Cambridge IGCSE Mathematics - Additional 0606 — 2009 Oct/Nov Paper 1 · Variant 1

This document consists of 5 printed pages and 3 blank pages. DC (SLM) 11532/2 © UCLES 2009 [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. * 8 1 6 0 8 6 3 2 1 0 * ADDITIONAL MATHEMATICS 0606/01 Paper 1 October/November 2009 2 hours Additional Materials: Answer Booklet/Paper Electronic calculator Mathematical tables www.theallpapers.com

2 0606/01/O/N/09 © UCLES 2009 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 0606/01/O/N/09 © UCLES 2009 [Turn over 1 Given that f (x) = 2x3 – 7x2 + 7ax + 16 is divisible by x – a, find (i) the value of the constant a, [2] (ii) the remainder when f (x) is divided by 2x + 1. [2] 2 Place Team 1st 2nd 3rd 4th Harriers 6 3 1 2 Strollers 3 2 4 3 Road Runners 2 5 5 0 Olympians 1 2 2 7 The table shows the results achieved by four teams in twelve events of an athletics match. In each event, 1st place scores 5 points, 2nd place scores 3 points, 3rd place scores 2 points and 4th place scores 1 point. (i) Write down two matrices whose product shows the total number of points scored by each team. [2] (ii) Evaluate this product of matrices. [2] 3 Find the values of k for which the equation x 2 – 2(2k + 1) x + (k + 2) = 0 has two equal roots. [4] 4 Solve the simultaneous equations x + 3y = 13, x 2 + 3y2 = 43. [5] 5 B A C 3 + 2 3 – 2 The diagram shows a triangle ABC, where angle B is a right angle, the length of AB = 3 + 2 and the length of BC = 3 – 2 . (i) Find the length of AC in the form k , where k is an integer. [2] (ii) Find tan A in the form a b c + 2 , where a, b and c are integers. [3] www.theallpapers.com

4 0606/01/O/N/09 © UCLES 2009 6 Set A is such that A = {x : 3x2 – 10x – 8  0}. (i) Find the set of values of x which define the set A. [3] Set B is such that B = {x : 7 – 2x  1}. (ii) Find the set of values of x which define the set A  B. [2] 7 A committee of 8 people is to be selected from 7 teachers and 6 students. Find the number of different ways in which the committee can be selected if (i) there are no restrictions, [2] (ii) there are to be more teachers than students on the committee. [4] 8 The number, N, of bacteria present in an experiment, t minutes after measurements begin, is given by N = 1000e–kt, where k is a constant. (i) State the number of bacteria when t = 0. [1] When t = 0, the number of bacteria is decreasing at the rate of 20 per minute. Find (ii) the value of k, [3] (iii) the time taken for the number of bacteria to decrease by 50%. [3] 9 Differentiate, with respect to x, (i) (1 – 2x)20, [2] (ii) x2ln x, [3] (iii) tan(2x + 1) __________ x . [3] 10 A curve has equation y = 3x 3 – 2x 2 + 2x. (i) Show that the equation of the tangent to the curve at the point where x = 1 is y = 7x – 4. [4] (ii) Find the coordinates of the point where this tangent meets the curve again. [5] www.theallpapers.com

5 0606/01/O/N/09 © UCLES 2009 11 (a) Show that tan + cot = cosec sec. [3] (b) Solve the equation (i) tan x = 3 sin x for 0 ° ฀x  360 °, [4] (ii) 2cot2 y + 3 cosec y = 0 for 0 ฀y  2π฀radians. [5] 12 Answer only one of the following two alternatives. EITHER A solid circular cylinder has radius r cm and height h cm. The volume of the cylinder is 1000 cm3. (i) Find an expression for h in terms of r. [2] (ii) Hence show that the total surface area, A cm2, of the cylinder is given by A = 2π r2 + 2000 r . [2] (iii) Given that r varies, find, correct to 2 decimal places, the value of r when A has a stationary value. [4] (iv) Find this stationary value of A and determine its nature. [3] OR y x O A B y = x + cos 2x The diagram shows part of the curve y = x + cos 2x. The curve has a maximum point at A and a minimum point at B. (i) Find the x-coordinate of the point A and of the point B. [6] (ii) Find, in terms of π, the area of the shaded region. [5] www.theallpapers.com

