Cambridge IGCSE Mathematics - Additional 0606 — 2008 May/June Paper 2 · Variant 1

This document consists of 5 printed pages and 3 blank pages. SPA (NF/DT) T56699 © UCLES 2008 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. 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 soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Write your answers on the separate Answer Booklet/Paper provided. 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. * 0 1 3 2 6 3 6 6 8 3 * ADDITIONAL MATHEMATICS 0606/02 Paper 2 May/June 2008 2 hours Additional Materials: Answer Booklet/Paper Electronic calculator Mathematical tables www.XtremePapers.com

2 0606/02/M/J/08 © UCLES 2008 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.

3 0606/02/M/J/08 © UCLES 2008 [Turn over 1 The equation of a curve is given by y = x2 + ax + 3, where a is a constant. Given that this equation can also be written as y = (x + 4)2 + b, find (i) the value of a and of b, [2] (ii) the coordinates of the turning point of the curve. [1] 2 (a) Illustrate the following statements using a separate Venn diagram for each. (i) A ∩ B = ∅, (ii) (C ∪ D) ⊂ E. [2] (b) Y X
Express, in set notation, the set represented by the shaded region. [2] 3 Find the coordinates of the points where the straight line y = 2x – 3 intersects the curve x2 + y2 + xy + x = 30. [5] 4 (i) Sketch, on the same diagram, the graphs of y = x – 3 and y =
2x – 9
. [3] (ii) Solve the equation
2x – 9
= x – 3. [2] 5 Find the coefficient of x3 in the expansion of (i) (1 + 3x)8, [2] (ii) (1 – 4x)(1 + 3x)8. [3] 6 (a) Given that sin x = p, find an expression, in terms of p, for sec2 x. [2] (b) Prove that sec A cosec A – cot A ≡ tan A . [4]

4 0606/02/M/J/08 © UCLES 2008 7 y x O P (x, y) y = 4 2 x2 The diagram shows part of the curve y x = 4 2 2 . The point P (x, y) lies on this curve. (i) Write down an expression, in terms of x, for (OP)2. [1] (ii) Denoting (OP)2 by S, find an expression for dS dx . [2] (iii) Find the value of x for which S has a stationary value and the corresponding value of OP. [3] 8 Solve the equation (i) 22x + 1 = 20, [3] (ii) 5 25 125 25 4 1 3 2 y y y y − + − = . [4] 9 Given that A =  4 1 2 3, B =  3 –5 0 2 and C =  4 1, calculate (i) AB, [2] (ii) BC, [2] (iii) the matrix X such that AX = B. [4] 10 (a) Find (i)  12 (2x – 1)4 dx, [2] (ii) x(x –1)2dx. [3] (b) (i) Given that y x x = − + 2 5 4 ( ) , show that d d y x x x = + + 3 1 4 ( ) . [3] (ii) Hence find  ( ) x x + + 1 4 dx. [2]

5 0606/02/M/J/08 © UCLES 2008 11 The function f is defined by f(x) = (x + 1)2 + 2 for x
–1. Find (i) the range of f, [1] (ii) f 2(1), [1] (iii) an expression for f –1(x) . [3] The function g is defined by g(x) = 20 x + 1 for x
0. Find (iv) g–1 (2), [2] (v) the value of x for which fg(x) = 38. [4] 12 Answer only one of the following two alternatives. EITHER The diagram shows the curve y = 4x – x2, which crosses the x-axis at the origin O and the point A. The tangent to the curve at the point (1, 3) crosses the x-axis at the point B. (i) Find the coordinates of A and of B. [5] (ii) Find the area of the shaded region. [5] OR Solutions to this question by accurate drawing will not be accepted. The points A (–2, 2), B ( 4, 4) and C (5, 2) are the vertices of a triangle. The perpendicular bisector of AB and the line through A parallel to BC intersect at the point D. Find the area of the quadrilateral ABCD. [10]

8 0606/02/M/J/08 BLANK PAGE 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.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2008 question paper 0606 ADDITIONAL MATHEMATICS 0606/02 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. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2008 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.XtremePapers.com

Page 2 Mark Scheme Syllabus Paper IGCSE – May/June 2008 0606 02 © UCLES 2008 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.

Page 3 Mark Scheme Syllabus Paper IGCSE – May/June 2008 0606 02 © UCLES 2008 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.

