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

This document consists of 14 printed pages and 2 blank pages. DC (NH/SW) 42198/1 © UCLES 2011 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 1 6 6 8 6 0 3 2 4 3 * ADDITIONAL MATHEMATICS 0606/21 Paper 2 May/June 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. 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. For Examiner’s Use 1 2 3 4 5 6 7 8 9 10 11 Total www.theallpapers.com

2 0606/21/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 0606/21/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use 1 Without using a calculator, express (5 + 2 3 )2 2 + 3 in the form p + q 3 , where p and q are integers. [4] 2 (i) Find the coefficient of x3 in the expansion of 1 – x 2 12 . [2] (ii) Find the coefficient of x3 in the expansion of (1 + 4x) 1 – x 2 12 . [3] www.theallpapers.com

4 0606/21/M/J/11 © UCLES 2011 For Examiner’s Use 3 Relative to an origin O, the position vectors of the points A and B are i – 4j and 7i + 20j respectively. The point C lies on AB and is such that → AC = 2 3 → AB . Find the position vector of C and the magnitude of this vector. [5] www.theallpapers.com

5 0606/21/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use 4 Find the set of values of k for which the line y = 2x – 5 cuts the curve y = x2 + kx + 11 in two distinct points. [6] www.theallpapers.com

6 0606/21/M/J/11 © UCLES 2011 For Examiner’s Use 5 The expression x3 + 8x2 + px – 25 leaves a remainder of R when divided by x – 1 and a remainder of –R when divided by x + 2. (i) Find the value of p. [4] (ii) Hence find the remainder when the expression is divided by x + 3. [2] www.theallpapers.com

7 0606/21/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use 6 (a) A shelf contains 8 different travel books, of which 5 are about Europe and 3 are about Africa. (i) Find the number of different ways the books can be arranged if there are no restrictions. [2] (ii) Find the number of different ways the books can be arranged if the 5 books about Europe are kept together. [2] (b) 3 DVDs and 2 videotapes are to be selected from a collection of 7 DVDs and 5 videotapes. Calculate the number of different selections that could be made. [3] www.theallpapers.com

8 0606/21/M/J/11 © UCLES 2011 For Examiner’s Use 7 The variables x and y are related so that when lg y is plotted against lg x a straight line graph passing through the points (4, 12) and (6, 17) is obtained. (4,12) lg y lg x O (6,17) (i) Express y in terms of x, giving your answer in the form y = axb. [6] (ii) Find the value of x when y = 300. [2] www.theallpapers.com

9 0606/21/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use 8 The temperature, T ° Celsius, of an object, t minutes after it is removed from a heat source, is given by T = 55e–0.1t + 15. (i) Find the temperature of the object at the instant it is removed from the heat source. [1] (ii) Find the temperature of the object when t = 8. [1] (iii) Find the value of t when T = 25. [3] (iv) Find the rate of change of T when t = 16. [3] www.theallpapers.com

10 0606/21/M/J/11 © UCLES 2011 For Examiner’s Use 9 A coastguard station receives a distress call from a ship which is travelling at 15 km h–1 on a bearing of 150°. A lifeboat leaves the coastguard station at 15 00 hours; at this time the ship is at a distance of 30 km on a bearing of 270°. The lifeboat travels in a straight line at constant speed and reaches the ship at 15 40 hours. (i) Find the speed of the lifeboat. [5] (ii) Find the bearing on which the lifeboat travelled. [3] www.theallpapers.com

11 0606/21/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use 10 (i) Solve the equation 3 sin x + 4 cos x = 0 for 0° < x < 360°. [3] (ii) Solve the equation 6 cos y + 6 sec y = 13 for 0° < y < 360°. [5] (iii) Solve the equation sin(2z – 3) = 0.7 for 0 < z < π radians. [3] www.theallpapers.com

12 0606/21/M/J/11 © UCLES 2011 For Examiner’s Use 11 Answer only one of the following two alternatives. EITHER 1.8 rad C A B E O D 12 cm The diagram shows an isosceles triangle AOB and a sector OCDEO of a circle with centre O. The line AB is a tangent to the circle. Angle AOB = 1.8 radians and the radius of the circle is 12 cm. (i) Show that the distance AC = 7.3 cm to 1 decimal place. [2] (ii) Find the perimeter of the shaded region. [6] (iii) Find the area of the shaded region. [4] OR y x y = x sin x Q P O The diagram shows part of the curve y = x sin x and the normal to the curve at the point P  π 2 , π 2. The curve passes through the point Q(π, 0). (i) Show that the normal to the curve at P passes through the point Q. [4] (ii) Given that d dx (x cos x) = cos x – x sin x, find x sin xdx. [3] (iii) Find the area of the shaded region. [5] www.theallpapers.com

13 0606/21/M/J/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

14 0606/21/M/J/11 © UCLES 2011 For Examiner’s Use Continue your answer here if necessary. … … … … … … … … … … … … … … … … … … … … … … … … … … … www.theallpapers.com

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16 0606/21/M/J/11 © UCLES 2011 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

