Cambridge IGCSE Mathematics - Additional 0606 — 2012 May/June Paper 1 · Variant 3

This document consists of 16 printed pages. DC (SLM) 57492 © UCLES 2012 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 1 2 3 7 3 0 4 5 4 5 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 May/June 2012 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 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. For Examiner’s Use 1 2 3 4 5 6 7 8 9 10 11 Total www.theallpapers.com

2 0606/13/M/J/12 © UCLES 2012 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 0606/13/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use 1 (i) Sketch the graph of y = 2x – 5, showing the coordinates of the points where the graph meets the coordinate axes. [2] y x O (ii) Solve 2x – 5 = 3 . [2] www.theallpapers.com

4 0606/13/M/J/12 © UCLES 2012 For Examiner’s Use 2 The expression 2x3 + ax2 + bx – 30 is divisible by x + 2 and leaves a remainder of –35 when divided by 2x – 1. Find the values of the constants a and b. [5] www.theallpapers.com

5 0606/13/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use 3 Find the set of values of k for which the line y = 2x + k cuts the curve y = x2 + kx + 5 at two distinct points. [6] www.theallpapers.com

6 0606/13/M/J/12 © UCLES 2012 For Examiner’s Use 4 (a) Arrangements containing 5 different letters from the word AMPLITUDE are to be made. Find (i) the number of 5-letter arrangements if there are no restrictions, [1] (ii) the number of 5-letter arrangements which start with the letter A and end with the letter E. [1] (b) Tickets for a concert are given out randomly to a class containing 20 students. No student is given more than one ticket. There are 15 tickets. (i) Find the number of ways in which this can be done. [1] www.theallpapers.com

7 0606/13/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use There are 12 boys and 8 girls in the class. Find the number of different ways in which (ii) 10 boys and 5 girls get tickets, [3] (iii) all the boys get tickets. [1] www.theallpapers.com

8 0606/13/M/J/12 © UCLES 2012 For Examiner’s Use 5 (i) Find the equation of the tangent to the curve y = x3 + 2x2 – 3x + 4 at the point where the curve crosses the y-axis. [4] (ii) Find the coordinates of the point where this tangent meets the curve again. [3] www.theallpapers.com

9 0606/13/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use 6 (i) Given that 15cos2 θ + 2sin2 θ = 7, show that tan2 θ = 8 5. [4] (ii) Solve 15cos2 θ + 2sin2 θ = 7 for 0  θ  π radians. [3] www.theallpapers.com

10 0606/13/M/J/12 © UCLES 2012 For Examiner’s Use 7 The table shows values of variables x and y. x 1 3 6 10 14 y 2.5 4.5 0 –20 –56 (i) By plotting a suitable straight line graph, show that y and x are related by the equation y = Ax + Bx2, where A and B are constants. [4] www.theallpapers.com

11 0606/13/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use (ii) Use your graph to find the value of A and of B. [4] 8 (a) Find the value of x for which 2lg x – lg(5x + 60) = 1 . [5] (b) Solve log5 y = 4logy 5 . [4] www.theallpapers.com

12 0606/13/M/J/12 © UCLES 2012 For Examiner’s Use 9 Find the values of the positive constants p and q such that, in the binomial expansion of ( p + qx)10, the coefficient of x5 is 252 and the coefficient of x3 is 6 times the coefficient of x2. [8] www.theallpapers.com

13 0606/13/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use 10 Variables x and y are such that y = e2x + e–2x . (i) Find dy dx. [2] (ii) By using the substitution u = e2x , find the value of y when dy dx = 3. [4] (iii) Given that x is decreasing at the rate of 0.5 units s–1, find the corresponding rate of change of y when x = 1. [3] www.theallpapers.com

