Cambridge IGCSE Mathematics - Additional 0606 — 2004 Oct/Nov Paper 2 · Variant 1

This document consists of 5 printed pages and 3 blank pages. (KN) S69811 © UCLES 2004 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/02 Paper 2 October/November 2004 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/O/N/04 © UCLES 2004 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/O/N/04 [Turn over 1 Given that A =
 , find A–1 and hence solve the simultaneous equations 2x + 3y + 4 = 0 –5x + 4y + 13 = 0. [4] 2 Given that , where a and b are integers, find, without using a calculator, the value of a and of b. [4] 3 The diagram shows part of the curve y = 3sin 2x + 4cos x. Find the area of the shaded region, bounded by the curve and the coordinate axes. [5] 4 Find the values of k for which the line y = x + 2 meets the curve y2 + (x + k)2 = 2. [5] 5 Solve the equation log16 (3x – 1) = log4 (3x) + log4 (0.5). [6] 6 Given that x = 3sinθ – 2cosθ and y = 3cosθ + 2sinθ, (i) find the value of the acute angle θ for which x = y, [3] (ii) show that x2 + y2 is constant for all values of θ. [3] 7 Given that 6x3 + 5ax – 12a leaves a remainder of –4 when divided by x – a, find the possible values of a. [7] 8 A motor boat travels in a straight line across a river which flows at 3 ms–1 between straight parallel banks 200m apart. The motor boat, which has a top speed of 6 ms–1 in still water, travels directly from a point A on one bank to a point B, 150m downstream of A, on the opposite bank. Assuming that the motor boat is travelling at top speed, find, to the nearest second, the time it takes to travel from A to B. [7] y x O π 2 a b + = + 3 13 4 3 2 3 –5 4 © UCLES 2004

4 0606/02/O/N/04 © UCLES 2004 9 In order that each of the equations (i) y = abx, (ii) y = Axk, (iii) px + qy = xy, where a, b, A, k, p and q are unknown constants, may be represented by a straight line, they each need to be expressed in the form Y = mX + c, where X and Y are each functions of x and/or y, and m and c are constants. Copy the following table and insert in it an expression for Y, X, m and c for each case. [7] 10 The function f is defined by f: x  x2 – 8x + 7  for the domain 3
x
8. (i) By first considering the stationary value of the function x x2 – 8x + 7, show that the graph of y = f(x) has a stationary point at x = 4 and determine the nature of this stationary point. [4] (ii) Sketch the graph of y = f(x). [2] (iii) Find the range of f. [2] The function g is defined by g: x  x2 – 8x + 7  for the domain 3
x
k. (iv) Determine the largest value of k for which g–1 exists. [1] 11 The diagram shows a trapezium OABC, where O is the origin. The equation of OA is y = 3x and the equation of OC is y + 2x = 0. The line through A perpendicular to OA meets the y-axis at B and BC is parallel to AO. Given that the length of OA is units, calculate the coordinates of A, of B and of C. [10] 250 y x A B C y = 3x y + 2x = 0 O Y X m c y = abx y = Axk px + qy = xy

© UCLES 2004 5 0606/02/O/N/04 12 Answer only one of the following two alternatives. EITHER A particle, travelling in a straight line, passes a fixed point O on the line with a speed of 0.5 ms–1. The acceleration, a ms–2, of the particle, ts after passing O, is given by a = 1.4 – 0.6t. (i) Show that the particle comes instantaneously to rest when t = 5. [4] (ii) Find the total distance travelled by the particle between t = 0 and t = 10. [6] OR Each member of a set of curves has an equation of the form y = ax + , where a and b are integers. (i) For the curve where a = 3 and b = 2, find the area bounded by the curve, the x-axis and the lines x = 2 and x = 4. [4] Another curve of this set has a stationary point at (2, 3). (ii) Find the value of a and of b in this case and determine the nature of the stationary point. [6] b x2

8 0606/02/O/N/04 BLANK PAGE Every reasonable effort has been made to trace all copyright holders. The publishers would be pleased to hear from anyone whose rights we 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 November 2004 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 November 2004 question papers for most IGCSE and GCE Advanced Level syllabuses. www.XtremePapers.com

Grade thresholds taken for Syllabus 0606 (Additional Mathematics) in the November 2004 examination. minimum mark required for grade: maximum mark available A C E Component 2 80 60 28 17 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 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) CWO Correct Working Only – often written by a ‘fortuitous' answer 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.

