Cambridge IGCSE Mathematics - Additional 0606 — 2011 Oct/Nov Paper 2 · Variant 1

This document consists of 18 printed pages and 2 blank pages. DC (NF/CGW) 48828 © UCLES 2011 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 1 7 4 1 3 5 0 3 6 4 * ADDITIONAL MATHEMATICS 0606/21 Paper 2 October/November 2011 2 hours Candidates answer on the Question Paper. Additional Materials: Electronic calculator READ THESE INSTRUCTIONS FIRST 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 pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. 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. For Examiner’s Use 1 2 3 4 5 6 7 8 9 10 11 12 Total www.theallpapers.com

2 0606/21/O/N/11 © UCLES 2011 For Examiner’s Use 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 0606/21/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 1 Solve the equation |4x – 5| = 21. [3] 2 Given that the straight line y = 3x + c is a tangent to the curve y = x2 + 9x + k, express k in terms of c. [4] www.theallpapers.com

4 0606/21/O/N/11 © UCLES 2011 For Examiner’s Use 3 3 2 2+√3 θ Without using a calculator, find the value of cosθ, giving your answer in the form a + b 3 c , where a, b and c are integers. [5] www.theallpapers.com

5 0606/21/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 4 (i) Given that y = 1 x2 + 3 , show that dy dx = kx (x2 + 3)2 , where k is a constant to be found. [2] (ii) Hence find  6x (x2 + 3)2 dx and evaluate  3 1 6x (x2 + 3)2 dx. [3] www.theallpapers.com

6 0606/21/O/N/11 © UCLES 2011 For Examiner’s Use 5 (a) The functions f and g are defined, for x , by f : x  2x + 3, g : x  x2 – 1. Find fg(4). [2] (b) The functions h and k are defined, for x > 0, by h : x  x + 4, k : x  x . Express each of the following in terms of h and k. (i) x  x + 4 [1] (ii) x  x + 8 [1] (iii) x  x2 – 4 [2] www.theallpapers.com

7 0606/21/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 6 Solutions to this question by accurate drawing will not be accepted. The points A(1, 4), B(3, 8), C(13, 13) and D are the vertices of a trapezium in which AB is parallel to DC and angle BAD is 90°. Find the coordinates of D. [6] www.theallpapers.com

8 0606/21/O/N/11 © UCLES 2011 For Examiner’s Use 7 (a) Given that tan x = p, find an expression, in terms of p, for cosec2x. [3] (b) Prove that (1 + secθ )(1 – cosθ ) = sinθ tanθ. [4] www.theallpapers.com

9 0606/21/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 8 A X B P b a O In the diagram OA = a, OB = b and AP = 2 5 AB . (i) Given that OX = μOP, where μ is a constant, express OX in terms of μ, a and b. [3] (ii) Given also that AX = λ OB, where λ is a constant, use a vector method to find the value of μ and of λ. [5] www.theallpapers.com

10 0606/21/O/N/11 © UCLES 2011 For Examiner’s Use 9 The table shows experimental values of two variables x and y. x 1 2 3 4 5 y 3.40 2.92 2.93 3.10 3.34 It is known that x and y are related by the equation y = a x + bx, where a and b are constants. (i) Complete the following table. x x y x [1] (ii) On the grid on page 11 plot y x against x x and draw a straight line graph. [2] (iii) Use your graph to estimate the value of a and of b. [3] (iv) Estimate the value of y when x is 1.5. [1] www.theallpapers.com

11 0606/21/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 2 1 2 3 4 5 6 7 8 O 4 6 8 10 12 x √ x y √ x www.theallpapers.com

12 0606/21/O/N/11 © UCLES 2011 For Examiner’s Use 10 It is given that A =  3 2 1 –5 and B =  –1 4 –2 3. (i) Find 2A – B. [2] (ii) Find BA. [2] www.theallpapers.com

13 0606/21/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use (iii) Find the inverse matrix, A–1. [2] (iv) Use your answer to part (iii) to solve the simultaneous equations 3x + 2y = 23, x – 5y = 19. [2] www.theallpapers.com

14 0606/21/O/N/11 © UCLES 2011 For Examiner’s Use 11 (a) (i) Solve 52x+3 252x = 252–x 125x . [3] (ii) Solve 1g y + 1g(y – 15) = 2. [4] www.theallpapers.com

15 0606/21/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use (b) Without using a calculator, and showing each stage of your working, find the value of 2 log12 4 – 1 2 log12 81 + 4 log12 3. [3] www.theallpapers.com

16 0606/21/O/N/11 © UCLES 2011 For Examiner’s Use 12 Answer only one of the following two alternatives. EITHER y x A P (1,1n 2) y =1n (x+1) – 1n x C O B D The diagram shows part of the curve y = 1n (x +1) – 1n x. The tangent to the curve at the point P (1, 1n 2) meets the x-axis at A and the y-axis at B. The normal to the curve at P meets the x-axis at C and the y-axis at D. (i) Find, in terms of 1n 2, the coordinates of A, B, C and D. [8] (ii) Given that Area of triangle BPD Area of triangle APC = 1 k, express k in terms of 1n 2. [3] OR A curve has equation y = xex. The curve has a stationary point at P. (i) Find, in terms of e, the coordinates of P and determine the nature of this stationary point. [5] The normal to the curve at the point Q (1, e) meets the x-axis at R and the y-axis at S. (ii) Find, in terms of e, the area of triangle ORS, where O is the origin. [6] www.theallpapers.com

17 0606/21/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use Start your answer to Question 12 here. Indicate which question you are answering. EITHER OR … … … … … … … … … … … … … … … … … … … … … … … … www.theallpapers.com

