Cambridge IGCSE Mathematics - Additional 0606 — 2010 Oct/Nov Paper 2 · Variant 2

This document consists of 6 printed pages and 2 blank pages. DC (SLM) 34222 © UCLES 2010 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 9 3 3 7 1 6 4 2 8 8 * ADDITIONAL MATHEMATICS 0606/22 Paper 2 October/November 2010 2 hours Additional Materials: Answer Booklet/Paper Electronic calculator 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. www.theallpapers.com

2 0606/22/O/N/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 0606/22/O/N/10 © UCLES 2010 [Turn over 1 Solve the equation 2x + 10 = 7. [3] 2 The expression x3 + ax2 – 15x + b has a factor x – 2 and leaves a remainder of 75 when divided by x + 3. Find the value of a and of b. [5] 3 A number, N0, of fish of a particular species are introduced to a lake. The number, N, of these fish in the lake, t weeks after their introduction, is given by N = N0e–kt, where k is a constant. Calculate (i) the value of k if, after 34 weeks, the number of these fish has fallen to 1 2 of the number introduced, [2] (ii) the number of weeks it takes for the number of these fish to have fallen to 1 5 of the number introduced. [3] 4 Students take three multiple-choice tests, each with ten questions. A correct answer earns 5 marks. If no answer is given 1 mark is scored. An incorrect answer loses 2 marks. A student’s final total mark is the sum of 20% of the mark in test 1, 30% of the mark in test 2 and 50% of the mark in test 3. One student’s responses are summarized in the table below. Test 1 Test 2 Test 3 Correct answer 7 6 5 No answer 1 3 5 Incorrect answer 2 1 0 Write down three matrices such that matrix multiplication will give this student’s final total mark and hence find this total mark. [5] 5 Find the set of values of m for which the line y = mx – 2 cuts the curve y = x2 + 8x + 7 in two distinct points. [6] www.theallpapers.com

4 0606/22/O/N/10 © UCLES 2010 6 A 4-digit number is formed by using four of the seven digits 1, 3, 4, 5, 7, 8 and 9. No digit can be used more than once in any one number. Find how many different 4-digit numbers can be formed if (i) there are no restrictions, [2] (ii) the number is less than 4000, [2] (iii) the number is even and less than 4000. [2] 7 45 cm 60 cm x cm x cm x cm A rectangular sheet of metal measures 60 cm by 45 cm. A scoop is made by cutting out squares, of side x cm, from two corners of the sheet and folding the remainder as shown. (i) Show that the volume, V cm3, of the scoop is given by V = 2700x – 165x2 + 2x3. [2] (ii) Given that x can vary, find the value of x for which V has a stationary value. [4] 8 Solve the equation (i) lg(5x + 10) + 2 lg3 = 1 + lg(4x + 12), [4] (ii) 92y 37–y = 34y+3 27y–2 . [3] www.theallpapers.com

5 0606/22/O/N/10 © UCLES 2010 [Turn over 9 A plane, whose speed in still air is 250 kmh–1, flies directly from A to B, where B is 500 km from A on a bearing of 060°. There is a constant wind of 80 kmh–1 blowing from the south. Find, to the nearest minute, the time taken for the flight. [7] 10 Solutions to this question by accurate drawing will not be accepted. y x O B(6,5) A(1,4) C D The diagram shows a quadrilateral ABCD in which A is the point (1, 4) and B is the point (6, 5). Angle ABC is a right angle and the point C lies on the x-axis. The line AD is parallel to the y-axis and the line CD is parallel to BA. Find (i) the equation of the line CD, [5] (ii) the area of the quadrilateral ABCD. [4] 11 Solve the equation (i) 5 sin x – 3 cos x = 0, for 0°  x  360°, [3] (ii) 2 cos2 y – sin y – 1 = 0, for 0°  y  360°, [5] (iii) 3 sec z = 10, for 0  z  6 radians. [3] www.theallpapers.com

6 0606/22/O/N/10 © UCLES 2010 12 Answer only one of the following two alternatives. EITHER The functions f and g are defined, for x  1, by f(x) = (x + 1)2 – 4, g(x) = 3x + 5 x – 1 . Find (i) fg(9), [2] (ii) expressions for f –1(x) and g–1(x), [4] (iii) the value of x for which g(x) = g–1(x). [4] OR A particle moves in a straight line so that, at time t s after passing a fixed point O, its velocity is v ms–1, where v = 6t + 4 cos 2t. Find (i) the velocity of the particle at the instant it passes O, [1] (ii) the acceleration of the particle when t = 5, [4] (iii) the greatest value of the acceleration, [1] (iv) the distance travelled in the fifth second. [4] www.theallpapers.com

