Cambridge IGCSE Mathematics - Additional 0606 — 2013 May/June Paper 1 · Variant 2

This document consists of 17 printed pages and 3 blank pages. DC (LEO/SW) 57088/3 © UCLES 2013 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 6 1 1 1 6 3 9 0 6 5 * ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 2013 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. www.theallpapers.com

2 0606/12/M/J/13 © UCLES 2013 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 0606/12/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 1 A B The Venn diagram shows the universal set , the set A and the set B. Given that n(B ) = 5, n(A) = 10 and n( ) = 26, find (i) n( ) A B + , [1] (ii) n(A), [1] (iii) n( ) B A + l . [1] www.theallpapers.com

4 0606/12/M/J/13 © UCLES 2013 For Examiner’s Use 2 A 4-digit number is to be formed from the digits 1, 2, 5, 7, 8 and 9. Each digit may only be used once. Find the number of different 4-digit numbers that can be formed if (i) there are no restrictions, [1] (ii) the 4-digit numbers are divisible by 5, [2] (iii) the 4-digit numbers are divisible by 5 and are greater than 7000. [2] www.theallpapers.com

5 0606/12/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 3 Show that (1 – cos θ – sin θ )2 – 2(1 – sin θ )(1 – cos θ ) = 0. [3] www.theallpapers.com

6 0606/12/M/J/13 © UCLES 2013 For Examiner’s Use 4 Find the set of values of k for which the curve y = 2x2 + kx + 2k – 6 lies above the x-axis for all values of x. [4] www.theallpapers.com

7 0606/12/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 5 The line 3x + 4y = 15 cuts the curve 2xy = 9 at the points A and B. Find the length of the line AB. [6] www.theallpapers.com

8 0606/12/M/J/13 © UCLES 2013 For Examiner’s Use 6 The normal to the curve y + 2 = 3 tan x, at the point on the curve where x = 4 3r , cuts the y-axis at the point P. Find the coordinates of P. [6] www.theallpapers.com

9 0606/12/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 7 It is given that f(x) = 6x3 – 5x2 + ax + b has a factor of x + 2 and leaves a remainder of 27 when divided by x – 1. (i) Show that b = 40 and find the value of a. [4] (ii) Show that f(x) = (x + 2)(px2 + qx + r), where p, q and r are integers to be found. [2] (iii) Hence solve f(x) = 0. [2] www.theallpapers.com

10 0606/12/M/J/13 © UCLES 2013 For Examiner’s Use 8 (a) Given that the matrix A = 4 3 2 5 – c m, find (i) A2, [2] (ii) 3A + 4I, where I is the identity matrix. [2] www.theallpapers.com

11 0606/12/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (b) (i) Find the inverse matrix of 6 9 1 3 – c m. [2] (ii) Hence solve the equations 6x + y = 5, –9x + 3y = 2 3. [3] www.theallpapers.com

12 0606/12/M/J/13 © UCLES 2013 For Examiner’s Use 9 (i) Given that n is a positive integer, find the first 3 terms in the expansion of x 1 2 1 n + c m in ascending powers of x. [2] (ii) Given that the coefficient of x2 in the expansion of (1 – x) x 1 2 1 n + c m is 4 25, find the value of n. [5] www.theallpapers.com

13 0606/12/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 10 (a) (i) Find d x x 2 5 - y . [2] (ii) Hence evaluate d x x 2 5 3 15 - y . [2] www.theallpapers.com

14 0606/12/M/J/13 © UCLES 2013 For Examiner’s Use (b) (i) Find d d ln x x x 3 ^ h. [2] (ii) Hence find d ln x x x 2 y . [3] www.theallpapers.com

15 0606/12/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 11 (a) Solve cos2x + 2sec2x + 3 = 0 for x 0 360 G G c c. [5] (b) Solve sin y 6 2 2 r - ` j = 1 for 0 y G G r. [4] www.theallpapers.com

16 0606/12/M/J/13 © UCLES 2013 For Examiner’s Use 12 A particle P moves in a straight line such that, t s after leaving a point O, its velocity v m s–1 is given by v = 36t –3t2 for t H 0. (i) Find the value of t when the velocity of P stops increasing. [2] (ii) Find the value of t when P comes to instantaneous rest. [2] (iii) Find the distance of P from O when P is at instantaneous rest. [3] www.theallpapers.com

17 0606/12/M/J/13 © UCLES 2013 For Examiner’s Use (iv) Find the speed of P when P is again at O. [4] www.theallpapers.com

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20 0606/12/M/J/13 © UCLES 2013 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. BLANK PAGE www.theallpapers.com

