Cambridge IGCSE Mathematics - Additional 0606 — 2016 May/June Paper 1 · Variant 3

This document consists of 15 printed pages and 1 blank page. DC (SLM) 122174 © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 5 8 7 0 4 3 8 4 3 5 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 May/June 2016 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.

2 0606/13/M/J/16 © UCLES 2016 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/16 © UCLES 2016 [Turn over 1 Find the value of k for which the curve y x x k 2 3 2 = - + (i) passes through the point ( , ) 4 7 - , [1] (ii) meets the x-axis at one point only. [2] 2 (a) Solve the equation 16 8 x x 3 1 2 = - + . [3] (b) Given that a b a b 3 3 2 2 1 3 1 2 1 - - ` j = a p b q, find the value of each of the constants p and q. [2]

4 0606/13/M/J/16 © UCLES 2016 3 Find the equation of the normal to the curve ( ) n y x 1 2 7 2 = - at the point where the curve crosses the positive x-axis. Give your answer in the form ax b c y 0 + + = , where a, b and c are integers. [5]

5 0606/13/M/J/16 © UCLES 2016 [Turn over 4 (a) Given the matrices A 1 3 2 0 = - J L KK N P OO and B 3 1 0 2 = J L KK N P OO , find A B 2 2 - . [3] (b) Using a matrix method, solve the equations x y 4 1 + = , x y 10 3 1 + = . [4]

6 0606/13/M/J/16 © UCLES 2016 5 Do not use a calculator in this question. (i) Show that d d e e e x x px 4 x x x 4 4 4 - = J L KK N P OO , where p is an integer to be found. [4] (ii) Hence find the exact value of e d x x n x 4 0 1 2 y , giving your answer in the form n a c b 1 2 + , where a, b and c are integers to be found. [4]

7 0606/13/M/J/16 © UCLES 2016 [Turn over 6 The function f is defined by ( ) f x x 2 5 = - + for x 5 0 1 G - . (i) Write down the range of f. [2] (ii) Find ( ) f x 1 - and state its domain and range. [4] The function g is defined by ( ) g x x 4 = for x 5 1 1 G - - . (iii) Solve ( ) fg x 0 = . [3]

8 0606/13/M/J/16 © UCLES 2016 7 75 m 2.4 ms–1 30 m B A The diagram shows a river with parallel banks. The river is 75 m wide and is flowing with a speed of 2.4 ms–1. A speedboat travels in a straight line from a point A on one bank to a point B on the opposite bank, 30 m downstream from A. The speedboat can travel at a speed of 4.5 ms–1 in still water. (i) Find the angle to the bank and the direction in which the speedboat is steered. [4]

9 0606/13/M/J/16 © UCLES 2016 [Turn over (ii) Find the time the speedboat takes to travel from A to B. [4]

10 0606/13/M/J/16 © UCLES 2016 8 Solutions to this question by accurate drawing will not be accepted. Three points have coordinates ( , ) A 8 6 - , ( , ) B 4 2 and ( , ) C 1 7 - . The line through C perpendicular to AB intersects AB at the point P. (i) Find the equation of the line AB. [2] (ii) Find the equation of the line CP. [2] (iii) Show that P is the midpoint of AB. [3]

11 0606/13/M/J/16 © UCLES 2016 [Turn over (iv) Calculate the length of CP. [1] (v) Hence find the area of the triangle ABC. [2]

12 0606/13/M/J/16 © UCLES 2016 9 (i) Show that cos cot cot cos x x x x 2 1 2 + = + can be written in the form ( )( ) cos cos sin a x x x b 0 - - = , where a and b are constants to be found. [4] (ii) Hence, or otherwise, solve cos cot cot cos x x x x 2 1 2 + = + for x 0 1 1 r . [3]

13 0606/13/M/J/16 © UCLES 2016 [Turn over 10 (i) Given that ( ) f x x kx p 4 3 = + + is exactly divisible by x 2 + and ( ) f x l is exactly divisible by x 2 1 - , find the value of k and of p. [4] (ii) Using the values of k and p found in part (i), show that ( ) ( )( ) f x x ax bx c 2 2 = + + + , where a, b and c are integers to be found. [2] (iii) Hence show that ( ) f x 0 = has only one solution and state this solution. [2]

14 0606/13/M/J/16 © UCLES 2016 11 B O 2θ rad A r cm The diagram shows a circle, centre O, radius r cm. The points A and B lie on the circle such that angle AOB 2i = radians. (i) Find, in terms of r and i, an expression for the length of the chord AB. [1] (ii) Given that the perimeter of the shaded region is 20 cm, show that sin r 10 i i = + . [2]

