Cambridge IGCSE Mathematics - Additional 0606 — 2011 Oct/Nov Paper 1 · Variant 3

This document consists of 16 printed pages. DC (NF/CGW) 48825 © UCLES 2011 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 8 2 0 8 2 8 1 3 1 1 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2011 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. Answer all 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 12 Total www.theallpapers.com

2 0606/13/O/N/11 © UCLES 2011 For Examiner’s Use 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/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 1 Given that 6x 3 2y 4 5 4 2x 1 2y–1 = axp yq, find the values of the constants a, p and q. [3] 2 Express  in the form k cos θ, where k is a constant to be found. [4] 1 – cos2 θ 4 sec2 θ – 4 www.theallpapers.com

4 0606/13/O/N/11 © UCLES 2011 For Examiner’s Use 3 (i) Given that A =  4 3 – 8 –2 , find A–1. [2] (ii) Hence find the matrix M such that  4 3 – 8 –2 M =  1 4 2 3 . [3] www.theallpapers.com

5 0606/13/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 4 (a) Sets A and B are such that n(A) = 11, n(B) = 13 and n(A B) = 18. Find n(A B). [2] (b) Sets , X and Y are such that  = {θ:0  θ  2π}, X = {θ:sin θ = – 0.5}, Y = θ:sec2 θ = 4 3 . (i) Find, in terms of π, the elements of the set X. [1] (ii) Find, in terms of π, the elements of the set Y. [2] (iii) Use set notation to describe the relationship between the sets X and Y. [1] www.theallpapers.com

6 0606/13/O/N/11 © UCLES 2011 For Examiner’s Use 5 It is given that lg p3 q = 10a and lg p q2 = a. (i) Find, in terms of a, expressions for lg p and lg q. [5] (ii) Find the value of logp q. [1] www.theallpapers.com

7 0606/13/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 6 A curve has equation y = 6 cos x 2 + 4 sin x 2 , for 0  x  2π radians. (i) Find the x-coordinate of the stationary point on the curve. [5] (ii) Determine the nature of this stationary point. [2] www.theallpapers.com

8 0606/13/O/N/11 © UCLES 2011 For Examiner’s Use 7 x y B C A O M (1,3) (– 4,2) The figure shows a right-angled triangle ABC, where the point A has coordinates (– 4,2) , the angle B is 90° and the point C lies on the x-axis. The point M(1,3) is the midpoint of AB. Find the area of the triangle ABC. [7] www.theallpapers.com

9 0606/13/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 8 Vectors a and b are such that a =  3 + m 5 – 2n and b =  4 – 2n 10 + 3m . (i) Given that 3a + b =  1 + n –5  , find the value of m and of n. [4] (ii) Show that the magnitude of b is k √⎯5, where k is an integer to be found. [2] (iii) Find the unit vector in the direction of b. [1] www.theallpapers.com

10 0606/13/O/N/11 © UCLES 2011 For Examiner’s Use 9 The function f is defined, for 0°  x  360°, by f(x) = 2 sin 3x –1. (i) State the amplitude and period of f. [2] (ii) State the maximum value of f and the corresponding values of x. [3] (iii) Sketch the graph of f. [2] www.theallpapers.com

11 0606/13/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 10 (a) Differentiate tan (3x + 2) with respect to x. [2] (b) Differentiate (√⎯x + 1) 2 3 with respect to x. [3] (c) Differentiate ln (x3 – 1) 2x + 3 with respect to x. [3] www.theallpapers.com

12 0606/13/O/N/11 © UCLES 2011 For Examiner’s Use 11 A particle moves in a straight line so that, t s after leaving a fixed point O, its velocity v ms–1 is given by v = 3e2t + 4t. (i) Find the initial velocity of the particle. [1] (ii) Find the initial acceleration of the particle. [3] www.theallpapers.com

13 0606/13/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use (iii) Find the distance travelled by the particle in the third second. [4] www.theallpapers.com

