Cambridge IGCSE Mathematics - Additional 0606 — 2021 May/June Paper 2 · Variant 1

This document has 16 pages. Any blank pages are indicated. DC (RW/SG) 202074/2 © UCLES 2021 [Turn over * 9 7 2 6 8 6 3 0 4 8 * ADDITIONAL MATHEMATICS 0606/21 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 0606/21/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 0606/21/M/J/21 © UCLES 2021 [Turn over 1 (a) Write the expression x x 6 1 2 - + in the form ( ) x a b 2 + + , where a and b are constants. [2] (b) Hence write down the coordinates of the minimum point on the curve y x x 6 1 2 = - + . [1] 2 Variables x and y are such that, when lny is plotted against lnx, a straight line graph passing through the points (6, 5) and (8, 9) is obtained. Show that y x e p q = where p and q are integers. [4]

4 0606/21/M/J/21 © UCLES 2021 3 (a) Solve the inequality x 4 1 9 2 - . [3] (b) Solve the equation x x 2 11 12 0 - + = . [3]

5 0606/21/M/J/21 © UCLES 2021 [Turn over 4 The graph of tan y a bx 2 = + , where a and b are constants, passes through the point (0, 4 - ) and has period 480°. (a) Find the value of a and of b. [3] (b) On the axes, sketch the graph of y for values of x between 0° and 480°. [2] x y 0 480°

6 0606/21/M/J/21 © UCLES 2021 5 The curves y x2 = and y x 27 2 = intersect at O(0, 0) and at the point A. Find the equation of the perpendicular bisector of the line OA. [8]

7 0606/21/M/J/21 © UCLES 2021 [Turn over 6 Variables x and y are such that cos y x x 2 e x 2 = + , where x is in radians. Use differentiation to find the approximate change in y as x increases from 1 to h 1+ , where h is small. [6]

8 0606/21/M/J/21 © UCLES 2021 7 Find the exact values of the constant k for which the line y x 2 1 = + is a tangent to the curve y x kx k 4 2 2 = + + - . [6]

9 0606/21/M/J/21 © UCLES 2021 [Turn over 8 In this question, a, b, c and d are positive constants. (a) (i) It is given that ( ) ( ) log log y x x 3 2 1 a a = + + - . Explain why x must be greater than 2 1. [1] (ii) Find the exact solution of the equation ( ) log log y 3 6 2 a a + = . [3] (b) Write the expression ( )( ) log log log b a 9 9 a a b + in the form log c d 9 a + , where c and d are integers. [4]

10 0606/21/M/J/21 © UCLES 2021 9 A curve is such that r sin x y x 6 2 d d 2 2 = - b l. Given that x y 2 1 d d = at the point r r , 4 12 13 b l on the curve, find the equation of the curve. [7]

11 0606/21/M/J/21 © UCLES 2021 [Turn over 10 Relative to an origin O, the position vectors of the points A, B, C and D are OA 6 5 = - e o, OB 10 3 = e o, OC x y = e o and OD 12 7 = e o. (a) Find the unit vector in the direction of AB. [3] (b) The point A is the mid‑point of BC. Find the value of x and of y. [2] (c) The point E lies on OD such that OE : OD is 1 : 1 m + . Find the value of m such that BE is parallel to the x‑axis. [3]

12 0606/21/M/J/21 © UCLES 2021 11 The 2nd, 8th and 44th terms of an arithmetic progression form the first three terms of a geometric progression. In the arithmetic progression, the first term is 1 and the common difference is positive. (a) (i) Show that the common difference of the arithmetic progression is 5. [5] (ii) Find the sum of the first 20 terms of the arithmetic progression. [2]

13 0606/21/M/J/21 © UCLES 2021 [Turn over (b) (i) Find the 5th term of the geometric progression. [2] (ii) Explain whether or not the sum to infinity of this geometric progression exists. [1]

14 0606/21/M/J/21 © UCLES 2021 12 x B A O y ( )( ) y x x 9 3 = - - y k 3 = - ( , ) C k k 3 - The diagram shows part of the curve ( )( ) y x x 9 3 = - - and the line y k 3 = - , where k 3 2 . The line through the maximum point of the curve, parallel to the y‑axis, meets the x‑axis at A. The curve meets the x‑axis at B, and the line y k 3 = - meets the curve at the point C(k, k 3 - ). Find the area of the shaded region. [9]

15 0606/21/M/J/21 © UCLES 2021 Continuation of working space for Question 12.

