Cambridge IGCSE Mathematics - Additional 0606 — 2008 Oct/Nov Paper 2 · Variant 1

This document consists of 6 printed pages and 2 blank pages. SPA (CM/DT) T56701 © 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. * 6 2 8 3 1 1 7 4 5 3 * ADDITIONAL MATHEMATICS 0606/02 Paper 2 October/November 2008 2 hours Additional Materials: Answer Booklet/Paper Graph Paper (1 sheet) Electronic calculator Mathematical tables www.XtremePapers.com

2 0606/02/O/N/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/O/N/08 © UCLES 2008 [Turn over 1 Given that A =
13 6 7 4, find the inverse matrix A–1 and hence solve the simultaneous equations 13x + 6y = 41, 7x + 4y = 24. [4] 2 Variables x and y are connected by the equation y = (2x – 9)3. Given that x is increasing at the rate of 4 units per second, find the rate of increase of y when x = 7. [4] 3 Find the set of values of m for which the line y = mx + 2 does not meet the curve y = x2 – 5x + 18. [5] 4 (i) Differentiate x ln x with respect to x. [2] (ii) Hence find ln x dx. [3] 5 Solve the equation (i) 4x 25–x = 24x 8x–3 , [3] (ii) lg (2y + 10) + lg y = 2. [3] 6 (a) A sports team of 3 attackers, 2 centres and 4 defenders is to be chosen from a squad of 5 attackers, 3 centres and 6 defenders. Calculate the number of different ways in which this can be done. [3] (b) How many different 4-digit numbers greater than 3000 can be formed using the six digits 1, 2, 3, 4, 5 and 6 if no digit can be used more than once? [3]

4 0606/02/O/N/08 © UCLES 2008 7 Q P 1.4 ms –1 48 m The diagram shows a river with parallel banks. The river is 48 m wide and is flowing with a speed of 1.4 ms–1. A boat travels in a straight line from a point P on one bank to a point Q which is on the other bank directly opposite P. Given that the boat takes 10 seconds to cross the river, find (i) the speed of the boat in still water, [4] (ii) the angle to the bank at which the boat should be steered. [2] 8 The function f is defined, for 0
x
2π, by f(x) = 3 + 5 sin 2x. State (i) the amplitude of f, [1] (ii) the period of f, [1] (iii) the maximum and minimum values of f. [2] Sketch the graph of y = f(x). [3] 9 The line y = 2x – 9 intersects the curve x2 + y2 + xy + 3x = 46 at the points A and B. Find the equation of the perpendicular bisector of AB. [8]

5 0606/02/O/N/08 © UCLES 2008 [Turn over 10 y x O y = x − 8x + 16x 3 2 The diagram shows part of the curve y = x3 – 8x2 + 16x. (i) Show that the curve has a minimum point at (4, 0) and find the coordinates of the maximum point. [4] (ii) Find the area of the shaded region enclosed by the x-axis and the curve. [4] 11 The table shows experimental values of two variables x and y. x 2 4 6 8 y 2.25 0.81 0.47 0.33 (i) Using graph paper, plot xy against 1 x and draw a straight line graph. [3] (ii) Use your graph to express y in terms of x. [5] (iii) Estimate the value of x and of y for which xy = 4. [3] [Question 12 is printed on the next page.]

6 0606/02/O/N/08 © UCLES 2008 12 Answer only one of the following two alternatives. EITHER O D C B A 6 cm 0.6 rad 2 cm The diagram shows a sector AOB of a circle with centre O and radius 6 cm. Angle AOB = 0.6 radians. The point D lies on OB such that the length of OD is 2 cm. The point C lies on OA such that OCD is a right angle. (i) Show that the length of OC is approximately 1.65 cm and find the length of CD. [4] (ii) Find the perimeter of the shaded region. [3] (iii) Find the area of the shaded region. [3] OR A particle moves in a straight line so that t seconds after passing a fixed point O its acceleration, a ms–2, is given by a = 4t – 12. Given that its speed at O is 16 ms–1, find (i) the values of t at which the particle is stationary, [5] (ii) the distance the particle travels in the fifth second. [5]

