Cambridge IGCSE Mathematics - Additional 0606 — 2020 May/June Paper 1 · Variant 2

* 5 8 1 4 8 1 2 4 8 5 * DC (NF/CB) 196732/3 R © UCLES 2020 [Turn over This document has 16 pages. Blank pages are indicated. ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 2020 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 0606/12/M/J/20 © UCLES 2020 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x a b b ac 2 4 2 ! = - - Binomial Theorem ( ) … … a b a n a b n a b n r a b b 1 2 n n n n n r r n 1 2 2 + = + + + + + + - - - J L KK J L KK J L KK N P OO N P OO N P OO where n is a positive integer and ( )! ! ! n r n r r n = - J L KK N P OO Arithmetic series u a n d S n a l n a n d 1 2 1 2 1 2 1 n n − − = + = + = + ^ ^ ^ h h h # - Geometric series u ar S r a r r S r a r 1 1 1 1 1 n n n n 1 1 ! − − − = = = 3 − ^ ^ ^ h h h 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A Formulae for ∆ABC sin sin sin A a B b C c = = a2 = b2 + c2 – 2bc cos A ∆ = 2 1 bc sin A

3 0606/12/M/J/20 © UCLES 2020 [Turn over 1 On the axes below, sketch the graph of ( )( )( ) y x x x 2 1 2 = - + + showing the coordinates of the points where the curve meets the axes. [3] O y x

4 0606/12/M/J/20 © UCLES 2020 2 The volume, V, of a sphere of radius r is given by V r 3 4 3 r = . The radius, r cm, of a sphere is increasing at the rate of 0.5 cms-1. Find, in terms of r, the rate of change of the volume of the sphere when . r 0 25 = . [4]

5 0606/12/M/J/20 © UCLES 2020 [Turn over 3 (a) Find the first 3 terms in the expansion of x 4 16 6 - b l in ascending powers of x. Give each term in its simplest form. [3] (b) Hence find the term independent of x in the expansion of x x x 4 16 1 6 2 - - b b l l . [3]

6 0606/12/M/J/20 © UCLES 2020 4 (a) (i) Find how many different 5-digit numbers can be formed using the digits 1, 2, 3, 5, 7 and 8, if each digit may be used only once in any number. [1] (ii) How many of the numbers found in part (i) are not divisible by 5? [1] (iii) How many of the numbers found in part (i) are even and greater than 30 000? [4] (b) The number of combinations of n items taken 3 at a time is 6 times the number of combinations of n items taken 2 at a time. Find the value of the constant n. [4]

7 0606/12/M/J/20 © UCLES 2020 [Turn over 5 : ( ) x x 2 3 f 2 7 + for x 0 2 (a) Find the range of f. [1] (b) Explain why f has an inverse. [1] (c) Find f-1. [3] (d) State the domain of f-1. [1] (e) Given that : ( ) ln x x 4 g 7 + for , x 0 2 find the exact solution of ( )x 49 fg = . [3]

8 0606/12/M/J/20 © UCLES 2020 6 O y x D (1, 0) C A B x y 2 5 + =- xy 3 0 + = The diagram shows the straight line x y 2 5 + =- and part of the curve xy 3 0 + = . The straight line intersects the x-axis at the point A and intersects the curve at the point B. The point C lies on the curve. The point D has coordinates ( , ) 1 0 . The line CD is parallel to the y-axis. (a) Find the coordinates of each of the points A and B. [3]

9 0606/12/M/J/20 © UCLES 2020 [Turn over (b) Find the area of the shaded region, giving your answer in the form ln p q + , where p and q are positive integers. [6]

10 0606/12/M/J/20 © UCLES 2020 7 (a) Given that y x x 1 5 2 2 = - + ` j , show that x y x Ax Bx C 2 5 2 d d 2 = + + + , where A, B and C are integers. [5]

11 0606/12/M/J/20 © UCLES 2020 [Turn over (b) Find the coordinates of the stationary point of the curve y x x 1 5 2 2 = - + ` j , for x 0 2 . Give each coordinate correct to 2 significant figures. [3] (c) Determine the nature of this stationary point. [2]

12 0606/12/M/J/20 © UCLES 2020 8 O R B Q P A The diagram shows a triangle OAB such that OA a = and OB b = . The point P lies on OA such that OP OA 4 3 = . The point Q is the mid-point of AB. The lines OB and PQ are extended to meet at the point R. Find, in terms of a and b, (a) AB, [1] (b) PQ. Give your answer in its simplest form. [3]

13 0606/12/M/J/20 © UCLES 2020 [Turn over It is given that nPQ QR = and , BR kb = where n and k are positive constants. (c) Find QR in terms of n, a and b. [1] (d) Find QR in terms of k, a and b. [2] (e) Hence find the value of n and of k. [3]

