Cambridge IGCSE Mathematics - Additional 0606 — 2021 Oct/Nov Paper 1 · Variant 2

This document has 16 pages. Any blank pages are indicated. Cambridge IGCSE™ DC (CE/CB) 212422/4 © UCLES 2021 [Turn over ADDITIONAL MATHEMATICS 0606/12 Paper 1 October/November 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 [ ]. * 5 9 8 2 1 2 6 7 1 2 *

2 0606/12/O/N/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/O/N/21 © UCLES 2021 [Turn over 1 x 0 y - 3 - 5 1 5 The diagram shows the graph of the cubic function ( ) y x f = . The intercepts of the curve with the axes are all integers. (a) Find the set of values of x for which ( )x 0 f 1 . [1] (b) Find an expression for ( )x f . [3]

4 0606/12/O/N/21 © UCLES 2021 2 (a) Given that xy xz z zy x y z a b c 3 3 2 = - ` ` j j , find the exact values of the constants a, b and c. [3] (b) Solve the equation ( ) ( ) 5 2 17 2 3 0 p p 2 1 - + = + . [4]

5 0606/12/O/N/21 © UCLES 2021 [Turn over 3 (a) Write lg lg a b 3 2 4 + - as a single logarithm to base 10. [4] (b) Solve the equation log log a 3 4 2 7 a 4 + = . [5]

6 0606/12/O/N/21 © UCLES 2021 4 Solve the equation , cot x 2 3 3 0 r + - = b l where x 1 1 r r - radians. Give your answers in terms of r. [4] 5 Find the possible values of the constant c for which the line y c = is a tangent to the curve . sin y x 5 3 4 = + [3]

7 0606/12/O/N/21 © UCLES 2021 [Turn over 6 DO NOT USE A CALCULATOR IN THIS QUESTION. The polynomial ( )x x ax x b 10 10 p 3 2 = + - + , where a and b are integers, is divisible by x 2 1 + . When ( )x p is divided by x 1 + , the remainder is 24 - . (a) Find the value of a and of b. [4] (b) Find an expression for ( )x p as the product of three linear factors. [4] (c) Write down the remainder when ( )x p is divided by x. [1]

8 0606/12/O/N/21 © UCLES 2021 7 (a) O A B C a b c The diagram shows triangle OAC, where OA a = , OB b = and OC c = . The point B lies on the line AC such that : : AB BC m n = , where m and n are constants. (i) Write down AB in terms of a and b. [1] (ii) Write down BC in terms of b and c. [1] (iii) Hence show that ( ) n m m n a c b + = + . [2] (b) Given that ( ) ( ) 2 1 1 4 7 1 4 2 m n m + - - = + - e e e o o o, find the value of each of the constants m and n. [4]

9 0606/12/O/N/21 © UCLES 2021 [Turn over 8 (a) A 5-digit number is made using the digits 0, 1, 4, 5, 6, 7 and 9. No digit may be used more than once in any 5-digit number. Find how many such 5-digit numbers are even and greater than 50 000. [3] (b) The number of combinations of n objects taken 4 at a time is equal to 6 times the number of combinations of n objects taken 2 at a time. Calculate the value of n. [5]

10 0606/12/O/N/21 © UCLES 2021 9 M A O B D C N 6 cm 4 cm 12 cm The diagram shows a circle, centre O, radius 12 cm, and a rectangle ABCD. The diagonals AC and BD intersect at O. The sides AB and AD of the rectangle have lengths 6 cm and 4 cm respectively. The points M and N lie on the circumference of the circle such that MAC and NDB are straight lines. (a) Show that angle AOD is 1.176 radians correct to 3 decimal places. [2] (b) Find the perimeter of the shaded region. [4]

11 0606/12/O/N/21 © UCLES 2021 [Turn over (c) Find the area of the shaded region. [3]

12 0606/12/O/N/21 © UCLES 2021 10 x A B y - 1 2 0 ( ) y x 2 1 3 2 = + + ( ) x 2 + The diagram shows the graph of the curve y x 2 1 3 2 = + + ( ) x 2 + ` j for x 2 2- . The points A and B lie on the curve such that the x-coordinates of A and of B are -1 and 2 respectively. (a) Find the exact y-coordinates of A and of B. [2] (b) Find the area of the shaded region enclosed by the line AB and the curve, giving your answer in the form ln q p r - , where p, q and r are integers. [6]

13 0606/12/O/N/21 © UCLES 2021 [Turn over Additional working space for Question 10(b).