15 0606/13/M/J/16 © UCLES 2016 (iii) Given that r and i can vary, find the value of d dr i when i = 6 r . [4] (iv) Given that r is increasing at the rate of 15 cm s–1, find the corresponding rate of change of i when i = 6 r . [3]

16 0606/13/M/J/16 © UCLES 2016 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 the registered trademark of Cambridge International Examinations. This document consists of 7 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/13 Paper 1 May/June 2016 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 2016 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2016 0606 13 © Cambridge International Examinations 2016 Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied www without wrong working Question Answer Marks Guidance 1 (i) 27 − B1 (ii) 9 8 0 k − = 9 8 k = M1 A1 for use of discriminant with a complete method to get to k = Or d 4 3 d y x x = − when d 0 d y x = , 3 4 x = so 9 8 k = M1 A1 for a complete method to get to k = Or completing the square 2 3 9 2 4 8 y x k   = − + −     9 8 k = M1 A1 for a complete method to get to k = 2 (a) ( ) ( ) 4 3 1 3 2 2 2 x x − + = or ( ) ( ) 3 2 2 3 1 2 4 4 x x + −= or ( ) 4 3 1 2 3 8 8 x x − + = or ( ) 3 2 3 1 4 16 16 x x + −= leading to 10 9 x = cao B1 M1 A1 B1 for a correct statement for equating indices (b) 5 3 p = 2 q = − B1 B1

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2016 0606 13 © Cambridge International Examinations 2016 Question Answer Marks Guidance 3 On x-axis, 2 2 7 1 x − = 2 x = 2 d 4 d 2 7 y x x x = − When 2, x = d 8 d y x = Gradient of normal = 1 8 − Equation of normal ( ) 1 2 8 y x = − − Required form 8 2 0 x y + − = M1 A1 B1 M1 A1 for equating to 1 for attempt at perpendicular through their ( ) 2, 0 , must be using 0 y = must be equated to zero with integer coefficients 4 (a) 2 7 2 3 6 −   =   −   A 2 1 2 2 5 2 −   − =   −   A B B1 M1 A1 for their 2 2 − A B (b) 4 1 1 10 3 1 x y     =         so 3 1 1 1 10 4 1 2 x y −     =     −     leading to 1 3 x y     =     −     1 3 x y = = − M1 DM1 A1 A1 for pre-multiplication by their inverse matrix DM1for attempt at matrix multiplication Allow in matrix form 5 (i) ( ) ( ) 4 4 4 4 4 d e e e 4e e d 4 x x x x x x x x   − = − × +     4 4 e x x = − B1 M1 A1 A1 for 4 4 d e e d 4 x x x  =     for attempt to differentiate a product for a correct product for correct final answer (ii) ln2 4 ln2 4 4 0 0 1 e e d e 4 4 x x x x x x   = − −     ∫ = 1 16 1 16ln 2 4 4 4     − − −         15 4ln 2 16 = − B1FT B1 M1 A1 FT for use of their 4 4 1 e e 4 x x x p   × −     , must be numerical p, but 0 ≠ for 4ln 2 e 16 = for correct use of limits, must be an integral of the correct form

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2016 0606 13 © Cambridge International Examinations 2016 Question Answer Marks Guidance 6 (i) ( ) 2 5 f 2 x − < ≤ B2 B1 for 2 ≤ B1 for 2 5< − or awrt –0.24 Must be using f, f(x) or y, 2 5< − , if not then B1 max (ii) ( ) ( ) 2 1 f 2 5 x x − = − − Domain 2 5< 2 x − ≤ Range y or ( ) 1 5 f 0 x − − < ≤ M1 A1 B1 B1 for a correct method to find the inverse Must be using the correct variables for the B marks (iii) ( ) 4 fg f x x   =     4 2 5 x = − + leading to 4 x = − M1 DM1 A1 for correct order of functions for solution of equation 7 (i) Finding an angle of o 68.2 or o 21.8 4.5 2.4 sin68.2 sinα = leading to o 29.7 α = (allow 0.1 ± ) Direction is o 82.1 to the bank, upstream(allow o 0.1 ± ) B1 B1 B1 B1 for the sine rule (ii) 4.5 2.4 sin68.2 sin 29.7 sin82.1 rv = = leading to 4.8 rv = time taken = 80.78 16.8 4.8 = B1 B1 M1 A1 for the sine rule for resultant velocity for attempt to find AB and hence the time taken Alternative method: Finding an angle of o 68.2 or o 21.8 ( ) 2 2 2 4.5 2.4 2 2.4 cos68.2 r r v v = + − × × leading to 4.8 rv = Use of sine rule to obtain angle and direction to obtain direction is o 82.1 to the bank, upstream Use of time taken = 80.78 16.8 4.8 = B1 B1 B1 B1 B1 B1 M1 A1 for correct use of the cosine rule for resultant velocity for use of the sine rule for o 29.7 α = for o 82.1 for attempt to find AB and hence the time taken

