Cambridge IGCSE Mathematics - Additional 0606 — 2011 Oct/Nov Paper 2 · Variant 3

This document consists of 16 printed pages. DC (LEO/CGW) 48835 © UCLES 2011 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 9 2 7 3 0 6 8 1 3 6 * ADDITIONAL MATHEMATICS 0606/23 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 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 12 Total www.theallpapers.com

2 0606/23/O/N/11 © UCLES 2011 For Examiner’s Use 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 0606/23/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 1 Solve the inequality x(2x – 1)  15. [3] 2 (i) Given that y = (12 – 4x)5, find dy dx . [2] (ii) Hence find the approximate change in y as x increases from 0.5 to 0.5 + p, where p is small. [2] www.theallpapers.com

4 0606/23/O/N/11 © UCLES 2011 For Examiner’s Use 3 (i) Find the coefficient of x3 in the expansion of (1 – 2x)7. [2] (ii) Find the coefficient of x3 in the expansion of (1 + 3x2)(1 – 2x)7. [3] www.theallpapers.com

5 0606/23/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 4 Without using a calculator, find the positive root of the equation (5 – 2 √⎯2)x2 – (4 + 2 √⎯2)x – 2 = 0, giving your answer in the form a + b √⎯2, where a and b are integers. [6] www.theallpapers.com

6 0606/23/O/N/11 © UCLES 2011 For Examiner’s Use 5 A school council of 6 people is to be chosen from a group of 8 students and 6 teachers. Calculate the number of different ways that the council can be selected if (i) there are no restrictions, [2] (ii) there must be at least 1 teacher on the council and more students than teachers. [3] After the council is chosen, a chairperson and a secretary have to be selected from the 6 council members. (iii) Calculate the number of different ways in which a chairperson and a secretary can be selected. [1] www.theallpapers.com

7 0606/23/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 6 (i) In the space below sketch the graph of y = (2x + 3)(2x – 7). [4] (ii) How many values of x satisfy the equation (2x + 3)(2x – 7) = 2x ? [2] www.theallpapers.com

8 0606/23/O/N/11 © UCLES 2011 For Examiner’s Use 7 (1,3) O (3,–1) x x y The variables x and y are related in such a way that when y x is plotted against x a straight line is obtained, as shown in the graph. The line passes through the points (1, 3) and (3, –1). (i) Express y in terms of x. [4] (ii) Find the value of x and of y such that y x = – 9. [2] www.theallpapers.com

9 0606/23/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 8 A sector of a circle, of radius r cm, has a perimeter of 200 cm. (i) Express the area, A cm2, of the sector in terms of r. [3] (ii) Given that r can vary, find the stationary value of A. [3] www.theallpapers.com

10 0606/23/O/N/11 © UCLES 2011 For Examiner’s Use 9 An aircraft, whose speed in still air is 350 kmh–1, flies in a straight line from A to B, a distance of 480 km. There is a wind of 50 kmh–1 blowing from the north. The pilot sets a course of 130°. (i) Calculate the time taken to fly from A to B. [5] (ii) Calculate the bearing of B from A. [3] www.theallpapers.com

11 0606/23/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 10 The line y = 2x + 10 intersects the curve 2x2 + 3xy – 5y + y2 = 218 at the points A and B. Find the equation of the perpendicular bisector of AB. [9] www.theallpapers.com

12 0606/23/O/N/11 © UCLES 2011 For Examiner’s Use 11 (i) Solve 4 cot 1 2 x = 1, for 0°  x  360°. [3] (ii) Solve 3(1 – tan y cos y) = 5 cos2 y – 2, for 0°  y  360°. [5] www.theallpapers.com

13 0606/23/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use (iii) Solve 3 sec2 z = 4, for 0  z  2π radians. [3] www.theallpapers.com

14 0606/23/O/N/11 © UCLES 2011 For Examiner’s Use 12 Answer only one of the following two alternatives. EITHER O y x (1, 0) Q P (e, 1) y=ln x The diagram shows part of the curve y = ln x cutting the x-axis at the point (1, 0). The normal to the curve at the point P(e, 1) cuts the x-axis at the point Q. (i) Show that Q is the point e + 1 e , 0 . [4] (ii) Show that d dx (x ln x) = 1 + ln x. [1] (iii) Hence find ln xdx and the area of the shaded region. [5] OR y x O B A (0,1) y = e x cos x (̟ , 0) 2 The diagram shows part of the curve y = ex cos x, cutting the x-axis at the point  π 2, 0 . The normal to the curve at the point A(0, 1) cuts the x-axis at the point B. (i) Find the coordinates of B. [4] (ii) Show that d dx [ex (cos x + sin x)] = 2ex cos x. [2] (iii) Hence find ex cos xdx and the area of the shaded region. [4] www.theallpapers.com

15 0606/23/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use Start your answer to Question 12 here. Indicate which question you are answering. EITHER OR … … … … … … … … … … … … … … … … … … … … … … … … … www.theallpapers.com