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 0606/21 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 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 21 © 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 21 © 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 21 © University of Cambridge International Examinations 2011 1 ( ) 3 20 37 3 2 5 2 + = + B1 ( ) 3 2 3 2 3 2 3 20 37 − − × + + M1 3 3 14 + A1 + A1√ [4] 2 (i) 220 or 8 1 ± B1 (ii) –27.5 oe 16.5(x2) Correct method for collecting terms (66 + (i)) 38.5 oe B1 B1 M1 A1√ [5] 3 AB = 6i + 24j (or AC = 4i + 16j) B1 AB OA OC 3 2 + = ( )      + + − j i j i 24 6 3 2 4 M1 OC = 5i + 12j A1 2 2 12 5 + = OC M1 13 A1 [5] 4 Eliminates y x2 + kx – 2x + 16 (= 0) Uses b2 – 4ac k2 – 4k – 60*0 or (k – 2)*± 8 k = –6 or 10 k < –6 or k > 10 M1 A1 M1 A1 A1 A1 [6] 5 (i) (ii) f(1) = 1 + 8 + p – 25 (= p – 16) f(–2) = –8 + 32 – 2p – 25 (= –2p – 1) p – 16 = 2p + 1 oe p = –17 Evaluates f(–3) or divides by (x + 3) to remainder 71 (= 20 – 3p) B1 B1 M1 A1 M1 A1√ [6] 6 (a) (i) Evidence of 8, 7, 6, 5, 4, 3, 2, 1 or 8! 40320 (ii) Evidence of 5! (120) or 4! (or 24) 2880 M1 A1 B1 B1 (b) ) 35 ( )1 ( 2 3 5 6 7 = × × × × and ) 10 ( )1 ( 2 4 5 = × × B1 Multiply 350 M1 A1 [7] 7 (i) (ii) m = 2.5 c = 2 lg y = 2.5lg x + 2 2 = lg100 or lg102 2.5lg x = lg x2.5 y = 100 x2.5 Solve 2.5lg x = lg3 or x2.5 = 3 correctly 1.55 B1 B1 M1 B1√ B1√ A1 M1 A1 [8] www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 21 © University of Cambridge International Examinations 2011 8 (i) (ii) (iii) 70 39.7 55e–0.1t = 25 – 15 oe B1 B1 B1 0.1t = ln       10 55 oe M1 17(.0) A1 (iv)       t T d d = ke–0.1t M1 k = –5.5 oe –1.11 A1 A1 [8] 9 (i) Either B1 10 or 45 found Uses cosine rule D2 = 102 + 302 – 2 × 10 × 30 × cos60 or V2 = 152 + 452 – 2 × 15 × 45 × cos60 39.7 or 39.8 or 7 15 B1 M1 A1 A1 (ii) V D / 60 sin 15 / 10 sin = α (or D 60 sin 30 sin = β and use β ) M1 α = 19.1 or β = 101 251 A1 A1√ [8] 9 (i) Or B1 10 Dsinα = 10sin60 and Dcosα = 25 or Vsinα = 15sin60 and Vcosα = 37.5 Solve equations V = 39.7 or 39.8 B1 B1 M1 A1 (ii) 25 60 sin 10 tan = α M1 α = 19.1 251 A1 A1√ [8] 30/45 D/V 10/15 60 β α 30/45 10/15 60 D/V β α www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 21 © University of Cambridge International Examinations 2011 10 (i) tanx = –1.33 126.9 306.9 B1 B1 B1√ (ii) 13 cos 6 cos 6 = + y y or 13 sec 6 sec 6 = + y y B1 (iii) Forms quadratic in cosy or secy (6cos2y – 13cosy + 6 = 0) Solve 3 term quadratic 48.2 311.8 2z – 3 = 0.775 (or 2.37) radians Solves for z using radians 1.89 and 2.68 M1 M1 A1 A1√ B1 M1 A1 [11] 11 (i) EITHER 9.0 cos 12 = OA M1 (ii) AC = 19.3 – 12 = 7.3 Complete method for major arc (2π – 1.8) × 12 53.8 AB = 2 × 12tan0.9 or cosine rule 30.2 Complete plan (53.8 + 30.2 + 2 × 7.3) 98.6 A1 M1 A1 M1 A1 M1 A1 (iii) Complete method for major sector ( ) 8.1 2 12 2 1 2 − × × π M1 8.1 sin 3. 19 2 1 2 × × or 12 2. 30 2 1 × × M1 323 or 181 504 A1 A1 [12] 11 (i) OR Uses product rule       + = x x x x y cos sin d d M1 At x = 2 π gradient = 1 A1 Uses m1m2 = –1 M1 Correctly reaches conclusion. e.g.       − − = − 2 1 2 π π x y with y = 0 A1 (ii) x x x x x x x cos d sin d cos = −∫ ∫ x x x x x x x cos d cos d sin − = ∫ ∫ M1 M1 sinx – xcosx A1 (iii) Uses limits of π and 2 π M1 2.14 or π – 1 A1 Area triangle = 2 2 2 1 π π × × or 8 2 π or 1.23 B1 Subtracts area of triangle 0.908 (allow 0.906 or 0.907) or 0.91 or π − 1 – 8 2 π M1 A1 [12] www.theallpapers.com