Cambridge IGCSE Mathematics - Additional 0606 — 2011 May/June Paper 1 · Variant 1

This document consists of 16 printed pages. DC (SM/DJ) 42191/2 © UCLES 2011 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education 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 a pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, 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. * 6 0 6 1 0 6 0 0 4 0 * ADDITIONAL MATHEMATICS 0606/11 Paper 1 May/June 2011 2 hours Candidates answer on the Question Paper. Additional Materials: Electronic calculator For Examiner’s Use 1 2 3 4 5 6 7 8 9 10 11 12 Total www.theallpapers.com

2 0606/11/M/J/11 © UCLES 2011 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 © UCLES 2011 [Turn over For Examiner’s Use 0606/11/M/J/11 1 Show that 1 1 – cosθ + 1 1 + cosθ = 2cosec2θ. [3] 2 Express lg a + 3lg b – 3 as a single logarithm. [3] www.theallpapers.com

4 © UCLES 2011 For Examiner’s Use 0606/11/M/J/11 3 (a) Shade the region corresponding to the set given below each Venn diagram. A (A∪B)∩Cʹ B C A (A∪B∪C)ʹ B C A (A∩B)∪(B∩C)∪(C∩A) B C    [3] (b) Given that P = {p : tan p = 1 for 0º  p  540º}, find n(P). [1] 4 (a) Solve the equation 163x–2 = 82x. [3] www.theallpapers.com

5 © UCLES 2011 [Turn over For Examiner’s Use 0606/11/M/J/11 (b) Given that a b 4 3 2 5 2 5 2 5 – – a b 1 3 3 5 = apbq, find the value of p and of q. [2] 5 (i) y 4 5 6 3 2 1 0 45° 90° 135° 180° –1 –2 –3 –4 x On the diagram above, sketch the curve y = 1 + 3sin2x for 0º  x  180º. [3] (ii) y 4 5 6 3 2 1 0 45° 90° 135° 180° –1 –2 –3 –4 x On the diagram above, sketch the curve y = 1 + 3sin2x for 0º  x  180º. [1] (iii) Write down the number of solutions of the equation 1 + 3sin2x = 1 for 0º  x  180º. [1] www.theallpapers.com

6 © UCLES 2011 For Examiner’s Use 0606/11/M/J/11 6 The curves y = x2 and 3y = –2x2 + 20x – 20 meet at the point A. y x A O y = x2 3y = –2x2 + 20x – 20 (i) Show that the x-coordinate of A is 2. [1] (ii) Show that the gradients of the two curves are equal at A. [3] (iii) Find the equation of the tangent to the curves at A. [1] www.theallpapers.com

7 © UCLES 2011 [Turn over For Examiner’s Use 0606/11/M/J/11 7 The points A and B have coordinates (–2, 15) and (3, 5) respectively. The perpendicular to the line AB at the point A (–2, 15) crosses the y-axis at the point C. Find the area of the triangle ABC. [6] www.theallpapers.com

8 © UCLES 2011 For Examiner’s Use 0606/11/M/J/11 8 (a) The matrices A, B and C are given by A =  2 1 1 3 2 5, B =  2 1 3 4 1 5 6 7 and C =  9 10. Write down, but do not evaluate, matrix products which may be calculated from the matrices A, B and C. [2] (b) Given that X =  2 4 3 5 and Y =  2x 3y x 4y, find the value of x and of y such that X-1Y =  –12x + 3y 6 –7x + 3y 6. [6] www.theallpapers.com

9 © UCLES 2011 [Turn over For Examiner’s Use 0606/11/M/J/11 9 A body moves in a straight line such that, t s after passing through a fixed point O, its displacement from O is s m. The velocity v ms–1 of the body is such that v = 5cos4t. (i) Write down the velocity of the body as it passes through O. [1] (ii) Find the value of t when the acceleration of the body is first equal to 10 ms–2. [4] (iii) Find the value of s when t = 5. [4] www.theallpapers.com

10 © UCLES 2011 For Examiner’s Use 0606/11/M/J/11 10 (a) A curve is such that dy dx = ae1–x – 3x2, where a is a constant. At the point (1, 4), the gradient of the curve is 2. (i) Find the value of a. [1] (ii) Find the equation of the curve. [5] www.theallpapers.com

11 © UCLES 2011 [Turn over For Examiner’s Use 0606/11/M/J/11 (b) (i) Find (7x + 8) 1 3dx. [2] (ii) Hence evaluate 8 0 (7x + 8) 1 3dx. [2] www.theallpapers.com

