Cambridge IGCSE Mathematics - Additional 0606 — 2022 May/June Paper 1 · Variant 2

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 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 [ ]. * 9 8 9 6 9 9 4 4 4 4 * DC (CJ/FC) 303920/4 © UCLES 2022

2 0606/12/M/J/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/M/J/22 © UCLES 2022 [Turn over 1 y x 0 – 8 2 180° – 180° The diagram shows the graph of sin y a bx c = + , where a, b and c are integers, for ° ° x 180 180 G G - . Find the values of a, b and c. [3]

4 0606/12/M/J/22 © UCLES 2022 2 Given that sec x 2i = and cot y 2 2i + = , find y in terms of x. [4]

5 0606/12/M/J/22 © UCLES 2022 [Turn over 3 Variables x and y are such that, when g( ) l y 2 1 + is plotted against x2, a straight line graph passing through the points (1, 1) and (2, 5) is obtained. (a) Find y in terms of x. [4] (b) Find the value of y when x 2 3 = . [1] (c) Find the value of x when y 2 = . [2]

6 0606/12/M/J/22 © UCLES 2022 4 (a) Find the unit vector in the same direction as 15 8 - e o. [2] (b) Given that a b a b b a 2 5 3 4 12 4 2 - + - = + - e e e o o o, find the values of a and b. [3]

7 0606/12/M/J/22 © UCLES 2022 [Turn over 5 The first three terms, in ascending powers of x, in the expansion of ( ) x x 1 6 2 3 12 3 + - b l can be written in the form px qx 8 2 + + , where p and q are constants. Find the values of p and q. [8]

8 0606/12/M/J/22 © UCLES 2022 6 The polynomial p( )x x ax x b 6 6 3 2 = + + + , where a and b are integers, is divisible by x 2 1 - . When p( )x is divided by x 2 - , the remainder is 120. (a) Find the values of a and b. [4] (b) Hence write down the remainder when p( )x is divided by x. [1] (c) Find the value of p ( ) 0 ll . [2]

9 0606/12/M/J/22 © UCLES 2022 [Turn over 7 (a) Show that ( ) x x x 2 3 2 1 1 1 1 2 + - - + - can be written as ( ) ( ) x x x 1 2 3 8 3 2 - + - . [2] (b) Find ( ) ( ) d x x x x 1 2 3 8 3 a 2 2 - + - y where a 2 2 . Give your answers in the form ln c d + , where c and d are functions of a. [6]

10 0606/12/M/J/22 © UCLES 2022 8 (a) A team of 6 people is to be chosen from 10 people. Two of the people are sisters who must not be separated. Find the number of different teams that can be formed. [3] (b) A 6-character password is to be chosen from the following characters. Digits 2 4 8 Letters x y z Symbols * # ! No character may be used more than once in any password. Find the number of different passwords that may be chosen if (i) there are no other restrictions, [1] (ii) the password starts with two letters and ends with two digits. [3]

11 0606/12/M/J/22 © UCLES 2022 [Turn over 9 The normal to the curve ( ) ln y x x 1 3 2 2 = + + , at the point A on the curve where x 0 = , meets the x-axis at point B. Point C has coordinates ( , ) ln 0 3 2 . Find the gradient of the line BC in terms of ln2. [9]

12 0606/12/M/J/22 © UCLES 2022 10 (a) Given the simultaneous equations g , l lg x y 2 1 + = , x y 3 13 2 - = (i) show that x x 13 30 0 2 - - = . [4] (ii) Solve these simultaneous equations, giving your answers in exact form. [2]

13 0606/12/M/J/22 © UCLES 2022 [Turn over (b) Solve the equation log log x a 3 4 a x + = , where a is a positive constant, giving x in terms of a. [5]

14 0606/12/M/J/22 © UCLES 2022 11 In this question all lengths are in kilometres and time is in hours. A particle P moves in a straight line such that its displacement, s, from a fixed point at time t is given by ( )( ) s t t 2 5 2 = + - , for t 0 H . (a) Find the values of t for which the velocity of P is zero. [4] (b) On the axes, draw the displacement–time graph for P for t 0 6 G G , stating the coordinates of the points where the graph meets the coordinate axes. [2] s t 0 6 5 4 3 2 1