14 0606/12/O/N/21 © UCLES 2021 11 (a) t s v ms-1 60 30 10 0 25 35 55 The diagram shows the velocity–time graph for a particle P, travelling in a straight line with velocity v ms 1 - at a time t seconds. P accelerates at a constant rate for the first 10 s of its motion, and then travels at constant velocity, 30ms 1 - , for another 15 s. P then accelerates at a constant rate for a further 10 s and reaches a velocity of 0 6 ms 1 - . P then decelerates at a constant rate and comes to rest when t 55 = . (i) Find the acceleration when t 12 = . [1] (ii) Find the acceleration when t 50 = . [1] (iii) Find the total distance travelled by the particle P. [2]

15 0606/12/O/N/21 © UCLES 2021 (b) A particle Q travels in a straight line such that its velocity, v ms 1 - , at time t s after passing through a fixed point O is given by cos v t 4 3 4 = - . (i) Find the speed of Q when t 9 5r = . [2] (ii) Find the smallest positive value of t for which the acceleration of Q is zero. [3] (iii) Find an expression for the displacement of Q from O at time t. [2]

16 0606/12/O/N/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 0606/12 Paper 1 October/November 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 October/November 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 October/November 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.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 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

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 4 of 10 Question Answer Marks Guidance 1(a) 3 1 −< < x 5 > x B1 1(b) ( )( )( ) 1 3 1 5 3 − + − − x x x 3 B1 for a negative cubic function B1 for a cubic function multiplied by 1 3 B1 for ( )( )( ) 3 1 5 + − − x x x 2(a) 10 1 or 3 3 3 = a B1 7 1 or 2 3 3 = b B1 9 1 or 4 or 4.5 2 2 = c B1 2(b) ( ) ( ) 2 10 2 17 2 3 0 − + = p p ( ) ( ) ( ) ( ) 5 2 1 2 2 3 0 − − = p p 1 3 2 , 2 5 2 = = p p M1 For recognition of a quadratic in 2 p , attempt to factorise and solve for 2 p 1 ln ln1.5 5 or ln 2 ln 2 = = p p oe M1 For correct attempt to deal with 2 = p k 2.32 − A1 0.585 A1 3(a) 2 4 1000 lg a b 4 B1 for 3 lg1000 = B1 for use of power rule once B1 for use of addition or subtraction rule once B1 All correct

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 5 of 10 Question Answer Marks Guidance 3(b) Either 4 3 3log 4 log = a a B1 ( ) 2 4 4 2 log 7log 3 0 − + = a a ( )( ) 4 4 2log 1 log 3 0 − − = a a 4 4 1 log or log 3 2 = = a a M1 For obtaining a quadratic equation and solution 1 3 2 4 or 4 = = a a M1 Dep For dealing with the logarithm correctly once, may be implied by a correct solution 64 A1 2 A1 Or 4 2 2log log 4 = a a (B1) ( ) 2 3 log 4 7log 4 2 0 − + = a a ( )( ) 3log 4 1 log 4 2 0 − − = a a 1 log 4 or log 4 2 3 = = a a (M1) For obtaining a quadratic equation and solution 1 2 3 4 or 4 = = a a (M1) Dep For dealing with the logarithm correctly once, may be implied by a correct solution 64 (A1) 2 (A1) 4 π 1 tan 2 3 3   + =     x B1 7 5 11 , , , 12 12 12 12 π π π π = − − x 3 M1 for using correct order of operations A1 for two correct solutions A1 for two further correct solutions and no other solutions in range

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 6 of 10 Question Answer Marks Guidance 5 Either Maximum when sin 1 3 = x or minimum when sin 1 3 = − x M1 For recognition that value of maximum or minimum is necessary 9 = c A1 1 = − c A1 or d 5cos d 3 3 = y x x When d 0 d = y x , sin 1 3 = + x or 1 − (M1) For differentiation, equating to zero to obtain values for sin 3 x 9 = c (A1) 1 = − c (A1) 6(a) 5 0 5 4 4 = − + + + a b M1 For use of the factor theorem 24 10 10 − = − + + + a b M1 For use of the remainder theorem 4 15 + = − a b 24 + = − a b leading to M1 Dep on both previous M marks for solution of their equations without using a calculator 27 = − a , 3 = b A1 6(b) ( )( ) 2 2 1 5 … + + x x their b M1 Allow for observation or algebraic long division. Their a and b must be integers. ( )( ) 2 2 1 5 16 3 + − + x x x A1 ( )( )( ) 2 1 5 1 3 + − − x x x 2 M1 for attempt to factorise their 3-term quadratic A1 all correct from fully correct working 6(c) 3 B1 FT on their (integer) b 7(a)(i) − b a B1 7(a)(ii) − c b B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 7 of 10 Question Answer Marks Guidance 7(a)(iii) =   nAB mBC M1 For substitution of their (i) and (ii) into =   nAB mBC ( ) + = + n m m n a c b A1 For correct manipulation to obtain the given answer 7(b) 2 4 4 4 4 λ μ λ − + = + or 7 7 2 2 λ μ λ + − = − − M1 For equating like components at least once, allow unsimplified M1 Dep for solving their equations to obtain both λ and μ 5 μ = A1 10 λ = − A1 8(a) Either Starting with a 6: 120 ways B1 May be implied by final answer Starting with 5, 7 or 9: 540 ways B1 May be implied by final answer Total 660 B1 Or Alternative 1 Ending with a 6: 180 ways (B1) May be implied by final answer Ending with 0 or 4: 480ways (B1) May be implied by final answer Total 660 (B1) Or Alternative 2 11 ways of obtaining even 5-digit numbers which start with 5, 6, 7, 9 (B1) For 11 × k May be implied by final answer 5P3 ways of arranging remaining 3 digits: 60 (B1) For m × 60 where m is from an attempt to list all cases for first and last digits May be implied by final answer 11 60 660 × = (B1) Or Alternative 3 Total arrangements 7P5 minus (all odds + evens starting with 1 + evens starting with 0 or 4) = 2520 – (1440 + 180 + 240) (B2) For 2520 – (1440 + 180 + 240) 660 (B1)

