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

This document consists of 16 printed pages. DC (LEO/CGW) 48824 © UCLES 2011 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 6 8 9 0 0 6 4 0 6 4 * ADDITIONAL MATHEMATICS 0606/12 Paper 1 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 Total www.theallpapers.com

2 0606/12/O/N/11 © UCLES 2011 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/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 1 Show that 1 tan θ + cot θ = sin θ cos θ. [3] www.theallpapers.com

4 0606/12/O/N/11 © UCLES 2011 For Examiner’s Use 2 Find the coordinates of the points where the line 2y = x – 1 meets the curve x2 + y2 = 29. [5] www.theallpapers.com

5 0606/12/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 3 (i) Express logx2 in terms of a logarithm to base 2. [1] (ii) Using the result of part (i), and the substitution u = log2x, find the values of x which satisfy the equation log2x = 3 – 2 logx2. [4] www.theallpapers.com

6 0606/12/O/N/11 © UCLES 2011 For Examiner’s Use 4 A curve has equation y = (3x2 + 15) 2 3. Find the equation of the normal to the curve at the point where x = 2. [6] www.theallpapers.com

7 0606/12/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 5 Variables x and y are such that, when y2 is plotted against 2x, a straight line graph is obtained. This line has a gradient of 5 and passes through the point (16,81). O (16,81) 2 x y 2 (i) Express y2 in terms of 2x. [3] (ii) Find the value of x when y = 6. [3] www.theallpapers.com

8 0606/12/O/N/11 © UCLES 2011 For Examiner’s Use 6 (i) Given that (3 + x)5 + (3 – x)5 = A + Bx2 + Cx4, find the value of A, of B and of C. [4] (ii) Hence, using the substitution y = x2, solve, for x, the equation (3 + x)5 + (3 – x)5 = 1086. [4] www.theallpapers.com

9 0606/12/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 7 (i) Show that 4 – x  2 x can be written in the form px – 1 2 + q + rx 1 2, where p, q and r are integers to be found. [3] (ii) A curve is such that dy dx = 4 – x  2 x for x > 0. Given that the curve passes through the point (9, 30), find the equation of the curve. [5] www.theallpapers.com

10 0606/12/O/N/11 © UCLES 2011 For Examiner’s Use 8 The line CD is the perpendicular bisector of the line joining the point A (–1, –5) and the point B (5,3). (i) Find the equation of the line CD. [4] www.theallpapers.com

11 0606/12/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use (ii) Given that M is the midpoint of AB, that 2CM = MD, and that the x-coordinate of C is – 2, find the coordinates of D. [3] (iii) Find the area of the triangle CAD. [2] www.theallpapers.com

12 0606/12/O/N/11 © UCLES 2011 For Examiner’s Use 9 (i) Given that y = x sin 4x, find dy dx . [3] (ii) Hence find  x cos 4x dx and evaluate 0 π 8 x cos 4x dx. [6] www.theallpapers.com

13 0606/12/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use 10 (i) Solve 2 sec2 x = 5 tan x + 5, for 0° < x < 360°. [5] (ii) Solve 2 sin  y 2 + π 3 = 1, for 0 < y < 4π radians. [5] www.theallpapers.com

14 0606/12/O/N/11 © UCLES 2011 For Examiner’s Use 11 Answer only one of the following two alternatives. EITHER A curve has equation y = e–x (Acos 2x + Bsin 2x). At the point (0, 4) on the curve, the gradient of the tangent is 6. (i) Find the value of A. [1] (ii) Show that B = 5. [5] (iii) Find the value of x, where 0 < x < π 2 radians, for which y has a stationary value. [5] OR A curve has equation y = 1n(x2 – 1) x2 – 1 , for x > 1. (i) Show that dy dx = k x(1 – 1n(x2 –1)) (x2 – 1)2 , where k is a constant to be found. [4] (ii) Hence find the approximate change in y when x increases from 5 to 5 + p, where p is small. [2] (iii) Find, in terms of e, the coordinates of the stationary point on the curve. [5] Start your answer to Question 11 here. Indicate which question you are answering. EITHER OR … … … … … … … … … … … www.theallpapers.com

