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

This document has 16 pages. Cambridge IGCSE™ * 3 6 1 7 7 6 5 8 5 9 * ADDITIONAL MATHEMATICS 0606/12 Paper 1 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 [ ]. DC (LK/FC) 202077/1 © UCLES 2021 [Turn over

2 0606/12/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 0606/12/M/J/21 © UCLES 2021 [Turn over 1 Write ( ) p r q pqr r 2 1 3 2 3 1 - - ` j in the form p q r a b c, where a, b and c are constants. [3]

4 0606/12/M/J/21 © UCLES 2021 2 (a) On the axes, sketch the graph of y x 4 3 = - , stating the intercepts with the coordinate axes. [2] y O x (b) Solve the inequality x 4 3 7 H - . [3]

5 0606/12/M/J/21 © UCLES 2021 [Turn over 3 O P B A C c b a The diagram shows the quadrilateral OABC such that a OA = , b OB = and c OC = . The lines OB and AC intersect at the point P, such that : : AP PC 3 2 = . (a) Find OP in terms of a and c. [3] (b) Given also that : : OP PB 2 3 = , show that b c a 2 3 2 = + . [2]

6 0606/12/M/J/21 © UCLES 2021 4 A curve is such that - ( ) d d x y x 3 2 2 2 3 1 = + . The curve has gradient 4 at the point (2, 6.2). Find the equation of the curve. [6]

7 0606/12/M/J/21 © UCLES 2021 [Turn over 5 (a) Given that log log log log p 5 4 20 a a a a + - = , find the value of p. [2] (b) Solve the equation ( ) 3 8 3 3 0 x x 2 1 + - = + . [3] (c) Solve the equation 2 4log 2 log 4. y y + = [3]

8 0606/12/M/J/21 © UCLES 2021 6 DO NOT USE A CALCULATOR IN THIS QUESTION. A curve has equation y x x 3 5 8 5 60 2 = + - + ` j . (a) Find the x-coordinate of the stationary point on the curve, giving your answer in the form a b 5 + , where a and b are integers. [4]

9 0606/12/M/J/21 © UCLES 2021 [Turn over (b) Hence find the y-coordinate of this stationary point, giving your answer in the form c 5, where c is an integer. [3]

10 0606/12/M/J/21 © UCLES 2021 7 (a) A six-character password is to be made from the following eight characters. Digits 1 3 5 8 9 Symbols * $ # No character may be used more than once in a password. Find the number of different passwords that can be chosen if (i) there are no restrictions, [1] (ii) the password starts with a digit and finishes with a digit, [2] (iii) the password starts with three symbols. [2] (b) The number of combinations of 5 objects selected from n objects is six times the number of combinations of 4 objects selected from n 1 - objects. Find the value of n. [3]

11 0606/12/M/J/21 © UCLES 2021 [Turn over 8 Variables x and y are such that y Axb = , where A and b are constants. When lgy is plotted against lgx, a straight line graph passing through the points (0.61, 0.57) and (5.36, 4.37) is obtained. (a) Find the value of A and of b. [5] Using your values of A and b, find (b) the value of y when x 3 = , [2] (c) the value of x when y 3 = . [2]

12 0606/12/M/J/21 © UCLES 2021 9 (a) The first three terms of an arithmetic progression are , , 4 8 20 - . Find the smallest number of terms for which the sum of this arithmetic progression is greater than 2000. [4]

13 0606/12/M/J/21 © UCLES 2021 [Turn over (b) The 7th and 9th terms of a geometric progression are 27 and 243 respectively. Given that the geometric progression has a positive common ratio, find (i) this common ratio, [2] (ii) the 30th term, giving your answer as a power of 3. [2] (c) Explain why the geometric progression 1, sini, sin2i, … for r r 2 2 1 1 i - , where i is in radians, has a sum to infinity. [2]

14 0606/12/M/J/21 © UCLES 2021 10 (a) Solve the equation sin cosec cos sec 0 2 2 a a a a + = for r r 1 1 a - , where a is in radians. [4]

15 0606/12/M/J/21 © UCLES 2021 [Turn over (b) (i) Show that sin cos cos sin sec 1 1 2 i i i i i - + - = . [4] (ii) Hence solve the equation sin cos cos sin 1 3 3 3 1 3 4 z z z z - + - = for ° ° 0 180 G G z . [4] Question 11 is printed on the next page.