8 0606/02/O/N/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 October/November 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 October/November 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 – October/November 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 – October/November 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 – October/November 2008 0606 02 © UCLES 2008 1       − − = − 13 7 6 4 10 1 1 A B1+B1 evaluate       − 24 41 1 A M1 x = 2 , y = 2.5 A1 [4] 2 ( ) 2 2 9 k x − M1 ( ) 2 6 2 9 x − A1 substitute d 7 and 4 d x x t = = into t x x y t y d d d d d d × = M1 600 A1 [4] 3 eliminate y M1 use 2 4 b ac − or 2 4 b ac ∗ DM1 2 10 39 0 m m + − ∗ or ( ) 2 5 64 m + ∗ A1 factorise 3 term quadratic in m or take square root M1 13 3 m − < < A1 [5] 4 (i) ( ) d 1 ln d x x x = B1 1 ln x + B1 (ii) ( ) 1 ln d ln ( ) x x x x c + = + ∫ M1 ln d ln 1 d ( ) x x x x x c = − + ∫ ∫ M1 ( ) ln x x x c − + A1 [5] 5 (i) express as powers of 2 (or 4 or 8) M1 applies rules of indices [ ] 2 (5 ) 4 3( 3) x x x x − − = − − DM1 7 A1 (ii) ( ) ( ) { } lg 2 10 lg lg 2 10 y y y y + + = + or 2 lg100 = B1 2 2 10 100 y y + = oe B1 5 only B1 [6]

Page 5 Mark Scheme Syllabus Paper IGCSE – October/November 2008 0606 02 © UCLES 2008 6 (a) 10, 3 and 15 B1 multiply 3 values M1 450 A1 (b) ( ) 4 5 4 3 × × × B1+B1 240 B1 [6] 7 (i) speed of travel = 4.8 or distance downstream = 14 B1 draw right angle triangle with 1.4 and (4.8) at o 90 B1 2 2 1.4 (4.8) + M1 5 A1 (ii) 1 (4.8) tan 1.4 − oe M1 73.7 or 1.29 radians A1 [6] 8 (i) 5 B1 (ii) 180 or π B1 (iii) 8 and 2 − B1+B1 correct start and endpoints B1 2 cycles in 0 to 2π B1 correct max and min points B1 [7] (4.8) 1.4 OR (4.8) 1.4

Page 6 Mark Scheme Syllabus Paper IGCSE – October/November 2008 0606 02 © UCLES 2008 9 eliminate y (or x) M1 2 7 42 35 0 x x − + = (or 2 7 42 49 0 y y + − = )oe A1 solve 3 term quadratic M1 x = 1 and 5 (or y = –7 and 1) A1 find second coordinates M1 find mid-point M1 use AB m , 1 2 1 m m = − and coordinates of a point M1 1 3 ( 3) 2 y x + = − − or 2 3 0 x y + + = or 1 3 2 2 y x = − − A1 [8] 10 (i) 16 16 3 d d 2 + − = x x x y B1 equate to 0 and solve 3 term quadratic M1 4, 0 x y = = A1 AG 27 256 or 27 13 9 3 4 = = y x or 9.48 or 9.5 A1 (ii) integrate M1 4 3 2 8 8 4 3 x x x − + A1 use limits of 4 (and 0) DM1 3 1 21 or 21.3 A1 [8] 11 (i) plot xy against 1/x with linear scales M1 xy 4.5 3.24 2.82 2.64 1/x 0.5 0.25 0.17 0.125 A2, 1, 0 (ii) attempt at gradient using plotted points DM1 5 0.2 ± A1 intercept 2 0.1 ± B1 (or A1 if calculated from y mx c = + ) use Y mX c = + in correct way M1 2 5 2 y x x = + or 2 5 2x y x + = or 1 5 2 y x x   = +     A1√ (iii) read from graph or substitute in formula to find x M1 2.5 0.2 x = ± A1 1.6 0.1 y = ± A1 [11]

Page 7 Mark Scheme Syllabus Paper IGCSE – October/November 2008 0606 02 © UCLES 2008 12 EITHER (i) cos0.6 2 OC = or 2cos0.6 OC = or 2 sin 0.97 sin 2 OC π = M1 1.65 A1 2sin 0.6 CD = or 2 2 CD OD OC = − M1 1.13 A1 (ii) 6 0.6 × B1 complete plan CD + 4 + rθ +(6 – 1.65) M1 13.1 A1 (iii) 2 1 6 0.6 2 × × B1 complete plan 2 1 1 2 2 r OC CD θ − × × M1 9.87 A1 [10] OR (i) 2 2 12 16 t t − + B1+B1+B1 equate to 0 and solve quadratic for 2 values M1 2 and 4 A1 (ii) d s v t = ∫ M1 3 2 2 6 16 3 t t t − + A 2, 1, 0√ use limits and subtract DM1 3 2 2 or 2.67 A1 [10]