Cambridge IGCSE Mathematics - Additional 0606 — 2009 May/June Paper 1 · Variant 1

This document consists of 6 printed pages and 2 blank pages. SP (SM) V06774 © UCLES 2009 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. 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 a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Write your answers on the separate Answer Booklet/Paper provided. 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. * 6 9 4 2 3 1 1 6 4 2 * ADDITIONAL MATHEMATICS 0606/01 Paper 1 May/June 2009 2 hours Additional Materials: Answer Booklet/Paper Graph paper (2 sheets) Electronic calculator Mathematical tables www.theallpapers.com

2 0606/01/M/J/09 © UCLES 2009 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. www.theallpapers.com

3 0606/01/M/J/09 © UCLES 2009 [Turn over 1 15 cm 15 cm 12 cm B A O The diagram shows a sector AOB of a circle, centre O, radius 15 cm. The length of the arc AB is 12 cm. (i) Find, in radians, angle AOB. [2] (ii) Find the area of the sector AOB. [2] 2 The equation of a curve is y = x3 – 8. Find the equation of the normal to the curve at the point where the curve crosses the x-axis. [4] 3 Show that 1 – cos2
sec2
– 1 = 1 – sin2
. [4] 4 The line y = 5x – 3 is a tangent to the curve y = kx2 – 3x + 5 at the point A. Find (i) the value of k, [3] (ii) the coordinates of A. [2] 5 (a) Solve the equation 92x – 1 = 27x. [3] (b) Given that a b – 1 2 2 3 a b 3 – 2 3 = a pbq, find the value of p and of q. [2] 6 Solve the equation 2x3 + 3x2 – 32x + 15 = 0. [6] 7 (i) Find d–dx(xe3x – e3x ––– 3 ) . [3] (ii) Hence find ∫xe3xdx. [3] www.theallpapers.com

4 0606/01/M/J/09 8 A curve has equation y = 2x x2 + 9 . (i) Find the x-coordinate of each of the stationary points of the curve. [4] (ii) Given that x is increasing at the rate of 2 units per second, find the rate of increase of y when x = 1. [3] 9 At 10 00 hours, a ship P leaves a point A with position vector (– 4i + 8j) km relative to an origin O, where i is a unit vector due East and j is a unit vector due North. The ship sails north-east with a speed of 10√2 km h–1. Find (i) the velocity vector of P, [2] (ii) the position vector of P at 12 00 hours. [2] At 12 00 hours, a second ship Q leaves a point B with position vector (19i + 34j) km travelling with velocity vector (8i + 6j) km h–1. (iii) Find the velocity of P relative to Q. [2] (iv) Hence, or otherwise, find the time at which P and Q meet and the position vector of the point where this happens. [3] © UCLES 2009 www.theallpapers.com

5 0606/01/M/J/09 © UCLES 2009 10 Solutions to this question by accurate drawing will not be accepted. C y x Q O A (– 4, 0) P (1, 10) B (8, 9) The diagram shows the line AB passing through the points A(– 4, 0) and B(8, 9). The line through the point P(1, 10), perpendicular to AB, meets AB at C and the x-axis at Q. Find (i) the coordinates of C and of Q, [7] (ii) the area of triangle ACQ. [2] 11 The table shows experimental values of variables s and t. t 5 15 30 70 100 s 1305 349 152 55 36 (i) By plotting a suitable straight line graph, show that s and t are related by the equation s = kt n, where k and n are constants. [4] (ii) Use your graph to find the value of k and of n. [4] (iii) Estimate the value of s when t = 50. [2] [Turn over www.theallpapers.com

6 0606/01/M/J/09 12 Answer only one of the following two alternatives. EITHER (i) State the amplitude of 1 + sin ( x–3). [1] (ii) State, in radians, the period of 1 + sin ( x–3). [1] y x x 3 O A B y = 1.5 y = 1 + sin (–) The diagram shows the curve y = 1 + sin ( x–3) meeting the line y = 1.5 at points A and B. Find (iii) the x-coordinate of A and of B, [3] (iv) the area of the shaded region. [6] OR A particle moves in a straight line such that t s after passing through a fixed point O, its velocity, v m s–1, is given by v = k cos 4t, where k is a positive constant. Find (i) the value of t when the particle is first instantaneously at rest, [1] (ii) an expression for the acceleration of the particle t s after passing through O. [2] Given that the acceleration of the particle is 12 m s–2 when t = 3π 8 , (iii) find the value of k. [2] Using your value for k, (iv) sketch the velocity-time curve for the particle for 0  t  π, [2] (v) find the displacement of the particle from O when t = π 24 . [4] © UCLES 2009 www.theallpapers.com

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8 0606/01/M/J/09 BLANK PAGE 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. University of 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. www.theallpapers.com