16 0606/12/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. 11 The normal to the curve ln y x x 2 3 2 2 = - + ` j at the point where x 2 = meets the y-axis at the point P. Find the coordinates of P. [7]

This document consists of 11 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 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.

0606/12 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.

0606/12 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 4 of 11 Question Answer Marks Guidance 1 2 0 5 3 − − p q r 3 B1 for 0 = a B1 for 5 = − b B1 for 2 3 = − c 2(a) 2 B1 for symmetrical V shape in the correct quadrant, touching the x-axis. Must have straight lines. B1 for 4 3 = x and 4 = y only, either seen or stated on a modulus graph. 2(b) 1 − x ≤ , 11 3 x≥ or 3.67 or better 3 B1 for –1 from a correct method. B1 for 11 3 or 3.67 or better, from a correct method. 3(a) = −  AC c a B1 May be implied 3 5 = +   OP AC a or 2 5 −  AC c M1 Maybe implied, for correct use of ratio ( ) 3 5 = +   OP theirAC a or ( ) 2 5 −  theirAC c 2 3 5 5 = +  OP a c A1 Allow unsimplified 3(b) 2 5 =  OP b oe B1 2 2 3 5 5 5 = + b a c 2 2 3 = + b a c B1 Dep on previous B mark for equating vectors and rearrangement to obtain AG Alternative 2 3 3 5 5 5 = + + b a c b (B1) Need a clear indication of the method used, in the form of a correct unsimplified statement. 2 2 3 = + b a c (B1) Dep for simplification to obtain AG

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 5 of 11 Question Answer Marks Guidance 4 ( ) ( ) 2 3 d 1 3 2 d 2   = + +     y x c x M1 For ( ) 2 3 1 3 2 + k x where 1k a constant. 4 2 = + c M1 Dep for use of 4 and 2 = x in their d d y x to obtain c ( ) 2 3 d 1 3 2 2 d 2   = + +     y x x A1 May be implied by subsequent integration or by 2 = c ( ) ( ) 5 3 1 3 2 2 10 = + + + y x x d M1 For ( ) 5 3 2 3 2 + k x where 2k is a constant. ( ) 1 6.2 32 4 10 = + + d M1 Dep on previous M1 for use of 2 = x and 6.2 = y in their y ( ) 5 3 1 3 2 2 1 10 = + + − y x x A1 Must be an equation 5(a) 16 = p 2 B1 for 5 log log 20 4 = a a p oe B1 for 16, nfww 5(b) ( ) ( )( ) 3 3 1 3 3 0 − + = x x M1 For recognition of a correct quadratic in 3x and attempt to factorise or use quadratic formula 1 3 3 = x 1 = − x 2 M1 dep for a correct attempt to solve 3 , 0 = > x k k A1 for one solution only, must be from a correct solution.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 6 of 11 Question Answer Marks Guidance 5(c) 2 1 log 2 log = y y or 2 1 log log 2 = y y or log 2 log 2 log = a y a y and 2 log log log 2 = a a y y B1 May be implied ( ) ( ) 2 4 log 2 4 log 2 1 0 − + = y y ( ) 2 2log 2 1 0 − = y , 1 log 2 2 = y or ( ) ( ) 2 2 2 log 4 log 4 0 − + = y y ( ) 2 2 log 2 0 − = y , 2 log 2 = y M1 For obtaining a 3 term quadratic equation in either log 2 y or 2 log y and solving to obtain log 2 = y k or 2 log = y k , may be implied or equivalent using an alternative base. or ( ) ( )( ) 2 log 4 log 2 log 4 log − a a a a y y ( ) 2 4 log 2 0 + = a ( ) 2 log 2log 2 0 − = a a y log 2log 2 = a a y 4 = y A1 nfww 6(a) ( ) d 2 3 5 8 5 d = + − y x x or ( ) 8 5 2 3 5 = + x M1 Either For differentiation must have one correct term. or for use of 2 ' 4 0' − = b ac , so 2 = −b x a at the stationary point. ( ) ( ) 3 5 4 5 3 5 3 5 − = × + − x oe leading to 12 5 20 4 − oe, this is the minimum acceptable working for this method. M1 Dep for equating their d d y x to zero with attempt to solve and rationalisation using a two term factor, or rationalisation of 2 = −b x a , using a two term factor with sufficient detail to imply no use of a calculator. Allow multiple equivalents. Allow one numerical slip or sign error. 5 3 5 = −+ x 2 A1 for 5 − A1 for 3 5

