Cambridge IGCSE Mathematics - Additional 0606 — 2007 Oct/Nov Paper 2 · Variant 1

This document consists of 6 printed pages and 2 blank pages. SP (SC) T44479/1 © UCLES 2007 [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 9 7 5 2 9 0 3 5 8 * ADDITIONAL MATHEMATICS 0606/02 Paper 2 October/November 2007 2 hours Additional Materials: Answer Booklet/Paper Electronic calculator Mathematical tables www.XtremePapers.com

2 0606/02/O/N/07 © UCLES 2007 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/07 © UCLES 2007 [Turn over 1 The two variables x and y are related by the equation yx2 = 800. (i) Obtain an expression for dy dx in terms of x. [2] (ii) Hence find the approximate change in y as x increases from 10 to 10 + p, where p is small. [2] 2 Solve the equation 3sin ( x–2 – 1) = 1 for 0 < x < 6π radians. [5] 3 (i) Express 9 x + 1 as a power of 3. [1] (ii) Express 3 272x as a power of 3. [1] (iii) Express 54 × 3 272x ––––––––––––––– 9x + 1 + 216(32x – 1) as a fraction in its simplest form. [3] 4 A cycle shop sells three models of racing cycles, A, B and C. The table below shows the numbers of each model sold over a four-week period and the cost of each model in $. Model Week A B C 1 8 12 4 2 7 10 2 3 10 12 0 4 6 8 4 Cost ($) 300 500 800 In the first two weeks the shop banked 30% of all money received, but in the last two weeks the shop only banked 20% of all money received. (i) Write down three matrices such that matrix multiplication will give the total amount of money banked over the four-week period. [2] (ii) Hence evaluate this total amount. [4] 5 (i) Expand (1 + x)5. [1] (ii) Hence express (1 + 2)5 in the form a + b 2, where a and b are integers. [3] (iii) Obtain the corresponding result for (1 – 2)5 and hence evaluate (1 + 2)5 + (1 – 2)5. [2]

4 0606/02/O/N/07 © UCLES 2007 6 Two circular flower beds have a combined area of 29π ––– 2 m2. The sum of the circumferences of the two flower beds is 10π m. Determine the radius of each flower bed. [6] 7 The position vectors of points A and B, relative to an origin O, are 2i + 4j and 6i + 10j respectively. The position vector of C, relative to O, is ki + 25j, where k is a positive constant. (i) Find the value of k for which the length of BC is 25 units. [3] (ii) Find the value of k for which ABC is a straight line. [3] 8 Given that x

and that
= {x : 2 < x < 10} , A = {x : 3x + 2 < 20} and B = {x : x2 < 11x – 28} , find the set of values of x which define (i) A
B, (ii) (A  B). [7] 9 A particle travels in a straight line so that, t s after passing through a fixed point O, its speed, v ms–1, is given by v = 8cos ( t–2). (i) Find the acceleration of the particle when t = 1. [3] The particle first comes to instantaneous rest at the point P. (ii) Find the distance OP. [4]

5 0606/02/O/N/07 © UCLES 2007 [Turn over 10 y x O M X The diagram shows part of the curve y = 4 x – x. The origin O lies on the curve and the curve intersects the positive x-axis at X. The maximum point of the curve is at M. Find (i) the coordinates of X and of M, [5] (ii) the area of the shaded region. [4] 11 Solutions to this question by accurate drawing will not be accepted. y x O C B A (6, –3) The diagram shows a triangle ABC in which A is the point (6, –3). The line AC passes through the origin O. The line OB is perpendicular to AC. (i) Find the equation of OB. [2] The area of triangle AOB is 15 units2. (ii) Find the coordinates of B. [3] The length of AO is 3 times the length of OC. (iii) Find the coordinates of C. [2] The point D is such that the quadrilateral ABCD is a kite. (iv) Find the area of ABCD. [2]

