Cambridge IGCSE Mathematics - Additional 0606 — 2010 May/June Paper 2 · Variant 2

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. * 0 3 1 6 7 4 6 5 1 3 * ADDITIONAL MATHEMATICS 0606/22 Paper 2 May/June 2010 2 hours Additional Materials: Answer Booklet/Paper Electronic calculator Graph paper (2 sheets) This document consists of 6 printed pages and 2 blank pages. DC (LEO/KN) 25706 © UCLES 2010 [Turn over www.theallpapers.com

2 0606/22/M/J/10 © UCLES 2010 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 0606/22/M/J/10 © UCLES 2010 [Turn over 1 Find ∫ (2 + 5x – 1 –––––– (x – 2)2) dx. [3] 2 (a) A B C  Copy the diagram and shade the region which represents the set A ∪ (B ∩ C). [1] (b) X Y  Express, in set notation, the set represented by the shaded region. [1] (c) The universal set  and the sets P and Q are such that n() = 30, n(P) = 18 and n(Q) = 16. Given that n(P ∪ Q) = 2, find n(P ∩ Q). [2] 3 The volume V cm3 of a spherical ball of radius r cm is given by V = 4–3 πr3. Given that the radius is increasing at a constant rate of 1–π cm s–1, find the rate at which the volume is increasing when V = 288π. [4] www.theallpapers.com

4 0606/22/M/J/10 © UCLES 2010 4 A θ B C 16 –– 2 7 3 The diagram shows a right-angled triangle ABC in which the length of AB is 16 –– 2 , the length of BC is 7 3 and angle BCA is θ. (i) Find tan θ in the form a 6 –––– b , where a and b are integers. [2] (ii) Calculate the length of AC, giving your answer in the form c d , where c and d are integers and d is as small as possible. [3] 5 Solve the equation 2x3 – 3x2 – 11x + 6 = 0. [6] 6 R Q(x, y) P O y = 12 – 2x y x The diagram shows part of the line y = 12 – 2x. The point Q (x, y) lies on this line and the points P and R lie on the coordinate axes such that OPQR is a rectangle. (i) Write down an expression, in terms of x, for the area A of the rectangle OPQR. [2] (ii) Given that x can vary, find the value of x for which A has a stationary value. [3] (iii) Find this stationary value of A and determine its nature. [2] www.theallpapers.com

5 0606/22/M/J/10 © UCLES 2010 [Turn over 7 (i) Sketch the graph of y = ⏐3x + 9⏐ for –5 < x < 2, showing the coordinates of the points where the graph meets the axes. [3] (ii) On the same diagram, sketch the graph of y = x + 6. [1] (iii) Solve the equation ⏐3x + 9⏐= x + 6. [3] 8 (a) (i) Write down the first 4 terms, in ascending powers of x, of the expansion of (1 – 3x)7. [3] (ii) Find the coefficient of x3 in the expansion of (5 + 2x)(1 – 3x)7. [2] (b) Find the term which is independent of x in the expansion of x2 + 2––x  9. [3] 9 (i) Given that y = x + 2 ––––––––– (4x + 12)½, show that dy –– dx = k(x + 4) ––––––––– (4x + 12) 3/2, where k is a constant to be found. [5] (ii) Hence evaluate ∫ 1 13 x + 4 ––––––––– (4x + 12) 3/2 dx. [3] 10 (a) Given that logp X = 6 and logp Y = 4, find the value of (i) logp  X2 ––– Y , [2] (ii) logY X. [2] (b) Find the value of 2z, where z = 5 + log23. [3] (c) Express 512 as a power of 4. [2] 11 (a) Solve, for 0 < x < 3 radians, the equation 4 sin x – 3 = 0, giving your answers correct to 2 decimal places. [3] (b) Solve, for 0° < y < 360°, the equation 4 cosec y = 6 sin y + cot y. [6] www.theallpapers.com

6 0606/22/M/J/10 © UCLES 2010 12 Answer only one of the following two alternatives. EITHER It is given that f(x) = 4x2 + kx + k. (i) Find the set of values of k for which the equation f(x) = 3 has no real roots. [5] In the case where k = 10, (ii) express f(x) in the form (ax + b)2 + c, [3] (iii) find the least value of f(x) and the value of x for which this least value occurs. [2] OR The functions f, g and h are defined, for x ∈ , by f(x) = x2 + 1, g(x) = 2x – 5, h(x) = 2x. (i) Write down the range of f. [1] (ii) Find the value of gf(3). [2] (iii) Solve the equation fg(x) = g–1 (15). [5] (iv) On the same axes, sketch the graph of y = h(x) and the graph of the inverse function y = h–1(x), indicating clearly which graph represents h and which graph represents h–1. [2] www.theallpapers.com

