Cambridge IGCSE Mathematics - Additional 0606 — 2013 May/June Paper 2 · Variant 2

This document consists of 15 printed pages and 1 blank page. DC (NH/SW) 57090/3 © UCLES 2013 [Turn over * 5 8 9 9 8 3 2 6 4 7 * ADDITIONAL MATHEMATICS 0606/22 Paper 2 May/June 2013 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. DO NOT WRITE IN ANY BARCODES. 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. UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education www.theallpapers.com

2 0606/22/M/J/13 © UCLES 2013 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/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 1 (4, 18) (1, 3) x2 O y Variables x and y are such that when y is plotted against x2 a straight line graph passing through the points (1, 3) and (4, 18) is obtained. Express y in terms of x. [4] www.theallpapers.com

4 0606/22/M/J/13 © UCLES 2013 For Examiner’s Use 2 (a) Solve the equation 3p+1 = 0.7 , giving your answer to 2 decimal places. [3] (b) Express (4 ) y y x 8 3 3 2 # in the form 2a × xb × yc, where a, b and c are constants. [3] www.theallpapers.com

5 0606/22/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 3 (a) O y x O y A x O y x O y x O y x O y x B C D E F (i) Write down the letter of each graph which does not represent a function. [2] (ii) Write down the letter of each graph which represents a function that does not have an inverse. [2] (b) x O 5 5 y y = f(x) The diagram shows the graph of a function y = f(x). On the same axes sketch the graph of y = f –1(x). [2] www.theallpapers.com

6 0606/22/M/J/13 © UCLES 2013 For Examiner’s Use 4 The position vectors of the points A and B, relative to an origin O, are 4i – 21j and 22i – 30j respectively. The point C lies on AB such that AB AC 3 = . (i) Find the position vector of C relative to O. [4] (ii) Find the unit vector in the direction OC. [2] www.theallpapers.com

7 0606/22/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 5 Calculators must not be used in this question. A B D X x C 5 4 + 7 2 The diagram shows a trapezium ABCD in which AD = 7 cm and AB = 4 5 cm + ^ h . AX is perpendicular to DC with DX = 2 cm and XC = x cm. Given that the area of trapezium ABCD is 15 5 2 cm2 + ^ h , obtain an expression for x in the form a + b 5, where a and b are integers. [6] www.theallpapers.com

8 0606/22/M/J/13 © UCLES 2013 For Examiner’s Use 6 O r › 3 r A B The shaded region in the diagram is a segment of a circle with centre O and radius r cm. Angle AOB = 3 r radians. (i) Show that the perimeter of the segment is r 3 3 r + c m cm. [2] (ii) Given that the perimeter of the segment is 26 cm, find the value of r and the area of the segment. [5] www.theallpapers.com

9 0606/22/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 7 Differentiate, with respect to x, (i) (3 − 5x)12, [2] (ii) x2 sin x, [2] (iii) 1 tanx e x 2 + . [4] www.theallpapers.com

10 0606/22/M/J/13 © UCLES 2013 For Examiner’s Use 8 Solutions to this question by accurate drawing will not be accepted. The points A(− 6, 2), B(2, 6) and C are the vertices of a triangle. (i) Find the equation of the line AB in the form y = mx + c . [2] (ii) Given that angle ABC = 90°, find the equation of BC. [2] www.theallpapers.com

11 0606/22/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (iii) Given that the length of AC is 10 units, find the coordinates of each of the two possible positions of point C. [4] www.theallpapers.com

12 0606/22/M/J/13 © UCLES 2013 For Examiner’s Use 9 (a) The graph of y = k(3x) + c passes through the points (0, 14) and (− 2, 6). Find the value of k and of c. [3] (b) The variables x and y are connected by the equation y = ex + 25 − 24e–x. (i) Find the value of y when x = 4. [1] (ii) Find the value of ex when y = 20 and hence find the corresponding value of x. [4] www.theallpapers.com

13 0606/22/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 10 (a) The function f is defined, for 0° x 360°, by f(x) = 1 + 3 cos 2x. (i) Sketch the graph of y = f(x) on the axes below. [4] O –5 5 y x 90° 180° 270° 360° (ii) State the amplitude of f. [1] (iii) State the period of f. [1] (b) Given that cos x = p , where 270° < x < 360°, find cosec x in terms of p. [3] www.theallpapers.com

14 0606/22/M/J/13 © UCLES 2013 For Examiner’s Use 11 A curve has equation y = 3x + ( ) x 4 1 3 - . (i) Find x y d d and x y d d 2 2 . [4] (ii) Show that the coordinates of the stationary points of the curve are (5, 16) and (3, 8). [2] (iii) Determine the nature of each of these stationary points. [2] www.theallpapers.com

