for 2 (2 6) 2 x    soi 7(b)(ii) 7 3 1 2 2 x x x   oe B1   8 4 1 ln 16 4 2 x x x c    oe or   8 4 1 ln 2 16 4 2 x x x c    oe B2 B1 for any two correct

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 9 of 11 Question Answer Marks Partial Marks 8(a)(i) [Domain f-1] 0 ⩽ x ⩽ 2.25 oe B2 B1 for either end correct or for 0 and 2.25 in an incorrect inequality [Range f-1 ] 0 ⩽ 1 f ⩽ 3 B1 8(a)(ii) x = 1.6 oe or x = 0 2 B1 for each 8(a)(iii) 0 x y 3 3 2.25 2.25 2 B1 for attempt at correct graph of inverse function drawn over correct domain soi B1 for correct shape with intersection in approximately correct location 8(b)(i) For a complete method to find the inverse, including changing the subject and swapping the variables M1 3 1 3 3 g ( ) 8 x x        oe mark final answer A1 8(b)(ii) [ k =] 0 1 8(b)(iii) 3 12 8e 3 x  mark final answer 1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 10 of 11 Question Answer Marks Partial Marks 9(a) 2 1 24 432 2     M1 3 2  rads soi A1 24  their  M1 36 cao A1 Alternative method soi and 1 432 2 s r r s     (B1) 1 24 432 2 s    (M1) 432 2 24 s   oe (M1) [s =] 36 (A1) 9(b)(i) [OB =] 2 cos y oe B1 9(b)(ii) ( 2 cos sin 2 their y y      2 1 cos 2 their y      oe M2 M1 for either area correct completion to   2 cos 2sin cos 2 y      A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 11 of 11 Question Answer Marks Partial Marks 10 [Term independent of x:] 9 6 3 3 C a b   or 6 3 84 a b   B1 6 3 145152 84 a b   M1 dep on B1   3 2 1728 a b  leading to 2 12 a b  or 2 3 1728 12 a b    A1 9 8 1 C a b   or 8 9 a b   B1 Correctly solves correct equations simultaneously 8 9 6912 a b    and 2 12 a b  as far as 6 a =… or 3 b =….. M1 Must be solving correct equations a = 2, b = 3 and no other values nfww B2 B1 for each nfww dep on previous B1B1 11 Eliminates one variable 2 2 ( 3 ) 2 9 0 k y y y      M1 2 2 10 (2 6 ) ( 9) 0 y k y k      soi A1 Uses 2 4 *0 b ac  with their 3-term quadratic:   2 2 (2 6 ) 4(10)( 9) *0 k k    M1 * can be = or any inequality sign   2 4 24 364 *0 k k    A1 Factorises 2 4 24 364 k k    or solves their 2 4 24 364 0 k k     M1 k = 7 only A1 Uses their k in 2 2 10 (2 6 ) ( 9) 0 y k y k      oe M1 y = 2 only A1 x = 1 only A1