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2009 question paper for the guidance of teachers 0606 ADDITIONAL MATHEMATICS 0606/01 Paper 1, 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2009 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.theallpapers.com

Page 2 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2009 0606 01 © UCLES 2009 Mark Scheme Notes Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Accuracy 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 generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. • The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously “correct” answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2, 1, 0 means that the candidate can earn anything from 0 to 2. www.theallpapers.com

Page 3 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2009 0606 01 © UCLES 2009 The following abbreviations may be used in a mark scheme or used on the scripts: AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) CAO Correct Answer Only (emphasising that no “follow through” from a previous error is allowed) ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) Penalties MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become “follow through √ ” marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. OW –1,2 This is deducted from A or B marks when essential working is omitted. PA –1 This is deducted from A or B marks in the case of premature approximation. S –1 Occasionally used for persistent slackness – usually discussed at a meeting. EX –1 Applied to A or B marks when extra solutions are offered to a particular equation. Again, this is usually discussed at the meeting. www.theallpapers.com

Page 4 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2009 0606 01 © UCLES 2009 1 (i) 12 15θ = , 0.8 θ = rads M1, A1 [2] M1 for use of s rθ = (ii) Area = ( ) 2 115 0.8 2 leading to 90 (cm2) M1 A1 [2] M1 for use of 2 1 2 A r θ = 2 x3 = 8, leading to x = 2 2 3 d d x x y = leading to grad of 12 1 − for normal ( ) 2 12 1 0 − − = − x y       + − = 6 1 12 1 x y B1 M1 DM1 A1 [4] B1 for finding where curve crosses the x axis M1 for attempt to differentiate and use of m1m2 = –1 DM1 for attempt at equation of normal Allow unsimplified 3 2 2 2 2 1 cos sin sec 1 tan θ θ θ θ − = − M1 M1 M1 for use of 2 2 1 cos sin θ θ − = M1 for use of 2 2 sec 1 tan θ θ −= 2 cos θ = M1 M1 for attempt to simplify = 2 1 sin θ − A1 [4] Alt Scheme 2 2 2 2 2 1 cos sin 1 cos sec 1 cos θ θ θ θ θ − = − − 2 2 2 sin cos sin θ θ θ = 2 cos θ = = 2 1 sin θ − M1 M1 M1 A1 M1 for use of 2 2 1 cos sin θ θ − = M1 for attempting to get all in terms of cos M1 for attempt to simplify 4 (i) 5 3 3 5 2 + − = − x kx x M1 M1 for equating line and curve equations 0 8 8 2 = + −x kx using 0 4 2 = −ac b , k = 2 (Alt scheme: 5 2 3 kx = − , 4 x k = 20 16 12 3 5 k k k − = − + leading to 2 k = ) DM1, A1 [3] DM1 for use of 2 4 b ac − on resulting quadratic (Alt scheme: M1 for attempt to differentiate quadratic and equate to 5 DM1 for simplification and solution using resulting quadratic (ii) leading to x = 2, y = 7 M1, A1 [2] M1 for obtaining x and y coords www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2009 0606 01 © UCLES 2009 5 (a) ( ) 2 2 1 3 3 3 x x −= 4 2 3 x x − = 2 x = (b) 2 a b − or 2 b a (allow here) 1 2, = − = q p B1 B1 B1 [3] B1 B1 [2] B1 for ( ) 2 2 1 3 x− B1 for 33 x B1 for 2 x = B1 for each 6 f (3), f ( 5) or f (0.5) 0 − = spotted Either ( )( ) 15 2 1 2 2 − + − x x x Or ( )( ) 3 7 2 5 2 + − + x x x Or ( )( ) 5 9 2 3 2 − + − x x x B1 M1 A1 M1 B1 for spotting one root M1 for attempt to obtain quadratic factor A1 all correct M1 for solution of quadratic x = 3, –5, 0.5 A2,1,0 [6] A2 for all 3 solutions (–1 each error) Correct factors only – lose 1 A mark 7 (i) 3 3 3 3 e e e x x x x + − = 3 3 e x x M1, A1, B1 [3] M1 for attempt to differentiate a product. A1 for correct product. B1 for 3e x − (ii) 3 3 3 1 e e d e 3 3 x x x x x x   = −     ∫ DM1 DM1 A1 [3] DM1 for recognition of the ‘reverse’ to (i) DM1 for dealing with ‘3’ A1 all correct (condone omission of c) 8 (i) ( ) ( ) ( ) 2 2 2 9 2 2 2 9 d d + − + = x x x x x y ( ) 2 2 2 9 2 18 + − = x x , turning points, x = ± 3 B2,1,0 M1 A1 [4] Attempt to differentiate a quotient –1 each error M1 for correct attempt to find the turning points. A1 for both (ii) d 2 d x t = d 16 2 d 100 y t   = ×    = 0.32 or 8 25 B1 M1 A1 [3] B1 for use of d 2 d x t = M1 for use of rates of change www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2009 0606 01 © UCLES 2009 9 (i) 1 1 10 2 10 10 2 2   + = +     i j i j M1 A1 [2] M1 for attempt at a correct direction vector A1 all correct (ii) ( ) ( ) 4 8 20 20 16 28 − + + + = + i j i j i j M1 A1 [2] M1 for valid attempt A1 all correct (iii) ( ) ( ) 10 10 8 6 2 4 + − + = + i j i j i j M1 A1 [2] M1 for attempt at vector difference A1 condone negative (iv) displacement of ( ) ( ) 19 34 16 28 3 6 + − + = + i j i j i j time =1330 hours (accept 1.5 hours) at 31 43 + i j Alternative scheme: ( ) ( ) ( ) ( ) 19 34 8 6 16 28 10 10 t t + + + = + + + i j i j i j i j or equivalent M1 A1 A1 [3] M1 for displacement and attempt to obtain time A1 for correct time A1 for correct position vector M1 for attempt to equate like vectors A marks as above 10 (i) mAB = 0.75 line AB ) 4 ( 75 .0 0 + = − x y mPQ = 3 4 − line PQ ( ) 4 10 1 3 y x − = − − intersection at C (4, 6) Q (8.5 0) (ii) AC = 10, CQ = 7.5 Area = 37.5 M1 A1 M1 A1 M1 A1 √ B1 [7] M1 A1 [2] M1 for attempt at mAB and line AB M1 for use of ‘m1m2 = –1’ and attempt at line PQ M1 for attempt at solving simultaneous equations Ft on their line PQ M1 for attempt at lengths and area www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2009 0606 01 © UCLES 2009 11 (i) k t n s ln ln ln + = ln t 1.6 2.7 3.4 4.2 4.6 ln s 7.2 5.9 5 4 3.6 Plot s ln against t ln (ii) grad n = –1.2 (–1.4 to –1.0) Intercept = ln k, leading to k = 7900 – 10 000 (iii) when t = 50, ln t = 4.4 leading to s = 80 (72 – 92) Alternative method (i) k t n s lg lg lg + = lg t 0.7 1.2 1.5 1.8 2 lg s 3.1 2.5 2.2 1.7 1.6 M1, A1 M1 A1 [4] M1, A1 M1, A1 [4] M1 A1 [2] M1for attempt to take logs A1 for correct form M1 for attempt to plot correct graph A1 for a reasonable straight line M1 for use of grad = n M1 for use of intercept = ln k M1 for attempt to obtain s Same scheme applies 1 2 3 4 5 1 2 3 4 5 6 7 8 9 10 ln t ln s www.theallpapers.com