14 0606/13/O/N/11 © UCLES 2011 For Examiner’s Use 12 Answer only one of the following two alternatives. EITHER A function f is such that f(x) = ln (5x – 10), for x  2. (i) State the range of f. [1] (ii) Find f –1 (x). [3] (iii) State the range of f –1. [1] (iv) Solve f(x) = 0. [2] A function g is such that g(x) = 2x – ln 2, for x  . (v) Solve gf(x) = f(x2). [5] OR A function f is such that f(x) = 4e–x + 2, for x  . (i) State the range of f. [1] (ii) Solve f(x) = 26. [2] (iii) Find f –1(x). [3] (iv) State the domain of f –1. [1] A function g is such that g(x) = 2ex – 4, for x  . (v) Using the substitution t = ex or otherwise, solve g(x) = f(x). [5] Start your answer to Question 12 here. Indicate which question you are answering. EITHER OR … … … … … … … … www.theallpapers.com

15 0606/13/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use Continue your answer to Question 12 here. … … … … … … … … … … … … … … … … … … … … … … … … … … … www.theallpapers.com

16 0606/13/O/N/11 © UCLES 2011 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 October/November 2011 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 October/November 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 – October/November 2011 0606 13 © 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 – October/November 2011 0606 13 © 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 – October/November 2011 0606 13 © University of Cambridge International Examinations 2011 1 5 21 2 11 648 y x B1 B1 B1 [3] B1 for each correct 2 θ θ 2 2 tan 4 sin B1 B1 B1 for numerator B1 for denominator = θ θ tan 2 sin M1 M1 for rearrangement = 2 cosθ , k = 0.5 A1 [4] OR 4 cos 4 cos 1 2 2 − − θ θ B1 = θ θ θ 2 2 2 cos cos 4 4 cos 1 − − B1 = 2 cosθ M1 A1 3 (i) A–1 =       − − 4 8 3 2 16 1 B2, 1, 0 –1 each error (ii) M =             − − 3 2 4 1 4 8 2 2 16 1 M1 DM1 M1 for pre-multiplication DM1 for attempt to multiply matrices, at least one element correct =       − − 44 16 17 8 16 1 A1 [5] A1 all correct 4 (a) 11 – x + x + 13 – x = 18 x = 6 M1 A1 M1 for a valid method (b) (i) X : 6 7π , 6 11π B1 B1 for both (ii) Y: 6 π , 6 5π , 6 7π , 6 11π B1, B1 B1 for each pair (iii) Y X ⊂ , Y X ∩ = X or Y X ∪ = Y B1 [6] www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 13 © University of Cambridge International Examinations 2011 5 (i) lg p3 + lg q = 10a or lg p – lg q2 = a 3lg p + lg q = 10a lg p – 2lg q = a lg p = 3lg q leading to log p = 3a and log q = a M1 A1 A1 M1 A1 M1 for attempt to simplify logs as shown A1 for each of any 2 correct M1 for attempt to solve simultaneous equations A1 for both (ii) 3 1 log log = p q √B1 [6] ft on their log p and log q, both need to be simple functions of a 6 (i) 2 cos 2 2 sin 3 d d x x x y + − = M1 A1, A1 M1 for attempt to differentiate A1 for each correct term When 0 d d = x y , 3 2 2 tan = x M1 M1 for their 0 d d = x y x = 1.18 A1 A1 correct solution only (ii) 2 sin 2 cos 2 3 d d 2 2 x x x y − − = M1 M1 for a valid method – needs to be seen, When x = 1.18 2 2 d d x y is –ve (–1.8) Maximum A1 [7] 7 B (6, 