Cambridge IGCSE Mathematics - Additional 0606 — 2009 Oct/Nov Paper 2 · Variant 1

This document consists of 5 printed pages and 3 blank pages. DC (CW) 15823 © 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. * 3 9 9 3 0 3 2 0 0 0 * ADDITIONAL MATHEMATICS 0606/02 Paper 2 October/November 2009 2 hours Additional Materials: Answer Booklet/Paper Graph paper (2 sheets) Electronic calculator Mathematical tables www.theallpapers.com

2 0606/02/O/N/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 0606/02/O/N/09 © UCLES 2009 [Turn over 1 A function f is defined by f: x ฀e x–1, where x  0. (i) State the range of f. [1] (ii) Find an expression for f –1. [2] (iii) State the domain of f –1. [1] 2 (i) Find the first four terms, in ascending powers of x, in the expansion of 2 – x 2 6 . [4] (ii) Find the coefficient of x3 in the expansion of (l + x)2 2 – x 2 6 . [2] 3 The table shows experimental values of the variables x and y which are related by the equation y = a x2 + b x , where a and b are constants. x 2 4 6 8 10 y 6.24 2.82 1.79 1.33 1.05 (i) Using graph paper, plot x2y against x and draw a straight line graph. [3] (ii) Use your graph to estimate the value of a and of b. [4] 4 Find the coordinates and the nature of the stationary points of the curve y = x3 + 3x2 – 45x + 60. [7] 5 Relative to an origin O, the position vectors of points A and B are  7 24 and  10 20 respectively. Find (i) the length of OA, [2] (ii) the length of AB . [2] Given that ABC is a straight line and that the length of AC is equal to the length of OA, find (iii) the position vector of the point C. [3] 6 (i) Given that y = x x 4 +12 , show that dy dx = k x x ( + 2) 4 +12 , where k is a constant to be found. [4] (ii) Hence evaluate 6 –2 3 + 6 4 +12 x x dx . [3] www.theallpapers.com

4 0606/02/O/N/09 © UCLES 2009 7 (i) Using graph paper, draw the curve y = sin 2x for 0°  x ฀360 °. [3] In order to solve the equation 1 + sin 2x = 2cos x another curve must be added to your diagram. (ii) Write down the equation of this curve and add this curve to your diagram. [3] (iii) State the number of values of x which satisfy the equation 1 + sin 2x = 2cos x for 0 °  x ฀360 °. [1] 8 It is given that A =  2 –1 4 3, B =  1 –3 2 0 and C =  5 –2. Find (i) AB, [2] (ii) BC, [2] (iii) A–1, and hence find the matrix X such that AX = B. [4] 9 A particle moves in a straight line so that, t seconds after passing through a fixed point O, its velocity, v ms–1, is given by v = 20 (2t + 4)2 . Find (i) the velocity of the particle at O, [1] (ii) the acceleration of the particle when t = 3, [3] (iii) the distance travelled by the particle in the first 8 seconds. [4] 10 (a) Solve lg(7x – 3) + 2 lg5 = 2 + lg(x + 3). [4] (b) Use the substitution u = 3x to solve the equation 3x+1 + 32– x = 28 . [5] www.theallpapers.com

5 0606/02/O/N/09 © UCLES 2009 11 Answer only one of the following two alternatives. EITHER 3 cm 3 cm 3 cm Q B P D C A 3 cm π 3 _ In the diagram, ACB is an arc of a circle with centre P, and ADB is an arc of a circle with centre Q. Angle AQB = π 3, AQ = BQ = 3 cm and AP = BP = 3 cm. (i) Show that angle APB = 2π 3 . [2] (ii) Find the perimeter of the shaded region. [3] (iii) Find the area of the shaded region. [5] OR Solutions to this question by accurate drawing will not be accepted. A B(8, 1) (–2, 1) E F y x O C(6, 9) D The diagram shows a parallelogram with vertices A(–2, 1), B(8, 1), C(6, 9) and D. (i) Find the coordinates of D. [2] The point E lies on the diagonal DB such that DE = 1 4 DB . (ii) Find the coordinates of E. [2] The point F is such that EF is parallel to AB. The area of trapezium AEFB is 11 2  (the area of parallelogram ABCD). (iii) Find the coordinates of F. [6] www.theallpapers.com

