Cambridge IGCSE Mathematics - Additional 0606 — 2015 May/June Paper 1 · Variant 3

This document consists of 15 printed pages and 1 blank page. DC (ST/SW) 105918 © UCLES 2015 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 7 5 5 2 6 5 9 0 8 8 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 May/June 2015 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 an HB pencil for any diagrams or graphs. Do not use staples, paper clips, 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.

2 0606/13/M/J/15 © UCLES 2015 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x a b b ac 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

3 0606/13/M/J/15 © UCLES 2015 [Turn over 1 (i) State the period of sin x 2 . [1] (ii) State the amplitude of cos x 1 2 3 + . [1] (iii) On the axes below, sketch the graph of (a) sin y x 2 = for ° ° x 0 180 G G , [1] (b) cos y x 1 2 3 = + for ° ° x 0 180 G G . [2] O –2 45° 90° 135° 180° 2 4 –4 y x (iv) State the number of solutions of sin cos x x 2 2 3 1 - = for ° ° x 0 180 G G . [1]

4 0606/13/M/J/15 © UCLES 2015 2 Do not use a calculator in this question. 4 3 2 + ^ h C B cm cm A 5 2 8 + ^ h rad i The diagram shows the triangle ABC where angle B is a right angle, AB 4 3 2 = + ^ h cm, BC 8 5 2 = + ^ h cm and angle BAC i = radians. Showing all your working, find (i) tan i in the form a b 2 + , where a and b are integers, [2] (ii) sec2i in the form c d 2 + , where c and d are integers. [3]

5 0606/13/M/J/15 © UCLES 2015 [Turn over 3 (i) Find the first 4 terms in the expansion of x 2 2 6 + ^ h in ascending powers of x. [3] (ii) Find the term independent of x in the expansion of x x 2 1 3 2 6 2 2 + - ^ c h m . [3]

6 0606/13/M/J/15 © UCLES 2015 4 (a) Given that the matrix k 2 4 0 X = - J L KK N P OO, find X2 in terms of the constant k. [2] (b) Given that the matrix a b 1 5 A = J L KK N P OO and the matrix 6 5 3 2 6 1 3 1 A 1 = - - - J L K KK N P O OO, find the value of each of the integers a and b. [3]

7 0606/13/M/J/15 © UCLES 2015 [Turn over 5 The curve y xy x 4 2 = + - intersects the line y x 3 1 = - at the points A and B. Find the equation of the perpendicular bisector of the line AB. [8]

8 0606/13/M/J/15 © UCLES 2015 6 The polynomial x ax x bx 15 2 f 3 2 = - + - ^ h has a factor of x 2 1 - and a remainder of 5 when divided by x 1 - . (i) Show that b 8 = and find the value of a. [4] (ii) Using the values of a and b from part (i), express x f^ h in the form x 2 1 - ^ h x g^ h, where x g^ h is a quadratic factor to be found. [2] (iii) Show that the equation x 0 f = ^ h has only one real root. [2]

9 0606/13/M/J/15 © UCLES 2015 [Turn over 7 The point A, where x 0 = , lies on the curve ln y x x 1 4 3 2 = - + ^ h. The normal to the curve at A meets the x-axis at the point B. (i) Find the equation of this normal. [7] (ii) Find the area of the triangle AOB, where O is the origin. [2]

10 0606/13/M/J/15 © UCLES 2015 8 It is given that x 3 f e x 2 = ^ h for x 0 H , x x 2 5 g 2 = + + ^ ^ h h for x 0 H . (i) Write down the range of f and of g. [2] (ii) Find g 1 - , stating its domain. [3] (iii) Find the exact solution of x 41 gf = ^ h . [4]

11 0606/13/M/J/15 © UCLES 2015 [Turn over (iv) Evaluate ln4 f l^ h. [2]