15 0606/22/M/J/13 © UCLES 2013 For Examiner’s Use (iv) Find ∫ ( ) x x 3 4 1 3 + - c m dx. [2] (v) Hence find the area of the region enclosed by the curve, the line x = 5, the x-axis and the line x = 6 . [2] www.theallpapers.com

16 0606/22/M/J/13 © UCLES 2013 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

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2013 series 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. www.theallpapers.com

Page 2 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 22 © Cambridge International Examinations 2013 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 Syllabus Paper IGCSE – May/June 2013 0606 22 © Cambridge International Examinations 2013 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 Syllabus Paper IGCSE – May/June 2013 0606 22 © Cambridge International Examinations 2013 1 1 4 3 18 − − = m or 5 soi Y – 3 = their 5(X – 1) or Y – 18 = their 5(X – 4) or 3 = their 5 + c or 18 = their 5 × 4 + c y = (their m) x2 + (their c) or y = (their m) (x2 – 1) + 3 or y = (their m) (x2 – 4) + 18 y = (5x2 – 2)2 or y = (5(x2 – 1) + 3)2 or y = (5(x2 – 4) + 18)2 cao, isw M1 M1 M1 A1 or 18 = 4m + c and 3 = m + c subtracting/substituting to solve for m or c, condone one error or using their m or their c to find their c or their m, without further error their m and c must be validly obtained 2 (a) (p + 1) ln 3 = ln 0.7 1 3 ln 7.0 ln − = p or 1 3 lg 7.0 lg − = p –1.32 cao M1 M1 A1 or p + 1 = log3 0.7 or       = 3 7.0 ln 3 ln p or p = log3 0.7 – 1 or 3 ln 3 7.0 ln 3 ln ÷       = p allow M2 for p = log3       3 7.0 correct answer only scores B3 (b) 2 1 6 2 5 2 − × × y x or 2 5 = a , 6 = b , 2 1 − = c B3 B1 for each component 3 (a) (i) A and E B2 1 mark for each B1 for 1 extra, B0 if 2 or more extras (ii) C and D B2 1 mark for each B1 if 1 extra, B0 if 2 or more extras (b) 5 5 y x B2 (–1, 0), (1, 3), (3, 4) or B1 for two points correct and joined or for three points correct but clearly not joined www.theallpapers.com

Page 5 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 22 © Cambridge International Examinations 2013 4 (i) OC OA AC = + uuur uuur uuur or ( ) 3 OB OA OC OA − = − uuur uuur uuur uuur soi ( ) 18 9 i j ± − o.e. or 2 1 3 3 OC OA OB = + uuur uuur uuur ( )) 9 18 ( 3 1 21 4 j i j i − + − their o.e. or 2 1 (4 21 (22 30 3 3 − + − i j) i j) 10 24 − i j cao B1 B1 M1 A1 or 1 2 3 3( 4) 3( 21) AC c c = − + + uuur i j o.e. soi or 3(c1 – 4) = their 18 and 3(c2 + 21) = their (–9) (ii) 2 2 ) 24 ( 10 − + = their their OC soi ( ) 1 5 12 13 − i j or ( ) 1 10 24 26 − i j isw M1 A1 FT condone 2 2 ) 24 ( 10 their their OC + = FT their xi + yj o.e. 5 AX = 45 AX = 5 3 ( ) 1 4 5 2 45 2 + + + × x their soi ) 2 5 ( 15 + = ( ) 1 4 5 2 45 2 + + + × x their or better Correctly divide their equation by their 5 or their 45 and rationalise denominator completion to 4 3 5 + www B1 B1 M1 M1 M1 A1 may be implied by 3 5 may be seen later may be implied by e.g. summation of rectangle and two triangles or correctly multiply both sides of their equation by their 5 or their 45 and obtain a rational coefficient of x soi answer only does not score www.theallpapers.com

Page 6 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 22 © Cambridge International Examinations 2013 6 (i) arc AB = 3 r π       chord AB = r with justification and summation and completion to given answer B1 B1       + 3 3 π r (ii) r = 12.7             − × × 3 sin 3 2 1 2 π π r their awrt 14.6 B1 M3 A1 must be seen; accept awrt 12.7 may be implied for example 84.45…– 69.84… or M1 for 3 2 1 2 π × × r their or 84.45... and M1 for 3 sin 2 1 2 π × × r their o.e. or 69.84… and M1 for Area Sector – Area triangle attempted 7 (i) 11 ) 5 3 ( x k − 11 ) 5 3 ( 12 5 x − × or better, isw M1 A1 (ii) x2(their cos x) + (their 2x) sin x x2 cos x + 2x sin x isw M1 A1 clearly applies correct form of product rule (iii) Quotient rule attempt: ( ) 2 d tan sec d x x x = ( ) 2 2 d 1 e 2e d x x x + = clearly applies correct form of quotient rule 2 2 2 2 2 ) e 1( tan ) e 2 ( ) sec )( e 1( x x x x their x their + − + 2 2 2 2 2 ) e 1( tan e 2 sec ) e 1( x x x x x + − + isw B1 B1 M1 A1 Product rule attempt: ( ) 2 d tan sec d x x x = 2 2 2 1 2 ) e 1( e 2 ) e 1( d d − − + − = + x x x x tan x (their – 2e2x (1 + e2x)–2) + (1 + e2x)–1(their sec2x) tan x (–2e2x(1 + e2x)–2) + (1 + e2x)–1(sec2x) www.theallpapers.com