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8 0606/02/O/N/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 October/November 2009 question paper for the guidance of teachers 0606 ADDITIONAL MATHEMATICS 0606/02 Paper 2, 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 October/November 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 – October/November 2009 0606 02 © 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 – October/November 2009 0606 02 © 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 – October/November 2009 0606 02 © UCLES 2009 1 (i) > e–1 or > 0.37 B1 (ii) Uses ln function properly M1 1 + ln x A1 (iii) > e–1 B1√ [4] 2 (i) 64 – 96x + 60x2 – 20x3 B1+B1+B1+B1 (ii) 1 × (–20) + 2 × (60) + 1 × (–96) M1 –20 + 120 – 96 = 4 A1 [6] 3 (i) Plots x2y against x with linear scale. M1 x 2 4 6 8 10 x2y 24.96 45.12 64.44 85.12 105 A2,1,0 (ii) x2y = bx + a B1 Calculates gradient M1 b = 10 ± 0.4 A1 a = 5 ± 2 from intercept or substitution B1 [7] (ii) Alternative last 3 marks Equates intercept to a(5 ± 2) B1 Uses a to find b M1 b = 10 ± 0.4 A1 4 45 6 3 d d 2 − +       = x x x y B2, 1, 0 Equates x y d d to 0 and solves 3 term quadratic M1 x = 3 and x = –5 A1 (3, –21) and (–5, 235) A1 Complete method for max/min M1 minimum when x = 3 and maximum when x = –5 A1 [7] 5 (i) 2 2 24 7 + M1 25 = OA A1 (ii)       − = 4 3 AB B1 5 = AB B1 www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2009 0606 02 © UCLES 2009 (iii)       − = = 20 15 5AB AC M1 AC OA OC + = used DM1       4 22 A1 [7] 6 (i) Uses product rule M1 2 1 ) 12 4 ( 4 2 1 12 4 − + × + + x x x A1 Expresses with common denominator M1 k = 6 A1 (ii) 12 4 3 + x x k M1 Uses limits of 6 and –2 in 12 4 + x Cx M1 20 A1√ [7] 7 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 (i) Attempt at sine curve M1 Correct position at multiples of 45° A2, 1,0 (ii) 2cos x – 1 B1 Attempt at cosine curve M1 (0, 1), (90, –1), (180, –3), (270, –1), (360, 1), A1 (iii) 2 B1√ [7] 8 (i) Matrix multiplication M1       − − 12 10 6 0 A1 (ii) Matrix multiplication M1       10 11 A1 www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2009 0606 02 © UCLES 2009 (iii) A–1 =       − 2 4 1 3 10 1 B1+B1 X = A–1B stated M1       − 12 0 9 5 10 1 A1 [8] 9 (i) 1.25 B1 (ii) 3) 4 2 ( d d + = = t k t v a M1 Substitutes 3 into t v d d M1 –0.08 A1 (iii) s = 4 2 d + = ∫ t k t v M1 4 2 10 + − t A1 Correct use of limits of 0 and 8 only on attempt at ∫ t vd or finds c from s = 0, t = 0 and substitutes t = 8 M1 2 A1 [8] 10 (a) 2 lg 5 = lg25 or lg52 B1 2 = lg 100 or lg102 B1 Uses rules of logs correctly (lg(175x – 75) = lg(100x + 300)) M1 5 A1 (b) Substitutes and express as equation in u M1 3u2 – 28u + 9 = 0 A1 Solves 3 term quadratic M1 u = 3 1 and 9 A1 x = –1 and 2 A1 [9] www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2009 0606 02 © UCLES 2009 11 EITHER (i) AB = 3 or 3 6 sin 3 sin π = ∠APQ B1 Correct use of trigonometry to APB = 3 2π B1 (ii) Uses s = rθ M1 3.14 (π) or 3.63       3 3 2 π A1 6.77       + 3 3 2 π π A1 (iii) Uses θ 2 2 1 r or rs 2 1 M1 Uses 2 2 1 r sin θ or area kite M1 Either 4.71 (1.5π) and 3.14 (π), or 3.90       4 3 9 and 1.30       4 3 3 or 5.20 ( ) 3 3 A1 Complete plan DM1 2.65 to 2.66 ( ) 3 3 5.2 − π A1 [10] OR (i) Method for D M1 (–4, 9) A1 (ii) Method for E M1 (–1,7) A1 (iii) Finds area parallelogram (= 80) M1 Area trapezium = 120 A1 Height trapezium = 6 B1 Uses Area = ½ × (6) × (AB + EF) M1 EF = 30 A1 F (29, 7) A1 [10] (iii) alternative last 4 marks Array method complete (with only one variable) M1 F (k, 7) A1 3k + 33 = 120 oe A1 F (29, 7) A1 www.theallpapers.com