12 © UCLES 2011 For Examiner’s Use 0606/11/M/J/11 11 (a) The function f is such that f(x) = 2x2 – 8x + 5. (i) Show that f(x) = 2(x + a)2 + b, where a and b are to be found. [2] (ii) Hence, or otherwise, write down a suitable domain for f so that f–1 exists. [1] (b) The functions g and h are defined respectively by g(x) = x2 + 4, x  0, h(x) = 4x – 25, x  0. (i) Write down the range of g and of h–1. [2] www.theallpapers.com

13 © UCLES 2011 [Turn over For Examiner’s Use 0606/11/M/J/11 (ii) On the axes below, sketch the graphs of y = g(x) and y = g–1(x), showing the coordinates of any points where the curves meet the coordinate axes. [3] y x O (iii) Find the value of x for which gh(x) = 85. [4] www.theallpapers.com

14 © UCLES 2011 For Examiner’s Use 0606/11/M/J/11 12 Answer only one of the following two alternatives. EITHER The equation of a curve is y = (x – 1)(x2 – 6x + 2). (i) Find the x-coordinates of the stationary points on the curve and determine the nature of each of these stationary points. [6] (ii) Given that z = y2 and that z is increasing at the constant rate of 10 units per second, find the rate of change of y when x = 2. [2] (iii) Hence find the rate of change of x when x = 2. [2] OR The diagram shows a cuboid with a rectangular base of sides x cm and 2x cm. The height of the cuboid is y cm and its volume is 72 cm3. y cm x cm 2x cm (i) Show that the surface area A cm2 of the cuboid is given by A = 4x2 + 216 x . [3] (ii) Given that x can vary, find the dimensions of the cuboid when A is a minimum. [4] (iii) Given that x increases from 2 to 2 + p, where p is small, find, in terms of p, the corresponding approximate change in A, stating whether this change is an increase or a decrease. [3] www.theallpapers.com

15 © UCLES 2011 For Examiner’s Use 0606/11/M/J/11 Start your answer to Question 12 here. Indicate which question you are answering. EITHER OR … … … … … … … … … … … … … … … … … … … … … … … … … www.theallpapers.com

16 © UCLES 2011 0606/11/M/J/11 For Examiner’s Use 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. Continue your answer here if necessary. … … … … … … … … … … … … … … … … … … … … … … … … … www.theallpapers.com