12 0606/13/M/J/15 © UCLES 2015 9 O C D B y x y = x3 – 5x2 + 3x + 10 y = 3x + 10 A The diagram shows parts of the line y x 3 10 = + and the curve y x x x 5 3 10 3 2 = - + + . The line and the curve both pass through the point A on the y-axis. The curve has a maximum at the point B and a minimum at the point C. The line through C, parallel to the y-axis, intersects the line y x 3 10 = + at the point D. (i) Show that the line AD is a tangent to the curve at A. [2] (ii) Find the x-coordinate of B and of C. [3]

13 0606/13/M/J/15 © UCLES 2015 [Turn over (iii) Find the area of the shaded region ABCD, showing all your working. [5]

14 0606/13/M/J/15 © UCLES 2015 10 (a) Solve sin cosec x x 4 = for ° ° x 0 360 G G . [3] (b) Solve tan sec y y 3 2 3 2 0 2 - - = for ° ° y 0 180 G G . [6]

15 0606/13/M/J/15 © UCLES 2015 (c) Solve tan z 3 3 r - = ` j for z 0 2 G G r radians. [3]

16 0606/13/M/J/15 © UCLES 2015 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. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. 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

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME for the May/June 2015 series 0606 ADDITIONAL MATHEMATICS 0606/13 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 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 2015 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 13 © Cambridge International Examinations 2015 Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied www without wrong working 1 (i) 180° or π radians or 3.14 radians ( or better) B1 (ii) 2 B1 (iii) (a) (b) B1 B1 B1 x y 2 sin = all correct for either ↑↓↑starting at their highest value and ending at their lowest value Or a curve with highest value at 3 = y and lowest value at 1 − = y completely correct graph (iv) 3 B1 2 (i) ( )( ) ( )( ) 2 3 4 2 3 4 2 3 4 2 5 8 tan − + − + = θ 18 16 30 2 20 2 24 32 − − + − = 2 2 1+ = cao M1 A1 attempt to obtain θ tan and rationalise. Must be convinced that no calculators are being used 45 90 135 180 −4 −3 −2 −1 1 2 3 4 x y

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 13 © Cambridge International Examinations 2015 (ii) θ θ 2 2 tan 1 sec + = ( ) 2 2 2 1 1 + − + = 8 2 4 1 1 + − + = 2 4 10 − = Alternative solution: ( ) ( ) 2 2 2 2 5 8 2 3 4 + + + = AC 2 104 148 + = ( ) 2 2 2 3 4 2 104 148 sec + + = θ ( ) 2 24 34 2 24 34 2 3 4 2 104 148 2 − − × + + = 2 4 10 − = M1 DM1 A1 M1 DM1 A1 attempt to use θ θ 2 2 tan 1 sec + = , with their answer to (i) attempt to simplify, must be convinced no calculators are being used. Need to expand ( ) 2 2 2 1+ − as 3 terms 3 (i) 6 4 2 160 240 192 64 x x x + + + B3,2,1,0 –1 each error (ii) ( )       + − + + 4 2 4 2 9 6 1 240 192 64 x x x x Terms needed ) 9 240 ( ) 6 192 ( 64 × + × − = 1072 B1 M1 A1 expansion of 2 2 3 1      −x attempt to obtain 2 or 3 terms using their (i)