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2013 series 0606 ADDITIONAL MATHEMATICS 0606/12 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. www.theallpapers.com

Page 2 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 12 © Cambridge International Examinations 2013 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 Syllabus Paper IGCSE – May/June 2013 0606 12 © Cambridge International Examinations 2013 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 Syllabus Paper IGCSE – May/June 2013 0606 12 © Cambridge International Examinations 2013 1 (i) n(A∩B) = 5 B1 (ii) n(A) = 16 B1 (iii) n (B´∩A) B1 2 (i) 6 × 5 × 4 × 3 = 360 or 6P4 = 360 B1 B1 unsimplified/evaluated (ii) Position 1 2 3 4 Number of ways 5 4 3 1 or 6 1 (i) or 5P3 or 5C3 × 6C1 Number of 4 digit numbers = 60 M1 A1 M1 for a correct attempt unsimplified (iii) Position 1 2 3 4 Number of ways 3 4 3 1 or 3P1 × 4P2 Number of 4 digit numbers = 36 M1 A1 M1 for a correct attempt unsimplified 3 EITHER 1 – 2sinθ – 2cosθ + sin2θ + cos2θ + 2sinθ cosθ B1 B1 for correct expansion of (1 – cosθ – sinθ )2 Use of sin2θ + cos2θ = 1 in simplification = 0 M1 A1 M1 for use of sin2θ + cos2θ = 1 in this form A1 must be convinced as AG OR (1 – cosθ – sinθ )2 = 1 – 2sinθ – 2cosθ + sin2θ + cos2θ + 2sinθ cosθ [B1 B1 for correct expansion of (1 – cosθ – sinθ )2 = 2 – 2sinθ – 2cosθ + 2sinθ cosθ M1 M1 for use of sin2θ + cos2θ = 1 in this form = 2 (1 – sinθ ) (1 – cosθ ) A1] A1 for simplification and factorising www.theallpapers.com

Page 5 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 12 © Cambridge International Examinations 2013 4 EITHER 2x2 + kx + 2k – 6 = 0 has no real roots k2 – 16k + 48 I 0 (k – 4) (k – 12) I 0 M1 DM1 M1 for attempted use of b2 – 4ac DM1 for attempt to obtain critical values from a 3 term quadratic Critical values 4 and 12 4 I k I 12 or k K=4 and k I=12 A1 A1 A1 for both critical values A1 for correct final answer OR 0 3 16 4 2 2 = − + −       + k k k x [M1] M1 for attempting to complete the square and obtain a 3 term quadratic 3 16 2 − + − k k K 0 so k2 – 16k + 48 I 0= Then as EITHER OR k x x y + = 4 d d When 0 d d = x y , x k 4 − = By substitution x2 + 4x + 3 I 0 leading to x = –1, k = 4 [M1 DM1 M1 for differentiation, equating to zero and obtaining a quadratic equation in x DM1 for attempt to obtain critical values of k from a 3 term quadratic in x followed by substitution to obtain a value for k and x = –3, k = 12 4 I k I 12 or k K 4 and k I 12 A1 A1] A1 for both critical values A1 for correct final answer OR k x x y + = 4 d d [M1] M1 for differentiation, equating to zero and obtaining a quadratic equation in k When 0 d d = x y , x = 4 k − leading to k2 – 16k + 48 I 0 Then as EITHER www.theallpapers.com

Page 6 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 12 © Cambridge International Examinations 2013 5 9 3 4 15 2 =       − y y or 9 4 3 15 2 =       −x x M1 M1 for attempt to obtain equation in one variable 8y2 – 30y + 27 = 0 or 3x2 – 15x + 18 = 0 (4y – 9) (2y – 3) = 0 or (x – 3) (x – 2) = 0 DM1 DM1 for attempt to solve a 3 term quadratic in that variable x = 2, y = 4 9 and x = 3, y = 2 3 A1, A1 A1 for each ‘pair’, x values must be simplified to single integer form AB2 = 12 + (0.75)2, AB = 1.25 M1, A1 M1 for a correct attempt to find AB, must have non zero differences and be using points calculated previously. 6 x y d d = 3sec2x When x = 4 3π , x y d d = 6 y = 5 B1 B1 B1 B1 for 3sec2x B1 for x y d d = 6, may be implied by later work B1 for y Perpendicular gradient = 6 1 − M1 M1 for perpendicular gradient from x y d d Equation of normal y + 5 =       − − 4 3 6 1 π x M1 M1 for attempt at the normal using their y value correctly and x = 4 3π and substitution of x = 0 When x = 0, y = 8 π – 5 o.e. or –4.61 or –4.6 but not –4.60 A1 A1 for obtaining y value www.theallpapers.com