15 0606/12/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use Continue your answer to Question 11 here. … … … … … … … … … … … … … … … … … … … … … … … … … … … www.theallpapers.com

16 0606/12/O/N/11 © UCLES 2011 For Examiner’s Use 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. Continue your answer here if necessary. … … … … … … … … … … … … … … … … … … … … … … … … … 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 0606/12 Paper 1, 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 12 © 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 12 © 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 12 © University of Cambridge International Examinations 2011 1 1 sin cos cos sin θ θ θ θ + M1 M1 for adding fractions in terms of sin/cos/tan/cot correctly 2 2 sin cos sin cos θ θ θ θ = + M1 M1 for use of correct identity sin cos θ θ = A1 [3] A1 for correct solution only OR 2 2 tan cot or tan 1 cot 1 θ θ θ θ + + M1 M1 for adding fractions in terms of tan/cot correctly = 2 2 tan cot or sec cosec θ θ θ θ M1 M1 for use of correct identity sin cos θ θ = A1 A1 for correct solution only 2 ( ) 2 2 2 1 29 y y + + = (or 2 5 2 115 0 x x − − = ) M1 M1 for attempt to get an equation in terms of one variable only leading to 2 5 4 28 0 y y + − = DM1 DM1 for obtaining a 3 term quadratic equation (or 2 2 1 29 2 x x −   + =     ) DM1 DM1 for attempt to solve quadratic equation 23 14 , 5 5 x y = − = − and A1 A1 for a pair of values 5, 2 x y = = (5, 2) spotted gets B1 A1 [5] www.theallpapers.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 12 © University of Cambridge International Examinations 2011 3 (i) 2 1 log x or 2 2 log 2 log x B1 (ii) 2 3 2 0 u u − + = ( )( ) 1 2 0 u u − − = M1 M1 for a correct attempt to obtain and solve a quadratic equation in terms of u or 2 log x 1, 2 u = A1 M1 A1 for 1, 2 u = M1 for attempt to solve an equation of the form 2 log x k = leading to 2k x = 2 x = and 4 x = A1 [5] A1 for both 4 When x = 2, y = 9 B1 B1 for y = 9 ( ) 1 2 3 d 2.6 . 3 15 d 3 y x x x − = + B1, B1 B1 for 2.6 3 x , B1 for ( ) 1 2 3 3 15 x − + when x = 2, d 8 d 3 y x = ∴grad of normal = 3 8 − normal : ( ) 3 9 2 8 y x − = − − M1 M1 M1 for use of 1 2 1 m m = − M1 for attempt to find equation of normal, must be using gradient of a perpendicular line 8 3 78 y x + = A1 [6] A1 allow unsimplified 5 (i) 2 2x y m c = + M1 M1 for use of straight line equation as given 81 = 80 + c c = 1 M1 M1 for use of (16, 81) ( ) 2 5 2 1 x y = + A1 A1 – do not allow if subsequent incorrect work is seen (ii) ( ) 36 5 2 1 x = + M1 M1 substitution of y = 6 into their equation in terms of 2y and 2x 1eading to 7 2x = DM1 DM1 for correct solution of equation of the form 2x k = x = 2.81 A1 [6] www.theallpapers.com

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 12 © University of Cambridge International Examinations 2011 6 (i) 2 3 4 5 243 405 270 90 15 x x x x x + + + + + 2 3 4 5 243 405 270 90 15 x x x x x − + − + − 2 4 486 540 30 x x + + (ii) 2 30 540 600 0 y y + − = B1 B1 B1 B1 M1 DM1 Assuming correct terms in x B1 for 486 B1 for 540 B1 for 30 B1 for all terms correct and no extra terms M1 for attempt to obtain a 3 term quadratic equation. DM1 for correct attempt at solution of quadratic 1.05 y = A1 A1 need both solutions leading to 1.02 x = ± A1 [8] A1 need both solutions 7 (i) 1 1 2 2 16 8 x x −− + B1, B1, B1 B1 for each correct term (ii) ( ) 1 3 2 2 2 32 8 3 y x x x c = − + + M1 A2, 1, 0 M1 for attempt to integrate a 3 term expression –1 for each error When x = 9 and y = 30, c = –12 M1 A1 [8] M1 for attempt to find c, must have attempted integration A1 for c = –12 8 (i) M (2, 4) B1 Allow in (ii) Grad AB = 2 6 −, ⊥ grad = 3 M1 M1 for attempt to find gradient of perpendicular CD: ( ) 4 3 2 y x − = − DM1 A1 DM1 for straight line equation (ii) C (–2, 2) D (10, -7) B1 B1, B1 B1 for y coordinate of A B1 for each (iii) Area = 37.5 M1, A1 [9] M1 for a valid attempt at area Special case: (ii) allow D( ) 6,5 − (iii) area 12.5 B1, B1 M1, A1 www.theallpapers.com