Page 8 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2009 0606 01 © UCLES 2009 12 EITHER (i) amplitude = 1 (ii) period = 6π, 18.8 (iii) 2 1 3 sin =      x , 2 5 , 2 π π = x (iv) Area under curve x x d 3 sin 1 2 π 5 2 π∫    + = 2 5π 2 π 3 cos 3    − x x B1 [1] B1 [1] M1 A1, A1 [3] M1 B1, B1 M1 for attempt to solve correctly A1 for each (allow degrees here) M1 for attempt to integrate B1 for x, B1 for 3cos 3 x − leading to 2π + 3 3 Area of rectangle = 2 3 2 π 2 π 5 ×       − = 3π Shaded area = 3 3 – π (2.05) DM1 M1 A1 [6] DM1 for correct use of limits M1 for attempt at rectangle plus subtraction – must be working in radians Alternative solution: Shaded area x x d 5.0 3 sin 2 π 5 2 π∫     − = 5π 2 π 2 0.5 3cos 3 x x − −       M1 M1 B1, B1 DM1, A1 M1 for subtraction (must be using radians) M1 for attempt to integrate B1 for 0.5x − , B1 for 3cos 3 x − DM1 for correct use of limits www.theallpapers.com

Page 9 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2009 0606 01 © UCLES 2009 OR (i) π 8 t = (ii) 4 sin 4 a k t = − (iii) 3π 12 4 sin 2 k = − leading to 3 k = (iv) 1 2 3 –4 –3 –2 –1 1 2 3 4 t v (v) ∫ = 24 π 0 d. 4 cos 3 t t s π 24 0 3 sin 4 4 t   =     leading to 3 8 B1 [1] M1, A1 [2] M1 A1 [2] B1 √ B1 [2] M1, √ A1 DM1, A1 [4] M1 for attempt to differentiate M1 for attempt to substitute into their acceleration equation B1 for correct shape B1 ft on their value for k M1 for attempt to integrate Ft on their value for k DM1 for application of limits or equivalent www.theallpapers.com