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 8 of 10 Question Answer Marks Guidance 8(b) ( ) ( ) ! 6 ! 4 !4! 2 !2! = − − n n n n B1 ( )( ) 2 3 72 − − = n n 2 B1 for ( )( ) 2 3 − − n n B1 for 72 11 = n only 2 M1 for correct attempt to form and solve a quadratic equation A1 for 11 = n only 9(a) 1 2 2 tan 3 −  = ×     AOD M1 For correct method to find AOD 1.1760... = AOD 1.176 = AOD [to 3dp] A1 Need to see 4 dp or more to justify 3 dp answer 9(b) Major arc ( ) 2 1.176 12 π = − MN B1 ND or MA =12 13 − B1 Perimeter = major arc MN + MA + ND + 16 oe B1 For their values in a correct plan, may be implied by a correct answer Perimeter = 94.1 B1 9(c) Minor sector area = 2 1 1.176 12 2 × × or Major sector area = ( ) 2 1 2 1.176 12 2 π × − × B1 Area = major sector area – remainder of rectangle or Area = area of circle – minor sector area – remainder of rectangle or Area = circle – rectangle – minor sector + triangle AOD B1 For their values in a correct plan, may be implied by a correct answer Area = 350 B1 Allow greater accuracy 10(a) At A 4 = y B1 At B 13 16 = y or 0.8125 B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 9 of 10 Question Answer Marks Guidance 10(b) Either Area of trapezium = 231 32 B1 Allow unsimplified ( ) 2 2 1 1 3 d 2 2 − + + +  x x x ( ) 2 1 1 3ln 2 2 −   = − + +   +   x x 2 B1 for 1 2 − + x B1 for ( ) 3ln 2 + x ( ) 1 3ln 4 1 4     − + −−         M1 For correct use of limits in their integral, but must have at least one of the two preceding B marks Area = 207 ln64 32 − 2 A1 for 207 32 A1 for ln 64 − Or ( ) 2 2 1 17 47 1 3 d 16 16 2 2 −− + − − + +  x x x x ( ) 2 2 1 17 47 1 3ln 2 32 16 2 x x x x −     − + + − +     +     (3) B1 for 2 17 47 32 16 − + x x B1 for ( ) ( ) 2 1 1 d 2 2 x x x = − + +  B1 for ( ) 3 d 3ln 2 2 x x x = + +  17 47 1 17 47 3ln 4 1 8 8 4 32 16     − + + − −− − +         (M1) For correct use of limits in their integral, but must have at least one of the two preceding B marks Area = 207 ln64 32 − (2) A1 for 207 32 A1 for ln 64 − 11(a)(i) 0 B1 11(a)(ii) 3 − B1 11(a)(iii) ( ) ( ) 1 1 1 25 15 30 30 60 10 20 60 2 2 2       + × + + × + × ×             M1 For an unsimplified expression for the required area allowing at most one incorrect length Total distance = 1650 A1 11(b)(i) 5π 4cos 4 3 = − v 2 = − M1 Speed = 2 A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 10 of 10 Question Answer Marks Guidance 11(b)(ii) 12sin3 = − a t B1 sin3 0 = t 3 π = t Leading to M1 For equating to zero and attempt to solve to obtain t, allow if in degrees π 3 = t A1 11(b)(iii) ( ) sin3 4 = − + s k t t c M1 4sin3 4 3 = − s t t A1