4) B1 grad AM = 5 1 x grad BC = –5 M1 M1 for attempt at gradient of BC BC equation: y – 4 = –5(x – 6) M1 A1 M1 for attempt at straight line equation A1 for correct equation in any form When y = 0, x = 6.8 √B1 Ft on their BC equation Area = 20.8 M1,A1 [7] M1 for a correct method for area of triangle www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 13 © University of Cambridge International Examinations 2011 8 (i) 12 + 3m = 3n 30 + 3m = 6n leading to m = 2, n = 6 M1 A1 M1, A1 M1 for equating like vectors A1 for both correct equations M1 for solution of equations (ii) b =      − 16 8 , b = 5 8 M1, A1 M1 for attempt at magnitude (iii)      − 16 8 5 8 1 √B1 [7] Ft on their b and its magnitude Allow decimals 9 (i) Amplitude = 2 Period = 120º B1 B1 (ii) Max Value = 1 Occurs when sin3x = 1 x = 30º, 150º, 270º B1 B1,B1 (iii) Sketch B1 B1 [7] B1 one cycle correct – ignore x values B1 all correct 10 (a) 3sec2(3x + 2) B1, B1 B1 for 3, B1 for 3sec2(3x + 2) (b) 3 1 2 1 2 1 1 2 1 3 2 − −         + × x x B1 B1 B1 B1 for 3 2 , B1 for 2 1 2 1 − x B1 for 3 1 2 1 1 −         + x (c) ( )( ) ( ) ( )2 3 3 2 3 2 1 ln 2 1 3 3 2 + − − − − x x x x x M1, B1 A1 [8] M1 for differentiation of a quotient B1 for differentiation of ln term A1 for everything else correct www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 13 © University of Cambridge International Examinations 2011 11 (i) 3 B1 (ii) a = 6e2t + 4 When t = 0, a = 10 M1, A1 A1 M1 for attempt at differentiation (iii) ( ) c t x t + + = 2 2 2 e 2 3 M1, A1 M1 for attempt at integration x3 = 623.14 + c, x2 = 89.90 + c or 621.64 or 88.40 Distance travelled = 533 M1 A1 [8] M1 for attempt to find difference between 3x and 2x – condone omission of or incorrect c 12 EITHER (i) or equivalent B1 (ii) 10 5 e − = x y , 5 10 e + y = x M1 M1 rearrangement to x in terms of y f–1(x) = 5 10 + x e DM1 A1 DM1 for interchange of x and y A1 for correct form (iii) f–1(x) > 2 or y > 2 B1 (iv) 1 = 5x – 10 x = 2.2 B1 B1 (v) g(ln(5x – 10)) = ln(5x2 – 10) 2ln(5x – 10) – ln2 = ln(5x2 –10) 25x2 – 100x – 100 = 10x2 –20 3x2 – 20x + 24 = 0, leading to x = 5.10 only M1 M1 A1 M1 A1 [12] M1 for correct order gf M1 for dealing with x2 correctly A1 correct quadratic– allow unsimplified M1 for correct attempt at solution of a 3 term quadratic A1 for valid solution only www.theallpapers.com

Page 8 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 13 © University of Cambridge International Examinations 2011 12 OR (i) f(x) > 2 B1 (ii) 26 = 4e–x + 2 B1 6 = e–x so x = –ln6, 6 1 ln or –1.79 B1 (iii) ( ) 4 2 − y = e–x, ( ) 4 2 ln − y = –x M1 M1 rearrangement to x in terms of y f–1(x) = 2 4 ln − x or 4 2 ln − − x M1 A1 M1 for interchange of x and y A1 for correct form (iv) f–1(x) or y > 2 B1 (v) 2ex – 4 = 4e–x + 2 (2t – 4 = 4t–1 + 2) e2x – 3ex – 2 = 0 (t2 – 3t – 2 = 0) ex = 3.56 so x = 1.27 M1 A1 M1 M1 A1 [12] M1 for attempt to deal with t–1 or e–x A1 for correct quadratic equation M1 for solution of quadratic M1 for correct attempt to obtain x A1 for 1 solution only www.theallpapers.com