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2016 0606 13 © Cambridge International Examinations 2016 Question Answer Marks Guidance 8 (i) ( ) 4 6 8 12 y x − = − + ( 3 10 y x + = ) M1 A1 for a correct method allow unsimplified (ii) ( ) 7 3 1 y x − = + ( ) 3 10 y x = + DM1 A1 for attempt at a perpendicular line using ( ) 1, 7 − allow unsimplified (iii) point of intersection ( ) 2, 4 − which is the midpoint of AB M1 M1 A1 for attempt to find the point of intersection using simultaneous equations for attempt to find midpoint for all correct Alternative method: Midpoint ( ) 2, 4 − Verification that this point lies on CP. M1 M1 A1 for attempt to find midpoint for full verification for all correct (iv) 10 CP = or 3.16 B1 (v) Area = 1 10 4 10 2 × × = 20 M1 A1 for correct method using CP for 19.9 20.1 −

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2016 0606 13 © Cambridge International Examinations 2016 Question Answer Marks Guidance 9 (i) 2cos cot cot 2cos x x x x = + cos cos 2cos 1 2cos sin sin x x x x x x + = + 2 2cos sin cos 2cos sin x x x x x + = + 2 2cos 2cos sin cos sin x x x x x − = − ( ) 2cos cos sin cos sin x x x x x − = − ( )( ) 2cos 1 cos sin 0 x x x − − = M1 DM1 DM1 A1 for use of cos cot sin x x x = for both terms for multiplication throughout by sin x for attempt to factorise for completely correct solution www Alternative method: 2 cos cos sin cos sin 0 a x a x x b x b x − − + = cos cot cos cot 0 a x x a x b x b − − + = 2, 1 a b = = M1 DM1 DM1 A1 for expansion of RHS for division by sin x for comparing like terms to obtain both a and b for both correct www (ii) ( )( ) 2cos 1 cos sin 0 x x x − − = 1 cos 2 x = , tan 1 x = π 3 x = , π 4 x = M1 A1,A1 for either A1 for each, penalise extra solutions within the range by withholding the last A mark Alternative method: ( )( ) 2cos 1 cot 1 0 x x − − = Leading to 1 cos 2 x = , tan 1 x = π 3 x = , π 4 x = M1 A1,A1 for attempt to factorise the original equation and attempt to solve A1 for each, penalise extra solutions within the range by withholding the last A mark 10 (i) ( ) f 2 32 2 0 k p − = − − + = 1 12 f 0 2 4 k   ′ = + =     leading to 3 k = − and 26 p = M1 M1 A1,A1 for attempt at ( ) f 2 − for attempt at 1 f 2   ′    A1 for each (ii) ( )( ) 2 2 4 8 13 x x x + − + B1FT B1 FT for their 2 p all correct (iii) Showing that 2 4 8 13 0 x x − + = has no real roots so 2 x = − only www M1, A1 M1 for a valid attempt at solution of equation leading to no solution or consideration of the discriminant

Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2016 0606 13 © Cambridge International Examinations 2016 Question Answer Marks Guidance 11 (i) 2 sin AB r θ = or 2 2 2 2 cos2 r r r θ + − or sin 2 π sin 2 r θ θ   −     or sin 2 cos r θ θ B1 (ii) 2 sin 2 20 r r θ θ + = 10 sin r θ θ = + M1 A1 for use of (i) + arc length = 20, oe must be convinced (iii) ( ) ( ) 2 10 1 cos d d sin r θ θ θ θ + = − + When π 6 θ = , d 17.8 d r θ = − M1 A2,1,0 A1 for a correct attempt to differentiate –1 each error allow awrt 17.8 − (iv) d 15 d r t = d d d d d d r r t t θ θ = ÷ d 0.842 dt θ = − B1 M1 A1 may be implied for use of 15 (iii) their allow 0.84 − or 0.843 −