Cambridge IGCSE Mathematics - Additional 0606 — 2017 May/June Paper 1 · Variant 3

This document consists of 15 printed pages and 1 blank page. DC (KN/CGW) 129468/2 © UCLES 2017 [Turn over * 8 2 6 2 8 4 1 6 1 7 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 May/June 2017 2 hours Candidates answer on the Question Paper. Additional Materials: Electronic calculator READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 80. Cambridge International Examinations Cambridge International General Certificate of Secondary Education

2 0606/13/M/J/17 © UCLES 2017 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)n = an + ( n 1)an–1 b + ( n 2)an–2 b2 + … + ( n r)an–r br + … + bn, where n is a positive integer and ( n r) = n! (n – r)!r! 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A Formulae for ∆ABC a sin A = b sin B = c sin C a2 = b2 + c2 – 2bc cos A ∆ = 1 2 bc sin A

3 0606/13/M/J/17 © UCLES 2017 [Turn over 1 (a) On the Venn diagram below, shade the region which represents A B C B + , + l l ^ ^ h h. [1] A B C (b) Complete the Venn diagram below to show the sets Y and Z such that Z X Y 1 1 . [1] X 2 Given that cos y x 3 4 9 = + , write down (i) the amplitude of y, [1] (ii) the period of y. [1]

4 0606/13/M/J/17 © UCLES 2017 3 (i) On the axes below, sketch the graph of sin y x 3 2 = - for 0 360 G G i c c. [3] 6 y x 4 2 0 90° 180° 270° 360° –2 –4 –6 (ii) Given that sinx k 0 3 2 G G - for x 0 360 G G c c, write down the value of k. [1]

5 0606/13/M/J/17 © UCLES 2017 [Turn over 4 In this question, all dimensions are in centimetres. A B C 3 + 2√5 M 4 + 6√5 The diagram shows an isosceles triangle ABC, where AB = AC. The point M is the mid-point of BC. Given that AM 3 2 5 = + and BC 4 6 5 = + , find, without using a calculator, (i) the area of triangle ABC, [2] (ii) tan ABC, giving your answer in the form c a b 5 + where a, b and c are positive integers. [3]

6 0606/13/M/J/17 © UCLES 2017 5 The normal to the curve y x 4 9 = + , at the point where x 4 = , meets the x- and y-axes at the points A and B. Find the coordinates of the mid-point of the line AB. [7]

7 0606/13/M/J/17 © UCLES 2017 [Turn over 6 (a) Given that A = 3 2 1 4 J L KK N P OO, B = 5 2 1 1 4 0 - J L K KK N P O OO and C = 5 3 2 1 - J L KK N P OO, find (i) A + 3C, [2] (ii) BA. [2] (b) (i) Given that X = 1 4 3 2 - - J L KK N P OO, find X–1. [2] (ii) Hence find Y, such that XY = 5 15 10 20 - J L KK N P OO. [3]

8 0606/13/M/J/17 © UCLES 2017 7 (a) Show that cos sec tan sin tan sin 2 2 i i i i i i + + = . [4]

9 0606/13/M/J/17 © UCLES 2017 [Turn over (b) Given that sin x 3 z = and cos y 3 z = , find the numerical value of y x y 9 2 2 2 - . [3]

10 0606/13/M/J/17 © UCLES 2017 8 It is given that ( )x x ax x b 2 4 p 3 2 = + + + , where a and b are constants. It is given also that x 2 1 + is a factor of ( )x p and that when ( )x p is divided by x 1 - there is a remainder of 12 - . (i) Find the value of a and of b. [5] (ii) Using your values of a and b, write ( )x p in the form ( ) ( ) x x 2 1 q + , where ( )x q is a quadratic expression. [2] (iii) Hence find the exact solutions of the equation ( )x 0 p = . [2]

11 0606/13/M/J/17 © UCLES 2017 [Turn over 9 It is given that x 5 6 15e e d x x k k 5 5 - = - - ^ h y . (i) Show that 3 e e k k 5 5 - = - . [5] (ii) Hence, using the substitution y e k 5 = , or otherwise, find the value of k. [3]