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8 0606/22/O/N/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 October/November 2010 question paper for the guidance of teachers 0606 ADDITIONAL MATHEMATICS 0606/22 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, 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 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 – October/November 2010 0606 22 © 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 – October/November 2010 0606 22 © 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 – October/November 2010 0606 22 © UCLES 2010 1 –1.5 Solve 2x + 10 = –7 or (2x + 10)2 = 49 –8.5 B1 M1 A1 [3] 2 Find f(2) or f(–3) or long division to remainder 8 + 4a – 30 + b = 0 or 4a + b = 22 –27 + 9a + 45 + b = 75 or 9a + b = 57 Solve simultaneous equations a = 7, b = –6 M1 A1 A1 M1 A1 [5] 3 (i) Solve 0.5 = e–34k using ln or log correctly k = 0.0204 or 2 ln 34 1 M1 A1 (ii) ekt = 5 or e–kt = 0.2 with k numerical 5 ln 1 k t = with k numerical 79 B1 M1 A1 [5] 4 ( )                     − 5.0 3.0 2.0 0 1 2 5 3 1 5 6 7 2 1 5 or ( )           −           2 1 5 0 5 5 1 3 6 2 1 7 5.0 3.0 2.0 Matrix multiplication using 3 × 3 matrix ( ) 30 31 32 or           7.0 6.3 7.5 (or transposed) Matrix multiplication of 1 × 3 with 3 × 1 30.7 B1 M1 A1 DM1 A1 [5] 5 Eliminate y x2 + (8 – m)x + 9 = 0 Use b2 * 4ac Reach (8 – m) * ±6 or solves m2 – 16m + 28 * 0 (* is either > or =) m = 2 and 14 m < 2, m > 14 M1 A1 DM1 DDM1 A1 A1 [6] www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2010 0606 22 © UCLES 2010 6 (i) 7 × 6 ×5 × 4 840 B1 B1 (ii) 2 × 6 × 5 × 4 or ( ) 840 7 2 × 240 M1 A1 (iii) 2 × 5 × 4 × 2 or ( ) 240 6 2 × or clear indication of method 80 M1 A1 [6] 7 (i) ( )( )x x x V − − = 60 2 45 Correctly reach 3 2 2 165 2700 x x x V + − = M1 A1 ag (ii) 2 6 330 2700 d d x x x V + −       = Solve 3 term quadratic expression for 0 d d = x V . 10 only B2,1,0 M1 A1 [6] 8 (i) 9 lg 3 lg 2 = or 2 3 lg 3 lg 2 = 10 lg 1 = Use lg rules correctly to eliminate logs (e.g. 9(5x + 10) = 10(4x + 12)) x = 6 B1 B1 M1 A1 (ii) Express in powers of 3       = − + − ) 2 ( 3 3 4 7 4 3 3 3 3 y y y y Correctly use rules of indices y = 4 M1 M1 A1 [7] www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2010 0606 22 © UCLES 2010 9 250 60 sin 80 sin = α α = 16.1 β = 104 β cos 250 80 2 250 80 2 2 2 × × × − + = v or      = = α β sin 80 60 sin 250 sin v v = 280(.2…) v t 500 = 1 hour 47 minutes or 107 mins M1 A1 A1√ DM1 A1 DM1 A1 [7] 10 (i) 5 1 = AB m Use m1m2 = –1 in equation for BC [y – 5 = –5(x – 6) or 5x + y = 35] C (7,0) Use mCD = mAB and point C in equation of line CD: y(–0) = 5 1 (x – 7) or x – 5y = 7 B1 M1 A1 M1 A1 (ii) At D x = 1 At D y = –1.2 Method for area not involving measuring 28.6 M1 A1 M1 A1 [9] 11 (i) tan x = 0.6 31(.0) or 30.96(…) 211 (= 31 + 180) B1 B1 B1√ (ii) Use cos2 y = 1 – sin2 y 2sin2 y + sin y – 1 = 0 Solve 3 term quadratic for sin y 30 and 150 270 M1 A1 M1 A1 B1 (iii) cos z = 0.3 1.27 5.02 or 5.01 (= 2π – 1.27) B1 B1 B1√ [11] OR www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2010 0606 22 © UCLES 2010 12E (i) fg(9) = f(4) evaluated or fg(x) 4 1 1 5 3 2 −       + − + = x x 21 M1 A1 (ii) Method for f–1(x) f–1(x) 1 4 − + = x Put 1 5 3 − + = x x y and rearrange g–1(x) 3 5 − + = x x M1 A1 M1 A1 (iii) Rearrange two of x x x x x = − + = − + 3 5 1 5 3 to quadratic equation 2(x2 – 4x – 5) = 0 Solve 3 term quadratic 5 only M1 A1 M1 A1 [10] 12O (i) 4 B1 (ii) Differentiate v to find an expression for a 6 – 8 sin 2t Substitute t = 5 10.3 to 10.4 M1 A1 DM1 A1 (iii) 14 B1 (iv) Integrate v to find an expression for s s = 3t2 + 2 sin 2t Use limits 4 and 5 23.9 M1 A1 DM1 A1 [10] www.theallpapers.com