November 2004 INTERNATIONAL GCSE MARK SCHEME MAXIMUM MARK: 80 SYLLABUS/COMPONENT: 0606/02 ADDITIONAL MATHEMATICS PAPER 2

Page 1 Mark Scheme Syllabus Paper IGCSE EXAMINATIONS – NOVEMBER 2004 0606 2 © University of Cambridge International Examinations 2005 1 [4]       − = − 2 5 3 4 1 A × 23 1       − =       − −       − =       2 1 13 4 2 5 3 4 23 1 y x B1 B1 M1 A1 2 [4] 3 4 3 4 3 4 3 4 13 − = − − × + or 3 8 19 169 3 4 13 2 + =       + ( ) 3 8 19 3 4 2 − = − or 3 8 19 3 8 19 3 8 19 3 8 19 169 − = − − × + OR ( )( )    = + = + ⇒ = + + 169 24 19 0 8 19 169 3 8 19 3 b a a b b a solve M1    − = = ⇒ 8 19 b a A1 M1 A1 M1 A1 3 [5] Integrate –3/2 cos 2x + 4 sin x [ ] ( ) 7 5 . 1 5 . 5 2 / 0 = − − = π Must use both limits properly, not assume cos0 = 0, not use 2 π degrees. M1 A1 A1 M1 A1 4 [5] Eliminate y → (x + 2)2 + (x + k)2 or x → x2 + (y – 2 + k)2 2x2 + (4 + 2k)x + (2 + k2) = 0 or 2y2 + (2k – 4)y + (k2 – 4k + 2) = 0 Apply “b2 – 4ac” ⇒ 16k – 4k2 ⇒ 4 0 4 0 ≤ ≤ = k k or OR       ≤ ≥ 1 4 1 0 B B k k Solving quadratic in k to 2 solutions – condone < M1 M1 A1 M1 A1 5 [6] log4 (3x) + log4 (0.5) = log4 (1.5x) log16 (3x – 1) = ( ) 16 log 1 3 log 4 4 − x For change of base – also to base 10,16, 2 ½ log4 (3x – 1) = log4 1 3 − x or 2log4 (1.5x) = log4 (2.25x2) Changing k log z to log zk 3x – 1 = 2.25x2 9x2 – 12x + 4 = 0 ⇒ (3x – 2)2 = 0 ⇒ x = 3 2 Solving 3 term quadratic Accept 0.66 or 0.67 or better B1 M1 M1 A1 M1 A1

Page 2 Mark Scheme Syllabus Paper IGCSE EXAMINATIONS – NOVEMBER 2004 0606 2 © University of Cambridge International Examinations 2005 6 [6] (i) 3sinθ – 2cosθ = 3cosθ + 2sinθ ⇒ sinθ = 5cosθ ⇒ tanθ = 5 OR, squaring + Pythagoras ⇒ sinθ = 26 5 or cosθ = 26 1 for M1 θ = 78.7° or 1.37 rad or better (acute angle only accepted) (ii) x2 + y2 = (9 sin2θ - 12sinθ cosθ + 4 cos2θ) + (9cos2θ + 12sinθ cosθ + 4 sin2θ) = 13 sin2θ + 13cos2θ = 13 Pythagoras M1 A1 A1 B1 M1 A1 c.s.o 7 [7] Put x = a ⇒ 6a3 + 5a2 – 12a = –4 or divide by x – a to remainder Search 6(–2)3 + 5(–2)2 = 12(–2) + 4 = 0 ⇒ a = –2 (at least 2, if unsuccessful, for M1) similarly, if a = 2 1 or 3 2 is found 6a3 + 5a2 – 12a + 4 ≡ (a + 2) (6a2 – 7a + 2) OR, finding 2nd root 6a2 – 7a + 2 ≡ (3a – 2) (2a – 1) = 0 ⇒ a = 2 1 , 3 2 OR, finding 3rd root M1 M1 A1 M1 A1 M1 A1 8 [7] BAX = tan-1 200/150 = tan-1 4/3 ≈ 53.13°, or 36.87°, or 250 ABX = sin-1 {(3sinBAX) ÷ 6} = sin-1 0.4 ≈ 23.58° Incorrect obtuse-angled ∆s – allow M1 for use of sine or cosine rule) AXB = 180° – (53.13° + 23.58° ) = 103.29° V = (6sin103.29°) ÷ sin 53.13° = 7.3 [or via cosine rule] [or VACROSS = 6sin 76.71° ≈ 5.84 or VDOWN = 3 + 6cos76.71° ≈ 4.38] Time = 250 ÷ 7.3 [or 200 ÷ 5.8 or 150 ÷ 4.4] ≈ 34 s (accept 34 ∼ 34.5) 2 stages can be combined by applying cosine rule to velocities: 36 = V 2 + 9 – 6V cos 53.13° M1 ⇒ 10 V 2 – 36V – 270 = 0 A1 Solve M1 V = 2.3 A1 3 stages can be combined by applying cosine rule to displacements: (6t)2 = (250)2 + (3t)2 – 6t cos 53.13° M2 ⇒27t 2 + 900t – 62500 = 0 A2 Solve DM1 t = 34.3 A1 First 5 marks by vector method: V = (150i + 200j )/t B1 VBOAT = (150i + 200j )/t – 3i M1A1 B1 M1 A1 M1 A1 DM1 A1