Page 7 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 22 © Cambridge International Examinations 2013 8 (i) ( ) 6 2 2 6 2 6 y x −   − = +   +   o.e. soi 5 2 1 + = x y isw M1 A1 or y – 6 = ( ) 2 6 2 2 6 −       + − x (ii) Use of m1m2 = –1 y – 6 = (their –2)(x – 2) or better, isw M1 A1 FT or y = (their – 2) x + c, c = their 10, isw (iii) ( ) ( ) 2 2 2 10 2 6 = − + + y x o.e. Substitute y = their (–2x + 10) Solve their quadratic (0, 10) and (4, 2) o.e. only B1 M1* M1 dep* A1 or (x – 2)2 + (y – 6)2 = ( 20 )2 o.e. or ( 80 )2 + ( ) 2 2 ) 6 ( ) 2 ( − + − y x = 102 or identifying one point by inspection from the length equation and testing it in the equation of BC or vice versa or identifying the second point by inspection from the length equation and testing it in the equation of BC or vice versa answer only does not score 9 (a) 14 = k + c and 6 = 9 k + c o.e. c = 5 k = 9 M1 A1 A1 for two equations in k and c; may be unsimplified; condone one slip in one equation (b) (i) 79.2 or 79.158574 … rot to 4 or more sf B1 (ii) e2x + 5ex – 24(= 0) or (ex)2 + 5ex – 24(= 0) o.e. factorise their 3 term quadratic ex = 3 x = ln 3 or 1.1(0) or 1.0986122 … rot to 3 or more sf as only answer from fully correct working M1 M1 A1 A1 condone one error, but must be three terms or correct/correct ft use of formula or completing the square ignore ex = –8 do not allow final mark if value given from ex = –8 if M0M0 then SC2 if ex = 3 is seen www and leads to x = ln3 or 1.1(0) or 1.0986122… rot to 3 or more sf www.theallpapers.com

Page 8 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 22 © Cambridge International Examinations 2013 10 (a) (i) 90 180 270 360 y x B1 B1 B1 B1 shape; cosine curve – ends must be approaching a turning point be centred on y = 1 clear intent to have min at –2 and max at 4 2 cycles (ii) 3 B1 (iii) 180 B1 (b) cosec x = x sin 1 soi 2 2 1 or cos 1 sin p x x − − = 2 1 1 p − − o.e. B1 B1 B1 or 1 + tan2 x = x 2 cos 1 or cosec2x = 1 + 2 2 1 1 p p − soi or – 2 2 1 1 p p − + or better www.theallpapers.com

Page 9 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0606 22 © Cambridge International Examinations 2013 11 (i) 4 ) 4 (3 3 d d − − − = x x y o.e. isw 2 2 d d x y = (their 12)(x – 4)their (–5) o.e. 2 2 d d x y = 12(x – 4)–5 o.e. isw B1 + B1 M1 A1 if M0 then SC1 for 12(x – 4)–5 + one other term (ii) Verifies x y d d = 0 when x = 3 and x = 5 or solves 0 ) 4 ( 3 3 4 = − − x to obtain 3 and 5 Shows that x = 3 ⇒y = 8 and x = 5⇒y = 16 M1 A1 if M0 then SC1 for verifying or correctly solving to find one x coordinate and showing that it gives rise to the corresponding y coordinate (iii) x = 5 2 2 d d x y (=12) > 0 ⇒ min or x = 3 2 2 d d x y (= –12) < 0 ⇒ max Both correct cao M1 A1 or, using first derivative e.g. x – 5 + x y d d 0 min at x = 5 or x – 3 + x y d d 0 max at x = 3 (iv) ) ( 2 ) 4 ( 2 3 2 2 c x x + − − − o.e. isw B1 + B1 may be unsimplified (v) their               − − −       − − 2 2 2 2 ) 4 5 ( 2 1 2 ) 5 ( 3 ) 4 6 ( 2 1 2 ) 6 ( 3 16.875 to 3 or more sf or 8 135 or 8 7 16 cao M1 A1 www.theallpapers.com