Cambridge IGCSE Mathematics - Additional 0606 — 2015 Feb/March Paper 2 · Variant 2

This document consists of 16 printed pages. DC (AC/FD) 99297/4 © UCLES 2015 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 8 4 7 5 1 4 8 9 1 2 * ADDITIONAL MATHEMATICS 0606/22 Paper 2 February/March 2015 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 0606/22/F/M/15 © UCLES 2015 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x b b ac a = − − 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 0606/22/F/M/15 © UCLES 2015 [Turn over 1 (i) State the amplitude of cosx 4 3 - . [1] (ii) State the period of cosx 4 3 - . [1] (iii) The function f is defined, for x 0 360 c c G G , by ( ) cos x x 4 3 f = - . Sketch the graph of ( ) y x f = on the axes below. [2] 90° 0 5 y x –5 180° 270° 360°

4 0606/22/F/M/15 © UCLES 2015 2 (a) Jean has nine different flags. (i) Find the number of different ways in which Jean can choose three flags from her nine flags. [1] (ii) Jean has five flagpoles in a row. She puts one of her nine flags on each flagpole. Calculate the number of different five-flag arrangements she can make. [1] (b) The six digits of the number 738925 are rearranged so that the resulting six-digit number is even. Find the number of different ways in which this can be done. [2]

5 0606/22/F/M/15 © UCLES 2015 [Turn over 3 Solve the simultaneous equations . , y x x xy y 2 4 3 2 16 2 2 - = - + = [5]

6 0606/22/F/M/15 © UCLES 2015 4 (i) Differentiate sin cos x x with respect to x, giving your answer in terms of sin x. [3] (ii) Hence find sin x x d 2 y . [3]

7 0606/22/F/M/15 © UCLES 2015 [Turn over 5 The position vectors of the points A and B relative to an origin O are −2i + 17j and 6i + 2j respectively. (i) Find the vector AB. [1] (ii) Find the unit vector in the direction of AB. [2] (iii) The position vector of the point C relative to the origin O is such that m OC OA OB = + , where m is a constant. Given that C lies on the x-axis, find the vector OC. [3]

8 0606/22/F/M/15 © UCLES 2015 6 B C O θ rad 20 cm A AOB is a sector of a circle with centre O and radius 20 cm. Angle AOB = θ radians. AOC is a straight line and triangle OBC is isosceles with OB = OC. (i) Given that the length of the arc AB is 15π cm, find the exact value of θ. [2] (ii) Find the area of the shaded region. [4]

9 0606/22/F/M/15 © UCLES 2015 [Turn over 7 It is given that A = 1 3 5 10 - - J L KK N P OO and B = 3 4 5 10 J L KK N P OO. (i) Find A2 + B. [2] (ii) Find det B. [1] (iii) Find the inverse matrix, B−1. [2] (iv) Find the matrix X, given that BX = A. [2]

10 0606/22/F/M/15 © UCLES 2015 8 Solutions to this question by accurate drawing will not be accepted. The points A and B have coordinates (2, −1) and (6, 5) respectively. (i) Find the equation of the perpendicular bisector of AB, giving your answer in the form ax by c + = , where a, b and c are integers. [4] The point C has coordinates (10, −2). (ii) Find the equation of the line through C which is parallel to AB. [2]

11 0606/22/F/M/15 © UCLES 2015 [Turn over (iii) Calculate the length of BC. [2] (iv) Show that triangle ABC is isosceles. [1]

12 0606/22/F/M/15 © UCLES 2015 9 y = 4 (2x + 1)2 + 2x y A O y = 4x x The diagram shows part of the curve ( ) y x x 2 1 4 2 2 = + + and the line y = 4x. (i) Find the coordinates of A, the stationary point of the curve. [5] (ii) Verify that A is also the point of intersection of the curve ( ) y x x 2 1 4 2 2 = + + and the line y = 4x. [1]

13 0606/22/F/M/15 © UCLES 2015 [Turn over (iii) Without using a calculator, find the area of the shaded region enclosed by the line y = 4x, the curve and the y-axis. [6]