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8 0606/22/M/J/10 © UCLES 2010 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 May/June 2010 question paper for the guidance of teachers 0606 ADDITIONAL MATHEMATICS 0606/22 Paper 22, 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 May/June 2010 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 2010 0606 22 © UCLES 2010 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 2010 0606 22 © UCLES 2010 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 2010 0606 22 © UCLES 2010 1 ( ) 2 5 1 2 2 2 x x c x + + + − oe B1 + B1 + B1 [3] 2 (a) B1 (b) Y X ∪ ' , )'' ( Y X ∩ , ) ( ' Y X X ∩ ∪ , )' ( Y X Y ∪ ∪ , or )' ' ( Y X Y ∩ ∪ oe B1 (c) 18 + 16 + 2 = 30 + x M1 6 A1 [4] 3 2 π 4 d d r r V = B1 r = 6 B1 Uses π 1 π 3 4 d d d d 3 ×       = r r t V M1 144 A1 [4] 4 (i) Evidence of rationalising 2 16 or 3 7 1 or 6 7 16 M1 21 6 8 oe A1 (ii) ( ) 2 2 3 7 2 16 +       M1 275 A1 11 5 A1 [5] A B C www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2010 0606 22 © UCLES 2010 5 Search for first root or factor M1 x = –2 or 2 1 or 3 or (x + 2) or (x – 3) or (2x – 1) A1 Attempt to factorise cubic M1 (x + 2)(2x2 – 7x + 3) or (x – 3)(2x2 + 3x – 2) A1 or (2x – 1)(x2 – x – 6) Solve 3 term quadratic M1 x = –2 and 2 1 and 3 A1 [6] 6 (i) A = xy M1 A = 12x – 2x2 oe A1 (ii) x x A 4 12 d d − = B1√ equate to 0 and solve M1 x = 3 A1√ (iii) A = 18 A1 Completely correct method and maximum B1√ [7] 7 (i) Idea of modulus correct M1 Shape and position completely correct A1 (0, 9) (–3, 0) indicated on graph A1 (ii) Straight line with +ve gradient and +ve y intercept, correct position B1 (iii) 3x + 9 = x + 6 ⇒ x = –1.5 B1 Solve –(3x + 9) = (x + 6) or (3x + 9)2 = (x + 6) 2 M1 x = –3.75 A1 [7] 8 (a) (i) 1 – 21x + 189x2 – 945x3 B3, 2, 1, 0 (ii) 2 × (189) and 5 × (–945) M1 –4347 A1√ (b) Identifies relevant (x2)3 6 2       x B1 Multiplies by 84 M1 5376 A1 [8] www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2010 0606 22 © UCLES 2010 9 (i) ( ) ( ) 4 12 4 2 1 12 4 d d 2 1 2 1 × + = + − x x x or ( ) ( ) 4 12 4 2 1 12 4 d d 2 3 2 1 × + − = + − − x x x B1 Uses quotient rule or product rule M1 ( ) ( )( ) 1 1 2 2 4 12 2 2 4 12 4 12 x x x x − + − + + + or ( ) ( )( ) 3 12 2 4 12 2 2 4 12 x x x − − + − + + A1 Express with common denominator of ( ) 4 12 n x + M1 ( ) ( ) 32 2 4 4 12 x x + + or 2 k = A1 (ii) ( ) 2 1 4 12 x k x + × + M1 uses both limits correctly on ( ) 2 4 12 C x x × + + M1 9 16 oe A1 [8] 10 (a) (i) 2log log p p X Y − or 8 log p p M1 8 A1 (ii) log log p p X Y A1 1.5 A1 (b) (i) 2 log 3 2 3 = or 5 2 3 z−= or 2 2 log 32 log 3 + B1 52 3 × or 2 log 96 M1 96 A1 (ii) ( ) 92 512 2 = or ( ) 92 512 4 = or log 512 log 4 x = B1 94 4 B1 [9] 11 (a) 0.85 B1 Subtract from π M1 2.29 A1 www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2010 0606 22 © UCLES 2010 (b) 4 cos 6sin sin sin y y y y = + B1 Uses 2 2 sin 1 cos y y = − to reach quadratic in cos y M1 Solve 3 term quadratic DM1 2 1 cos and cos 3 2 y y = = − A1 48.2 and 120 A1 311.8 and 240 A1√ [9] 12E (i) Rearranges to ( ) 2 0 ax bx c + + = and uses b2 * 4ac M1 k2 – 16k + 48 * 0 A1 Solve 3 term quadratic M1 4 and 12 k = A1 4 < < 12 k A1 (ii) 2 5 15 2 2 4 x   + +     or 2 a = , 52 b = , 15 or 3.75 4 c = B1 + B1 + B1 (iii) minimum 15 or 3.75 4 B1√ 54 x − = or 1.25 − B1√ [10] 12O (i) (f) 1 ≥ B1 (ii) gf(3)=g(10) or ( ) 2 gf ( ) 2 1 5 x x = + − M1 15 A1 (iii) g –1(15) = 10 B1 Finds fg(x) M1 ( ) 2 2 2 5 1 or 4 20 26 x x x − + − + A1 Solves ( ) 2 2 5 1 10 x − + = or 2 4 20 16 0 x x − + = M1 1 and 4 A1 (iv) complete, labelled exponential curve B1 complete logarithmic curve not cutting exponential curve B1 [10] www.theallpapers.com