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 13 © Cambridge International Examinations 2015 4 (a) X2       − − − = k k k 4 2 8 4 4 B2,1,0 –1 each incorrect element (b) Use of AA–1 = I       =           − −       1 0 0 1 3 1 3 2 6 1 6 5 5 1 b a Any 2 equations will give a = 2, b = 4 Alternative method 1:           − − =       − − 3 1 3 2 6 1 6 5 1 5 5 1 a b b a Compare any 2 terms to give a = 2, b = 4 Alternative method 2: Inverse of       =       − − 5 4 1 2 2 4 1 5 6 1 M1 A1,A1 M1 A1,A1 M1 A1,A1 use of AA-1 = I and an attempt to obtain at least one equation. correct attempt to obtain A-1 and comparison of at least one term. reasoning and attempt at inverse 5 4 )1 3 ( 1 3 2 − + − = − x x x x or 4 3 1 3 1 2 −       + +       + = y y y y 0 3 4 4 2 = − −x x or 0 35 4 4 2 = − −y y ( )( ) 0 1 2 3 2 = + − x x or ( )( ) 0 5 2 7 2 = + − y y leading to 2 1 , 2 3 − = = x x and 2 5 , 2 7 − = = y y Midpoint 1 1 , 2 2       Perpendicular gradient = 3 1 − Perp bisector:      − − = − 2 1 3 1 2 1 x y ) 0 2 3 ( = − + x y M1 DM1 A1 A1 B1 M1 M1 A1 equate and attempt to obtain an equation in 1 variable forming a 3 term quadratic equation and attempt to solve x values y values for midpoint, allow anywhere correct attempt to obtain the gradient of the perpendicular, using AB straight line equation through the midpoint; must be convinced it is a perpendicular gradient. allow unsimplified

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 13 © Cambridge International Examinations 2015 6 (i) 0 2 2 4 15 8 2 1 f = − + − =       b a leading to 46 4 = + b a 5 2 15 )1( f = − + − = b a leading to 22 = + b a giving 8 = b (AG), 14 = a M1 A1 M1,A1 correct use of either       2 1 f or f(1) paired correctly both equations correct (allow unsimplified) M1 for solution of equations A1 for both a and b. AG for b. (ii) ( )( ) 2 4 7 1 2 2 + − − x x x M1,A1 M1 for valid attempt to obtain g(x), by either observation or by algebraic long division. (iii) 0 2 4 7 2 = + −x x has no real solutions as ac b 4 2 < 56 16 < M1 A1 use of ac b 4 2 − correct conclusion; must be from a correct g(x) or 2g(x) www 7 (i) ( ) ( ) 2 2 2 )1 ( 3 4 ln 2 4 8 )1 ( d d − + − + − = x x x x x x y When x = 0, 3 ln − = y oe 3 ln d d − = x y so gradient of normal is 3 ln 1 (allow numerical equivalent) normal equation x y 3 ln 1 3 ln = + or 10 .1 910 .0 − = x y , or 10 11 11 10 − = x y cao (Allow 1.1 91 .0 − = x y ) M1 B1 A1 B1 M1 M1 A1 differentiation of a quotient (or product) correct differentiation of ( ) 3 4 ln 2 + x all else correct for y value valid attempt to obtain gradient of the normal attempt at normal equation must be using a perpendicular (ii) when x = 0, 3 ln − = y when y = 0, ( )2 3 ln = x Area = ±0.66 or ±0.67 or awrt these or ( )3 3 ln 2 1 M1 A1 valid attempt at area

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 13 © Cambridge International Examinations 2015 8 (i) Range for f: y ≥ 3 Range for g: y ≥ 9 B1 B1 (ii) 5 2 − + − = y x ( ) 5 2 g 1 − + − = − x x Domain of g–1: x ≥ 9 Alternative method: 0 9 4 2 = − + + x y y 4 16 4(9 ) 2 x y −+ − − = M1 A1 B1 M1 A1 attempt to obtain the inverse function Must be correct form for domain attempt to use quadratic formula and find inverse must have + not ± (iii) Need g( ) x 2 e 3 ( ) 41 5 2 e 3 2 2 = + + x or 0 32 e 12 e 9 2 4 = − + x x ( )( ) 0 8 e 3 4 e 3 2 2 = + − x x leading to 6 2 e 3 2 ± = + x so 3 4 ln 2 1 = x or 3 4 e2 = x so 3 4 ln 2 1 = x Alternative method: Using ) 41 ( g ) ( f 1 − = x , 4 ) 41 ( g 1 = − leading to 4 e 3 2 = x , so 3 4 ln 2 1 = x M1 DM1 M1 A1 M1 DM1 M1 A1 correct order correct attempt to solve the equation dealing with the exponential correctly in order to reach a solution for x Allow equivalent logarithmic forms correct use of 1 g− dealing with ) 41 ( g 1 − to obtain an equation in terms of x 2 e dealing with the exponential correctly in order to reach a solution for x Allow equivalent logarithmic forms (iv) x x 2 e 6 ) ( g = ′ 96 ) 4 (ln g = ′ B1 B1 B1 for each

Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 13 © Cambridge International Examinations 2015 9 (i) 3 10 3 d d 2 + − = x x x y When x = 0, for curve 3 d d = x y , gradient of line also 3 so line is a tangent. Alternate method: 10 3 5 10 3 2 3 + + − = + x x x x leading to 0 2 = x , so tangent at 0 = x M1 A1 M1 A1 for differentiation comparing both gradients attempt to deal with simultaneous equations obtaining 0 x = (ii) When 0 d d = x y , 0 ) 3 )( 1 3 ( = − − x x 3 1 = x , 3 = x M1 A1,A1 equating gradient to zero and valid attempt to solve A1 for each (iii) Area = 3 3 2 0 1 (10 19)3 5 3 10d 2 x x x x + − − + + ∫ 3 0 2 3 4 10 2 3 3 5 4 2 87       + + − − = x x x x 87 81 27 45 30 2 4 2   = − − + +     7. 24 = or 24.8 Alternative method: Area ( ) 3 3 2 0 (3 10) 5 3 10 d x x x x x = + − − + + ∫ 3 3 2 0 5 d x x x = − + ∫ 4 99 3 5 4 3 0 3 4 =       + − = x x B1 M1 A1 DM1 A1 B1 M1 A1 DM1 A1 area of the trapezium attempt to obtain the area enclosed by the curve and the coordinate axes, by integration integration all correct correct application of limits (must be using their 3 from (ii) and 0) correct use of ‘Y–y’ attempt to integrate integration all correct correct application of limits 10 (a) 4 1 sin2 = x 2 1 ) ( sin ± = x ° ° ° ° = 330 , 210 , 150 , 30 x M1 A1,A1 using 1 cosec sin x x = and obtaining = x sin … A1 for one correct pair, A1 for another correct pair with no extra solutions

Page 8 Mark Scheme Syllabus Paper Cambridge IGCSE – May/June 2015 0606 13 © Cambridge International Examinations 2015 (b) ( ) 0 2 3 sec 2 1 3 sec2 = − − − y y 0 3 3 sec 2 3 sec2 = − − y y 0 ) 3 3 )(sec 1 3 (sec = − + y y leading to 1 3 cos − = y , 3 1 3 cos = y 3 180 , 540 3 70.5 , 289.5 , 430.5 y y = ° ° = ° ° ° ° ° ° ° ° = 5. 143 , 5. 96 , 5. 23 , 180 , 60 y Alternative 1: 0 3 3 sec 2 3 sec2 = − − y y leading to 1 3 cos 2 3 cos 3 2 − + y y 0 )1 )(cos 1 cos 3 ( = + − y y Alternative 2: 0 2 cos 2 cos sin 2 2 = − − y y y ( ) 0 cos 2 cos 2 cos 1 2 2 = − − − x x x M1 M1 M1 A1,A1 A1 M1 M1 M1 M1 use of the correct identity attempt to obtain a 3 term quadratic equation in sec 3y and attempt to solve dealing with sec and 3y correctly A1 for a correct pair, A1 for a second correct pair, A1 for correct 5th solution and no other within the range use of the correct identity attempt to obtain a quadratic equation in cos 3y and attempt to solve dealing with 3y correctly A marks as above use of the correct identity, y y y cos sin tan = and y y cos 1 sec = , then as before (c) π π 4π , 3 3 3 z − = 3 π 5 , 3 π 2 = z or 2.09 or 2.1, 5.24 M1 A1,A1 correct order of operations A1 for a correct solution A1 for a second correct solution and no other within the range