14 0606/12/M/J/20 © UCLES 2020 9 (a) A particle P moves in a straight line such that its displacement, x m, from a fixed point O at time t s is given by sin x t 10 2 5 = - . (i) Find the speed of P when t r = . [1] (ii) Find the value of t for which P is first at rest. [2] (iii) Find the acceleration of P when it is first at rest. [2]

15 0606/12/M/J/20 © UCLES 2020 [Turn over (b) V t s v ms–1 0 5 10 15 20 25 The diagram shows the velocity–time graph for a particle Q travelling in a straight line with velocity v ms-1 at time t s. The particle accelerates at 3.5 ms-2 for the first 10 s of its motion and then travels at constant velocity, V ms-1, for 10 s. The particle then decelerates at a constant rate and comes to rest. The distance travelled during the interval t 20 25 G G is 112.5 m. (i) Find the value of V. [1] (ii) Find the velocity of Q when t 25 = . [3] (iii) Find the value of t when Q comes to rest. [3] Question 10 is printed on the next page.

16 0606/12/M/J/20 © UCLES 2020 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. 10 (a) Solve tan x 3 1 =- for x 2 2 G G r r - radians, giving your answers in terms of r. [4] (b) Use your answers to part (a) to sketch the graph of tan y x 4 3 4 = + for x 2 2 G G r r - radians on the axes below. Show the coordinates of the points where the curve meets the axes. O y x r 2 – r 4 – r 2 r 4 [3]

This document consists of 9 printed pages. © UCLES 2020 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 2020 MARK SCHEME Maximum Mark: 80 Published Students did not sit exam papers in the June 2020 series due to the Covid-19 global pandemic. This mark scheme is published to support teachers and students and should be read together with the question paper. It shows the requirements of the exam. The answer column of the mark scheme shows the proposed basis on which Examiners would award marks for this exam. Where appropriate, this column also provides the most likely acceptable alternative responses expected from students. Examiners usually review the mark scheme after they have seen student responses and update the mark scheme if appropriate. In the June series, Examiners were unable to consider the acceptability of alternative responses, as there were no student responses to consider. Mark schemes should usually be read together with the Principal Examiner Report for Teachers. However, because students did not sit exam papers, there is no Principal Examiner Report for Teachers for the June 2020 series. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the June 2020 series for most Cambridge IGCSE™ and Cambridge International A & AS Level components, and some Cambridge O Level components.

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 2 of 9 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 2020 © UCLES 2020 Page 3 of 9 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 2020 © UCLES 2020 Page 4 of 9 Question Answer Marks Partial Marks 1 B1 Shape B1 Correct x-coordinates B1 Correct y-coordinate and max in first quadrant 2 d 0.5 d r t = B1 2 d 4π d V r r = B1 d d d d d d V V r t r t = × 2 d π d V r t = M1 For attempt to use a correct form of the chain rule When 1 4 r = , d 0.125π d V t = A1 3(a) 4096 – 384x + 15x2 B1 For 4096 B1 For 384x − B1 For 2 15x 3(b) (4096 – 384x + 15x2) 2 2 1 2 x x   − +     B1 For 2 2 1 2 x x   − +     Term independent of x : ( ) 2 4096 15 − + M1 For use of 2 appropriate terms –8177 A1 4(a)(i) 720 B1 4(a)(ii) 600 B1 FT on their (i) 5 6 ×

