Cambridge IGCSE Mathematics - Additional 0606 — 2011 May/June Paper 1 · Variant 2

This document consists of 16 printed pages. DC (NH/CGW) 42186/1 © UCLES 2011 [Turn over 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 7 3 2 1 8 7 5 6 0 * ADDITIONAL MATHEMATICS 0606/12 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 UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education www.theallpapers.com

2 0606/12/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 0606/12/M/J/11 1 Find the value of k for which the x-axis is a tangent to the curve y = x2 + (2k + 10)x + k 2 + 5. [3] www.theallpapers.com

4 © UCLES 2011 For Examiner’s Use 0606/12/M/J/11 2 The coefficient of x3 in the expansion of (2 + ax)5 is 10 times the coefficient of x2 in the expansion of 1 + ax 3 4 . Find the value of a. [4] www.theallpapers.com

5 © UCLES 2011 [Turn over For Examiner’s Use 0606/12/M/J/11 3 (a) 5 4 3 2 1 π π x 2 y O The figure shows the graph of y = k + m sin px for 0  x  π, where k, m and p are positive constants. Complete the following statements. k = … m = … p = … [3] (b) The function g is such that g(x) = 1 + 5cos3x . Write down (i) the amplitude of g, [1] (ii) the period of g in terms of π. [1] www.theallpapers.com

6 © UCLES 2011 For Examiner’s Use 0606/12/M/J/11 4 You must not use a calculator in Question 4. In the triangle ABC, angle B = 90°, AB = 4 + 2 2 and BC = 1 + 2 . (i) Find tan C, giving your answer in the form k 2 . [2] (ii) Find the area of the triangle ABC, giving your answer in the form p + q 2, where p and q are integers. [2] (iii) Find the area of the square whose side is of length AC, giving your answer in the form s + t 2, where s and t are integers. [2] www.theallpapers.com

7 © UCLES 2011 [Turn over For Examiner’s Use 0606/12/M/J/11 5 (i) Show that 2x – 1 is a factor of 2x3 – 5x2 + 10x – 4. [2] (ii) Hence show that 2x3 – 5x2 + 10x – 4 = 0 has only one real root and state the value of this root. [4] www.theallpapers.com

8 © UCLES 2011 For Examiner’s Use 0606/12/M/J/11 6 The figure shows the graph of a straight line with 1g y plotted against x. The straight line passes through the points A (5,3) and B (15,5). A (5,3) B (15,5) x lg y O (i) Express lg y in terms of x. [3] (ii) Show that y = a (10bx) where a and b are to be found. [3] www.theallpapers.com

9 © UCLES 2011 [Turn over For Examiner’s Use 0606/12/M/J/11 7 A team of 6 members is to be selected from 6 women and 8 men. (i) Find the number of different teams that can be selected. [1] (ii) Find the number of different teams that consist of 2 women and 4 men. [3] (iii) Find the number of different teams that contain no more than 1 woman. [3] www.theallpapers.com

10 © UCLES 2011 For Examiner’s Use 0606/12/M/J/11 8 (i) Sketch the curve y = (2x – 5)(2x + 1) for –1  x  3, stating the coordinates of the points where the curve meets the coordinate axes. [4] (ii) State the coordinates of the stationary point on the curve. [1] (iii) Using your answers to parts (i) and (ii), sketch the curve y = ⏐(2x – 5)(2x + 1)⏐ for –1  x  3. [2] www.theallpapers.com

11 © UCLES 2011 For Examiner’s Use 0606/12/M/J/11 [Turn over 9 The figure shows a circle, centre O, radius r cm. The length of the arc AB of the circle is 9π cm. Angle AOB is θ radians and is 3 times angle OBA. θ rad A B O r cm 9π cm (i) Show that θ = 3π 5 . [2] (ii) Find the value of r. [2] (iii) Find the area of the shaded region. [3] www.theallpapers.com

12 © UCLES 2011 For Examiner’s Use 0606/12/M/J/11 10 Relative to an origin O, points A and B have position vectors  5 –6 and  29 –13 respectively. (i) Find a unit vector parallel to AB . [3] The points A, B and C lie on a straight line such that 2AC = 3 AB . (ii) Find the position vector of the point C. [4] www.theallpapers.com

13 © UCLES 2011 For Examiner’s Use 0606/12/M/J/11 [Turn over 11 Solve (i) 2cot2 x – 5cosec x – 1 = 0 for 0° < x < 180°, [5] (ii) 5cos 2y – 4sin 2y = 0 for 0° < y < 180°, [4] (iii) cos z + π 6  = – 1 2 for 0 < z < 2π radians. [3] www.theallpapers.com