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 7 of 11 Question Answer Marks Guidance 6(b) ( )( ) 2 3 5 3 5 5 = + − y ( ) 8 5 3 5 5 60 − − + ( )( ) 3 5 45 25 30 5 120 40 5 60 = + + − − + + 210 70 5 90 5 150 120 40 5 60 = + − − − + + M1 For substitution of their x and simplification with sufficient detail to imply no use of a calculator. Allow one numerical slip or sign error in the expansion of ( )( ) 2 3 5 3 5 5 + − or one sign error in the other terms. 20 5 = 2 A1 for all non surd terms = 0 A1 for 20 5 7(a)(i) 20160 B1 7(a)(ii) 7200 2 B1 for 6 4 P or ( ) 6 5 4 3 360 × × × = for ‘inner’ characters or 5 2 P or ( ) 4 5 20 × = for ‘outer’ characters Must be part of a product 7(a)(iii) 360 2 B1 for 3 3P or 3! or 6 for arrangements of symbols or 5 3P or 5 4 3 × × (= 60) for the digits Must be part of a product 7(b) ( ) ( ) ( ) ( ) 6 1 ! ! 5 !5! 1 4 !4! − = − − − n n n n B1 May be implied by simplification e.g. ( ) 1 ! ! 6 5! 4! − = n n or ( )( )( )( ) 1 2 3 4 5! − − − − n n n n n ( )( )( )( ) 6 1 2 3 4 4! − − − − = n n n n Simplification of either the numerical factorials or the algebraic factorials M1 30 = n A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 8 of 11 Question Answer Marks Guidance 8(a) lg lg lg = + y b x A B1 May be implied by subsequent work 4.37 5.36 lg = + b A 0.57 0.61 lg = + b A M1 For at least one correct equation 0.8 = b A1 ( ) lg 0.082 = A k 10 = k A M1 Dep for substitution to obtain lg = A k and hence A 1.21 = A A1 Alternative 1 lg lg lg = + y b x A (B1) May be implied by subsequent work Gradient = 4.37 0.57 5.36 0.61 − − (M1) 0.8 = b (A1) ( ) lg 0.082 = A k 10 = k A (M1) Dep for substitution into a correct equation to obtain lg = A k and hence A 1.21 = A (A1) Alternative 2 4.37 5.36 10 10 = × b A or 0.57 0.61 10 10 = × b A (B1) 3.8 4.75 = b (M1) For eliminating A correctly Must have B1. 0.8 = b (A1) ( ) 4.37 5.36 ( ) 10 − × = theirb A oe (M1) For a correct attempt to find A. Must have B1 A = 1.21 (A1) 8(b) ( ) 0.8 1.21 3 or lg 0.8lg3 0.082 = = + y y B1 FT for substitution into their equation y = awrt 2.9 B1 8(c) 0.8 3 1.21 or lg3 0.8lg 0.082 = = + x x B1 FT for substitution into their equation x = awrt 3.1 B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 9 of 11 Question Answer Marks Guidance 9(a) 12 = d B1 ( ) ( ) 8 1 12 2000 2 −+ − > n n 2 3 5 1000 0 − − > n n M1 For use of sum formula to obtain a three term quadratic inequality or equation 5 25 12000 6 ± + = n 19.1 = n M1 Dep for attempt at critical value(s) using their quadratic, may be using a calculator, so may be implied by a correct answer of 20. 20 = n A1 9(b)(i) 3 = r 2 M1 For 6 27 = ar and 8 243 = ar with an attempt to eliminate a to obtain 2 r . Allow other valid methods. 