Cambridge IGCSE Mathematics - Additional 0606 — 2020 May/June Paper 1 · Variant 3

Cambridge IGCSE™ DC (JC/CT) 196731/3 R © UCLES 2020 [Turn over This document has 16 pages. Blank pages are indicated. * 2 0 3 7 5 4 7 4 3 1 * ADDITIONAL MATHEMATICS 0606/13 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 [ ].

2 0606/13/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 0606/13/M/J/20 © UCLES 2020 [Turn over 1 ( ) e x 3 f x = + for x R ! ( )x x 9 5 g = - for x R ! (a) Find the range of f and of g. [2] (b) Find the exact solution of ( ) ( ) x x f g 1 = - l . [3] (c) Find the solution of ( )x 112 g2 = . [2]

4 0606/13/M/J/20 © UCLES 2020 2 (a) Given that log log x y 2 8 2 4 + = , find the value of xy. [3] (b) Using the substitution y 2x = , or otherwise, solve 2 2 2 1 0 x x x 2 1 1 - - + = + + . [4]

5 0606/13/M/J/20 © UCLES 2020 [Turn over 3 At time t s, a particle travelling in a straight line has acceleration ( ) t2 1 ms 2 2 1 + - - . When t 0 = , the particle is 4 m from a fixed point O and is travelling with velocity 8ms 1 - away from O. (a) Find the velocity of the particle at time t s. [3] (b) Find the displacement of the particle from O at time t s. [4]

6 0606/13/M/J/20 © UCLES 2020 4 (a) Write x x 2 3 4 2 + - in the form ( ) a x b c 2 + + , where a, b and c are constants. [3] (b) Hence write down the coordinates of the stationary point on the curve y x x 2 3 4 2 = + - . [2] (c) On the axes below, sketch the graph of y x x 2 3 4 2 = + - , showing the exact values of the intercepts of the curve with the coordinate axes. [3] O y x (d) Find the value of k for which x x k 2 3 4 2 + - = has exactly 3 values of x. [1]

7 0606/13/M/J/20 © UCLES 2020 [Turn over 5 ( )x x ax x b 6 12 p 3 2 = + + + , where a and b are integers. ( )x p has a remainder of 11 when divided by x 3 - and a remainder of 21 - when divided by x 1 + . (a) Given that ( ) ( ) ( ) x x Q x 2 p = - , find ( ) Q x , a quadratic factor with numerical coefficients. [6] (b) Hence solve ( )x 0 p = . [2]

8 0606/13/M/J/20 © UCLES 2020 6 (a) Find the unit vector in the direction of 5 12 - e o. [1] (b) Given that k r 4 1 2 3 10 5 + - = - e e e o o o, find the value of each of the constants k and r. [3]

9 0606/13/M/J/20 © UCLES 2020 [Turn over (c) Relative to an origin O, the points A, B and C have position vectors p, 3q p - and 9 5 q p - respectively. (i) Find AB in terms of p and q. [1] (ii) Find AC in terms of p and q. [1] (iii) Explain why A, B and C all lie in a straight line. [1] (iv) Find the ratio AB : BC. [1]

10 0606/13/M/J/20 © UCLES 2020 7 O A C D B rad i 12 cm 10 cm The diagram shows an isosceles triangle OAB such that OA OB 12cm = = and angle AOB i = radians. Points C and D lie on OA and OB respectively such that CD is an arc of the circle, centre O, radius 10 cm. The area of the sector OCD 35cm2 = . (a) Show that .0 7 i = . [1] (b) Find the perimeter of the shaded region. [4] (c) Find the area of the shaded region. [3]

11 0606/13/M/J/20 © UCLES 2020 [Turn over 8 (a) An arithmetic progression has a first term of 7 and a common difference of 0.4. Find the least number of terms so that the sum of the progression is greater than 300. [4] (b) The sum of the first two terms of a geometric progression is 9 and its sum to infinity is 36. Given that the terms of the progression are positive, find the common ratio. [4]

12 0606/13/M/J/20 © UCLES 2020 9 xy 2 = x 3 = y x 5 3 – = y x O B A C D The diagram shows part of the curve xy 2 = intersecting the straight line y x 5 3 = - at the point A. The straight line meets the x-axis at the point B. The point C lies on the x-axis and the point D lies on the curve such that the line CD has equation x 3 = . Find the exact area of the shaded region, giving your answer in the form ln p q + , where p and q are constants. [8]

13 0606/13/M/J/20 © UCLES 2020 [Turn over Additional working space for question 9.

14 0606/13/M/J/20 © UCLES 2020 10 (a) Given that y x x 2 = + , show that y x x Ax B 2 2 d d = + + , where A and B are constants. [5]

15 0606/13/M/J/20 © UCLES 2020 (b) Find the exact coordinates of the stationary point of the curve y x x 2 = + . [3] (c) Determine the nature of this stationary point. [2]

16 0606/13/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. BLANK PAGE

This document consists of 8 printed pages. © UCLES 2020 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/13 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.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 2 of 8 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.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 3 of 8 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