15 0606/12/M/J/22 © UCLES 2022 (c) On the axes below, draw the velocity–time graph for P for t 0 6 G G , stating the coordinates of the points where the graph meets the coordinate axes. [2] velocity t 0 6 5 4 3 2 1 (d) (i) Write down an expression for the acceleration of P at time t. [1] (ii) Hence, on the axes below, draw the acceleration–time graph for P for t 0 6 G G , stating the coordinates of the points where the graph meets the coordinate axes. [2] acceleration t 0 6 5 4 3 2 1

16 0606/12/M/J/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 10 printed pages. © UCLES 2022 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 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 May/June 2022 series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 4 of 10 Question Answer Marks Guidance 1 5  a B1 4  b B1 3  c B1 2 2 1 tan 2   y soi or 2 1 tan   x soi B1 Must be in terms of 2 tan  Use of 2 2 tan 1 sec    1 1 2   x y oe M1 For a valid attempt to eliminate  1 2 1    y x or 3 2 1    x y x oe 2 Dep M1 for attempt to rearrange to obtain in the required form A1 for a correct form Alternative 2 1 cos   x and 2 2 cos 2 sin     y soi (B1) 1 2 1 1    x y x oe (M1) For a valid attempt to eliminate , making use of 2 2 sin cos 1     1 2 1    y x or 3 2 1    x y x oe (2) Dep M1 for attempt to rearrange to obtain in the required form A1 for a correct form 3(a) Gradient = 4 soi B1 Intercept = 3  soi B1   2 lg 2 1 4 3    y x oe M1 For     2 lg 2 1    y their m x their c   2 4 3 1 10 1 2    x y or 2 4 10 1 1000 2   x y A1 3(b) y = 0 B1 Must have at least 3 marks from part (a)

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 5 of 10 Question Answer Marks Guidance 3(c)   2 4 3 1 2 10 1 2    x oe and attempt to obtain x = … M1 Dep on M mark in part (a) for use of 2  y in their   2 4 3 1 10 1 2    x y , 2 4 10 1 1000 2   x y or   2 lg 2 1 4 3    y x and attempt to obtain x = … x =  0.962 or better A1 4(a) 15 1 8 17        oe 2 B1 for 17 seen 4(b) 2 4 12 4 4     a b b a 5 3 4 8    a b M1 For equating like vectors to obtain at least one equation 5 2, 4   a b oe 2 Dep M mark for solving their 2 equations to obtain both a and b 5 12 2 11 1 1 2 6 6          x x x 2 B2 for 3 correct terms B1 for 2 correct terms   3 2 2 3 8 36 54 ...     x x x 2 B2 for 3 correct terms B1 for 2 correct terms Term in x: 36 16   x x M1 A correct method using their terms or coefficients, must be considering 2 terms 20  p soi A1 Term in 2 : x 2 2 2 88 54 72 6   x x x M1 A correct method using their terms or coefficients, must be considering 3 terms 10 3  q soi A1 6(a) 1 p : 4 15 0 2          a b oe B1 For 1 p 2       equated to zero  p 2 : 4 60   a b oe B1 For  p 2 equated to 120 17, 8   a b 2 Dep M1 on both previous B marks, for solving their equations to obtain a and b A1 for both

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 6 of 10 Question Answer Marks Guidance 6(b) 8  B1 FT on their integer b 6(c)  2 p 18 34 6     x x x  p 36 34    x x M1 For attempt to differentiate their p(x), may be implied by correct FT answer  p 0 34   A1 FT on 2 their a integer 7(a)          2 2 2 1 1 2 3 2 3 1 2 3         x x x x x x M1 Attempt at a fraction, allow with an extra   1  x term in each term of the numerator and the denominator =    2 8 3 1 2 3    x x x A1 AG – must see sufficient detail to justify the given result, if an extra  1  x term involved, it must be dealt with correctly 7(b)       2 1 ln 2 3 ln 1 1              a x x x 3 B1 for each correct term       1 ln 2 3 ln 1 ln7 1 1              a a a M1 Dep on at least one ln term from integration, for applying limits correctly in their integral   2 2 3 ln 1 7 1              a a a a oe 2 A1 for 2 1   a a or 1 1 1   a A1 for   2 3 ln 7 1           a a Alternative 1 final 2 marks     e 2 3 1 ln 1 7 1              a a a oe (2) A1 for 1 1   a A1 for     e 2 3 ln 7 1           a a Alternative 2 final 2 marks 2 ln7 1    a a +   2 3 ln 1           a a oe (2) A1 for 2 ln7 1    a a or 1 1 ln7 1    a Allow 1.946 or better for ln 7 A1 for   2 3 ln 1           a a