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 12 © University of Cambridge International Examinations 2011 9 (i) d .4cos4 sin 4 d y x x x x = + M1 B1, A1 M1 for differentiation of a product B1 for 4cos4x , A1 all else correct (ii) 1 sin 4 sin 4 d 4 I x x x x    = −      ∫ DM1 DM1 for realising integration is form reverse process of (i) – do not need 1 4       until last A1 = 1 1 sin 4 cos4 4 4 x x x       −−             A1, A1 B1 A1 for sin 4 x x , A1 for sin 4 d x x ∫ B1 for 1 cos4 4 x − For definite integral π 8 0 1 1 sin 4 cos4 4 4 x x x       −−             M1 M1 for correct application of limits = π 1 32 16 − , or 0.0357 A1 [9] 10 (i) 2 2tan 2 5tan 5 x x + = + M1 M1 for use of correct identity 2 2tan 5tan 3 0 x x − − = M1 M1 for solution of 3 term quadratic equation 1 tan ,tan 3 2 x x = − = M1 M1 for attempt to solve tanx = k from a 3 term quadratic equation o o o o 153.4 ,333.4 and 71.6 ,251.6 x = A1, A1 A1 for any pair (ii) π 1 sin 0.5 3 2 y   + =     M1 M1 for dealing with 2 correctly π π 3π 9π 0.5 , , 3 4 4 4 y + = M1 M1 M1 for correct order of operations M1 for correct order of operations and attempt to get a solution in the range leading to 5π 23π , 6 6 y = A1, A1 [10] Allow decimal equivalents 2.62 and 12.0 www.theallpapers.com

Page 8 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0606 12 © University of Cambridge International Examinations 2011 11 EITHER (i) A = 4 B1 (ii) ( ) d e 2 sin 2 2 cos2 d x y A x B x x − = − + − M1 B1 M1 for differentiation of a product B1 for ( 2 sin 2 2 cos2 )e x A x B x − − + ( ) e cos2 sin 2 x A x B x − + B1 B1 for ( ) e cos2 sin 2 x A x B x − − + when x = 0, 6 = 2B – A, B = 5 (verification acceptable ) M1 A1 M1 for substitution to find B (iii) when d d y x = 0, M1 M1 for their d d y x = 0 ( ) e cos2 sin 2 0 x p x q x − − = M1 M1 for attempt to simplify leading to tan 2 p x q = 6 13   =     M1 M1 for attempt to obtain tan 2 p x q = x = 0.216 M1 A1 [11] M1 for attempt to solve tan 2x k = 11 OR (i) ( )( ) ( ) ( ) 2 2 2 2 2 2 1 2 ln 1 1 d d 1 x x x x x y x x − − − − = − M1 B1 M1 for differentiation of a quotient B1 ( ) 2 2 1 x x − Rearranging to get k = 2 A1 A1 A1 for all else correct A1 for rearrangement to get k = 2 (ii) d d y y p x ∂= , M1 M1 for substitution of 5 x = and correct method leading to 0.108 y p ∂= − √A1 √A1 on their k (iii) when d d y x = 0, ( ) 2 1 ln 1 0 x − − = M1 M1 for d d y x = 0 and attempt to simplify ( ) 2 ln 1 1 x − = A1 A1 for ( ) 2 ln 1 1 x − = 2 1 e x −= or 2.72 A1 A1 for 2 1 e x −= or 2.72 leading to 1 x e = + A1 1 y e = A1 [11] www.theallpapers.com