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 4 of 8 Question Answer Marks Partial Marks 1(a) f 3 > B1 Allow y but not x g∈ B1 Allow y but not x 1(b) ( ) ln 3 x − B1 ( ) ln 3 9 x − = 9 3 e x − = M1 For attempt to equate to 9 and solve, must get rid of ln 9e 3 x = + A1 1(c) ( ) 9 9 5 5 112 x − − = M1 For correct order of operation 2 x = A1 2(a) Either 4 2 2log log y y = Or 2 4 log 2log x x = B1 Either 2 2 log log 8 x y + = leading to 2 log 8 xy = Or 4 4 2log 2log 8 x y + = leading to 4 log 4 xy = M1 For use of log law 256 xy = A1 2(b) 2 2 3 1 0 y y − + = B1 1 , 1 2 y = M1 For attempt to solve for y 1 x = − A1 0 x = A1 3(a) ( ) ( ) 1 2 2 1 v t c = + + B1 For ( ) 1 2 2 1 v t = + condone absence of c 8 1 , 7 c c = + = M1 For attempt to find c must have ( ) 1 2 2 1 k t + ( ) 1 2 2 1 7 v t = + + A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 5 of 8 Question Answer Marks Partial Marks 3(b) ( ) ( ) 3 2 1 2 1 7 3 s t t d = + + + B1 For ( ) 3 2 1 2 1 3 t + M1 For attempt to integrate their answer to (a), must have ( ) 1 2 2 1 k t + in (a) 1 4 3 d = + , 11 3 d = M1 Attempt to find d ( ) 3 2 1 11 2 1 7 3 3 s t t = + + + A1 4(a) 2 3 41 2 4 8 x   + −     B3 B1 for 2 B1 for 3 4 B1 for 41 8 − 4(b) 3 41 , 4 8   − −     B2 B1 for 3 4 − or FT on their b − B1 for 41 8 − or FT on their c 4(c) B1 For shape with max in 2nd quadrant B1 For x-intercepts 3 41 4 −± B1 For y-intercept of 4 and cusps 4(d) 41 8 B1 FT on their c

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 6 of 8 Question Answer Marks Partial Marks 5(a) ( ) p 3 : 162 9 36 11 a b + + + = ( ) p 1 : 6 12 21 a b − − + − + = − M1 For attempt at ( ) p 3 and ( ) p 1 − 9 187 0 a b + + = 3 0 a b + + = A1 for both, may be implied by correct work later 23, 20 a b = − = M1 attempt to solve simultaneous equations A1 For both ( ) ( )( ) 2 p 2 6 11 10 x x x x = − − − M1 For attempt to factorise or use algebraic long division A1 For ( ) 2 6 11 10 x x − − 5(b) ( ) ( )( )( ) p 2 3 2 2 5 x x x x = − + − M1 For attempt to factorise or use quadratic formula – must be seen 2 5 2, , 3 2 − A1 For all three solutions 6(a) 5 1 12 13     −   B1 6(b) 4 2 10 k r − = − 1 3 5 k r + = M1 equating like vectors to obtain 2 equations 7 3 , 10 2 r k = − = − M1 Dep on previous M mark, for attempt to solve simultaneously A1 6(c)(i) 3 2 − q p B1 6(c)(ii) 9 6 − q p B1 6(c)(iii) A common point of A and the same direction vector B1 6(c)(iv) 1:2 B1 7(a) 2 1 10 35 2 θ × × = so 0.7 θ = B1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 7 of 8 Question Answer Marks Partial Marks 7(b) Arc length CD: 7 B1 ( ) 2 sin 0.35 12 AB = M1 For a complete method to find AB, could be using cosine rule 8.23(0) AB = A1 Perimeter = 7 4 8.23 19.2 + + = A1 7(c) Area of triangle 2 112 sin0.7 2 = M1 For complete attempt at triangle area, may use equivalent method Area of triangle 46.4 = A1 Shaded area 11.4 = A1 Follow through on their area of the triangle 8(a) ( ) ( ) 14 1 0.4 2 n n + − B1 ( ) ( ) 14 1 0.4 300 2 n n + − > 2 0.4 13.6 600 0 n n + − > M1 Attempt to form a 3 term inequality and find the positive critical value Positive critical value 25.29 A1 26 terms A1 8(b) 9 a ar + = B1 36 1 a r = − B1 ( )( ) 36 1 1 9 r r + − = M1 attempt at solution of simultaneous equations 3 2 r = A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 8 of 8 Question Answer Marks Partial Marks 9 ( ) 5 3 2 x x − = 2 5 3 2 0 x x − − = M1 attempt at a 3-term quadratic equation in one variable with solution 1 x = , 2 5 x = − A1 Allow if 2 5 x = − not seen ( ) 1, 2 A A1 3 , 0 5 B       B1 Area of triangle 2 5 = M1 Using their A and B Area under curve: [ ] 3 3 1 1 2 d 2ln x x x =  B1 For [ ] 3 1 2ln x 2ln3 = M1 For use of limits Total area 2 ln9 5 = + A1 10(a) ( ) ( ) 1 1 2 2 d 1 2 2 d 2 y x x x x − = + + + B1 For ( ) 1 2 1 2 2 x − + M1 For differentiation of a product A1 ( ) ( ) 1 2 d 1 2 2 2 d 2 y x x x x − = + + +    M1 For attempt to simplify d 3 4 d 2 2 y x x x + = + A1 10(b) 3 4 0 x + = M1 For setting their numerator in (a) to zero and attempt to solve 4 3 x = − A1 4 6 9 y = − oe A1 10(c) Using the gradient method or inspection of y-coordinates either side of stationary point. Allow use of second derivative M1 complete method Minimum A1 Must be from correct work