Cambridge IGCSE Mathematics - Additional 0606 — 2011 Oct/Nov Paper 2 · Variant 2

This document consists of 16 printed pages. DC (NF/CGW) 48833 © UCLES 2011 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 8 1 3 4 1 8 0 7 4 8 * ADDITIONAL MATHEMATICS 0606/22 Paper 2 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 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 0606/22/O/N/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 0606/22/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 1 (a) The universal set  and the sets A and B shown in the Venn diagram below are such that n(A) = 15, n(B) = 20, n(A B) = 6 and n() = 30. In the Venn diagram below insert the number of elements in the set represented by each of the four regions. [4] A  B (b) In the Venn diagram below shade the region that represents (P Q) R. [1] P R Q  www.theallpapers.com

4 0606/22/O/N/11 © UCLES 2011 For Examiner’s Use 2 (i) On the grid below, draw on the same axes, for 0°  x  180°, the graphs of y = sin x and y = 1 + cos 2x. [3] (ii) State the number of roots of the equation sin x = 1 + cos 2x for 0°  x  180°. [1] (iii) Without extending your graphs state the number of roots of the equation sin x = 1 + cos 2x for 0°  x  360°. [1] www.theallpapers.com

5 0606/22/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 3 It is given that 2x – 1 is a factor of the expression 4x3 + ax2 – 11x + b and that the remainder when the expression is divided by x + 2 is 25. Find the remainder when the expression is divided by x + 1. [6] www.theallpapers.com

6 0606/22/O/N/11 © UCLES 2011 For Examiner’s Use 4 It is given that A =  3 –2 1 –5, B =  1 –4 2 –3 5 0 and C =  4 2 –7. (i) Calculate AB. [2] (ii) Calculate BC. [2] (iii) Find the inverse matrix, A–1. [2] www.theallpapers.com

7 0606/22/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 5 Four boys and three girls are to be seated in a row. Calculate the number of different ways that this can be done if (i) the boys and girls sit alternately, [2] (ii) the boys sit together and the girls sit together, [2] (iii) a boy sits at each end of the row. [2] www.theallpapers.com

8 0606/22/O/N/11 © UCLES 2011 For Examiner’s Use 6 The length of a rectangular garden is x m and the width of the garden is 10 m less than the length. (i) Given that the perimeter of the garden is greater than 140 m, write down a linear inequality in x. [1] (ii) Given that the area of the garden is less than 3000 m2, write down a quadratic inequality in x. [1] (iii) By solving these two inequalities, find the set of possible values of x. [4] www.theallpapers.com

9 0606/22/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 7 D E C B A O F 8 cm 20 cm 6 cm θ In the diagram AD and BE are arcs of concentric circles centre O, where OA = 6 cm and AB = 8 cm. The area of the region ABED is 32 cm2. The triangle OCF is isosceles with OC = OF = 20 cm. (i) Find the angle  in radians. [3] (ii) Find the perimeter of the region BCFE. [5] www.theallpapers.com

10 0606/22/O/N/11 © UCLES 2011 For Examiner’s Use 8 A particle travels in a straight line so that, t s after passing through a fixed point O, its velocity, v ms–1, is given by v = 12cos ( t 3). (i) Find the value of t when the velocity of the particle first equals 2 ms–1. [2] (ii) Find the acceleration of the particle when t = 3. [3] (iii) Find the distance of the particle from O when it first comes to instantaneous rest. [4] www.theallpapers.com

11 0606/22/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 9 It is given that f(x) = 2x2 – 12x + 10. (i) Find the value of a, of b and of c for which f(x) = a(x + b)2 + c. [3] (ii) Sketch the graph of y = |f(x)| for –1  x  7. [4] (iii) Find the set of values of k for which the equation |f(x)| = k has 4 distinct roots. [2] www.theallpapers.com

12 0606/22/O/N/11 © UCLES 2011 For Examiner’s Use 10 y x P O Q y = x3 – 9x2 + 24x + 2 The diagram shows part of the curve y = x3 – 9x2 + 24x + 2 cutting the y-axis at the point P. The curve has a minimum point at Q. (i) Find the coordinates of the point Q. [4] www.theallpapers.com

13 0606/22/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use (ii) Find the area of the region enclosed by the curve and the line PQ. [6] www.theallpapers.com

14 0606/22/O/N/11 © UCLES 2011 For Examiner’s Use 11 Answer only one of the following two alternatives. EITHER O B X T A S In the diagram above ⎯→ OA = a, ⎯→ OB = b, ⎯→ OS = 3 5 ⎯→ OA and ⎯→ OT = 7 5 ⎯→ OB. (i) Given that ⎯→ AX = l ⎯→ AB, where l is a constant, express ⎯→ OX in terms of l, a and b. [2] (ii) Given that ⎯→ SX = k ⎯→ ST , where k is a constant, express ⎯→ OX in terms of k, a and b. [4] (iii) Hence evaluate l and k. [4] OR O C D R Q P In the diagram above ⎯→ OC = c and ⎯→ OD = d. The points P and Q lie on OC and OD produced respectively, so that OC : CP = 1 : 2 and OD : DQ = 2 : 1. The line CD is extended to R so that CD = DR. (i) Find, in terms of c and/or d, the vectors ⎯→ OP, ⎯→ OQ and ⎯→ OR. [5] (ii) Show that the points P, Q and R are collinear and find the ratio PQ : QR. [5] www.theallpapers.com

