Cambridge IGCSE Mathematics - Additional 0606 — 2005 May/June Paper 2 · Variant 1

This document consists of 6 printed pages and 2 blank pages. (KN) S76935 © UCLES 2005 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/02 Paper 2 May/June 2005 2 hours Additional Materials: Answer Booklet/Paper Electronic calculator Graph paper Mathematical tables 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 on both sides of the paper. 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. 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. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. www.XtremePapers.com

2 0606/02/M/J/05 © UCLES 2005 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, . Binomial Theorem (a + b)n = an + an–1 b + an–2 b2 + … + an–r br + … + bn, where n is a positive integer and = . 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1. sec2 A = 1 + tan2 A. cosec2 A = 1 + cot2 A. Formulae for ∆ABC = = . a2 = b2 + c2 – 2bc cos A. ∆= bc sin A. 1 2 c sin C b sin B a sin A n! (n – r)!r! ) n r( ) n r( ) n 2( ) n 1( x b b ac a = −± − 2 4 2

3 0606/02/M/J/05 © UCLES 2005 [Turn over 1 A curve has the equation y = . (i) Find an expression for . [3] (ii) Given that y is increasing at a rate of 0.2 units per second when x = – 0.5, find the corresponding rate of change of x. [2] 2 A flower show is held over a three-day period – Thursday, Friday and Saturday. The table below shows the entry price per day for an adult and for a child, and the number of adults and children attending on each day. (i) Write down two matrices such that their product will give the amount of entry money paid on Thursday and hence calculate this product. [2] (ii) Write down two matrices such that the elements of their product give the amount of entry money paid for each of Friday and Saturday and hence calculate this product. [2] (iii) Calculate the total amount of entry money paid over the three-day period. [1] 3 The diagram shows a square ABCD of area 60m2. The point P lies on BC and the sum of the lengths of AP and BP is 12 m. Given that the lengths of AP and BP are x m and ym respectively, form two equations in x and y and hence find the length of BP. [5] A B D C P dy ––– dx 8 ––––– 2x – 1 Thursday Friday Saturday Price ($) – Adult 12 10 10 Price ($) – Child 5 4 4 Number of adults 300 180 400 Number of children 40 40 150

4 0606/02/M/J/05 © UCLES 2005 4 The functions f and g are defined by f : x sin x, 0
x
, g : x 2x – 3, x ∈
. Solve the equation g–1f(x) = g2(2.75). [5] 5 (i) Differentiate xln x – x with respect to x. [2] (ii) The diagram shows part of the graph of y = ln x. Use your result from part (i) to evaluate the area of the shaded region bounded by the curve, the line x = 3 and the x –axis. [4] 6 A curve has the equation y = , for 0 < x < π. (i) Find and show that the x-coordinate of the stationary point satisfies 2 sinx – cos x = 0. [4] (ii) Find the x-coordinate of the stationary point. [2] 7 Solve, for x and y, the simultaneous equations 125x = 25(5y), 7x ÷ 49y = 1. [6] dy ––– dx e2x –––– sin x y x O y = ln x 3 π–2

5 0606/02/M/J/05 © UCLES 2005 [Turn over 8 The Venn diagram above represents the sets
= {homes in a certain town}, C = {homes with a computer}, D = {homes with a dishwasher}. It is given that n(C ∩D) = k, n(C) = 7 × n(C ∩D), n(D) = 4 × n(C ∩D), and n(
) = 6 × n(C′ ∩ D′). (i) Copy the Venn diagram above and insert, in each of its four regions, the number, in terms of k, of homes represented by that region. [5] (ii) Given that there are 165 000 homes which do not have both a computer and a dishwasher, calculate the number of homes in the town. [2] 9 A plane, whose speed in still air is 300 km h–1, flies directly from X to Y. Given that Y is 720km from X on a bearing of 150° and that there is a constant wind of 120 km h–1 blowing towards the west, find the time taken for the flight. [7] 10 (a) Solve, for 0°< x < 360°, 4 tan2x + 15 secx = 0. [4] (b) Given that y > 3, find the smallest value of y such that tan (3y – 2) = –5. [4] 11 (a) (i) Expand (2 + x)5. [3] (ii) Use your answer to part (i) to find the integers a and b for which (2 + )5 can be expressed in the form a + b . [3] (b) Find the coefficient of x in the expansion of
x –  7 . [3] 4–x 3 3 C D