6 0606/02/O/N/07 © UCLES 2007 12 Answer only one of the following two alternatives. EITHER The function f is defined, for x > 0, by f : x  ln x. (i) State the range of f. [1] (ii) State the range of f –1. [1] (iii) On the same diagram, sketch and label the graphs of y = f(x) and y = f –1(x). [2] The function g is defined, for x > 0, by g : x  3x + 2. (iv) Solve the equation fg (x) = 3. [2] (v) Solve the equation f –1g–1 (x) = 7. [4] OR (i) Find the values of k for which y = kx + 2 is a tangent to the curve y = 4x2 + 2x + 3. [4] (ii) Express 4x2 + 2x + 3 in the form a(x + b)2 + c, where a, b and c are constants. [3] (iii) Determine, with explanation, whether or not the curve y = 4x2 + 2x + 3 meets the x-axis. [2] The function f is defined by f : x  4x2 + 2x + 3 where x  p. (iv) Determine the smallest value of p for which f has an inverse. [1]

8 0606/02/O/N/07 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 2007 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 2007 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 2007 0606 2 © UCLES 2007 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 2007 0606 2 © UCLES 2007 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 2007 0606 2 © UCLES 2007 1 [4] (i) dy/dx = x -3 × ( – 1600 ) oe (ii) δy = x y d d δx = – p 1000 1600 = – 1.6p M1 A1 M1 A1√ 2 [5] sin(x/2-1) =1/3 0.34 ( 19.5° ) or 2.8(0) or π - 0.34 ( 160.5° ) Using radians, + 1, ×2 ⇒ 2.68 or 7.6(0) ( both, no other values < 15 ) or 15.2~3 ( no other values between 15 and 20 ) B1 B1√ M1 A1 A1 3 [5] (i) 9x+1 ⇒ 32x+2 (ii) 3 2 27 x ⇒ 32x (iii) Substitute (i) and (ii) into expression Correctly cancel powers involving x simplify ⇒ 2 / 3 B1 B1 B1√ M1 A1 4 [6] (i)( )                       800 500 300 4 8 6 0 12 10 2 10 7 4 12 8 2 .0 2 .0 3 .0 3 .0 or ( )                       2 .0 2 .0 3 .0 3 .0 4 0 2 4 8 12 10 12 6 10 7 8 800 500 300 (ii) 1st product ( ) 6 . 2 6 . 10 7 . 7 or             9000 9000 8700 11600 or transpose of either 2nd product [ 2310+5300 +2080 or 3480 +2610+1800+1800 ] ⇒ $9690 for 9690 with no matrices allow sc 1 M1 A1 M1 A1 M1 A1 5 [6] (i) ( 1 + x )5 = 1 + 5x + 10x2 + 10x3 + 5x4 + x5 (ii) (√2)3 = 2√2 (√2)5 = 4√2 ( 1 + √2 )5 = 41 + 29√2 (iii) ( 1 – √2 )5 = 41 – 29√2 ( 1 + √2 )5 + ( 1 – √2 )5 = 82 B1 B1 B1 B1 B1√ B1√

Page 5 Mark Scheme Syllabus Paper IGCSE – October/November 2007 0606 2 © UCLES 2007 6 [6] Uses x,y or equivalent correctly in either equation πx2 + πy2 = 29π / 2 and 2πx + 2πy = 10π Eliminate one variable [x2 + ( 5 – x )2 = 14.5] Correctly substitute into formula or factorise 3 term quadratic [ 4x2 – 20x + 21 = 0 ] 1.5, 3.5 M1 A1 A1 M1 M1 A1 7 [6] (i) BC = ( k – 6 )i + 15j (may be implied) ( k – 6 )2 + 152 = 252 ⇒ k = 26 ( ignore – 14 ) (ii) AB = 4i + 6j [ or AC = ( k – 2 )i + 21j ] or grad 6/4 or ratio 6:4       =       − 6 4 15 6 λ k or       =       − 6 4 21 2 µ k or       − =       − 15 6 21 2 k k ν λ = 2.5 or µ = 3.5 or ν = 1.4 or x step =10 or solve sim equations from y=mx+c k = 16 B1 M1 A1 B1 M1 A1 8 [7] A : x < 6 x2 – 11x + 28 = 0 ⇒ ( x – 4 )( x – 7 ) = 0 ⇒ x = 4, 7 B : 4 < x < 7 (i) 4 < x < 6 (ii) 7 ≤ x < 10 If integers used allow sc1 for (i) {5} and sc1 for (ii) {7, 8, 9} B1 M1 A1 M1A1 M1A1 9 [7] (i) ( a = ) sin ( t / 2 ) × ( – 4 ) When t = 1, a ≈ – 1.92 (ii) (s = ) sin ( t / 2 ) × 16 At P, v = 0, t = π OP = 16 2 sin 16 π 0 =          t M1 A1 A1√ M1 A1 B1 A1√ 10 [9] (i) At X, y = 0, x = 16 dy / dx = x -1/2 ×4×½ – 1 ( dy/dx ) = 0 ⇒ at M, x = 4, y = 4 (ii) x xd 4∫ = x 3/ 2 ×4×⅔ or ×4 ÷ 3/2 ( ) 2 / d 2 x x x − = − ∫ 16 0 2 2 / 3 2 3 8         −x x = 42⅔ B1 cso M1 A1 M1 A1 M1 A1 B1 A1

