Cambridge IGCSE Mathematics - Additional 0606 — 2011 Oct/Nov Paper 1 · Variant 1

This document consists of 14 printed pages and 2 blank pages. DC (NF/CGW) 48822 © UCLES 2011 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 6 4 0 9 7 8 2 6 1 2 * ADDITIONAL MATHEMATICS 0606/11 Paper 1 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 Total www.theallpapers.com

2 0606/11/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 0606/11/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 1 (a) Sets A and B are such that n(A) = 15 and n(B) = 7. Find the greatest and least possible values of (i) n(A B), [2] (ii) n(A B). [2] (b) On a Venn diagram draw 3 sets P, Q and R such that P Q = ∅ and P R = P. [2] www.theallpapers.com

4 0606/11/O/N/11 © UCLES 2011 For Examiner’s Use 2 The function f is such that f(x) = 4x3 – 8x2 + ax + b, where a and b are constants. It is given that 2x – 1 is a factor of f(x) and that when f(x) is divided by x + 2 the remainder is 20. Find the remainder when f(x) is divided by x – 1. [6] www.theallpapers.com

5 0606/11/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 3 Variables t and N are such that when 1g N is plotted against 1g t, a straight line graph passing through the points (0.45, 1.2) and (1, 3.4) is obtained. lg N lg t (0.45, 1.2) (1, 3.4) O (i) Express the equation of the straight line graph in the form 1g N = m 1g t + 1g c, where m and c are constants to be found. [4] (ii) Hence express N in terms of t. [1] www.theallpapers.com

6 0606/11/O/N/11 © UCLES 2011 For Examiner’s Use 4 Six-digit numbers are to be formed using the digits 3, 4, 5, 6, 7 and 9. Each digit may only be used once in any number. (i) Find how many different six-digit numbers can be formed. [1] Find how many of these six-digit numbers are (ii) even, [1] (iii) greater than 500 000, [1] (iv) even and greater than 500 000. [3] www.theallpapers.com

7 0606/11/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 5 A particle moves in a straight line such that its displacement, x m, from a fixed point O at time t s, is given by x = 3 + sin 2t, where t  0. (i) Find the velocity of the particle when t = 0. [2] (ii) Find the value of t when the particle is first at rest. [2] (iii) Find the distance travelled by the particle before it first comes to rest. [2] (iv) Find the acceleration of the particle when t = 3π 4 . [2] www.theallpapers.com

8 0606/11/O/N/11 © UCLES 2011 For Examiner’s Use 6 (a) x y O (0, p) (q, 1) (– ̟ , –5) 12 The diagram shows part of the graph y = p + 3tan3x passing through the points (– π 12, –5), (0, p) and (q, 1). Find the value of p and of q. [4] (b) It is given that f(x) = a cos(bx)+c, where a, b and c are integers. The maximum value of f is 11, the minimum value of f is 3 and the period of f is 72 °. Find the value of a, of b and of c. [4] www.theallpapers.com

9 0606/11/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 7 The coefficient of x2 in the expansion of 1 + x 5 n , where n is a positive integer, is 3 5. (i) Find the value of n. [4] (ii) Using this value of n, find the term independent of x in the expansion of 1 + x 5 n 2 – 3 x 2 . [4] www.theallpapers.com

10 0606/11/O/N/11 © UCLES 2011 For Examiner’s Use 8 (a) Find (ex + 1)2 dx and hence evaluate  2 0 (ex + 1)2 dx. [6] (b) A curve is such that dy dx = (4x +1)– 1 2. Given that the curve passes through the point with coordinates (2, 4.5), find the equation of the curve. [5] www.theallpapers.com

11 0606/11/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 9 (i) Solve 1 + cot2 x = 8 sin x for 0°  x  360°. [5] (ii) Solve 4 sin(2y – 0.3) + 5 cos(2y – 0.3) = 0 for 0  y  π radians. [5] www.theallpapers.com