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 7 of 10 Question Answer Marks Guidance 7(b) Alternative 3 final 2 marks 1 1   a ln7  +     e 2 3 ln 1           a a oe 2 A1 for 1 1   a ln7  Allow 1.946 or better for ln 7 A1 for     e 2 3 ln 1           a a 8(a) With the sisters: 70 or 8 4 C oe B1 Without the sisters: 28 or 8 6 C oe B1 Total: 98 B1 8(b)(i) 60480 B1 8(b)(ii) The start of the password and the end of the password can each be chosen 6 ways B1 6 or 3 2 P oe seen twice The remaining characters can be chosen in 20 ways B1 20 or 5 2P oe seen Total number of ways: 720 B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 8 of 10 Question Answer Marks Guidance 9 When 0, ln 2   x y soi B1 May be implied in later work        2 2 2 6 1 ln 3 2 3 2 d d 1       x x x x y x x 3 B1 for   2 6 3 2  x x allow when seen M1 for attempt at a quotient or product A1 for all other terms apart from   2 6 3 2  x x correct When d 0, ln2 d   y x x M1 Dep on previous M mark for attempt to find the gradient using their d d y x Equation of normal: 1 ln2 ln2   y x M1 For attempt at a normal equation using their y (not 3 ln2) and   1 ln 2   their , must be from an attempt at differentiation When   2 0, ln 2   y x M1 For attempt to find the value of x when y = 0 using their normal equation Gradient BC =   2 3ln2 ln2 M1 Dep on both previous M marks 3 ln2 or   1 3 ln 2  A1 Must have correct exact working throughout

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 9 of 10 Question Answer Marks Guidance 10(a)(i) 2 xy soi B1 Simplification of the left-hand side of the first equation 1 lg10  soi B1 Simplification of right-hand side of equation 10 3 13         x x M1 For substitution of 2 y into linear equation oe and attempt to simplify 2 13 30 0    x x A1 AG – must see sufficient detail to justify the given answer Alternative   2 13 3  x y (B1)   13 lg lg 1 3    x x (M1) For attempt at substitution in the log equation   13 10 3   x x oe (B1) 2 13 30 0    x x (A1) AG – must see sufficient detail to justify the given answer 10(a)(ii) 15  x only B1 2 3  y or 6 3 or exact equivalent only B1 isw once exact value seen 10(b) 3 log log  a a x x or 1 3log log  x x a a B1 For an appropriate change of base   2 log 4log 3 0    a a x x or   2 3 log 4log 1 0   x x a a M1 For an attempt to obtain a 3-term quadratic equation in terms of loga x or logx a , equated to zero. log 3 log 1   a a x x or 1 log , log 1 3   x x a a M1 Dep on previous M mark for correct solution of their quadratic equation  x a A1 Must be from completely correct work 3  x a A1 Must be from completely correct work

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 10 of 10 Question Answer Marks Guidance 11(a) 2 d 3 16 5 d    s t t t oe 2 M1 for attempt at differentiation of a product or expansion and differentiation with at least two out of three terms of their expansion differentiated correctly A1 all correct, allow factorised 1, 5 3   t t 2 Dep M1 for attempt to solve their d 0 d  s t , must be in quadratic form A1 for both 11(b) 2 B1 for a correct curve in the first quadrant only. There must be an indication of a max point in the correct position and a min point at   5, 0 B1 for   0, 50 provided basic curve shape is correct 11(c) 2 B1 for a quadratic curve in the first and fourth quadrants B1 for   1 0, 5 , , 0 3       or   0.333, 0 marked and passing through   5, 0 on the x-axis 11(d)(i) Acceleration = 6 16  t B1 11(d)(ii) 2 B1 for a straight line with a positive gradient in the first and fourth quadrants, meeting the vertical axis Dep B1 for 8, 0 3           or 2.67, 0 and 0, 16 