Page 7 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 12 © Cambridge International Examinations 2013 7 (i) f (–2) leads to 68 = b – 2a M1 attempt at f (–2) = 0 allow unsimplified f(1) leads to 26 = a + b M1 attempt at f (1) = 27 allow unsimplified a = –14, b = 40 A1, B1 A1 for b = 40, B1 for a = –14 (ii) f (x) = (x + 2) (6x2 – 17x + 20) B2, 1, 0 –1 each error (iii) 6x2 – 17x + 20 = 0 has no real roots B1 B1 for dealing with quadratic factor either by use of formula, completing the square or use of b2 – 4ac to show that there are no real solutions x = –2 B1 8 (a) (i)       − − 31 3 2 22 B2, 1, 0 –1 each element error (ii)       −11 9 6 16 B2, 1, 0 –1 each element error (b) (i)       − + 6 9 1 3 9 18 1 B1, B1 B1 for 1 determinant (allow unsimplified), B1 for matrix (ii) ' . y x             − =       5 1 5 6 9 1 3 27 1 M1 M1 for correct use of inverse matrix, including correct multiplication to solve equation       = 54 5 13 27 1 . x = 0.5, y = 2 A1, A1 A1 for each www.theallpapers.com

Page 8 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 12 © Cambridge International Examinations 2013 9 (i) ( ) 2 2 2 1 2 1 2 1 1       − +       + =      + x n n x n x n B1, B1 B1 for 1 + second term, B1 for 3rd term Allow unsimplified (ii) ( ) ( )               − +       + − 2 2 2 1 2 1 1 x n n x n x M1 dealing with 2 terms involving x2 Multiply x and x n 2 to get ( ) 2 2 x n DM1 attempt to obtain one term Multiply 1 and ( ) 8 1 2 x n n − or ( ) 4 1 2 x n n − DM1 attempt to obtain a second term 4 25 2 8 2 = − − n n n n2 – 5n – 50 = 0 A1 correct quadratic equation n = 10 A1 A1 for n = 10 only 10 (a) (i) 3 1 (2x – 5) 2 3 B1, B1 B1 for k(2x – 5) 2 3 , B1 for ( )2 3 5 2 3 1 − x (ii) 3 124 3 1 3 125 = − Allow awrt 41.3 M1, A1 M1 for correct use of limits (b) (i) x x x x ln 3 1 2 3 + B1, B1 B1 for each term, allow unsimplified (ii) ∫ ∫ − = x x x x x x x d ln d ln 3 2 3 2 o.e. M1 for a use of answer to (i) ∫ = 3 d 3 2 x x x or A1 A1 for intergrating x2 or dividing by 3 ( ) ∫ ∫ − = x x x x x x x d ln 3 1 d ln 2 3 2 o.e. ∫       − = 3 ln 3 1 d ln 3 3 2 x x x x x x (+c) A1 www.theallpapers.com

Page 9 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 12 © Cambridge International Examinations 2013 11 (a) 0 3 2 cos 2 2 cos = + + x x M1 dealing with sec or cos leading to cos2 2x + 3 cos 2x + 2 = 0 2 sec2 2x + 3 sec 2x + 1 = 0 A1 simplification to correct 3 term quadratic in sec 2x or cos 2x (does not have to be equated to zero) (cos 2x + 2) (cos 2x + 1) = 0 or (2 sec 2x +1) (sec 2x + 1) = 0 M1 attempt to solve a 3 term quadratic, must obtain solutions in terms of cos 2x leading to cos 2x = –1 or sec 2x = –1 only 2x = 180°, 540° x = 90°, 270° A1, A1 (b) 2 1 6 π sin 2 =       − y so 2 1 6 π sin =       − y 4 π 3 , 4 π 6 π =       − y 12 π 11 , 12 5π = y Allow awrt 1.31, 2.88 M1 DM1 A1, A1 division by 2 and square root correct order of operation and attempt to solve 12 (i) t 6 36 d d − = t y M1 attempt to differentiate and equate to zero When ,0 d d = t y t = 6 A1 (ii) When v = 0, t = 12 M1, A1 M1 for equating v to zero and attempt to solve (iii) s = 18t2 – t3 (+c) When t = 12, s = 864 M1, A1 M1 for a correct attempt to integrate at least one term, allow unsimplified A1 for all correct A1 for s = 864 (iv) When s = 0, t = 18 v = –324 So speed is 324 M1 √A1 DM1 M1 for substitution of s = 0 into their s equation √A1 on their s DM1 for substitution of their t back into v equation A1 for 324 only www.theallpapers.com