6 0606/02/M/J/05 © UCLES 2005 12 Answer only one of the following two alternatives. EITHER Solutions to this question by accurate drawing will not be accepted. The diagram, which is not drawn to scale, shows a right-angled triangle ABC, where A is the point (6, 11) and B is the point (8, 8). The point D(5, 6) is the mid-point of BC. The line DE is parallel to AC and angle DEC is a right-angle. Find the area of the entire figure ABDECA. [11] OR The diagram, which is not drawn to scale, shows a circle ABCDA, centre O and radius 10 cm. The chord BD is 16 cm long. BED is an arc of a circle, centre A. (i) Show that the length of AB is approximately 17.9 cm. For the shaded region enclosed by the arcs BCD and BED, find (ii) its perimeter, (iii) its area. [11] B D C E O A 16 cm 10 cm y x O C E D (5, 6) A (6, 11) B (8, 8)

8 0606/02/M/J/05 BLANK PAGE Every reasonable effort has been made to trace all copyright holders where the publishers (i.e. UCLES) are aware that third-party material has been reproduced. The publishers would be pleased to hear from anyone whose rights they have unwittingly infringed. University of Cambridge International Examinations is part of the 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 June 2005 question paper 0606 ADDITIONAL MATHEMATICS 0606/02 Paper 2, maximum raw mark 80 This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination. 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 discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the June 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.XtremePapers.com

Grade thresholds taken for Syllabus 0606 (Additional Mathematics) in the June 2005 examination. minimum mark required for grade: maximum mark available A C E Component 2 80 61 35 25 Grade A* does not exist at the level of an individual component.

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.

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.

June 2005 IGCSE MARK SCHEME MAXIMUM MARK: 80 SYLLABUS/COMPONENT: 0606/02 ADDITIONAL MATHEMATICS PAPER 2

BLANK PAGE 4] UNIVERSITY of CAMBRIDGE International Examinations

Page 1 Mark Scheme Syllabus Paper IGCSE – June 2005 0606 2 © University of Cambridge International Examinations 2005 1 (i) dy/dx = (2x –1) –2 × ( – 8) × 2 B3, 2, 1 (ii) dy/dt = [dy/dx]x = −0.5 × dx/dt ⇒ 0.2 = – 4 × dx/dt ⇒ dx/dt = – 0.05 M1 A1 [5] 2 (i) ( )       5 12 40 300 or ( )       40 300 5 12 = ( 3800) B1 B1 (ii)             4 10 150 400 40 180 or ( )       150 40 400 180 4 10 =       4600 1960 or ( ) 4600 1960 M1 A1 (iii) 3800 + 1960 + 4600 = 10 360 B1 [5] 3 x + y = 12 AP2 – BP2 = AB2 ⇒ x2 – y2 = 60 Solve for y [via (12 – y)2 – y2 = 60 or using x – y = 5] BP = 3.5 B1 M1 A1 M1 A1 [5] 4 g2 (2.75) = g (2.5) or via [2(2x – 3) – 3]x=2.75 = 2 g-1(x) = 2 3 + x g-1f(x) = 2 3 sin + x x = π /2 M1 A1 M1 M1 A1 [5] 5 (i) x d d (x ln x – x) = ln x + (x × 1/x) – 1 (= ln x) M1 A1 (ii) ∫ − = x x x x x ln d ln y = 0, x = 1 [ ] 3 1 = (3ln3 – 3 ) – (0 – 1) ≈ 1.30 M1 B1 M1 A1 [6]