18 0606/21/O/N/11 © UCLES 2011 For Examiner’s Use Continue your answer here if necessary. … … … … … … … … … … … … … … … … … … … … … … … … … … … www.theallpapers.com

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20 0606/21/O/N/11 © UCLES 2011 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. BLANK PAGE www.theallpapers.com

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2011 question paper for the guidance of teachers 0606 ADDITIONAL MATHEMATICS 0606/21 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. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the October/November 2011 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 2011 0606 21 © University of Cambridge International Examinations 2011 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 2011 0606 21 © University of Cambridge International Examinations 2011 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 2011 0606 21 © University of Cambridge International Examinations 2011 1 6.5 4x – 5 = –21 or (4x – 5)2 = 212 –4 B1 M1 A1 [3] 2 Eliminates y x2 + 6x + k – c (= 0) Uses b2 = 4ac or completes square k = c + 9 M1 A1 M1 A1 [4] OR dx dy = 2x + 9 B1 Equate to 3 and solve for x (x = –3) M1 Substitute in both equations and equate M1 k = c + 9 A1 3 ( ) ( ) 3 2 4 9 3 2 4 cos 2 + − + + = θ or ( ) ( ) θ cos 3 2 4 3 2 4 9 2 + − + + = M1 ( ) 3 4 7 3 2 2 + = + B1 ( ) 3 2 4 3 4 2 + + A1 Multiply top and bottom by 2 – 3 M1 2 3 3 2 4 + − oe A1 [5] 4 (i) ( ) 2 2 3 + x kx k = –2 M1 A1 (ii) ( ) 3 1 2 6 2 + × − x M1 Correct use of limits in 3 2 + x C M1 0.5 A1 [5] 5 (a) f(15) evaluated or fg(x) = 2(x2 – 1) + 3 33 M1 A1 (b) (i) kh B1 (ii) h2 or hh B1 (iii) h–1k–1 or (kh)–1 B2 [6] www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 21 © University of Cambridge International Examinations 2011 6 mAB = 2 B1 Uses m1m2 = –1 and point A M1 AD: )1 ( 2 1 4 − − = − x y or x + 2y = 9 or 2 9 2 1 + − = x y A1 CD: y – 13 = 2(x – 13) or y = 2x – 13 B1 Solve equation AD with equation CD M1 (7,1) A1 [6] 7 (a) x x 2 2 tan 1 cot = B1 cosec2x = 1 + cot2x B1 2 1 1 p + = or 2 2 1 p p + B1 OR Draw triangle with 1, p and p2 + 1 correct B1 cosec x p p 1 2 + = B1 cosec2 x 2 2 1 p p + = B1 (b) θ θ cos 1 sec = B1 Multiply out and correct use Pythagoras M1 θ θ cos sin2 A1 θ θ θ θ θ tan sin cos sin sin = A1 [7] 8 (i) b a 5 2 5 3 + = OP oe M1 A1       + = b a 5 2 5 3 µ OX A1 (ii) b a λ OX + = or a b a −       + = 5 2 5 3 µ AX B1 Equates a components M1 3 5 = µ A1 Equates b components M1 3 2 = λ A1 [8] www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 21 © University of Cambridge International Examinations 2011 9 (i) x x 1 2.83 5.20 8 11.18 x y 3.40 4.13 5.07 6.20 7.47 B1 (ii) Plot points on graph B2, 1, 0 (iii) Calculates gradient M1 b = 0.4 ± 0.001 A1 a = 3 ± 0.1 B1 (iv) 3.05 B1 [7] 10 (i)       −13 4 0 5 B2, 1, 0 (ii) Matrix multiplication M1       − − − 19 3 18 7 A1 (iii)       − − − − 3 1 2 5 17 1 or       −3 1 2 5 17 1 B1+ B1 (iv) evaluate       19 23 M1 x = 9, y = –2 or       − =       2 9 y x A1 [8] 11 (a)(i)Express in powers of 5.       = − + x x x x 3 ) 2 ( 2 4 3 2 5 5 5 5 M1 Use rules of indices (2x + 3 – 4x = 2(2 – x) – 3x) M1 3 1 A1 (ii) LHS = lg y(y – 15) B1 2 = lg 100 B1 Solve 3 term quadratic M1 20 only A1 (b) log12 16 – log12 9 + log12 81 B1 Correctly combine 3 logarithms M1 2 A1 [10] www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 21 © University of Cambridge International Examinations 2011 12E (i) x x x y 1 1 1 d d − + = B1 gradient tangent 2 1 − = B1 y – ln 2 = – )1 ( 2 1 − x M1 ( ) 0,2 ln 2 1+ A A1       + 2 ln 2 1 ,0 B A1 Uses m1m2 = –1 in equation of line (y – ln 2 = 2(x – 1)) M1      − 0,2 ln 2 1 1 C A1 ( ) 2 ln 2 ,0 + − D A1 (ii) Valid method for area of triangle M1 1.25 or 1.25 × (ln 2)2 or 0.601 A1 k = (ln 2)2 A1 [11] 12O (i) Use product rule M1 (x + 1)ex A1 Solve 0 d d = x y M1       − − e 1 ,1 A1 Shows minimum B1 (ii) Gradient tangent = 2e B1 Use m1m2 = –1 in equation of line ( )      − − = − 1 e 2 1 e x y M1 ( ) 0 , e 2 1 2 + R A1       + e 2 e 2 1 ,0 2 S A1 Area of triangle = ( ) e 4 e 2 1 2 2 + M1 A1 [11] www.theallpapers.com