12 0606/13/M/J/17 © UCLES 2017 10 It is given that ( ) ( ) ln y x x 10 2 5 1 = + + . (i) Find x y d d . [4] (ii) Hence show that ( ) ( ) ( ) ln ln x x ax b x x c 5 1 5 5 1 d + + + - + = y , where a and b are integers and c is a constant of integration. [3]

13 0606/13/M/J/17 © UCLES 2017 [Turn over (iii) Hence find ( ) ln x x 5 1 d 0 5 1 + y , giving your answer in the form ln d f 5 + , where d and f are integers. [2]

14 0606/13/M/J/17 © UCLES 2017 11 A curve has equation y x x x 6 = - . (i) Find the coordinates of the stationary point of the curve. [4] (ii) Determine the nature of this stationary point. [2] (iii) Find the approximate change in y when x increases from 4 to h 4 + , where h is small. [3]

15 0606/13/M/J/17 © UCLES 2017 12 A particle moves in a straight line, such that its velocity, v ms 1 - , t s after passing a fixed point O, is given by sin v t t 3 2 2 6 = + + . (i) Find the acceleration of the particle at time t. [2] (ii) Hence find the smallest value of t for which the acceleration of the particle is zero. [2] (iii) Find the displacement, x m from O, of the particle at time t. [5]

16 0606/13/M/J/17 © UCLES 2017 16 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 International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. BLANK PAGE

® IGCSE is a registered trademark. This document consists of 8 printed pages. © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/13 Paper 1 May/June 2017 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 will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 2 of 8 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 2017 © UCLES 2017 Page 3 of 8 Question Answer Marks Partial Marks 1(a) 1 1(b) 1 2(i) 4 1 2(ii) o 40 or 2π 9 or 0.698 rad 1 3(i) 3 B1 for a complete cycle starting and ending at –2 B1 for max at y = 1 and min at y = –5 B1 for a completely correct graph 3(ii) 5 1 FT their min value for y 4(i) Area = ( )( ) 1 3 2 5 4 6 5 2 + + ( ) 1 12 26 5 60 2 = + + M1 use of correct formula and attempt to expand out the brackets 36 13 5 = + A1 4(ii) 3 2 5 2 3 5 + + B1 3 2 5 2 3 5 2 3 5 2 3 5 + − = × + − M1 6 5 5 30 4 45 − − = − 24 5 5 41 + = A1 for answer 90 180 270 360 −6 −4 −2 2 4 6 x y

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 4 of 8 Question Answer Marks Partial Marks 5 When 4, 5 x y = = B1 for y ( ) 1 2 d 1 4 4 9 d 2 y x x − = × + B1 for ( ) 1 2 2 4 9 x − + , allow unsimplified When d 2 4, d 5 y x x = = , so perp grad = 5 2 − M1 obtaining numerical gradient for normal Equation of normal ( ) 5 5 4 2 y x − = − − ( ) 2 30 5 y x = − M1 for equation of normal ( ) ( ) 6, 0 , 0, 15 A B A2 A1 for each Midpoint 15 3, 2       B1 FT on their x/y intercepts 6(a)(i) dealing with multiplication and addition M1 implied by 2 correct elements A + 3C = 12 7 11 7 −       A1 6(a)(ii) correct attempt to multiply M1 implied by 2 correct elements BA = 17 9 14 18 3 1         − −   A1 6(b)(i) 1 2 3 1 4 1 10 − −   =   −   X B2 B1 for 1 10 , B1 for 2 3 4 1 −     −   6(b)(ii) 2 3 5 10 1 4 1 15 20 10 a b c d − −      =      −      M1 pre-multiplication using matrix from (b)(i) = 3.5 8 0.5 6     −   A2 –1 for each incorrect element