14 0606/22/F/M/15 © UCLES 2015 10 (a) (i) Sketch the graph of y 5 ex = - on the axes below, showing the exact coordinates of any points where the graph meets the coordinate axes. [3] O x y (ii) Find the range of values of k for which the equation k 5 ex - = has no solutions. [1] (b) Simplify log log log 2 8 2 1 a a a + + J L KK N P OO, giving your answer in the form log p 2 a , where p is a constant. [2] (c) Solve the equation log log x x 4 1 3 9 - = . [4]

15 0606/22/F/M/15 © UCLES 2015 [Turn over 11 (a) A particle P moves in a straight line. Starting from rest, P moves with constant acceleration for 30 seconds after which it moves with constant velocity, k ms−1, for 90 seconds. P then moves with constant deceleration until it comes to rest; the magnitude of the deceleration is twice the magnitude of the initial acceleration. (i) Use the information to complete the velocity-time graph. [2] t seconds v ms –1 k O (ii) Given that the particle travels 450 metres while it is accelerating, find the value of k and the acceleration of the particle. [4] Question 11(b) is printed on the next page.

16 0606/22/F/M/15 © UCLES 2015 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. (b) A body Q moves in a straight line such that, t seconds after passing a fixed point, its acceleration, a ms−2, is given by a t3 6 2 = + . When t = 0, the velocity of the body is 5 ms−1. Find the velocity when t = 3. [5]

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME for the March 2015 series 0606 ADDITIONAL MATHEMATICS 0606/22 Paper 2 (Paper 22), maximum raw mark 80 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 March 2015 series for most Cambridge IGCSE®, components.

Page 2 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2015 0606 22 © Cambridge International Examinations 2015 1 (i) 4 B1 (ii) 360 B1 or 2π (iii) B2 Correct symmetrical shape; one cycle; both maximums at 1 and minimum at –7 2 (a) (i) ( ) 9 3 84 = C B1 (ii) ( ) 9 5 15120 = P B1 (b) !6 6 2 × or 5! + 5! oe 240 M1 A1 or clear indication of method 3 Eliminate x or y 0 8 2 3 2 = − + x x or 0 32 44 12 2 = + − y y oe Factorise 3 term quadratic oe 3 4 = x and –2 3 8 = y and 1 M1 A1 M1 A1 A1 correct method Or allow A1 A1 for each (x, y) pair If second M0 then SC1 for one (x, y) pair found by inspection i.e. with no method or with no incorrect method shown 90 180 270 360 O −5 5 x y

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2015 0606 22 © Cambridge International Examinations 2015 4 (i) ( ) ( ) ( ) sin sin cos cos x their x x their x − + x x 2 2 cos sin + − oe x 2 sin 2 1− oe M1 A1 A1 clearly applies correct form of product rule If M1 A0 A0 then allow SC1 for 1 sin 2 cos sin 2 2 2 − = − x x x (ii) ( ) ( ) ∫ + = − c x x x x cos sin d sin 2 1 2 ∫ ∫ − = − x x x x x d 1 cos sin d sin 2 2 x x x cos sin 2 1 2 − [+ c] oe isw M1 M1 A1 or ( ) ( ) 2 2 1 sin d 2sin 1 d 1d oe 2 x x x x x = − + − − ∫ ∫ ∫ ∫ ∫ − − − = x x x x x d 1 2 1 cos sin 2 1 d sin 2 5 (i) 6i + 2j – (−2i + 17j) = 8i − 15j B1 (ii) ( )2 2 15 8 − + their their ( ) 17 15 8 their their j i − M1 A1ft ft their AB uuur (iii) −2i + 17j + m(6i + 2j) leading to 17 + 2m = 0 m = −8.5 oe −53i M1 M1 A1 If M0, allow SC1 for 6m – 2 = 0 leading to 53 3 j 6 (i) 15π = 20θ 3 4 θ π = or exact equivalent form isw M1 A1 (ii) Sector plus triangle approach: 2 1 3 Area sector = 20 2 4 their π   × ×    soi 2 1 1 Area triangle 20 sin soi 2 4 their π   = × ×     their sector area + their triangle area 613 or 612.6(60254…) rot to 4 sig figs B1 B1 M1 A1 Semicircle less segment approach: 2 1 1 Area sector = 20 2 4 their π   × ×    soi ( ) 2 20 2 π – (their area sector – their area triangle) soi