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8 0606/01/O/N/09 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. www.theallpapers.com

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2009 question paper for the guidance of teachers 0606 ADDITIONAL MATHEMATICS 0606/01 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. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2009 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 2009 0606 01 © UCLES 2009 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 2009 0606 01 © UCLES 2009 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 2009 0606 01 © UCLES 2009 1 (i) 2a3 – 7a2 + 7a2 + 16 = 0 leading to a3 = –8, a = –2 (ii) 16 2 1 14 2 1 7 2 1 2 2 3 +      − −      − −      − = 21 M1 A1 [2] M1 A1 [2] M1 for use of x = a and equated to zero, maybe implied M1 for substitution of x = – 2 1 into their expression or f(x) 2 (i) (ii)             =                         22 35 32 43 1 2 3 5 7 2 2 1 0 5 5 2 3 4 2 3 2 1 3 6 B1, B1 [2] B2, 1, 0 [2] B1 for each matrix, must be in correct order –1 for each error 3 4(2k + 1)2 = 4(k + 2) 4k2 + 3k – 1 = 0 leading to k = 4 1 , –1 M1 A1 M1 A1 [4] M1 for use of ‘b2 – 4ac’ Correct quadratic expression M1 for correct attempt at solution A1 for both values 4 (13 – 3y)2 + 3y2 = 43 (or x2 + 3 ) 13 ( 2 x − = 43) 6(2y2 – 13y + 21) = 0 (or 2(2x2 – 13x + 20) = 0) (2y – 7)(y – 3) = 0 (or (2x – 5)(x – 4) = 0 y = 3 or       = 4 or 2 5 2 7 x (or x = 4 or       = 3 or 2 7 2 5 y ) M1 A1 DM1 A1,A1 [5] M1 for eliminating one variable A1 for correct quadratic DM1 for correct attempt at solving quadratic A1 for each correct pair 5 (i) ( ) ( ) 22 2 3 2 3 2 2 = − + + AC = 22 (ii) tan A = 2 3 2 3 + − ( )( ) ( )( ) 2 3 2 3 2 3 2 3 − + − − = 7 2 6 11− M1 A1 [2] M1 M1, A1 [3] M1 for use of Pythagoras Use of decimals M1, A0 M1 for correct ratio M1 for rationalising 2 term denominator www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE– October/November 2009 0606 01 © UCLES 2009 6 (i) 3x2 – 10x – 8 = 0 (3x + 2)(x – 4) = 0 critical values – 3 2 , 4 A = {x : – 3 2 Y x Y 4} (ii) B = {x : x [ 3} A ∩ B = {x : 3 Y x Y 4} M1 A1 √A1 [3] B1 B1 [2] M1 for attempt to solve quadratic A1 for critical values Follow through on their critical values. B1 for values of x that define B. B1 (beware of fortuitous answers) 7 (i) 13C8 = 1287 (ii) 7 teachers, 1 student : 6 6 teachers, 2 students 7C6 × 6C2 : 105 5 teachers, 3 students 7C5 × 6C3 : 420 531 M1, A1 [2] B1 B1 B1 B1 [4] M1 for correct C notation 8 (i) When t = 0, N = 1000 (ii) t N d d = –1000ke–kt when t = 0, t N d d = –20 leading to k = 50 1 (iii) 500 = 1000e–kt t = –50ln 2 1 leading to 34.7 mins B1 [1] M1 DM1 A1 [3] M1 M1 A1 [3] M1 for differentiation DM1 for use of t N d d = ±20 M1 for attempt to formulate equation using half life M1 for a correct attempt at solution (beware of fortuitous answers) 9 (i) 20 × –2(1 – 2x)19 (ii) x2 2 1 + 2x ln x ISW (iii) 2 2 )1 2 tan( )) 1 2 ( sec 2 ( x x x x + − + ISW B1,B1 [2] M1 B1 A1 [3] M1 B1 A1 [3] B1 for 20 and (1 – 2x)19 B1 for –2 provided (1 – 2x)19 is present M1 for attempt to differentiate a product. B1 for x 1 A1 all other terms correct M1 for attempt to differentiate a quotient. B1 for differentiation of tan (2x + 1) A1 all other terms correct www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE– October/November 2009 0606 01 © UCLES 2009 10 (i) x y d d = 9x2 – 4x + 2 at P grad = 7 tangent y – 3 = 7(x – 1) (ii) at Q, 7x – 4 = 3x3 – 2x2 + 2x leading to 3x3 – 2x2 – 5x + 4 = 0 (x – 1)(3x2 + x – 4) – 0 (x – 1)(3x + 4)(x – 1) = 0 leading to x = – 3 4 , y = – 3 40 M1 A1 DM1 A1 [4] M1 B1 DM1 DM1 A1 [5] M1 for differentiation A1 for gradient = 7 and y = 3 DM1 for attempt to find tangent equation. M1 for equating tangent and curve equations B1 for realising (x – 1) is a factor DM1 attempt to factorise cubic DM1 for attempt to solve quadratic A1 for both 11 (a) tan θ + cot θ = θ θ θ θ sin cos cos sin + = θ θ θ θ sin cos cos sin 2 2 + = θ θ sin cos 1 = cos ec θ sec θ (b) (i) tan x = 3sin x x x x sin 3 cos sin = sin x – 3sin x cos x = 0 leading to cos x = 3 1 , sin x = 0 x = 70.5°, 289.5° and x = 180° (ii) 2 cot2 y + 3 cosecy = 0 2(cosec2y – 1) + 3 cosecy = 0 2 cosec2y + 3 cosecy – 2 = 0 (2 cosecy – 1)(cosecy + 2) = 0 leading to sin y = – 2 1 , y = 6 π 11 , 6 π 7 allow y = 3.67, 5.76 B1 B1 B1 [3] M1 A1√A1 B1 [4] M1 M1 M1 A1,A1 [5] B1 for attempt to obtain one fraction B1 for use of an appropriate identity B1 for simplification Scheme follows for alternative proofs M1 for use of tan x = x x cos sin and correct attempt to solve √A1 on their x = 70.5° B1 for x = 180° M1 for use of correct identity M1 for attempt to solve quadratic M1 for dealing with cosec/cot Scheme follows for alternative solutions www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE– October/November 2009 0606 01 © UCLES 2009 12 EITHER (i) πr2h = 1000, leading to h = 2 π 1000 r (ii) A = 2πrh + 2πr2 leading to given answer A = 2πr2 + r 2000 (iii) r A d d = 4πr – 2 2000 r when r A d d = 0, 4πr = 2 2000 r leading to r = 5.42 (iv) 2 2 d d r A = 4π + 3 4000 r + ve when r = 5.42 so min value Amin = 554 M1 A1 [2] B1 A1 [2] M1 A1 DM1 A1 [4] M1 A1 B1 [3] M1 for attempt to use volume B1 for A = 2πrh + 2πr2 GIVEN ANSWER M1 for attempt to differentiate given A A1 all correct DM1 for solution = 0 M1 for second derivative method or gradient method’ A1 for minimum, must be from correct work 12 OR (i) y = x + cos 2x x y d d = 1 – 2 sin 2x when x y d d = 0, sin 2x = 2 1 leading to x = 12 5 , 12 π π (ii) Area = ∫ + 12 5 12 d. 2 cos π π x x x = 12 5 12 2 2 sin 2 1 2 π π       + x x = 12 2 π M1 A1 DM1 DM1 A1,A1 [6] M1 A1,A1 DM1 A1 [5] M1 for attempt to differentiate DM1 for setting to 0 and attempt to solve DM1 for correct order of operations M1 for attempt to integrate A1for each term correct DM1 for correct use of limits – must be in radians (Trig terms cancel out) www.theallpapers.com