Page 2 Mark Scheme Syllabus Paper IGCSE – June 2005 0606 2 © University of Cambridge International Examinations 2005 6 (i) d/dx (e2x) = 2e2x x x x x x x x x 2 2 2 2 sin cos e ) e 2 ( sin sin e d d − =       or (sin x)−1(2e2x) + e2x (sin x)-2 (-cos x) = 0 when 2sin x – cos x = 0 B1 B2, 1, 0√ B1 (ii) tan x = 0.5 ⇒ x ≈ 0.464 (26.6°) M1 A1 [6] 7 Express in powers of 5 ⇒ 53x = 52+y [or xlg125 = lg25 + ylg 5] Express in powers of 7 ⇒ 7x – 2y = 70 or 7x = 72y [or x lg7 – y lg 49 = 0] Solve 3x = 2 + y [or 2.10 x = 1.40 + 0.70 y ⇒ x = 0.8 x = 2y 0.845 x – 1.69 y = 0] y = 0.4 B2, 1, 0 B2, 1, 0 M1 A1 [6] 8 (i) Insert k in C∩D 6k in C∩D′ 3k in C′∩D n (C∪D) = 5/6 n(ε ) = 10k ⇒Insert 2k in (C′ ∩D′) B1 B1 B1 M1 A1√ (ii) 11k = 165 000 ⇒ n(ε ) = 12k = 180 000 M1 A1 [7] 9 X Correct ∆ of velocities α 300 V 60° β Y 120 sin α = (120sin 120°)/300 ⇒ α ≈ 20.3° [β ≈ 39.7°] V = 300 sin 39.7°/sin 120° [or 120 sin 39.7°/sin 20.3° or cos rule] ≈ 221 T ≈ 720/221 ≈ 3.24~6 [or via components 300 sin β = V cos 30°, 300 cos β = V cos 60° + 120 Square, add and solve for V, T = 12/(√22 –1)] B1 M1 A1 M1 A1 DM1 A1 [7]

Page 3 Mark Scheme Syllabus Paper IGCSE – June 2005 0606 2 © University of Cambridge International Examinations 2005 10 (a) Replace tan2 x by sec2 x –1 4sec2 x + 15sec x – 4 = 0 ⇒ (4sec x –1)(sec x + 4) = 0 (cos x = 4), cos x = – 0.25⇒ x = 104.5° or 255.5° B1 M1 A1 A1√ (b) tan-15 ≈ 1.37 (or 78.7°), or any correct value of tan−1(−5) Any 2 correct values, or the specific value, of tan−1 (−5) = 1.77 (101.3°), 4.91 (281.3°), 8.05 (461.3°) Add 2 [or 114.6°], divide by 3 [consistent] y ≈ 3.35 [one answer only] B1 B1 M1 A1 [8] 11 (a) (i) 32 + 80x + 80x2 + 40x3 + 10x4 + x5 All coefficients to be resolved B3, 2, 1 (ii) x = √3 ⇒ x3 = 3√3, x5 = 9√3 32 + 80√3 + 240 + 120√3 + 90 + 9√3 = 362 + 209√3 B1 B1 B1 (b) …+ x4(– 4/x )3… × 7C4 (or 7C3 ) = 35 = – 2240 M1 A1 A1 [9]

Page 4 Mark Scheme Syllabus Paper IGCSE – June 2005 0606 2 © University of Cambridge International Examinations 2005 12 E DC = BD [or D (5, 6) midpoint of C (x, y), B(8, 8)] ⇒ C is (2, 4) mDE = mAC = 7/4 mCE = –1/mAC = – 4/7 Equation of DE is y – 6 = 7/4 (x – 5) Equation of CE is y – 4 = – 4/7 (x – 2) Solve for E ⇒ x = 3.4, y = 3.2 Complete method for entire area → 15.6 M1 A1 B1√ B1√ A1√ M1(either) A1√ M1 A1 M1 A1 [11] 12 O (i) ∠ BOD = 2sin−1 0.8 ≈ 1.855 (106.3°) or ∠ BOE = 0.927 (53.1°) ∠BAD = ½ ∠BOD ≈ 0.927 (53.1°) or ∠ BAE = 0.464 (26.6°) [or O to BD = √( 102 – 82 ) = 6, ∠BAD = 2tan−1 8/16] AB = 8/sin(½ ∠BAD) ≈ 17.9 [or via √(82 + 162)] M1 A1 M1 A1 M1 A1 (ii) Perimeter = (10 × 1.855) + (17.89 × 0.927) [or degrees × π /180] ≈ 35.1 M1 A1 (iii) Use of ½ r2θ or ½ r2 ( θ – sinθ ) (radians or degrees × 180 π ) Complete plan Segment BCDB = ½ 102 × 1.855 – ½ 16 × 6 ≈ 44.75 Segment BEDB = ½ 17.92 × 0.927 – ½ 16 × 16 ≈ 20.3~5 Area ≈ 24.2~5 M1 M1 A1 [11]