Cambridge IGCSE Mathematics - Additional 0606 — 2021 May/June Paper 2 · Variant 3

This document has 16 pages. Any blank pages are indicated. DC (RW/SG) 202072/3 © UCLES 2021 [Turn over * 2 4 1 7 6 2 7 7 0 5 * ADDITIONAL MATHEMATICS 0606/23 Paper 2 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 [ ]. Cambridge IGCSE™

2 0606/23/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 0606/23/M/J/21 © UCLES 2021 [Turn over 1 DO NOT USE A CALCULATOR IN THIS QUESTION. Write 7 3 5 4 5 - - with a rational denominator, simplifying your answer. [3] 2 Given that ( ) ( ) y 2 7 3 7 19 x x 2 1 = - + + , find the value of x when y 30 = . [4]

4 0606/23/M/J/21 © UCLES 2021 3 (a) Write ( ) y x xy 81 27 5 4 3 3 5 in the form x y 3a b c # # where a, b and c are constants. [3] (b) (i) Find the value of a such that log 2 8 2 3 a = . [2] (ii) Write log a 3 ( ) a2 as a single logarithm to base a. [2]

5 0606/23/M/J/21 © UCLES 2021 [Turn over 4 Variables x and y are such that cos sin y x x = . Using differentiation, find the approximate change in y as x increases from r 4 - to r h 4 - , where h is small. [4] 5 (a) Solve the inequality x x 2 17 21 0 2 G - + . [3] (b) Hence find the area enclosed between the curve y x x 2 17 21 2 = - + and the x‑axis. [3]

6 0606/23/M/J/21 © UCLES 2021 6 The polynomial p is given by ( ) p x x x x 36 15 2 1 3 2 = - - + . (a) Show that . x 0 25 =- is a root of the equation ( ) p x 0 = . [1] (b) Show that the equation ( ) p x 0 = has a repeated root. [4]

7 0606/23/M/J/21 © UCLES 2021 [Turn over 7 (a) Sketch the graph of the curve ( ) ln y x 4 3 = - on the axes, stating the intercept with the x‑axis. [2] – 1 3 x y 0 (b) Find the equation of the tangent to the curve ( ) ln y x 4 3 = - at the point where x 2 = . [5]

8 0606/23/M/J/21 © UCLES 2021 8 (a) (i) Find r sin 3 d z z + e o y . [2] (ii) Find ( ) sin cos 5 5 d 2 2 i i i + y . [2] (b) Show that 1 e x x 1 1 1 3 1 d e e 2 + - = - eb l o y . [4]

9 0606/23/M/J/21 © UCLES 2021 [Turn over 9 (a) The function f is defined, for all real x, by ( ) f x x x 13 4 2 2 = - - . (i) Write ( ) f x in the form ( ) a b x c 2 + + , where a, b and c are constants. [3] (ii) Hence write down the range of f. [1] (b) The function g is defined, for x 1 H , by ( ) g x x x 2 1 2 = + - . (i) Given that ( )x g 1 - exists, write down the domain and range of g 1 - . [2] (ii) Show that ( )x px q 1 g 1 2 =- + + - , where p and q are integers. [4]

10 0606/23/M/J/21 © UCLES 2021 10 In this question all lengths are in centimetres. The volume and curved surface area of a cone of base radius r, height h and sloping edge l are rr h 3 1 2 and rrl respectively. y x x y 2 2 + The diagram shows a cone of base radius x, height y and sloping edge x y 2 2 + . The volume of the cone is 10r. (a) Find an expression for y in terms of x and show that the curved surface area, S, of the cone is given by r S x x 900 6 = + . [3]

11 0606/23/M/J/21 © UCLES 2021 [Turn over (b) Given that x can vary and that S has a minimum value, find the exact value of x for which S is a minimum. [5]

12 0606/23/M/J/21 © UCLES 2021 11 (a) The first three terms of an arithmetic progression are p 1 , q 1 , q 1 - . (i) Show that the common difference can be written as p 3 2 - . [3] (ii) The 10th term of the progression is p k , where k is a constant. Find the value of k. [2]

13 0606/23/M/J/21 © UCLES 2021 [Turn over (b) The sum to infinity of a geometric progression is 8. The second term of the progression is 2 3. Find the two possible values of the common ratio. [5]

14 0606/23/M/J/21 © UCLES 2021 12 A particle moves in a straight line such that its displacement, s metres, from a fixed point O at time t seconds, is given by cos s t t 2 2 = + - , for t 0 H . (a) Find the displacement of the particle from O at the time when it first comes to instantaneous rest. [5]

15 0606/23/M/J/21 © UCLES 2021 (b) Find the time when the particle next comes to rest. [1] (c) Find the distance travelled by the particle for r t 0 2 3 G G . [2]

16 0606/23/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 10 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/23 Paper 2 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.