12 0606/11/O/N/11 © UCLES 2011 For Examiner’s Use 10 Answer only one of the following two alternatives. EITHER A B G D E F C θ θ θ r cm r cm The figure shows a sector ABC of a circle centre C, radius 2r cm, where angle ACB is 3θ radians. The points D, E, F and G lie on an arc of a circle centre C, radius r cm. The points D and G are the midpoints of CA and CB respectively. Angles DCE and FCG are each θ radians. The area of the shaded region is 5 cm2. (i) By first expressing θ in terms of r, show that the perimeter, P cm, of the shaded region is given by P = 4r + 8 r . [6] (ii) Given that r can vary, show that the stationary value of P can be written in the form k 2 , where k is a constant to be found. [4] (iii) Determine the nature of this stationary value and find the value of θ for which it occurs. [2] OR 10 cm r cm θ θ O A B D C E F The figure shows a sector OAB of a circle, centre O, radius 10 cm. Angle AOB = 2θ radians where 0 < θ < π 2. A circle centre C, radius r cm, touches the arc AB at the point D. The lines OA and OB are tangents to the circle at the points E and F respectively. (i) Write down, in terms of r, the length of OC. [1] (ii) Hence show that r = 10 sin θ 1 + sin θ . [2] (iii) Given that θ can vary, find dr dθ when r = 10 3 . [6] (iv) Given that r is increasing at 2 cms–1, find the rate at which θ is increasing when θ = π 6 . [3] www.theallpapers.com

13 0606/11/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use Start your answer to Question 10 here. Indicate which question you are answering. EITHER OR … … … … … … … … … … … … … … … … … … … … … … … www.theallpapers.com

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

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16 0606/11/O/N/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 October/November 2011 question paper for the guidance of teachers 0606 ADDITIONAL MATHEMATICS 0606/11 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 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 11 © 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 11 © 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 11 © University of Cambridge International Examinations 2011 1 (a) (i) 7 and 0 (ii) 22 and 15 (b) 3 ‘sets’ enclosed in a rectangle B2 B2 [4] B1 B1 [2] B1 for each. B1 for each. B1 for set P and set Q separate B1 for set R contained within set P 2 ( ) f 2 : 2 84 a b − − + = M1 A1 M1 for substitution of a correct value of x 1 1 3 f : 2 2 2 a b   + =     A1 A1 for each correct equation (allow unsimplified) 33, 18 a b = − = M1, A1 M1 for solution to obtain a and b ( ) f 1 19 = − √B1 [6] √B1 on their a and b 3 (i) Gradient m = 4 lg 0.6 c = − B1 M1 M1 M1 for a valid attempt to obtain lgc M1 for attempt to deal with lgc 0.251 c = A1 [4] (ii) 4 0.251 N t = √B1 [1] √B1 on their m and c 4 (i) 6! = 720 (ii) 2 × 5! = 240 (iii) 4 × 5! = 480 (iv) Even first and last: 4! (24) Odd first and even last: 4 x 4! (144) Total: 7 × 4! = 168 B1 [1] B1 [1] B1 [1] B1 B1 B1 [3] www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 11 © University of Cambridge International Examinations 2011 5 (i) 2cos2 v t = when t = 0, v = 2 (ii) cos2 0, t = π 2 2 t = π 4 t = ( 0.785) (iii) when π 4 t = , x = 4 when t = 0, x = 3 distance moved = 1 M1 A1 [2] M1 A1 [2] B1 √B1 [2] M1 for attempt to differentiate M1 for attempt to solve and deal with 2t √B1 for ‘their 4’ –3 (iv) 4sin 2 a t = − M1 when 3π , 4 4 t a = = A1 [2] 6 (a)      − + = − 4 tan 3 5 π p M1 A1 M1 for use of       − − 5 , 12 π 12 1 3 tan 3 tan 3 ' ' 1 2 π q q q p p = = + = − = ∴ M1 A1 [4] M1for use of their p and (q, 1) (b) amplitude a = 4 b = 5 B1 B1 When f = 11, x = 0, so c = 7 Or when f = 3, 3 π x = , so c = 7 M1 A1 [4] M1 for use of either max and x = 0, or min and 3 π x = www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 11 © University of Cambridge International Examinations 2011 7 (i) ( ) 1 3 2 25 5 n n − = × B1 B1 for correct term 2 30 0 n n − − = or 15 2 n =   M1 M1 M1 equating 3rd term to 3 5 M1 attempt to solve quadratic or realising that 15 2 n =   when n = 6 n = 6 A1 [4] (ii) 2 2 3 12 9 1 4 5 nx x x x    + + − +       term : 4 B1 B1 for 4 12 5 n − (14.4) M1 M1 for 2nd term ( ) 2 0.18 n n − (5.4) M1 M1 for 3rd term = – 5 A1 [4] 8 (a) x x x d 1 e 2 e 2 2 0 + + ∫ M1 M1 for expansion 2 2 0 e 2e 2 x x x   + +     B1 B1 B1 B1 for each correct term = 41.6 M1, A1 [6] M1 for correct use of limits (b) ( ) 1 2 1 4 1 ( ) 2 y x c = + + M1 M1 for attempt to integrate A1 A1 for ( ) 1 2 4 1 x + A1 A1 for ( ) 1 2 1 4 1 2 x + when y = 4.5, x = 2, c = 3 M1 M1 for attempt to find c, must be from integration ( ) 1 2 1 4 1 3 2 y x = + + A1 [5] A1 for c = 3 www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 11 © University of Cambridge International Examinations 2011 9 (i) 2 cosec 8sin x x = M1 M1 for use of correct identity or equivalent 3 1 sin 8 x = M1 M1 for dealing with cosec or equivalent 1 sin 2 x = M1 M1 for attempt to solve