14 0606/13/M/J/12 © UCLES 2012 For Examiner’s Use Answer only one of the following two alternatives. 11 EITHER The diagram shows part of the curve y = 9x2 – x3, which meets the x-axis at the origin O and at the point A. The line y – 2x + 18 = 0 passes through A and meets the y-axis at the point B. y O B A x y = 9x2 – x3 y – 2x + 18 = 0 NOT TO SCALE (i) Show that, for x  0, 9x2 – x3  108. [4] (ii) Find the area of the shaded region bounded by the curve, the line AB and the y-axis. [6] OR The diagram shows part of the curve y = 2sin 3x . The normal to the curve y = 2sin 3x at the point where x = π 9 meets the y-axis at the point P. y O P π 9 x y = 2sin 3x NOT TO SCALE (i) Find the coordinates of P. [5] (ii) Find the area of the shaded region bounded by the curve, the normal and the y-axis. [5] www.theallpapers.com

15 0606/13/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use Start your answer to Question 11 here. Indicate which question you are answering. EITHER OR … … … … … … … … … … … … … … … … … … … … … … … … … www.theallpapers.com

16 0606/13/M/J/12 © UCLES 2012 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 2012 question paper for the guidance of teachers 0606 ADDITIONAL MATHEMATICS 0606/13 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 2012 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 2012 0606 13 © University of Cambridge International Examinations 2012 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 2012 0606 13 © University of Cambridge International Examinations 2012 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 2012 0606 13 © University of Cambridge International Examinations 2012 1 (i) −2 −1 1 2 3 4 −1 1 2 3 4 5 6 x y B1 B1 B1 for shape B1 for both intercepts (ii) 2x – 5 = ±3, leading to x = 1, 4 B1, B1 [4] 2 f (–2) : 4a – 2b = 46 1 f : 2 21 2 a b   + = −     a = 5, b = –13 M1 M1 M1 A1 A1 [5] M1 for substitution of x = –2 and equating to zero M1 for substitution of x = 0.5 and equating to – 35 M1 for solution of equations 3 x2 + x(k – 2) + (5 – k) = 0 Using 'b2 > 4ac', (k – 22) > 4(5 – k) k2 > 16 k > 4, k < –4 M1 DM1 A1 A1 A1, A1 [6] M1 for equating line and curve DM1 for use of b2 > 4ac b = k –2 and c = 5 – k Accept < = ≥ ≤ etc. A1 for each 4 (a) (i) 15120 (ii) 210 (b) (i) 15504 (ii) 12C10 × 8C5 = 3696 (iii) 56 B1 B1 B1 B1, B1 B1 B1 [7] 5 (i) (0, 4) 3 4 3 d d 2 − + = x x x y When x = 0, 3 d d − = x y y – 4 = –3x B1 M1 M1 A1 M1 for differentiation M1 for attempt at line equation (ii) 4 – 3x = x3 + 2x2 – 3x + 4 leading to, 0 = x3 + 2x2, (–2, 10) M1 M1 A1 [7] M1 for equating line and curve M1 for solution of cubic A1 need x and y www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2012 0606 13 © University of Cambridge International Examinations 2012 6 (i) θ θ θ 2 2 2 cos 7 cos sin 2 15 = + 15 + 2tan2 θ = 7 sec2 θ 15 + 2tan2 θ = 7(1 + tan2 θ) leading to 5 8 tan2 = θ or 15cos2 θ + 2sin2 θ = 7(cos2 θ + sin2 θ) 8cos2 θ = 5sin2 θ leading to 5 8 tan2 = θ M1 M1 M1 A1 [M1] [M1] [M1] [A1] M1 for dividing by cos2 θ M1 for θ θ 2 2 sec cos 1 = M1 