16 0606/23/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 0606/23 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 23 © 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 23 © 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 23 © University of Cambridge International Examinations 2011 1 Rearrange to ( ) 0 15 2 2 > − −x x and solve 3 term quadratic M1 3 , 2 5 − A1 2 5 − < x , x > 3 A1 [3] 2 (i) x y d d = k(12 – 4x)4 M1 –20(12 – 4x)4 A1 (ii) Uses ( ) p dx dy y x × = ∂ = 5.0 M1 –200000p A1√ [4] 3 (i) –8 or 35 B1 –280 B1 (ii) –14 (x) B1 (–280) + (–14) × 3 M1 –322 A1√ [5] 4 Correct use of quadratic formula M1 ( ) 8 2 16 16 2 2 4 2 + + = + or ( ) 2 16 24 + B1 ( ) ( ) 2 2 5 2 64 2 2 4 − + + A1 Multiply top and bottom by 2 4 10 + or 2 2 5 + M1 Numerator ( ) 2 17 34 + k or denominator 17k A1 2 2 + only A1 [6] 5 (i) 1 2 3 4 5 6 9 10 11 12 13 14 × × × × × × × × × × or !6 !8 ! 14 × M1 3003 A1 (ii) Both 5 students + 1 teacher and 4 students + 2 teachers M1 56 × 6 or 70 × 15 B1 1386 A1 (iii) 30 B1 [6] www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 23 © University of Cambridge International Examinations 2011 6 (i) Completely correct idea of modulus B1 quadratic shape B1 cusps B1 correct position B1 (ii) complete method M1 2 A1 [6] 7 (i) Gradient = –2 B1 y intercept = 5 B1 5 2 + − = x x y M1 x x y 5 2 2 + − = A1√ (ii) x = 7 B1 y = –63 B1 [6] 8 (i) s = 200 – 2r or 2r + rθ = 200 B1 Use rs A 2 1 = or θ 2 2 1 r A = M1 2 100 r r A − = oe A1 (ii) Differentiate non-linear function in r M1 Solve 0 d d = r A DM1 A = 2500 A1 [6] www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 23 © University of Cambridge International Examinations 2011 9 (i) B1 130 cos 350 50 2 350 50 2 2 2 × × − + = V M1 384 = V A1 V T 480 = M1 1.25 hours A1 (ii) V 130 sin 350 sin = α or V 130 sin 50 sin = β M1 3. 44 = α or 72 .5 = β A1 135.7 or 136 A1√ [8] OR (i) Diagram B1 ) 268 ( 50 sin 350 = = X ) 275 ( 50 cos 350 50 = + = Y 2 2 2 Y X V + = Finds components and uses Pythagoras M1 384 = V A1 V T 480 = M1 75 mins or 1.25 hours A1 (ii) X Y = α tan M1 3. 44 = α A1 135.7 or 136 A1√ 10 Eliminate x or y M1 ( ) 0 14 5 2 = − + x x k or ( ) 0 56 10 2 = − − y y k A1 Factorise 3 term quadratic M1 x = –7 and 2 A1 y = –4 and 14 B1√ Finds midpoint      − 5 , 2 5 M1 mAB = 2 (may be implied) B1 Uses m1m2 = –1 in equation of line M1       + − = − 2 5 2 1 5 x y or 4y + 2x = 15 or y = –0.5x + 3.75 A1 [9] 50 350 130 OR www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 23 © University of Cambridge International Examinations 2011 11 (i) 4 2 1 tan = x B1 76 2 1       = x B1 151.9 or 152 B1√ (ii) y y y sin cos tan = and y y 2 2 sin 1 cos − = B1 Solve quadratic M1 180 0 sin = ⇒ = y y A1 36.9 A1 143(.1) A1√ (iii) 2 3 cos ± = z or 3 1 tan ± = z or 2 1 sin ± = z B1 6 π and 6 11π or 6 π and 6 7π or 6 π and 6 5π B1 6 5π and 6 7π or 6 5π and 6 11π or 6 7π and 6 11π B1 [11] 12E (i) x y d d = x 1 B1 Uses m1m2 = –1 DM1 y – 1 = –e(x – e) A1 Leading to      + 0 , e 1 e Q A1ag (ii) x x x ln 1 + × B1ag (iii) ∫ ∫ − = x x x x x d 1 ln d ln M1 ∫ + − = ) ( ln d ln c x x x x x A1 Uses limits of 1 and e in correct order M1 Area under curve + Area triangle M1 e 2 1 1+ or 1.18 A1 [10] www.theallpapers.com

Page 8 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 23 © University of Cambridge International Examinations 2011 12O (i) Uses product rule M1 x x x y x x sin e cos e d d − = A1 Uses m1m2 = –1 DM1 B(1, 0) A1 (ii) Uses product rule M1 x x x x x x x x x x cos e 2 sin e cos e sin e cos e = + + − A1ag (iii) ( )( ) c x x x x x x + + = ∫ sin cos e 2 1 d cos e M1 Uses limits of 0 and 2 π in correct order M1 Area under curve – Area triangle M1 1.40 or 1.41 or 1 e 2 1 2 − π A1 [10] www.theallpapers.com