14 © UCLES 2011 For Examiner’s Use 0606/12/M/J/11 12 Answer only one of the following two alternatives. EITHER The tangent to the curve y = 3x3 + 2x2 – 5x + 1 at the point where x = –1 meets the y-axis at the point A. (i) Find the coordinates of the point A. [3] The curve meets the y-axis at the point B. The normal to the curve at B meets the x-axis at the point C. The tangent to the curve at the point where x = –1 and the normal to the curve at B meet at the point D. (ii) Find the area of the triangle ACD. [7] OR R Q y y = x (x – 3)2 x O P The diagram shows the curve y = x (x – 3)2 . The curve has a maximum at the point P and touches the x-axis at the point Q. The tangent at P and the normal at Q meet at the point R. Find the area of the shaded region PQR. [10] www.theallpapers.com

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

16 © UCLES 2011 For Examiner’s Use 0606/12/M/J/11 Continue your answer here if necessary. … … … … … … … … … … … … … … … … … … … … … … … … 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 May/June 2011 question paper for the guidance of teachers 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 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 12 © 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 12 © 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 12 © University of Cambridge International Examinations 2011 1 ( ) ( ) 2 2 2 10 5 0 x k x k + + + + = ( ) ( ) 2 2 2 10 4 5 k k + = + 2 k = − (or ( ) ( ) d 2 2 10 , 5 d y x k x k x = + + = − + ( ) ( )( ) 2 2 0 5 2 10 5 5 k k k k = + − + + + + leading to 2 k = −) (or ( ) ( ) 2 2 2 2 10 5 x A x k x k + = + + + + ( ) 2 2 5 , 5 A k A k = + = + ( ) 2 2 5 5, k k + = + leading to 2 k = −) (or by completing the square ( ) ( ) ( ) ( ) 2 2 2 5 5 5 y x k k k = + + − + + + ( ) 2 2 5 5 k k + = + leading to 2 k = −) M1 M1 A1 [3] M1 M1 A1 M1 M1 A1 M1 M1 A1 M1 for equating to zero and use of 2 4 b ac = M1 for solution M1 for differentiation and attempt to equate to zero. M1 for attempt to substitute in for x in terms of k, for y = 0 and for attempt at solution. M1 for approach M1 for equating and attempt at solution M1 for approach M1 for equating last 2 terms to zero and attempt to solve 2 2 5 2 3 4 3 2 2 (10) 9 a C a C = 1 6 a = B1B1 M1 A1 [4] B1 for 5 2 3 32 C a , B1 for 2 4 2 9 a C M1 for a relationship between the 2 coefficients and attempt to solve 3 (a) k = 2, m = 3, p = 1 (b) (i) 5 (ii) 2π 3 B3 B1 B1 [5] B1 for each There must be evidence of working without a calculator in all parts 4 (i) ( ) ( ) ( ) ( ) 4 2 1 2 2 2 1 2 1 2 + − = + − (ii) Area = ( ) ( ) 1 4 2 2 1 2 2 × + × + = 4 3 2 + (iii) Area = 2 AC ( ) ( ) 2 2 4 2 2 1 2 = + + + = 27 18 2 + M1A1 M1 A1 M1 A1 [6] M1 for attempt to rationalise and attempt to expand M1 for attempt at area using surd form and attempt to expand M1 for attempt at 2 AC or AC in surd form, with attempt to expand www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 12 © University of Cambridge International Examinations 2011 5 (i) 1 1 1 2 5 10 4 8 4 2       − + −             = 0 (ii) ( )( ) 2 2 1 2 4 x x x − − + For ( ) 2 2 4 x x − + , ‘ 2 4 b ac < ’ so only one real root of x = 0.5 M1 A1 M1A1 M1 A1 [6] M1 for substitution of x = 0.5 or attempt at long division M1 attempt to obtain quadratic factor A1 for correct quadratic factor M1 for correct use of discriminant or solution of quadratic equation = 0 A1, all correct with statement of root. 6 (i) ( ) 1 lg 3 5 5 y x − = − (ii) Either 1 5 b = 1 2 5 10 x y   +     = , 1 2 5 10 10 x = 100 a = Or lg lg lg10bx y a = + lg lg y a bx = + , lg 2 a = 100 a = 1 5 b = Or ( ) 5 3 10 10 b a = ( ) 