9(b)(ii) 26 3 2 B1 for 1 27 = a or 3 3− nfww 9(c) Common ratio or sinθ = r B1 May be implied by e.g. 1 1 sinθ − or 1 sin 1 sin θ θ − − n 1 sin 1 θ −< < or sin 1 θ < or 1 1 −< < r or 1 < r with no incorrect statements seen. B1 Dep on previous B1 10(a) ( ) 1 1 0 sin cos α α + = B1 For dealing correctly with 2 cosec α and 2 sec α to obtain an expression in sinα and cosα only tan 1 α = − or sin cos α α = − B1 For an equation in tanα , may be implied by a correct solution. π 4 α = − or – 0.785 3π 4 α = or 2.36 2 B1 for one correct solution B1 for a second correct solution and no extra solutions in the range.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 10 of 11 Question Answer Marks Guidance 10(b)(i) ( ) 2 2 cos 1 2sin sin cos 1 sin θ θ θ θ θ + − + − M1 For dealing with the fractions correctly and expansion of ( ) 2 1 sinθ − ( ) 1 1 2sin cos 1 sin θ θ θ + − − or better M1 Dep for use of identity, may be implied by ( ) ( ) 2 1 sin cos 1 sin θ θ θ − − ( ) ( ) 2 1 sin cos 1 sin θ θ θ − − M1 Dep on previous M mark for simplification 2 2sec cos θ θ = A1 Need to see this detail for A1 Need to have had θ in every trigonometric ratio. Alternative 1 cos 1 sin 1 sin 1 sin 1 sin cos θ θ θ θ θ θ + −   × +   − +   (M1) ( ) 2 cos 1 sin 1 sin cos cos θ θ θ θ θ + − + (M1) Dep for use of identity 1 sin 1 sin cos cos θ θ θ θ + − + (M1) Dep on previous M mark for simplification 2 2sec cos θ θ = (A1) Need to see this detail for A1 Need to have had θ in every trigonometric ratio. Alternative 2 ( ) ( ) ( ) 2 2 1 sin 1 sin cos 1 sin θ θ θ θ − + − − (M1) For dealing with the fractions and using 2 2 cos 1 sin θ θ = − . ( )( ) ( ) ( ) 2 1 sin 1 sin 1 sin cos 1 sin θ θ θ θ θ − + + − − (M1) Dep for factorising 2 1 sin θ − 1 sin 1 sin cos θ θ θ + + − (M1) Dep for simplification 2 2sec cos θ θ = (A1) Need to see this detail for A1 Need to have had θ in every trigonometric ratio.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 11 of 11 Question Answer Marks Guidance 10(b)(ii) 1 cos3 2 φ = B1 o o o 20 ,100 ,140 φ = 3 M1 for one correct solution of their cos3φ = k using a correct order of operations A1 for 2 correct solutions A1 for a third correct solution with no extra solutions in the range 11 ( ) ( ) ( ) 2 2 2 2 2 3 2ln 2 d 2 d 2 3 − − + + = − x x x y x x x 3 B1 for 2 2 2 + x x M1 for differentiation of a quotient When d 4 2, 2ln6 d 6 = = − y x x , –2.92 Gradient of normal = 0.3428 M1 For 1 d d − y their x When ( ) 2, ln6 or 1.79 176 = = x y B1 Equation of normal: ( ) 1 ln6 2 d d − = − − y x y their x or ( ) 1 ln6 2 d d = − × + c y their x M1 Dep for equation of normal using 1 d d − y their x and their y with 2. = x When 0, awrt 1.11 = = x y A1 Must be evaluated.