Page 4 Mark Scheme Syllabus Paper IGCSE – May/June 2008 0606 02 © UCLES 2008 1 (i) a = 8 B1 b = –13 B1 (ii) (–4, –13) B1√ [3] 2 (a) (i) E A B B1 (ii) E E C D oe B1 (b) ( ) ( ) Y X Y X ∩ ∪ ∩ ' ' B2, 1, 0 or ( ) ( ) Y X Y X ∪ ∩ ∩ ' or ( ) ( ) Y X Y X ∪ ∩ ∪' ' or ( ) ( ) ( )'' Y X Y X ∪ ∪ ∩ or ( ) ( ) ( ) ( ) Y Y X X Y X ∩ ∩ ∪ ∩ ∩ ' ' [4] 3 Eliminates x or y M1 0 21 14 7 2 = − − x x or 0 105 14 7 2 = − + y y oe A1 Solve 3 term quadratic M1 ( )( ) 3 1 − + x x A1 (3,3) and (–1,–5) A1 [or 2 16 2 ± = x M1 –1 and 3 A1 (3,3) and (–1,–5) A1] [5]

Page 5 Mark Scheme Syllabus Paper IGCSE – May/June 2008 0606 02 © UCLES 2008 4 (i) straight line, +ve gradient, – ve intercept B1 idea of modulus (V shape on axis) B1 meets axes in correct places DB1 (ii) 6 B1 4 B1 [5] 5 (i) evidence of 27 or 56 in correct place B1 1512 B1 (ii) 28 x 9 B1 complete plan M1 504 A1 [5] 6 (a) uses cos2 x = 1 – sin2 x or sec2 x = x 2 cos 1 or sec2 x = 1 + x x 2 2 cos sin B1 2 1 1 p − or ( )( ) p p − + 1 1 1 or 2 2 1 1 p p − + B1 (b) express LHS in terms of sine and cosine B1 uses common denominator B1 uses sin2 A + cos2 A = 1 in useful way B1 correct conclusion B1 [6]

Page 6 Mark Scheme Syllabus Paper IGCSE – May/June 2008 0606 02 © UCLES 2008 7 (i) x2 +       + =       4 2 2 2 32 2 4 x x x B1 (ii) 2x – 5 128 x B1 + B1 (iii) equates to 0 and attempts to solve M1 x = 2 A1 OP = 6 or 2.45 A1 [6] 8 (i) (2x + 1)log2 = log20 or 2xlog2 = log10 M1 attempt at valid solution M1 1.66 A1 (ii) express in powers of 5 (or 25 or 125) M1 y y y y 2 4 9 3 2 1 4 5 5 5 5 − + − = A1 4y – 1 – 2y = 3y + 9 – (4 – 2y) M1 –2 A1 [7] 9 (i) Matrix multiplication M1       − − 4 6 18 12 A1 (ii) Matrix multiplication M1       2 7 A1 (iii) 10 1 1 = − A       − − 4 2 1 3 B1 + B1 premultiply       − 2 0 5 3 M1       − − 1.8 0.6 1.7 0.9 A1 [8] 10 (a) (i) k(2x – 1)–3 +(c) M1 k = –2 A1 (ii) multiplies out and integrates M1 4 3 2 2 ( ) 4 3 2 x x x c − + + A2, 1, 0

Page 7 Mark Scheme Syllabus Paper IGCSE – May/June 2008 0606 02 © UCLES 2008 (b) (i) uses product rule M1 4 ) 5 ( 4 2 + − + + x x x oe A1 correct completion A1 (ii) ( ) 4 5 + − x x k M1 3 2 = k oe A1 [10] 11 (i) ≥ 2 B1 (ii) 51 B1 (iii) method for inverse M1 ( ) 1 2 − − x A2, 1, 0 (iv) solve 2 1 20 = + x or g –1 : x 1 20 − x a M1 9 A1 (v) finds expression for fg(x) DM1 2 1 1 20 2 +       + + x A1 equate to 38 and solve quadratic M1 3 A1 [or (v) g(x) = f –1 (38) M1 g(x) = 5 A1 solve 5 1 20 = + x M1 3 A1] [or (v) x = g–1 f –1 (38) M1 x = g–1 (5) A1 evaluate M1 3 A1] [11] 12 EITHER (i) A (4, 0) B1 x y d d = 4 – 2x M1 gradient = 2 A1 y – 3 = 2(x – 1) or y = 2x + 1 DM1 correctly reaches x = – 2 1 A1

Page 8 Mark Scheme Syllabus Paper IGCSE – May/June 2008 0606 02 © UCLES 2008 (ii) 2x2 – 3 3 x B1 uses limits of 4 and 1 M1 9 A1 area triangle = 9/4 B1 11.25 A1 [10] OR M(1, 3) B1 grad AB = ⅓ B1 uses m1m2 = –1 M1 ( )1 3 3 − − = − x y or 6 3 = + y x A1 grad BC = –2 B1 ( ) 2 2 2 + − = − x y or 2 2 − = + y x M1 solve equation of MD with equation of AD M1 x = 8, y = –18 A1 method for area M1 77 A1 [10]