15 0606/22/O/N/11 © UCLES 2011 [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 0606/22/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 0606/22 Paper 2, 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 22 © 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 22 © 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 22 © University of Cambridge International Examinations 2011 1 (a) 6 correctly positioned 14 correctly positioned 1 correctly positioned 9 correctly positioned B1 B1 B1 B1√ (b) B1 [5] 2 (i) -0.5 0 0.5 1 1.5 2 2.5 0 1 2 3 4 Sine curve through (0, 0), (90, 1)(180, 0) Cosine curve through (0, 2), (45, 1), (90, 0), (135, 1), (180, 2) B1 M1 A1 (ii) 2 B1√ (iii) 2√ B1√ [5] 3 0 2 1 f =       or f(–2) = 25 (or f(–1) found) M1 0 2 11 4 1 2 1 = + − + b a or 5 4 1 = + b a or a + 4b = 20 A1 –32 + 4a + 22+ b = 25 or 4a + b = 35 Solve simultaneous equations a = 8, b = 3 f(–1) = 18 A1 M1 A1 A1√ [6] 1 6 14 9 www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 22 © University of Cambridge International Examinations 2011 4 (i) Matrix multiplication       − − 2 6 29 22 16 9 M1 A1 (ii) Matrix multiplication       − − 2 18 M1 A1 (iii)       − − − 3 1 2 5 13 1 or       − − 3 1 2 5 13 1 B1+B1 [6] 5 (i) (ii) (iii) Evidence of 4 × 3 × 3 × 2 × 2(× 1 × 1) or 4 × 3 × 3 × 2 × 2(× 1 × 1) or 4! × 3! 144 4! (or 24) for boys and 3! (or 6) for girls 288 Evidence of 4 × (120) × 3 or (4) × 5! × (3) 1440 M1 A1 M1 A1 M1 A1 [6] 6 (i) (ii) (iii) 4x – 20 > 140 oe x(x – 10) < 3000 x > 40 Factorise 3 term quadratic –50 and 60 40 < x < 60 B1 B1 B1 M1 A1 A1 [6] 7 (i) 32 ) 6 14 ( 2 1 2 2 = − θ M1A1 0.4 A1 (ii) 2 CF = 20 sin 0.2 or CF2 = 202 + 202 – 2 × 20 × 20 cos 0.4 M1 (CF=)7.95 Uses s = rθ Complete plan including s = rθ 25.5 or 25.6 A1 M1 M1 A1 [8] www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 22 © University of Cambridge International Examinations 2011 8 (i) Solve 12 2 3 cos =      t M1 4.21 or 4.2(0) A1 (ii) 3 sin d d t k t v a = = M1 k = –4 –3.36 or –3.37 A1 A1√ (iii) Find t when v = 0 M1 3 sin 36 d t t v s = =∫ B1 substitute t into 3 sin t k M1 36 A1√ [9] 9 (i) (ii) (iii) a = 2, b = –3, c = –8 or 2(x – 3)2 – 8 graph with 3 distinct parts all above and touching x axis quadratic shape for middle section reflected in x axis cusp completely correct identifies 0 and 8 0 < k < 8 B1+B1+B1 B1 B1 B1 B1 M1 A1 [9] 10 (i) 3x2 – 18x + 24 Solves 3 term quadratic equals zero x = 4 (and x = 2) Q(4, 18) B1 M1 A1 cso A1 (ii) x x x x 2 12 3 4 2 3 4 + + − B2, 1, 0 Uses limits of 0 and xQ in integral (area under curve = 72) Area under PQ = (32 + 8) Plan for area including use of limits 32 M1 B1 M1 A1 [10] www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 22 © University of Cambridge International Examinations 2011 11E (i) OX = a + µ (b-a) or (1 – µ)a + µ b M1A1 (ii) a 5 3 = OS B1 b 5 7 = OT B1       − + = a b a 5 3 5 7 5 3 λ OX or b a 5 7 5 3 ) 1( λ λ OX + − = M1A1 (iii) Equate components or arrange to (α) a = (β) b and put α = β = 0 Solve simultaneous equations M1 DM1 2 1 = λ A1 10 7 = µ A1 [10] 11O (i) OP = 3c B1 d 2 3 = OQ B1 c d − = = CD DR B1 c d − = + = 2 DR OD OR M1A1 (ii) Finds two of PR QR PQ , , M1 Two of c d c d c d 4 2 , 2 1 , 3 2 3 − = − = − = PR QR PQ A1+A1 Express one vector as multiple of another 3:1 M1 A1 [10] www.theallpapers.com