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 5 of 9 Question Answer Marks Partial Marks 4(a)(iii) Starting with 8: 1 × 4 × 3 × 2 × 1 = 24 B1 Starting with 3, 5 or 7: 3 × 4 × 3 × 2 × 2 = 144 M1 May be considering each case separately, need all three cases for M1 A1 Total = 168 A1 4(a)(iii) Alternative Plan for adding numbers ending in 2 and numbers ending in 8 M1 Ending in 2: 1 4 720 96 6 5   × × =     B1 Allow unsimplified Ending in 8: 1 3 720 72 6 5   × × =     B1 Allow unsimplified Total = 168 A1 4(b) 3 2 6 n n C C = B1 ( )( ) 1 2 3! n n n − − ( )( ) ( ) 1 2 6 1 3! 2! n n n n n − − − = B1 ( ) 6 1 2! n n − ( ) ( ) 1 2 18 0 n n n −  − − =   M1 Valid attempt to solve, must have at least one previous B mark 20 n = A1 4(b) Alternative 3 2 6 n n C C = ( ) ( ) 2 !2! 3 !3! n n − = − B1 For dealing with ( ) 2 ! n − and ( ) 3 ! n − to obtain ( ) 2 n − ( ) 2 6 3 n − = × B1 For dealing with 2! and 3! To obtain 6 20 n = M1 Valid attempt to solve, must have at least one previous B mark A1 5(a) f > 9 B1 Allow y but not x 5(b) It is a one-one function because of the restricted domain B1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 6 of 9 Question Answer Marks Partial Marks 5(c) ( ) 2 2 3 x y = + or equivalent M1 For a correct attempt to find the inverse 3 2 x y − = M1 For correct rearrangement 1 3 f 2 x − − = A1 Must have correct notation 5(d) x > 9 B1 FT on their (a) 5(e) ( ) ( ) f ln 4 49 x + = M1 For correct order ( ) ( ) 2 2ln 4 3 49 x + + = ( ) ln 4 2 x + = M1 For correct attempt to solve, dep on previous M mark, as far as x = 2 e 4 x = − A1 6(a) 5 , 0 2 A   −     B1 ( ) 5 2 3 0 x x −− + = 2 2 5 3 0 x x + − = ( )( ) 2 1 3 0 x x − + = M1 For attempt to eliminate one variable, obtain a 3-term quadratic equation = 0 and attempt to solve 1 , 6 2 B   −     A1 Allow A1 if just the x-coordinates or just the y-coordinates are given 6(b) Area of triangle 1 5 1 6 2 2 2   = + ×     , = 9 M1 For attempt at triangle using their values [ ] 1 1 1 1 2 2 3d 3ln x x x − = −  M1 For attempt to integrate, must have ln 1 3ln 2 = M1 correct application of limits, dep on previous M mark 3ln 2 = − M1 realisation that value of integral is negative and making the adjustment M1 application of log law, dep on previous M mark Area = 9 + ln8 A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 7 of 9 Question Answer Marks Partial Marks 7(a) ( ) ( ) ( ) 1 1 2 2 2 d 5 1 5 2 2 5 2 d 2 y x x x x x − = − + + + B1 For ( ) 1 2 5 5 2 2 x − + M1 For differentiation of a product A1 ( ) ( ) ( ) ( ) 1 2 2 5 2 d 5 1 4 5 2 d 2 x y x x x x − + = − + + or equivalent M1 Dep on previous M mark for attempt to simplify 2 d 25 8 5 d 2 5 2 y x x x x + − = + A1 7(b) 2 25 8 5 0 x x + − = M1 Equating their numerator in (a) to zero and attempt to solve 0.315 x = A1 1.70 y = − A1 7(c) Consideration of gradient or y values either side of stationary point, remembering that x > 0. M1 Must be a complete method making use of their (a). Allow consideration of 2 25 8 5 x x + − as a ‘minimum curve’. Accept 2nd derivative method. Minimum A1 8(a) b – a B1 8(b) ( ) 1 1 4 2 + − a b a or ( ) 3 1 4 2 − + + a a b B1 For 1 4 a or 3 4 − a B1 For ( ) 1 2 − b a or ( ) 1 2 + a b 1 1 2 4 − b a B1 Correct and simplified 8(c) 1 1 2 4 n  −     b a B1 FT on their answer to (b) 8(d) ( ) 1 2 k − + b a b M1 For use of their (a) and kb A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 8 of 9 Question Answer Marks Partial Marks 8(e) ( ) 1 1 1 2 2 4 k n  − + −     b a b = b a 1 2 4 n − = − 1 2 2 n k + = M1 For equating their (c) and (d) and then equating like vectors to obtain 2 equations 2 n = A1 1 2 k = A1 9(a)(i) 20cos2 v t = when π, 20 t v = = B1 9(a)(ii) 20cos2 0 t = M1 Equating their (i) to zero, must be a cosine and attempt to solve π 4 t = A1 9(a)(iii) 40sin 2 a t = − M1 Attempt to differentiate their v, dep on previous M mark, and use their value for (ii) –40 A1 9(b)(i) 35 B1 9(b)(ii) ( ) 1 112.5 35 5 2 x = + × M1 Use of area under appropriate part of the graph A1 x = 10 A1 9(b)(iii) 25 10 5 't = M1 Using a ratio method or otherwise, find extra time to stop = 2s or equivalent ' 2 t = A1 27 A1

0606/12 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 9 of 9 Question Answer Marks Partial Marks 10(a) 5π π 3π 3 , 4 4 4 x = − − M1 For a correct attempt to solve, may be implied by one correct solution π 12 x = − A1 π 4 x = A1 5π 12 x = − A1 10(b) B1 Shape – must have three ‘parts’ with asymptotes B1 For correct x-coordinates B1 For correct y-coordinate