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 5 of 8 Question Answer Marks Partial Marks 7(a) LHS = 2 2 2 sin sin cos 1 cos cos θ θ θ θ θ + + M1 for obtaining all in terms of sinθ and cosθ = 2 2 2 2 2 sin sin cos cos cos 1 cos θ θ θ θ θ θ + + M1 for simplification using addition of fractions = ( ) ( ) 2 2 2 sin 1 cos cos cos 1 θ θ θ θ + + 2 sin cos θ θ = M1 for factorisation and subsequent cancelling of common term tan sin θ θ = RHS A1 correct final simplification Alternative 2 2 sec 1 cos 1 cos sec θ θ θ θ −− + + M1 use of correct identities = ( )( ) ( ) sec cos sec cos sec cos θ θ θ θ θ θ − + − sec cos θ θ = − M1 attempt to factorise and simplify 2 1 cos cos θ θ − = M1 simplification to obtain terms in sinθ and cosθ only 2 sin cos θ θ = tan sin θ θ = A1 for final simplification 7(b) 3 sin , cos 3 x y φ φ = = M1 for obtaining sinφ and cosφ in terms of x and y and attempt to use correct identity Using 2 2 sin cos 1 φ φ + = leads to 2 2 9 1 9 x y + = and hence 2 2 2 81 9 x y y + = M1 attempt at simplification 81 A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 6 of 8 Question Answer Marks Partial Marks Alternative method using substitution 2 2 2 9 9 9 9sin cos cos φ φ φ     × − ×         M1 attempt to substitute in for x and y = 2 2 2 81 81sin cos cos φ φ φ    −        M1 simplification of fractions = ( ) 2 2 81 1 sin cos φ φ − or ( ) 2 2 81 sec tan φ φ − leading to 81 A1 use of correct identity to obtain 81 8(i) 1 2 p 2 2 8 4 a b   − = − + − +     M1 for attempt at 1 p 2   −     leading to 4 9 a b + = oe A1 ( ) p 1 2 4 a b = + + + leading to 18 a b + = − oe B1 solution of simultaneous equations M1 27, 9 a b = − = A1 for both 8(ii) attempt at factorisation using either long division or observation M1 ( )( ) 2 2 1 14 9 x x x + − + A1 8(iii) attempt to solve q(x) = 0 M1 1 7 2 10, 2 x = ± − A1 for all 3 solutions 9(i) 5 5 3e e 6 k x x k − −   + =   B2 B1 for each term integrated correctly ( ) ( ) 5 5 5 5 3e e 3e e 6 k k k k − − + − + = M1 for use of limits with 5 5 e e x x a b − + 5 5 2e 2e 6 k k − − = A1 correct unsimplified 5 5 e e 3 k k − − = A1 correct simplification to obtain given answer

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 7 of 8 Question Answer Marks Partial Marks 9(ii) 2 3 1 0 y y − −= M1 for correct attempt to obtain a quadratic equation in terms of y or 5e x 3 9 4 2 y ± + = , 5e 3.303 k y = = only DM1 for attempt to solve quadratic equation and solve for k 0.239 k = A1 A0 if more than one solution is given 10(i) for attempt to differentiate a product M1 5 5 1 x + B2 B1 for 1 5 1 x + ( ) ( ) d 5 10 2 10ln 5 1 d 5 1 y x x x x = + × + + + A1 all else correct 10(ii) ( ) 5 10 2 10 5 1 x x + × = + B1 simplification to obtain 10, allow if seen in (i) ( ) ( ) ( ) 10 ln 5 1 d 10 2 ln 5 1 10 x x x x x + = + + − ∫ M1 use of result from part (i) ( ) ( ) ( ) ln 5 1 d 5 1 ln 5 1 5 x x x x x + + = + − ∫ A1 10(iii) ( ) ( ) 1 5 0 0.2 ln 5 1 x x x  + + −   M1 use of limits in result from (ii) = 1 2 ln 2 5 5 − + = 1 ln 4 5 −+ cao A1 11(i) attempt to differentiate M1 1 2 d 3 6 d 2 y x x = − A1 When d 0 d y x = M1 equating to zero and attempt to solve 16, x = 32 y = A1 both correct

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 8 of 8 Question Answer Marks Partial Marks 11(ii) 1 2 2 2 d 3 4 d y x x − = − B1 correct differentiation This is negative so a maximum point DB1 correct conclusion 11(iii) When x = 4, d 3 d y x = B1 d d y y h x ∂≈ × M1 use of small increases 3h ≈ A1 FT their (iii) 12(i) attempt to differentiate M1 6cos2 6 t + A1 12(ii) cos2 1 t = − M1 attempt to equate (i) to zero and solve π 2 t = A1 12(iii) attempt to integrate M1 ( ) 2 3 cos2 3 2 2 x t t t c = − + + + A2 –1 for each error When 0, 0 t x = = , so 3 2 c = M1 attempt to find c 2 3 3 cos2 3 2 2 2 x t t t = − + + A1