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

This document consists of 16 printed pages. DC (NH/SW) 42199/1 © UCLES 2011 [Turn over * 7 5 6 0 4 0 0 8 8 6 * ADDITIONAL MATHEMATICS 0606/22 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 UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education www.theallpapers.com

2 0606/22/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 0606/22/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use 1 (i) Given that y = sin 3x, find dy dx. [1] (ii) Hence find the approximate increase in y as x increases from π 9 to π 9 + p, where p is small. [2] 2 (a) An outdoor club has three sections, walking, biking and rock-climbing. Using  to denote the set of all members of the club and W, B and R to denote the members of the walking, biking and rock-climbing sections respectively, write each of the following statements using set notation. (i) There are 72 members in the club. [1] (ii) Every member of the rock-climbing section is also a member of the walking section. [1] (b) (i) X Y  On the diagram shade the region which represents the set X Y ʹ. [1] (ii) Using set notation express the set X Y ʹ in an alternative way. [1] www.theallpapers.com

4 0606/22/M/J/11 © UCLES 2011 For Examiner’s Use 3 (i) Given that A =  2 1 –2 5, find the inverse of the matrix A + I, where I is the identity matrix. [3] (ii) Hence, or otherwise, find the matrix X such that AX + X = B, where B =  14 4. [2] www.theallpapers.com

5 0606/22/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use 4 (a) Prove that 1 1 – sin x – 1 1 + sin x = 2 tan x sec x. [3] (b) An acute angle x is such that sin x = p. Given that sin 2x = 2 sin x cos x, find an expression, in terms of p, for cosec 2x. [3] www.theallpapers.com

6 0606/22/M/J/11 © UCLES 2011 For Examiner’s Use 5 (i) Given that y = x 2x + 15, show that dy dx = k(x + 5) ––––––– 2x + 15 , where k is a constant to be found. [3] (ii) Hence find  x + 5 ––––––– 2x + 15dx and evaluate 5 –3 x + 5 ––––––– 2x + 15dx. [3] www.theallpapers.com

7 0606/22/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use 6 The line y = 3x – 9 intersects the curve 49x2 – y2 + 42x + 8y = 247 at the points A and B. Find the length of the line AB. [7] www.theallpapers.com

8 0606/22/M/J/11 © UCLES 2011 For Examiner’s Use 7 A particle moves in a straight line so that, t s after passing through a fixed point O, its velocity, v ms–1, is given by v = 60 ––––––– (3t + 4)2 . (i) Find the velocity of the particle as it passes through O. [1] (ii) Find the acceleration of the particle when t = 2. [3] (iii) Find an expression for the displacement of the particle from O, t s after it has passed through O. [4] www.theallpapers.com

9 0606/22/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use 8 (a) (i) Solve 3x = 200, giving your answer to 2 decimal places. [2] (ii) Solve log5 (5y + 40) – log5 (y + 2) = 2. [4] (b) Given that (24z3)2 ––––––– 27 × 12z = 2a3bzc, evaluate a, b and c. [3] www.theallpapers.com

10 0606/22/M/J/11 © UCLES 2011 For Examiner’s Use 9 Solutions to this question by accurate drawing will not be accepted. y x D C x = 14 B (–2, –10) A (4, 2) M O The diagram shows the quadrilateral ABCD in which A is the point (4, 2) and B is the point (–2, –10). The points C and D lie on the line x = 14. The diagonal AC is perpendicular to AB and passes through the mid-point, M, of the diagonal BD. Find the area of the quadrilateral ABCD. [9] www.theallpapers.com

11 0606/22/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use Continue your answer here if necessary. www.theallpapers.com

12 0606/22/M/J/11 © UCLES 2011 For Examiner’s Use 10 (a) (i) Express 18 + 16x – 2x2 in the form a + b(x + c)2, where a, b and c are integers. [3] A function f is defined by f : x → 18 + 16x – 2x2 for x . (ii) Write down the coordinates of the stationary point on the graph of y = f(x). [1] (iii) Sketch the graph of y = f(x). [2] www.theallpapers.com

13 0606/22/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use (b) A function g is defined by g : x → (x + 3)2 – 7 for x > –3. (i) Find an expression for g–1 (x). [2] (ii) Solve the equation g–1 (x) = g(0). [3] www.theallpapers.com

14 0606/22/M/J/11 © UCLES 2011 For Examiner’s Use 11 Answer only one of the following two alternatives. EITHER (a) Using an equilateral triangle of side 2 units, find the exact value of sin 60° and of cos 60°. [3] (b) P Q x cm 60° 60° x cm y cm S R PQRS is a trapezium in which PQ = RS = x cm and QR = y cm. Angle QPS = angle RSP = 60° and QR is parallel to PS. (i) Given that the perimeter of the trapezium is 60 cm, express y in terms of x. [2] (ii) Given that the area of the trapezium is A cm2 , show that A = 3 (30x – x2) 2 . [3] (iii) Given that x can vary, find the value of x for which A has a stationary value and determine the nature of this stationary value. [4] OR r cm h cm For a sphere of radius r: Volume = 4 3 π r3 Surface area = 4π r2 The diagram shows a solid object in the form of a cylinder of height h cm and radius r cm on top of a hemisphere of radius r cm. Given that the volume of the object is 2880 π cm3, (i) express h in terms of r, [2] (ii) show that the external surface area, A cm2, of the object is given by A = 5 3 π r2 + 5760π r . [3] Given that r can vary, (iii) find the value of r for which A has a stationary value, [4] (iv) find this stationary value of A, leaving your answer in terms of π, [2] (v) determine the nature of this stationary value. [1] www.theallpapers.com