Page 3 Mark Scheme Syllabus Paper IGCSE EXAMINATIONS – NOVEMBER 2004 0606 2 © University of Cambridge International Examinations 2005 | VBOAT | =| (150 – 3t)i/t + 200j/t | =6 M1 ⇒ 27t 2 + 900t – 62500 = 0 A1 By scale drawing: Construct 53.13 with 200, 150 ⇒ Velocity ∆ M1A1 ⇒ V = 7.3 ± 0.1 M1A1 ⇒ T = 34 DM1A1 9 [7] (i) Y = log y, X = x m = log b, c = log a (ii) Y = log y, X = log x m = k, c = log A (iii) Y = 1/y, X = 1/x    − = = p m p c / 9 / 1 [Other valid alternatives acceptable Y y x y x y x x 1 X x y y y x x y 1 m q q 1 p p 1 q p − c p q p − q p q − q 1 ] B1 DB1 B1 DB1 M1 A1 A1

Page 4 Mark Scheme Syllabus Paper IGCSE EXAMINATIONS – NOVEMBER 2004 0606 2 © University of Cambridge International Examinations 2005 10 [9] (i) Let y = x2 – 8x + 7 dy/dx = 2x – 8 = 0 at x = 4 d2y/dx2 = 2 ∴min at x = 4 OR via completing the square: y = (x – 4)2 – 9 ⇒ min –9 at x = 4 ∴ f(x) has maximum at x = 4, corroborated by argument re reflection of –9 or by graph (ii) Judge by shape, unless values clearly incorrect. Ignore curve outside domain. Cusp needed at x-axis. Accept straight line for right-hand arm, but curvature, if shown, must be correct. (iii) 0 ≤ f (x) ≤ 9 [condone <] (iv) k = 4 M1 A1 B2, 1, 0 B2, 1, 0 B1 B1 B1

Page 5 Mark Scheme Syllabus Paper IGCSE EXAMINATIONS – NOVEMBER 2004 0606 2 © University of Cambridge International Examinations 2005 11 [10] Let A be (x, y) i.e. (x, 3x) Length of OA = 250 9 2 2 = + x x ⇒ x = 5, A is (5, 15) ( 250 2 2 = + y x enough for M1) Gradient of AB is – 3 1 Equation of AB is y – 15 = −3 1 (x – 5) ⇒ B is (0, 16 3 2 ) AND substitute x = 0 for M1 Decimals 16.6 or 16.7, – 1 p.a. Gradient of BC is 3 Equation of BC is y = 3x + 16 3 2 Meets y + 2x = 0 when –2x = 3x + 16 3 2 ⇒ x = –3 3 1 , C is (–3 3 1 , 6 3 2 ) but accept (–3.32, 6.64), (–3.34, 6.68) In essence, scheme is 3 marks for each of A, B, C. Possible to find B before A e.g. 1 3 2 16 1 565 . 71 sin / 250 1 565 . 71 3 tan ˆ 1 A M B ⇒ = = = − o o OB X O A Gradient of AB is – 3 1 B1 Solve y −16 3 2 = – 3 1 x with y = 3x M1 ⇒ (5,15) A1 B1 M1 A1 B1 M1 A1 B1 M1 M1 A1 12 [10] EITHER (i) v = ∫ + − = 5 . 0 3 . 0 4 . 1 2t t t a d At rest v = 0 ⇒ 3t 2 – 14t – 5 = 0 ⇒ (3t + 1) (t – 5) = 0 ⇒ t = 5 OR, by verifying [1.4t – 0.3t 2 + 0.5]t = 5 = 0 (ii) s = ∫ + − = t t t t v 5 . 0 1 . 0 7 . 0 3 2 d [s] t = 5 = 7.5 [s] t = 10 = –25 OR s 10 – s 5 = –32.5 Total distance = (2× 7.5) + 25 = 40 OR 7.5 + 32.5 M1 A2,1,0 A1 M1 A1√ A1√ A1M1A1c.s.o.

Page 6 Mark Scheme Syllabus Paper IGCSE EXAMINATIONS – NOVEMBER 2004 0606 2 © University of Cambridge International Examinations 2005 12 [10] OR (i) ∫ ∫ − =       + = x x x x x x y 2 2 3 d 2 3 d 2 2 accept x b ax − 2 2 One term correct sufficient for M1 [ ] ( ) 5 . 18 1 6 2 1 24 4 2 = − −       − = (ii) (2, 3) on curve ⇒ 3 = 2a + 4 b 3 2 d d x b a x y − = 0 4 0 d d 2 = − ⇒ =     = b a x y x Solving a = 1, b = 4 y = x + 4 2 2 3 2 24 d d 8 1 d d 4 x x y x x y x = ⇒ − = ⇒ > 0 when x = 2 ∴min [or any equivalent method] M1 A1 DM1 A1 B1 M1 M1 A1 M1 A1c.s.o