Cambridge IGCSE Mathematics - Additional 0606 — 2010 May/June Paper 1 · Variant 2

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. ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 2010 2 hours Additional Materials: Answer Booklet/Paper Electronic calculator Graph paper (2 sheets) This document consists of 6 printed pages and 2 blank pages. DC (SM/KN) 25700 © UCLES 2010 [Turn over * 6 3 1 6 6 1 6 0 8 9 * UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education www.theallpapers.com

2 0606/12/M/J/10 © UCLES 2010 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 0606/12/M/J/10 © UCLES 2010 [Turn over 1 Find the coordinates of the points of intersection of the curve y2 + y = 10x – 8x2 and the straight line y + 4x + 1 = 0. [5] 2 The expression 6x3 + ax2 – (a + 1)x + b has a remainder of 15 when divided by x + 2 and a remainder of 24 when divided by x + 1. Show that a = 8 and find the value of b. [5] 3 Given that J OA =  –17 25 and J OB =  4 5, find (i) the unit vector parallel to J AB, [3] (ii) the vector J OC, such that J AC = 3J AB. [2] 4 O y2 sec x × × (2.4, 1.6) (1.3, 3.8) Variables x and y are such that, when y2 is plotted against sec x, a straight line graph passing through the points (2.4, 1.6) and (1.3, 3.8) is obtained. (i) Express y2 in terms of sec x. [3] (ii) Hence find the exact value of cos x when y = 2. [2] www.theallpapers.com

4 0606/12/M/J/10 © UCLES 2010 5 y x O A B y = 6 – 3 3–x The diagram shows part of the curve y = 6 – 3–x which passes through the point A where x = 3. The normal to the curve at the point A meets the x-axis at the point B. Find the coordinates of the point B. [5] 6 (a) (i) On the same diagram, sketch the curves y = cos x and y = 1 + cos 2x for 0  x  2π. [3] (ii) Hence state the number of solutions of the equation cos 2x – cos x + 1 = 0 where 0  x  2π. [1] (b) The function f is given by f(x) = 5sin 3x. Find (i) the amplitude of f, [1] (ii) the period of f. [1] 7 The table shows values of the variables p and v which are related by the equation p = kvn, where k and n are constants. v 10 50 110 230 p 1412 151 53 19 (i) Using graph paper, plot lg p against lg v and draw a straight line graph. [3] Use your graph to estimate (ii) the value of n, [2] (iii) the value of p when v = 170. [2] www.theallpapers.com

5 0606/12/M/J/10 © UCLES 2010 [Turn over 8 Given that A =  4 3 1 2 and B =  –2 0 1 4, find (i) 3A – 2B, [2] (ii) A–1, [2] (iii) the matrix X such that XB–1 = A. [3] 9 O A B X Y 8 cm 3 cm The diagram shows a sector OXY of a circle centre O, radius 3 cm and a sector OAB of a circle centre O, radius 8 cm. The point X lies on the line OA and the point Y lies on the line OB. The perimeter of the region XABYX is 15. 5 cm. Find (i) the angle AOB in radians, [3] (ii) the ratio of the area of the sector OXY to the area of the region XABYX in the form p : q, where p and q are integers. [4] 10 A music student needs to select 7 pieces of music from 6 classical pieces and 4 modern pieces. Find the number of different selections that she can make if (i) there are no restrictions, [1] (ii) there are to be only 2 modern pieces included, [2] (iii) there are to be more classical pieces than modern pieces. [4] www.theallpapers.com

6 0606/12/M/J/10 © UCLES 2010 11 A particle moves in a straight line such that its displacement, x m, from a fixed point O on the line at time t seconds is given by x = 12{1n (2t + 3)}. Find (i) the value of t when the displacement of the particle from O is 48 m, [3] (ii) the velocity of the particle when t = 1, [3] (iii) the acceleration of the particle when t =1. [3] 12 Answer only one of the following two alternatives. EITHER y y = 5 x O A B , 7 C π––4 The diagram shows part of a curve for which dy –– dx = 8 cos 2x. The curve passes through the point B  π–4, 7. The line y = 5 meets the curve at the points A and C. (i) Show that the curve has equation y = 3 + 4 sin 2x. [3] (ii) Find the x-coordinate of the point A and of the point C. [4] (iii) Find the area of the shaded region. [5] OR A curve is such that dy –– dx = 6e3x – 12. The curve passes through the point (0, 1). (i) Find the equation of the curve. [4] (ii) Find the coordinates of the stationary point of the curve. [3] (iii) Determine the nature of the stationary point. [2] (iv) Find the coordinates of the point where the tangent to the curve at the point (0, 1) meets the x-axis. [3] www.theallpapers.com