for sec2 θ = 1 + tan2 θ A1 for rearrangement to get required result M1 for use of identity M1 for simplification M1 for use of θ θ θ cos sin tan = (ii) 5 8 tan ± = θ leading to θ = 0.902, 2.24 (also, sin θ = ± , 13 8 cos θ = ± 13 5 ) M1 A1, A1 [M1] [7] M1 for attempt to solve M1 for attempt to solve 7 (i) Bx A x y + = x 1 3 6 10 14 y/x 2.5 1.5 0 –2 –4 2 4 6 8 10 12 14 16 −5 −4 −3 −2 −1 1 2 3 4 x y (ii) Grad = B = –0.5 Intercept = A = 3 B1 M1 A2, 1, 0 M1, A1 M1, A1 [8] M1 for attempt to calculate points and plot –1 each error M1 for grad = B M1 for intercept = A 8 (a) 2lg x – lg(5x + 60) = lg10 10 60 5 2 = + x x leading to x2 – 50x – 600 = 0 x = 60 B1 B1 M1 DM1 A1 B1 for lg 10 B1 for dealing with ‘power’ M1 for dealing with division DM1 for attempt to solve quadratic (b) log5 y = y 5 5 log 5 log 4 (log5 y)2 = 4 log5 y = ±2 y = 25, 25 1 M1 M1 A1,A1 [9] M1 for change of base M1 for valid attempt to solve www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2012 0606 13 © University of Cambridge International Examinations 2012 9 120(p7q3) and (45)(p8q2) 120p7q3 = 270p8q2 252p5q5 = 252 pq = 1 and 4q = 9p leading to 2 3 , 3 2 = = q p B3,2,1,0 M1 B1 B1 A1, A1 [8] –1 per element (of 4) incorrect M1 for equating and multiplying by 6 B1 for 252p5q5 B1 for pq = 1 OR 4p = 9q A1 for each 10 (i) x x x y 2 2 e 2 e 2 d d − − = B1,B1 One per term (ii) 3 = 2e2x – 2e–2x 2e4x – 3e2x – 2 = 0 (2e2x + 1)(e2x – 2) = 0 e2x = 2, y = 2 5 M1 DM1 M1 A1 M1 for attempt to obtain in ‘quadratic’ form DM1 for attempt to solve M1 for attempt to solve for y (iii) ,5.0 d d − = t x ) 5.0 ( ) 2 2 ( d d 2 2 − × − = − e e t y = –7.25 M1, M1 A1 [9] M1 for substitution of x = 1 M1 for correct application of chain rule 11 EITHER (i) 2 3 18 d d x x x y − = When ) 6 ( 3 0 ,0 d d x x x y − = = Turning points when x = 0, 6 When x = 6, y = 108 M1 M1 M1 A1 M1 for differentiation M1 for equating to zero and attempt to solve M1 for finding y (ii) Area 9 0 4 3 4 3       −x x = 546.75 B(0, –18) Area of triangle = 81 Total Area = 628 Or: Area = ∫ + − − 9 0 3 2 18 2 9 x x x dx 628 18 4 3 9 0 2 4 3 =       + − − x x x x M1, A1 DM1,A1 B1 A1 [M1] [A3,2,1,0] [DM1,A1] [10] M1 for attempt to integrate DM1 for correct application of limits B1 for area of triangle www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2012 0606 13 © University of Cambridge International Examinations 2012 11 OR (i) x x y 3 cos 6 d d = When 3 ,3 d d , 9 = = = y x y x π Equation of normal      − − = − 9 3 1 3 π x y When x = 0, y = 1.85 B1 B1 M1 M1 A1 B1 for differentiation For y Use of m1m2 = –1 M1 for equation of normal and attempt to solve when x = 0 (ii) ( ) ∫ − + 9 0 d 3 sin 2 9 85 .1 3 2 1 π π x x 0.6251 9 0 3 cos 3 2 π    − − x 0.6251       −3 1 = 0.292 Alt method: Area = x x x d 3 sin 2 9 3 1 3 9 0 ∫ −      − − π π 292 .0 3 cos 3 2 27 6 3 9 0 2 =       + + − π π x x x x B1 M1 A1 M1, A1 [M1] [A2,1,0] [DM1] [A1] [10] B1 for trapezium – allow unsimplified M1 for attempt to integrate A1 correct integration M1 for correct application of limits www.theallpapers.com