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2011 question paper for the guidance of teachers 0606 ADDITIONAL MATHEMATICS 0606/11 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. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2011 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 2011 0606 11 © University of Cambridge International Examinations 2011 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 2011 0606 11 © University of Cambridge International Examinations 2011 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 2011 0606 11 © University of Cambridge International Examinations 2011 1 ( ) ( ) ( )( ) 1 cos 1 cos 1 cos 1 cos θ θ θ θ + + − + − = 2 2 1 cos θ − = 2 2 sin θ = 2 cosec2 θ M1 M1 A1 [3] M1 for attempt to deal with fractions M1 for attempt at simplification and use of 2 2 1 cos sin θ θ − = in denominator 2 3 lg lg1000 ab − = 3 lg1000 ab B1B1 B1 [3] B1 for 3 lg ab , B1 for lg1000 3 (a) (i) (ii) (iii) (b) n(P) = 3 B1 B1 B1 B1 [4] B1 for each region shaded correctly 4 (a) Powers of 2: ( ) ( ) 4 3 2 3 2 x x − = or equivalent for powers of 4, 8 or 16 4 3 x = , allow 1.33 (b) p = 1, q = – 5 4 M1 A1 A1 B1B1 [5] M1 for powers of 2, 4, 8 or 16 A1 for all powers correct A C B www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 11 © University of Cambridge International Examinations 2011 5 (i) 45 90 135 180 −2 −1 1 2 3 4 x y (ii) 45 90 135 180 −2 −1 1 2 3 4 x y (iii) 5 B1 B1 B1 √B1 √B1 [5] B1 for shape B1 for 1 cycle between 4 and –2 B1 all correct √B1 for modulus of (i) √ on their graph 6 (i) 2 2 3 2 20 20 x x x = − + − and verification. Or ( ) 2 2 0 x − = , x = 2 (ii) for OA, d 2 d y x x = when x = 2, grad = 4 for other curve , d 4 20 d 3 3 y x x = − + when x = 2, grad = 4 Or 4 20 2 3 3 x x = − + leading to x = 2 (iii) tangent ( ) 4 4 2 y x − = − B1 B1 B1 M1 A1 M1 M1 A1 B1 [5] substitution of x = 2 B1 for solution of equation B1 for grad at A from OA M1 for attempt to differentiate the other curve and substitute x = 2 M1 for differentiation of both M1 for equating and attempt to solve www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 11 © University of Cambridge International Examinations 2011 7 Grad of AB = –2, perp grad = 1 2 Eqn of perp ( ) 1 15 2 2 y x − = + C(0, 16) Area = 1 125 5 2 = 12.5 (or ( ) 2 3 0 2 1 1 38 13 15 5 16 15 2 2 − − = − ) B1M1 M1 A1 M1 A1 [6] B1 for grad AB M1 use of 1 2 1 m m = − M1 for correct attempt to find the equation of AC and hence to find C M1 for a valid method to find area 8 (a) AB, AC (b) Either: Y = X       + − + − 6 3 7 6 3 12 y x y x       + − + − =       48 24 71 36 18 52 4 3 2 y x y x y x y x leading to y = 12 and x = 4 Or       − − − 2 3 4 5 12 10 1 Y =       + − + − 6 3 7 6 3 12 y x y x       + − + − =       − − − − 6 3 7 6 3 12 4 6 2 1 y x y x y x y x leading to y = 12 and x = 4 B2,1,0 M1 M1A1 M1 A1A1 B1 B1 M1 M1 A1A1 [8] –1 each one incorrect or extra M1 for pre-multiplying by X M1 for multiplication of matrices A1 for correct product M1 for equating like elements B1 for determinant for inverse B1 for ‘matrix part’ of inverse M1 for multiplication of matrices M1 for equating like elements 9 (i) 5 (ii) 20sin 4 a t = − sin 4 0.5 t = − 7π 24 t = ( allow 0.916) (iii) 5 sin 4 ( ) 4 s t c = + When t = 5, s = 1.14 B1 M1A1 DM1 A1 M1A1 DM1 A1 [9] M1 for attempt to differentiate DM1 for attempt to solve for 4t M1 for attempt to integrate DM1 for substitution of t in radians www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 11 © University of Cambridge International Examinations 2011 10 (a) (i) 2 3 a = − , a = 5 (ii) 1 3 5 x y e x c − = − − + 10 c = 1 3 5 10 x y e x − = − − + (b) (i) ( ) 4 3 1 3 7 8 7 4 x + (ii) ( ) 8 4 3 0 3 7 8 28 x   +     =180 7 or 25.7 B1 √B1 B1B1 M1 A1 B1B1 M1A1 [10] √B1for first term using their a B1 for 3x − , B1 for c + M1 for attempt to find c B1 for 1 7 , B1 for ( ) 4 3 3 7 8 4 x + M1 for use of limits 11 (i) ( ) 2 2 2 3 x − − (ii) 2 x ≥ or equivalent (b) (i) ( ) g 4 x ≥ , ( ) -1 h 0 x ≥ (ii) Correct sketch (iii) ( ) g 4 25 85 x − = ( ) 2 4 25 4 85 x − + = 17 2 x = , x = 4 Discarding x = 4 B1B1 √B1 B1B1 B1 B1 B1 M1 DM1 A1 B1 [12] B1 for –2, B1 for –3 √ on their ‘–2’ B1 for each B1 for g(x) B1 for g-1(x) B1 for idea of symmetry M1 for correct order DM1 for attempt to solve A1 for both B1 for discarding x = 4 www.theallpapers.com

Page 8 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 11 © University of Cambridge International Examinations 2011 12 EITHER (i) 2 d 3 14 8 d y x x x = − + When d 0 d y x = , 2 ,4 3 x = 2 2 d 6 14 d y x x = − , 2 3 x = max, 4 x = min (ii) Use of d d d d d d y y z t z t = × , leading to d d y t = 5 6 − allow –0.833 (iii) Use of d d d d d d y y x t x t = × leading to d 5 d 48 x t = M1A1 M1A1 M1 A1 M1 A1 M1 √A1 [10] M1 for attempt to differentiate M1 for attempt to equate to zero and solve M1 for attempt to differentiate (or other valid method) A1 correct from correct working for both M1 for attempt to use rates of change M1 for attempt to use rates of change ft on d d y t 12 OR (i) 2 2 72 x y = , 2 4 6 A x xy = + leading to given answer (ii) 2 d 216 8 d A x x x = − When d 0 d A x = , 3 27 x = = 3 Dimensions are 3 by 6 by 4 (iii) Use of d d A A x x ∂ ≈ ×∂ leading to 38 A p ∂ = − , decrease B1M1 A1 M1A1 M1 A1 M1 A1√A1 [10] B1 for 2 2 150 x y = , M1 for 2 4 6 A x xy = + M1 for attempt to differentiate M1 for attempt to equate to zero and solve A1 for dimensions M1 for attempt to use small changes A1 for – 38 p, √A1 on their A ∂ www.theallpapers.com