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8 0606/12/M/J/10 © UCLES 2010 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 May/June 2010 question paper for the guidance of teachers 0606 ADDITIONAL MATHEMATICS 0606/12 Paper 12, 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 May/June 2010 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– May/June 2010 0606 12 © UCLES 2010 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– May/June 2010 0606 12 © UCLES 2010 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– May/June 2010 0606 12 © UCLES 2010 1 24x2 – 6x = 0 (or y2 + 3y + 2 = 0) leading to (0, 1) and       −2 , 4 1 M1 M1 DM1 A1,A1 [5] M1 for attempt to get an equation in one variable. M1 for attempt to get 2 or 3 term quadratic = 0 DM1 for attempt to solve A1 for each pair of values 2 6(–2)3 + a(–2)2 – (a + 1)(–2) + b = 15 6a + b = 61 when x = –1, 2a + b = 29 leading to a = 8 and b = 13 M1 A1 A1 M1 A1 [5] M1 for substitution of x = –2 or –1, or verification A1 for each correct (allow unsimplified) M1 for attempt to solve A1 for a = 8, b = 13 3 (i)      − −       = 25 17 5 4 B A r       − = 20 21 unit vector =       − 29 20 29 21 or equivalent (ii)       − =      − − 20 21 3 25 17 C O r       − = 35 46 C O r B1 M1, A1 [3] M1 A1 [2] B1 for B A r M1 for magnitude of B A r M1 for B A r 3 25 17 +      − 4 (i) gradient = –2 y2 = –2sec x + c leading to y2 = –2sec x + 6.4 (ii) when y = 2, cos x = 6 5 B1 M1 A1 [3] DM1 A1 [2] B1 for gradient M1 for correct attempt to link y2 and sec x DM1 for attempt to solve their equation using y = 2 5 2 3 d d x x y = , gradient at A = 3 1 , normal grad = –3 coords of A (3, 5) normal y – 5 = –3(x – 3) when y = 0, x = 3 14 M1 DM1 B1 DM1 A1 [5] M1 for attempt to differentiate DM1 for use of perp grads DM1 for attempt at normal www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE– May/June 2010 0606 12 © UCLES 2010 6 (a) (i) 2 4 6 −2 −1 1 2 3 x y (ii) 4 (b) (i) 5 (ii) 3 π 2 B1 B1 B1 [3] B1 [1] B1 [1] B1 [1] B1 for y = cos x B1 for either a translation of       1 0 or 2 cycles B1 for correct curve 7 (i) lgv 1 1.70 2.04 2.36 lgp 3.15 2.18 1.72 1.28 (ii) gradient = n = –1.37 (allow 1.32 to 1.42) (iii) p = 30 (allow 28 to 32) M1 A2,1,0 [3] M1 A1 [2] M1 A1 [2] M1 for attempt to take logs and plot graph –1 for each error either in table or on graph. M1 for use of gradient M1 for use of graph or their equation 8 (i)       −2 1 9 16 (ii)       − − − 4 1 3 2 3 8 1 (iii) X = AB =      − 8 0 12 5 B1 B1 [2] B1 B1 [2] M1 A2,1,0 [3] B1 at least 2 correct B1 all correct B1 for determinant B1 for matrix M1 for attempt at valid method –1 each error www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE– May/June 2010 0606 12 © UCLES 2010 9 (i) 5 + 5 + 3θ + 8θ = 15.5 θ = 0.5 (ii) ½(3)2 θ : ½(8)2 θ – ½(3)2 θ = 9 : 55 M1, DM1 A1 [3] M1 DM1 DM1, A1 [4] M1 for use of arc length DM1 for attempt to find perimeter M1 for a sector area M1 for attempt to find area of XABY M1 for attempt to obtain ratio 10 (i) 10C7 = 120 (ii) 6C5 × 4C2 = 36 (iii) Need (6C + 1M) + (5C + 2M) + (4C + 3M) 4 + (ii) + (6C4 × 4C3) = 100 B1 [1] B1, B1 [2] M1 B1, B1 A1 [4] B1 for 6C5 × 4C2, B1 for 36 M1 for a correct method B1 for 4, B1 for 60 11 (i) 48 = 12 ln (2t + 3) 2t + 3 = e4 t = 25.8 (ii) x = 12 ln (2t + 3) 3 2 24 + = t v when t = 1, v = 4.8 (iii) 2) 3 2 ( 48 + − = t a when t = 1, a = –1.92 M1 DM1 A1 [3] B1 B1 B1 [3] B1 √B1 B1 [3] M1 for attempt to deal with logs DM1 for attempt to solve B1 3 2 1 + t B1 24 B1 for 4.8 B1 for 2) 3 2 ( 1 + t √B1 on ‘24’ B1 for –1.92 www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE– May/June 2010 0606 12 © UCLES 2010 12 EITHER (i) y = 4 sin 2x + c passes through       7, 4 π , c = 3 (ii) 5 = 4 sin 2x + 3 0.5 = sin 2x x = 12 π 5 , 12 π (iii) ∫ 12 12 5 π π 4 sin 2x + 3dx [ ] 12 12 5 3 2 cos 2 π π x x + − = π + 2 3 Shaded area = π + 2 3 – 3 5π (= 1.37) M1 M1 A1 [3] M1 M1 A1 √A1 [4] M1 A1 DM1 M1 A1 [5] M1 for attempt to integrate M1 for attempt to get c provided a function of sin 2x is used M1 for attempt to equate to 5 and solve M1 for a correct method to find x √A1 on first solution M1 for attempt to integrate DM1 for correct use of limits M1 for area of rectangle 12 OR (i) y = 2e3x – 12x + c Passes through (0, 1), so c = –1 (ii) 6e3x – 12 = 0 leading to x = 3 1 ln 2 and y = 3 – 4 ln 2 (allow (0.231, 0.227) (iii) 2 2 d d x y = 18e3x, always +ve so min (iv) at (0, 1), gradient = –6 tangent : y – 1 = –6(x – 0) when y = 0, x = 6 1 M1, A1 M1, A1 [4] M1 A1, A1 [3] M1, A1 [2] M1 DM1 A1 [3] M1 for attempt to integrate, condone omission of c M1 for attempt to obtain c M1 for attempt to solve M1 for a complete, correct method M1 for attempt to get equation of tangent at (0, 1) DM1 for substitution of y = 0 www.theallpapers.com