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 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.

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 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

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 4 of 10 Question Answer Marks Partial Marks 1 4 5 7 3 5 7 3 5 7 3 5 − + × − + attempted M1 Correct expansion 28 12 5 7 5 15 49 45 + − − − M1 DEP condone one arithmetic or sign slip 13 5 5 4 + or simplified equivalent A1 2 Attempts to solve 2 2(7 ) 21(7 ) 11 0 − − = x x or uses u = 7x and attempts to solve 2 2 21 11 0 − − = u u B1 (2(7x ) + 1)( 7x − 11) or (2u + 1)(u – 11) M1 FT their 2 2(7 ) (7 ) 0 + + = x x b c or 2 2 0 + + = u bu c with b and c both non-zero [7x = 1 2 − or] 7x = 11 A1 7 log 11 = x or ln11 ln7 or lg11 lg7 isw or 1.23[227...] only A1 3(a) 34 × 8 3 x × 15 4 y B3 B1 for each correct power or M1 for ( ) 5 5 3 5 4 243 3 x x y y or better 3(b)(i) 3 2 64 = a or 3 4 8 = a oe M1 a = 16 A1 If 0 scored, SC1 for correctly finding a from log 8 = a k , where k ≠ 0.75 3(b)(ii) Correct change of base to a: 2 log 3 log a a a a oe M1 Simplifies denominator: 1 2 log (3 ) a a oe A1

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 5 of 10 Question Answer Marks Partial Marks 4 tan = y x B1 2 d sec d = y x x B1 Alternative method for first 2 marks: B1 for d d cos and sin d d = = − u v x x x x B1 for 2 2 2 cos ( sin ) cos −− x x x ; allow unsimplified π 4 δ d d =− = x y y their h x M1 2h A1 5(a) (2x – 3)(x – 7) M1 CV 1.5, 7 A1 1.5 ⩽ x ⩽ 7 nfww A1 FT their CVs 5(b) 7 2 1.5(2 17 21)d − +  their their x x x = 1.5 7 3 2 2 17 21 3 2   − +    their their x x x B1 F(their 7) – F(their 1.5) M1 FT their 7 and their 1.5 from (a); must have at least two terms correct [ 1331 24 − , therefore area =] 1331 24 isw or 55.5 or 55.4583333... rot to 4 or more sig figs; nfww A1 6(a) p(−0.25) = 36(−0.25)3 − 15(−0.25)2 − 2(−0.25) +1 = 0 oe B1

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 6 of 10 Question Answer Marks Partial Marks 6(b) (4x + 1)(9x2 – 6x + 1) oe B2 B1 for any two correct terms in the quadratic factor (4x + 1)(3x – 1)(3x – 1 ) nfww B1 dep on B2 States e.g. Repeated factor, so repeated root or finds the remaining roots as 1 1 , 3 3 = = x x or finds 1 3 = x and indicates e.g. twice B1 dependent on all previous marks Alternative method 2 p ( ) 108 30 2 ′ = − − x x x (B1) solving p ( ) 0 ′ = their x or factorising p ( ) ′ their x (B1) 1 1 , 3 18 = = − x x (B1) 3 2 1 1 1 1 p 36 15 2 1 0 3 3 3 3         = − − + =                 [x-axis tangential to turning point, therefore root is repeated oe] (B1) 7(a) Correct sketch O x y x = 0.75 1 B2 B1 for correct shape passing through (1, 0) B1 for attempt at correct shape with asymptote at x = 0.75 soi 7(b) d 4 d 4 3 = − y x x B2 B1 for d d 4 3 = − y k x x where k ≠ 4 or 0 2 d 4 d 4(2) 3 = = − x y x or 4 5 M1 FT their k; dep on at least B1 awarded for differentiation When x = 2, y = ln 5 B1 y – ln 5 = ( ) 4 2 5 − x oe, isw A1 FT their ln5 and their 0.8