Page 6 Mark Scheme Syllabus Paper IGCSE – October/November 2007 0606 2 © UCLES 2007 11 [9] (i) mAO = – 1/ 2 Equation of OB is y = 2x (ii) (Length of OA = √ 45 ) ⇒ Length of OB = 2√ 5 oe ( ) 5 2 2 2 2 = + x x ⇒ x = 2 B is ( 2, 4 ) (iii) 3 1 3 1 = = AO OC ( – 6, 3 ) = ( – 2, 1 ) (iv) Area COB = ⅓ × area AOB = 5 ⇒ Area ABCD = 40 [ Any valid method] B1 B1√ M1 M1 A1 M1 A1 M1 A1 12E [10] (i) – ∞ < (f )< ∞ ( i.e. all real values ) (ii) ( f -1 )> 0 ( i.e. all positive values ) (iii) Graph of f (x) = ln x Graph of f -1(x) ( as ex or reflection in y = x ) (iv) fg(x) = ln ( 3x + 2 ) = 3 ⇒ x = ( e3 – 2 ) / 3 [≈ 6.0(3)] (v) g -1(x) = ( x – 2 ) / 3 f -1(x) = ex [ can be earned anywhere] e ( x – 2 ) / 3 = 7 ⇒ x = 3ln 7 + 2 oe [≈ 7.84 condone 7.8] Or g -1(x) = ( x – 2 ) / 3 f -1g -1(x) = 7 ⇒ g -1(x) = f (7) ( x – 2 ) / 3 = ln 7 ⇒ x = 3ln 7 + 2 oe [≈ 7.84 condone 7.8] Or f -1g-1(x) = 7 ⇒ g -1(x) = f (7) ⇒ x = g f (7) x = g ( ln 7 ) = 3ln 7 + 2 oe [≈ 7.84 condone 7.8] B1 B1 B1 B1 M1 A1 B1 B1 M1 A1 B1 B1 M1 A1 B1 B1 M1 A1 12O [10] (i) Eliminate y and rearrange [4x2 + ( 2 – k)x + 1 = 0] Equal roots ⇒ [( 2 – k)2 = 16] Or dy/dx = k = 8x + 2 Use 3 equations to eliminate 2 of k,y,x Solve for 2 values of k k = – 2 or 6 (ii) 4( x + ¼ )2 + 2¾ or a = 4, b = ¼, c = 2¾ (iii) 4x2 + 2x + 3 = 0 ⇒ 22 – 4×4×3 < 0 ⇒ no real roots Or using (ii), minimum value of y = 2¾ [when x = – ¼ ] Or minimum y = 2¾ [when x = – ¼ ] via dy/dx = 8x + 2, And d2y/dx2 = 8 > 0 or indication of shape Or 4( x + ¼ )2 = -11/4 not possible (iv) f must be one-one ; hence, from (ii), (p =) – ¼ M1 M1 M1 M1 M1 A1cso B1 B1 B1 M1 A1 M1 A1 M1 A1 M1 A1 B1√