16 0606/21/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 11 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/21 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.

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 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 2021 © UCLES 2021 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. 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

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 4 of 11 Question Answer Marks Partial Marks 1(a) (x – 3)2 – 8 B2 B1 for (x – 3)2 + k where k ≠ −8 or a = −3 or (x + m)2 − 8 where m ≠ −3 or b = −8 1(b) (3, −8) B1 strict FT their a and b 2 9 5 8 6 − = − m oe M1 9 = their 2(8) + c oe or 5 = their 2(6) + c M1 lny = 2lnx – 7 A1 Correct completion to answer: 2 ln 7 7 2 e e − − = = x y x nfww A1 Alternative lny = p + qlnx soi (B1) 9 5 8 6 − = − m oe (M1) 9 = their 2(8) + c oe or 5 = their 2(6) + c (M1) 7 2 e− = y x (A1) 3(a) 4x – 1 *9 oe and 4x − 1*−9 oe OR 16x2 − 8x − 80*0 oe soi M1 where * could be = or any inequality sign x > 5 2 , x < −2 only; mark final answer A2 not from wrong working A1 for CV 5 2 , −2 oe If M0 then SC1 for any correct inequality with at most one extra inequality

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 5 of 11 Question Answer Marks Partial Marks 3(b) ( )( ) 2 3 4 − − x x or x = u2 and (2u – 3)(u – 4) oe soi M1 3, 4 2 = = x x oe A1 x = 9 4 , x = 16 B1 FT their x Alternative ( ) 2 1 2 2 2 12 11   + =     x x simplified to 2 4 73 144 0 − + = x x (M1) solves 3 term quadratic in x (M1) x = 9 4 , x = 16 (A1) 4(a) a = −4 B1 180 480 = b oe M1 b = 3 8 A1 4(b) Correct sketch 0° x y 480° 240° B2 correct tan shape, two branches starting and finishing on same negative y value asymptote implied at x = 240 root between 120 and 240 B1 for correct tan shape with exactly two branches plus one other correct property Maximum B1 if not fully correct