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 7 of 10 Question Answer Marks Partial Marks 8(a)(i) π 3cos ( ) 3 φ +   − +     c oe B2 B1 for π cos ( ) 3 φ +  +     k c where k < 0 or k = 3 8(a)(ii) 5dθ   =    5θ + c B2 B1 for 2 2 5sin 5cos 5 θ θ + = soi prior to integrating 8(b) 2 2 1 d   +      x x x soi B1 e 1 1 2ln 1 −   +   −   x x M1 FT 2 1 d   +      k x x x [ ] 1 2lne 2ln1 1 e   − − −     DM1 1 3e 1 2 1 e e − − + = A1 9(a)(i) 15 – 2(x + 1)2 isw B3 B1 for ( ) 2 1 + x B1 for a = 15 9(a)(ii) f ⩽ 15 B1 STRICT FT their a 9(b)(i) Domain: x ⩾ 2 B1 Range: g−1 ⩾ 1 B1 9(b)(ii) 2 2 2 ( 1 ) 0 + + −− = x x y or 2 2 2 ( 1 ) 0 + + −− = y y x B1 Correctly applies quadratic formula: [ ] 2 2 2 2 4(1)( 1 ) 2 −± − −− = y x or [ ] 2 2 2 2 4(1)( 1 ) 2 −± − −− = x y M1 FT their 2 2 2 ( 1 ) 0 + + −− = x x y or 2 2 2 ( 1 ) 0 + + −− = y y x with at most one sign error in the equation Justifies the positive square root at some point B1 Correct completion to 1 2 g ( ) 1 2 − = −+ + x x A1

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 8 of 10 Question Answer Marks Partial Marks 10(a) 2 30 = y x oe B1 2 2 2 30 π   = +     S x x their x M1 FT their 2 30 = y x providing 2 1 10π = π 3 x y was attempted Correct completion to given answer 6 π 900 + = x S x A1 10(b) 1 6 6 5 2 d([π] 900) 1 [π ] ( 900) 6 d 2 − + = × + × x x x x B2 B1 for 1 5 6 2 [π ] ( 900) − × + kx x , k ≠ 3 or 0 Applies correct form of quotient or product rule e.g.: ( ) 1 1 5 6 6 2 2 2 π 3 ( 900) π( 900) − + − + x x x x x or ( ) 1 1 2 6 5 6 2 2 π π ( 900) 3 ( 900) − − − + + + x x x x x M1 FT their 6 d([π] 900) d + x x their d d S x = 0 and attempt to solve M1 DEP x = 6 450 isw A1

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 9 of 10 Question Answer Marks Partial Marks 11(a)(i) 1 1 1 1 − = − − q p q q oe or 1 1 (3 1) (2 1) − − − = + − d d p p or 1 3 2 (3 1) 2   = + −     d p p M2 M1 for 1 1 [ ] = − d q p or 1 1 [ ] = − − d q q or 1 1 [2 ] = − − d q p or 1 1 (3 1) − = + − d q p or 1 1 (2 1) = + − d q p or 1 1 1 3 2 (3 1) 2   + − = + −     d p q q p correct completion to given answer 2 3 −p e.g. 1 1 2 3 3 3 − − = − p p p or 1 1 1 3 2 3 3 3 3 − = − = − p p p p p or makes d the subject of 1 1 (3 1) (2 1) − − − = + − d d p p or 1 3 2 (3 1) 2   = + −     d p p A1 11(a)(ii) 10 1 2 oe or 9 3   −   = +         k u p p p M1 k = −5 A1

0606/23 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 10 of 10 Question Answer Marks Partial Marks 11(b) ar = 1.5 and 8 1 = − a r oe, soi B1 Correctly eliminates a: 3 8(1 ) 2 = −r r oe M1 2 16 16 3 0 − + = r r oe A1 Attempts to solve their 3-term quadratic in r M1 Correct solutions r = 3 4 1 4 = r A1 Alternative method ar = 1.5 and 8 1 = − a r oe, soi (B1) Correctly eliminating r: 3 1 8 2   − =     a a oe (M1) 2 8 12 0 − + = a a (A1) Attempting to solve their 3-term quadratic in a and use the values of a to find r (M1) Correct solutions r = 3 4 1 4 = r (A1) 12(a) d 1 2sin d   = = +     s v t t soi B1 Puts their 1 2sin 0 + = t and solves for t M1 FT a + bsint where a and b are non-zero 7π 6 = t A1 7π 7π 2 2cos 6 6 = + − s M1 FT their t ≠ 0; dep on previous M1 7.4[0] or 7.397[24...] (metres) rot to 4 or more sig figs A1 12(b) t = 11π 6 B1 12(c) 7.3972... + (7.3972... – 6.7123...) M1 8.08[20...] (metres) A1