15 0606/22/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

16 0606/22/M/J/11 © UCLES 2011 For Examiner’s Use Continue your answer here if necessary. … … … … … … … … … … … … … … … … … … … … … … … … 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 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 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 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 – May/June 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 – May/June 2011 0606 22 © University of Cambridge International Examinations 2011 1 (i) d 3cos3 d y x x = B1 (ii) Uses x x y y ∂ ×       = ∂ d d at attempt with 9 x π = and x p ∂= . M1 1.5p A1√ on k [3] 2 (a) (i) n(E) = 72 or ( ) n W B R 72 ∪ ∪ = B1 (ii) R ⊂ W or R W R ∩ = or R W W ∪ = or / R W ∩ = ∅ B1 (b) (i) B1 (ii) ( )′ ∩ ′ Y X or ( ) Y Y X ′ ∪ ∩ or ( ) X Y X ∪ ′ ∪ or ( ) X Y X ∪ ′ ∩ ′ B1 [4] 3 (i)       − = + 6 2 1 3 I A B1       − = + 3 2 1 6 20 1 ) ( 1- I A B1 + B1√ (ii)       = + = 4 14 ) ( 1- I A X evaluated to matrix with 2 entries M1 4 2    A1 [5] www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 22 © University of Cambridge International Examinations 2011 4 (a) Expresses with common denominator M1 2 2sin cos x x A1 sin 1 2 cos cos x x x =2tan sec x x or 1 2tan cos x x = 2tan sec x x A1 ag (b) 2 cos 1 x p = − B1 1 cosec2 sin 2 x x = B1 2 1 2 1 p p − B1 [6] 5 (i) Uses product rule M1 2 15 2 15 x x x + + + A1 ( ) ( ) 3 5 3 2 15 x k x + ⇒ = + A1 (ii) 1 2 15 x x k + M1 Uses limits on 2 15 Cx x + M1 34 3 A1 [6] 6 Eliminates y (or x) M1 2 3 10 0 x x + − = (or 2 27 72 0 y y + + = ) oe A1 Factorises 3 term quadratic or solves using formula M1 5 and 2 x = − (or 24 and -3 y = − ) A1 24 and -3 y = − (or 5 and 2 x = − ) A1√ Uses Pythagoras M1 22.1 or 490 or 7 10 A1 [7] www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 22 © University of Cambridge International Examinations 2011 7 (i) 3.75 oe B1 (ii) 3) 4 3 ( d d + = = t k t v a M1 k = –360 oe A1 –0.36 oe A1√ (iii) ( ) 3 4 k s c t = + + M1 20 k = − oe A1 Substitutes t = 0, s = 0 into k(3t + 4)n M1 20 5 3 4 t − + or 15 3 4 t t + A1 [8] 8 (a) (i) log3 log 200 x = or 3 log 200 B1 4.82 B1 (ii) 2 log 25 = or 2 2 log5 = B1 ( ) ( ) 5 40 log 5 40 log 2 log 2 y y y y   + + − + =   +   oe M1 Deals with logs correctly and solves DM1 0.5 y = − A1 (b) a = 4 B1 b = –2 B1 c = 5 B1 [9] 9 2 AB m = B1 1 2 1 m m = − M1 AC: ( ) 1 2 4 2 y x − = − − or 1 4 2 y x = − + or 2 8 x y + = A1 C(14, –3) A1 Midpoint M(6, 1) B1 + B1 D(14, 12) B1 Complete method for area M1 150 √ on D y A1√ [9] www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 22 © University of Cambridge International Examinations 2011 10 (a) (i) a = 50 B1 b = –2 OR ( ) 2 50 2 4 x − − B1 c = –4 B1 (ii) (4, 50) B1√ (iii) Correct shape B1 Maximum and y intercept in correct quadrant, B1 (b) (i) Method for inverse M1 7 3 x + − A1 (ii) g(0) = 2 B1 Solves g–1(x) = 2 or solves x = g(g(0)) = g(2) M1 18 A1 [11] -20 -10 0 10 20 30 40 50 60 -2 3 8 www.theallpapers.com

Page 8 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0606 22 © University of Cambridge International Examinations 2011 11 EITHER (a) cos 0.5 x = B1 Uses Pythagoras (to find 3 ) M1 2 3 sin = x A1 (b) (i) PS x y = + B1 60 3 2 x y − = B1 (ii) height = 3 2 x B1 Substitutes height and y into ( ) height Area 2 PS y = × + M1 or height into ( ) height Area 60 2 2 x = × − Correctly reaches ( ) 2 3 30 2 x x − A1 ag (iii) ( ) d 3 30 2 d 2 A x x = − B1 Equates to 0 and solves M1 15 x = A1 Completely correct for method with 15 x = leading to maximum B1 [12] OR (i) 3 2 2 2880 3 r r h π π π = + B1 2 2880 2 3 h r r = − or 3 2 8640 2 3 r h r − = B1 (ii) 2 3 2 A r rh π π = + B1 substitute for h M1 correctly reaches 2 5 5760 3 A r r π π = + A1 ag (iii) 2 d 10 5760 d 3 A r r r π π = − B1 + B1 equate to 0 and solve M1 12 r = A1 (iv) substitute in formula for A DM1 720π A1 (v) Completely correct method with 12 r = leading to minimum. B1 [12] www.theallpapers.com