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2015 0606 22 © Cambridge International Examinations 2015 7 (i) A2 =       − − 85 27 45 14 seen       − − 95 23 50 11 M1 A1 condone one error (ii) 10 B1 (iii) 10 1 their or       − − 3 4 5 10 oe, seen       − − 3 4 5 10 10 1 oe isw B1 B1 (iv) X = B–1A soi       − 1 5.0 0 5.0 oe M1 A1ft ft their B-1 8 (i) (4, 2) mAB = 2 3 ⇒ mPerp = 3 2 − ( ) 4 3 2 2 − − = − x y oe 14 3 2 = + y x B1 M1 M1 A1 allow unsimplified allow arithmetic slips provided method is correct ft their mid-point and perpendicular gradient allow any correct equivalent form with integer a, b, c (ii) mAB used ( ) 10 2 − = + x m their y AB M1 A1ft (iii) ( ) ( ) ( )2 2 2 5 6 10 − − + − oe 65 or 8.0622577… rot to 3 or more sf M1 A1 any valid method (iv) ( ) ( ) ( )2 2 2 2 1 10 2 − − − + − = AC and AC2 = BC2 = 65 or showing C lies on the perpendicular bisector of AB or showing line from C to (4, 2) is perpendicular to AB B1 any valid method

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2015 0606 22 © Cambridge International Examinations 2015 9 (i) ( ) 3 1 2 − + x k ( ) 3 8 2 1 2 x − − + × oe + 2 0 d d = x y their and solves 1 , 2 2 x y = = M1 A1 B1 M1 A1 (ii) 2 2 1 4 = × = y B1 or equivalent correct method (iii) ( ) ∫       + + x x x d 2 1 2 4 2 ( ) 2 2 2 1 2 4 2 1 x x + − + × − or better ( ) 5.0 0 2 1 2 2 2 1 2 4 their x x their               + − + × − Substitution of correct limits seen, leading to 1 14 Shaded area = 1 1 14 2 − their their 3 4 M1 A1 M1 A1 M1 A1 Alternative method: M1 for ( ) ∫       − + + x x x x d 4 2 1 2 4 2 A1 for ( ) 2 2 1 2 2 2 2 1 2 4 x x x − + − + × − or better M1 for ( ) 5.0 0 2 1 2 2 2 1 2 4 their x x their               − − + × − M1 for subst of their limits into their genuine attempt at an integral A1 for subst of correct limits into correct expression A1 for for 3 4

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2015 0606 22 © Cambridge International Examinations 2015 10 (a)(i) B3 B1 correct shape B1 through (0, −4) B1 through (ln5, 0) (ii) k < −5 B1 (b) 2 log 2 log 3 2 log 2 1 a a a − + or ( ) 1 3 2 2 2 log 2 1 − × × a oe 2 log 2 1 2 a oe M1 A1 condone one error (c) 9 log 4 log 4 log 3 3 9 x x = or 3 log log log 9 9 3 x x = 1 2 4 log log 3 3 = − x x or 9 9 log log 4 1 1 2 x x − = ( ) 3 log 4 log 3 3 2 1 = x x or 9 log 4 log 9 2 9 = x x oe x = 36 B1 M1 M1 A1 soi −4 O ln5

Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2015 0606 22 © Cambridge International Examinations 2015 11 (a)(i) k v ms-1 O 30 120 135 t s B2 Horizontal line of correct length; deceleration correctly drawn; key times soi on horizontal axis (ii) k × × = 30 2 1 450 k = 30 30 30 their a = a = 1 [ms–2] M1 A1 M1 A1 (b) ( ) ∫ ∫ + = = t t t a v d 6 3 d 2 ( ) 3 6 5 v t t = + + When t = 3, v = ( ) 5 3 6 33 + + 50 [ms–1] M1 A2 M1 A1 A1 for two terms correct