150 , 30 = x A1, A1 [5] Withhold last A1 if extra solutions (ii) ( ) 5 tan 2 0.3 4 y − = − M1, A1 M1 for attempt to get in terms of tan 2y – 0.3 = 2.2455, 5.387 M1 M1 for dealing with order correctly y = 1.27, 2.84 (allow 1.28 and 2.85) A1, A1 [5] www.theallpapers.com

Page 8 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 11 © University of Cambridge International Examinations 2011 10 EITHER (i) ( ) ( ) 2 2 1 1 2 3 2 5 2 2 r r θ θ − = M1 M1 M1 for use of sector area M1 for attempt to equate area to 5 2 1 r θ = A1 ( ) 2 3 2 2 2 P r r r r θ θ = + + + M1 M1 for use of arc length leading to 8 4 P r r = + M1 M1 for attempt to get P in terms of r and θ (answer given) A1 [6] (ii) 2 d 8 4 d P r r = − + M1 M1 for attempt to differentiate and equate to zero. when d d P r = 0, 2 r = A1 8 2 P = M1 A1 [4] M1 for attempt to obtain P (iii) 2 2 3 d 16 d P r r = , + ve ∴minimum B1 B1 for correct method and conclusion when 2, r = 1 2 θ = B1 [2] www.theallpapers.com

Page 9 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 11 © University of Cambridge International Examinations 2011 10 OR (i) 10 OC r = − B1 [1] (ii) sin r OC θ = ,sin 10 r r θ = − M1 M1 for attempt to use sinθ leading to 10sin 1 sin r θ θ = + A1 [2] A1 for correct attempt to simplify to given answer (iii) ( ) 2 d 10cos d 1 sin r θ θ θ = + M1 A2, 1, 0 M1 for correct attempt to differentiate a quotient – 1 each error when 10 1 3 , sin , cos 3 2 2 r θ θ = = = M1 M1 M1 for attempt to find sin or cos M1 for substitution d 20 3 d 9 r θ ∴ = (3.85) A1 [6] (iv) d 2 d r t = , B1 when π 6 θ = , d 3 3 d 20 r θ = d d d d d d r t t r θ θ = × M1 M1 for correct use of rates of change leading to d dt θ = 3 3 10 (0.520) A1 [3] www.theallpapers.com