15 5 10 10 b a = 1 5 b = , 100 a = B1M1 A1 B1 M1 A1 [6] M1 A1 B1 M1 B1, A1 B1 for gradient, M1 for use of straight line equation B1 for 1 5 b = M1 for use of powers of 10 correctly to obtain a A1 for a M1 for use of logarithms correctly to obtain a A1 for a B1 for 1 5 b = M1 for simultaneous equations involving powers of 10 B1 for 1 5 b = , A1 for 100 a = 7 (i) 14 6 3003 C = (ii) 8 6 4 2 C C × 1050 = (iii) 8 8 6 5 6 364 C C + = B1 B1B1 B1 B1B1 B1 [7] B1 for 8 4 C or 6 2 C B1 for × by 6 2 C or 8 4 C B1 for 1050 B1 for 8 6 C or equivalent B1 for 8 5 6 C or equivalent B1 for 364 www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 12 © University of Cambridge International Examinations 2011 8 (i) −2 −1 1 2 3 4 −10 −5 x y (ii) ( ) 1, 9 − (iii) −2 −1 1 2 3 4 5 10 x y B1 B1 B1 B1 B1 √B1 B1 [7] B1 for x = – 0.5 B1 for x = 2.5 B1 for y = –5 B1 for shape √B1 on shape from (i) B1 for a completely correct sketch 9 (i) : 2 π 3 OBA θ θ   ∆ + =     (ii) 3π 9π 5 r = × r = 15 (iii) Area = 2 2 1 3π 1 3π 15 15 sin 2 5 2 5     × × − × ×         =105 M1 A1 M1 A1 M1M1 A1 [7] M1 for using angles in an isosceles triangle M1 for use of s rθ = M1 for use of 2 1 2 r θ or 1 2 rs M1 for use of 2 1 sin 2 r θ or other correct method 10 (i) 29 5 24 13 6 7       − =       − − −       Magnitude = 25, unit vector 24 1 7 25     −   (ii) 2 3 AC AB = uuur uuur or 2 2 3 AB BC AB + = uuur uuur uuur leading to 36 10.5 AC   =   −   uuur OC OA AC = + uuur uuur uuur or 2 2 OB OA OC OB − = − uuur uuur uuur uuur leading to 41 16.5 OC   =   −   uuur (equivalent methods acceptable) M1 M1 A1 M1 M1 A1 A1 [7] M1 for subtraction M1 for attempt to find magnitude of their vector M1 for attempt to find AC uuur – may be part of a larger method M1 for attempt to find OC uuur A1 for each www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 12 © University of Cambridge International Examinations 2011 11 (i) 2 2cosec 5cosec 3 0 x x − − = ( )( ) 2cosec 1 cosec 3 0 θ θ + − = leading to sin x = 3 1 , x = 19.5°, 160.5° (ii) 5 tan 2 4 y = 2y = 51.34°, 231.34° y = 25.7°, 115.7° (iii) π 2π 4π , 6 3 3 z   + =     2π π 3 6 z = − 4π π 3 6   −     π 7π , 2 6 z = allow 1.57, 3.67 M1A1 DM1 A1√A1 M1 M1 A1,√A1 M1 A1, A1 [12] M1 for use of correct identity or attempt to get in terms of sin x DM1 for attempt to solve √ o 180 – their x M1 for attempt to get in terms of tan M1 for dealing correctly with double angle √ 90° their y M1 for dealing with order correctly and attempt to solve 12 EITHER (i) 2 d 9 4 5 d y x x x = + − when x = –1, x y d d = 0 tangent y = 5, A (0, 5) (ii) B (0, 1) At B, d 5 d y x = − normal 1 1 5 y x −= C (–5, 0) At D 1 1 5 5 x + = , D (20, 5) Area = 1 20 5 2 × × , = 50 M1 DM1 A1 B1 M1A1 M1A1 M1 A1 [10] M1 for differentiation and substitution of x = –1 DM1 for attempt at equation of tangent and coordinates of A B1 for B M1 for attempt at normal and C, must be from differentiation and using correct point M1 for attempt to obtain D, equating normal and tangent equations M1 for valid attempt at area www.theallpapers.com

Page 8 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 12 © University of Cambridge International Examinations 2011 12 OR 2 d 3 12 9 d y x x x = − + When x y d d = 0, x = 1, 3 ( ) 1,4 P Area = 8 – ∫ 3 1 3 x – 6x2 + 9x dx = 3 4 2 3 1 9 8 2 4 2 x x x   − − +     27 11 8 4 4 = − + = 4 M1 M1 A1 A1 √B1M1 A2,1,0 DM1 A1 [10] M1 for differentiation and equating to 0, can be using a product M1 for attempt to solve A1 for both x values A1 for y coordinate √B1 on y coordinate for area of rectangle M1 for attempt to integrate –1 each error DM1 for application of limits www.theallpapers.com