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 6 of 11 Question Answer Marks Partial Marks 5 ( ) 2 2 27 = x x or 2 2 27   =     y y oe M1 if M0 then, for first 4 marks, SC4 if (3, 9) only stated and verified in both equations, ignore (0, 0) or SC2 for (3, 9) only stated with no working, ignore (0, 0) If first M1 then (3, 9) with no additional working award M1SC1 4 27 0 − = x x or 4 729 0 − = y y oe nfww A1 x(x3 − 27) = 0 or y(y3 − 729) = 0 oe M1 A(3, 9) oe only nfww A1 Mid-point = ( ) 1.5,4.5 B1 9 3 = OA m oe B1 3 9 ⊥= − m oe M1 y – 4.5 = 3 ( 1.5) 9 − − x oe isw A1 FT their mid-point and 1 9 3 − their 6 2 2 d(e ) 1 e d 2 = x x x B1 d(cos2 ) 2sin 2 d = − x x x soi B1 x × their(−2sin2x) + cos2x M1 2 d 1 e d 2   =     x y x −2xsin2x + cos2x A1 FT their 2 2 d e e d      = x x k x 1 δ d d = = x y y their h x M1 −1.41[03...]h nfww A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 7 of 11 Question Answer Marks Partial Marks 7 2 4 2 2 1 + + − = + x kx k x M1 2 4 ( 2) 3[*0] + − + − x k x k soi A1 * can be <, >, =, ≤, ≥ ( ) 2 2 4(4)( 3) − − − k k M1 2 20 52 0 − + ∗ k k A1 2 ( 20) ( 20) 4(1)(52) 2 −− ± − − = k M1 k = 10 48 ± oe isw A1 Alternative (using calculus) 2 8 = + x k oe (M1) ( ) 2 4 2 8 2 8 2 = + − + − − y x x x x or 2 4 6 = − − y x x (M1) 0 = 4x2 + 8x +1 (A1) 2 8 8 4(4)(1) 8 −± − = x (M1) 48 1 8 = −± x oe (A1) for k = 10 48 ± oe (A1) 8(a)(i) Valid explanation e.g. This ensures the argument of both logarithms is greater than 0 B1 8(a)(ii) 2 log 6 log ( 3) = + a a y oe B1 2 ( 3) 6 + = y M1 3 6 = −+ y oe only final answer A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 8 of 11 Question Answer Marks Partial Marks 8(b) Within a complete expression: Correct change of base to a: log 9 log 9 log = a b a a a b Correct use of power law: 1 log log 2 = a a b b Correct use of addition/multiplication law: log 9 log 9 log = + a a a a a Correct use of loga a = 1 M3 M2 for 2 or 3 correct steps within complete expression or M1 for 1 correct step within complete expression leading to 2 3log 9. + a nfww A1 9 π cos 6 π 2 sin 6 d 2 6   −       − = − +      x x x c B2 B1 for π π sin 6 d cos 6 2 2     − = − +          x x k x c where k < 0 or 1 6 = k or 1 π cos 6 6 2   − −     x 1 1 6π π cos 2 6 4 2   = − − +     c M1 FT their k provided B1 awarded π cos 6 1 2 d 6 3 π sin 6 1 2 36 3     −       − +       −     = − + +  x x x x A M2 FT π cos 6 2   − +     their k x their c provided at least B1 awarded M1 for π 1 sin 6 2 3     − + +         m x their x A where m < 0 or 1 36 = m 13π 1 6π π 1 π sin 12 36 4 2 3 4     = − − + +         A M1 FT their m and their c provided at least M1 awarded π sin 6 1 2 π 36 3   −     = − + + x y x oe cao A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 9 of 11 Question Answer Marks Partial Marks 9 Alternative sin6 cos6 d 6 − = − +  x x x c B2 B1 for cos6 d sin6 − = +  x x k x c where k < 0 or 1 6 = k or sin6 6 − x 1 1 3π sin 2 6 2 = − + c oe M1 FT their k provided B1 awarded sin6 1 d 6 3 cos6 1 36 3   − + =     + +  x x x x A M2 FT sin6 + their k x their c provided at least B1 awarded M1 for 1 cos6 3   + +     m x their x A where m > 0 or 1 36 = − m 3π cos 13π 1 π 2 12 36 3 4   = + +     A M1 FT their m and their c cos6 1 π 36 3 = + + x y x oe cao A1 10(a) 4 8  =    AB B1 2 2 4 8 + M1 FT their 4 8    4 1 8 80    oe isw A1 FT provided working shown 10(b) 6 10 1 5 3 2 +     =     − +     x y oe M1 x = 2, y = −13 A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 10 of 11 Question Answer Marks Partial Marks 10(c) 12 1 7 1 λ   =   +    OE oe seen B1 Solves their 7 3 1 λ = + M1 λ = 4 3 oe A1 Alternative 3  =    x OE 12 7 3 = x 36 7 = x (B1) 1 12 36 1 7 λ + = (M1) FT their x λ = 4 3 (A1) 11(a)(i) 1 + d, 1 + 7d, 1 + 43d soi B1 [ ] 1 7 1 43 1 1 7 + + = = + + d d r their their d d M2 FT their ratios of terms provided in terms of a and d M1 FT for either [ ] 1 7 1 + = + d r d or [ ] 1 43 1 7 + = + d r d Simplifies to 6d 2 – 30d = 0 oe nfww A1 Verifies that d = 5 by substitution or factorises and solves to obtain d = 5 only A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 11 of 11 Question Answer Marks Partial Marks 11(a)(i) Alternative 1 + d, 1 + 7d, 1 + 43d soi B1 2 7 6 43 42 − −  =     a a a a oe M2 M1 for 7 6 − a a or for 43 42 − a a oe 2 6a – 42a + 36 = 0 oe A1 Finds a = 6 and uses it to show that d = 5 only A1 11(a)(ii) { } 20 20 2[1] (20 1)(5) 2 = + − S M1 970 A1 11(b)(i) 7776 nfww B2 B1 for 6 × 65–1 11(b)(ii) Valid explanation e.g. The sum to infinity does not exist for this GP as the common ratio is greater than 1. B1 12 x-coordinate of A = 6 soi B1 x-coordinate of B = 9 soi B1 k – 3 = (9 – k)(k – 3) M1 k = 8 [therefore C(8, 5)] A1 ( ) 8 6 5 − × or 10 oe soi B1 ( ) 9 2 8 12 27 d − −  their their x x x 3 2 12 27 2 3 = − −x x x M2 M1 for 2 correct terms their10 + F(their 9) – F(their 8) M1 DEP on at least M1 for integration 38 3 or 2 12 3 or 12.7 or 12.66[66...] rot to 4 or more figs nfww A1