Cambridge IGCSE Mathematics - Additional 0606 — 2013 Oct/Nov Paper 1 · Variant 3

* 0 7 4 8 0 0 0 4 4 7 * This document consists of 16 printed pages. DC (NH/CGW) 67137/4 © UCLES 2013 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 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. www.theallpapers.com

2 0606/13/O/N/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 0606/13/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use 1 The coefficient of x2 in the expansion of px 2 6 + ^ h is 60. (i) Find the value of the positive constant p. [3] (ii) Using your value of p, find the coefficient of x2 in the expansion of x px 3 2 6 - + ^ ^h h . [3] www.theallpapers.com

4 0606/13/O/N/13 © UCLES 2013 For Examiner’s Use 2 Solve lg lg y y 2 5 60 1 - + = ^ h . [5] www.theallpapers.com

5 0606/13/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use 3 Show that tan sin sin sec 2 2 4 2 i i i i - = . [4] www.theallpapers.com

6 0606/13/O/N/13 © UCLES 2013 For Examiner’s Use 4 A curve has equation e y x 3 x 2 2 = + ^ h . (i) Show that d d e x y x A x 3 2 x 3 2 = + + ^ ^ h h, where A is a constant to be found. [4] (ii) Find the exact coordinates of the point on the curve where d d x y 0 = . [2] www.theallpapers.com

7 0606/13/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use 5 For x R ! , the functions f and g are defined by f x x 2 3 = ^ h , g x x x 4 5 2 = - ^ h . (i) Express f 2 1 2c m as a power of 2. [2] (ii) Find the values of x for which f and g are increasing at the same rate with respect to x. [4] www.theallpapers.com

8 0606/13/O/N/13 © UCLES 2013 For Examiner’s Use 6 Do not use a calculator in this question. The diagram shows part of the curve y x 4 2 = - . O 4 y x y = 4 – x 2 2 Show that the area of the shaded region can be written in the form p 2, where p is an integer to be found. [6] www.theallpapers.com

9 0606/13/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use 7 It is given that A t t t t 2 1 2 2 = - + c m. (i) Find the value of t for which det A = 1. [3] (ii) In the case when t 3 = , find A–1 and hence solve x y 3 5 + = , x y 7 3 11 + = . [5] www.theallpapers.com

10 0606/13/O/N/13 © UCLES 2013 For Examiner’s Use 8 The diagram shows two concentric circles, centre O, radii 4 cm and 6 cm. The points A and B lie on the larger circle and the points C and D lie on the smaller circle such that ODA and OCB are straight lines. A D 4 cm rad 2 cm B C O i (i) Given that the area of triangle OCD is 7.5 cm2, show that 1.215 i = radians, to 3 decimal places. [2] (ii) Find the perimeter of the shaded region. [4] www.theallpapers.com

11 0606/13/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use (iii) Find the area of the shaded region. [3] www.theallpapers.com

12 0606/13/O/N/13 © UCLES 2013 For Examiner’s Use 9 (a) (i) Solve sin cos x x 6 5 2 = + for x 0 180 ° ° 1 1 . [4] (ii) Hence, or otherwise, solve cos sin y y 6 5 2 = + for y 0 180 ° ° 1 1 . [3] www.theallpapers.com

13 0606/13/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use (b) Solve cot cot z z 4 3 0 2 - = for z 0 1 1 r radians. [4] www.theallpapers.com

14 0606/13/O/N/13 © UCLES 2013 For Examiner’s Use 10 The variables s and t are related by the equation t ksn = , where k and n are constants. The table below shows values of variables s and t. s 2 4 6 8 t 25.00 6.25 2.78 1.56 (i) A straight line graph is to be drawn for this information with lg t plotted on the vertical axis. State the variable which must be plotted on the horizontal axis. [1] (ii) Draw this straight line graph on the grid below. [3] O 1 2 1g t www.theallpapers.com

15 0606/13/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use (iii) Use your graph to find the value of k and of n. [4] (iv) Estimate the value of s when t = 4. [2] Question 11 is printed on the next page. www.theallpapers.com

16 0606/13/O/N/13 © UCLES 2013 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. 11 (i) Given that e e dx 2 2 5 4 3 x x k 2 2 0 – - = < c m , where k is a constant, show that 4e 12e 5 0 k k 4 2 - + = . [5] (ii) Using a substitution of e y k 2 = , or otherwise, find the possible values of k. [4] www.theallpapers.com

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2013 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 October/November 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 GCE O LEVEL – October/November 2013 0606 13 © 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 GCE O LEVEL – October/November 2013 0606 13 © Cambridge International Examinations 2013 1 (i) 6C2 (24) (px)2 or ( )2 4 2 2 6 px       2 240 60 p = 1 2 p = B1 M1 A1 [3] Seen or implied, unsimplified M1 for their coefficient of x2 = 60 and attempt to solve (ii) coefficients of the terms needed (–1) 6C1 (2)5 p + (3 × 60) = 84 M1 B1 A1 [3] M1 for realising that 2 terms are involved B1 for (–1) 6C1 (2)5 p or –192p, using their p. 2 lg 60 5 2 + y y = lg10 B1 B1 B1 for 2 lg y = lg y2 B1 for 1 = lg10 or equivalent, allow when seen Or lg y2 = lg10 (5y + 60) y2 – 50y – 600 = 0 leading to y = –10, 60 y must be positive so y = 60 M1 DM1 A1 [5] M1 for use of logA – logB = logA/B or logA + logB = log AB DM1 for forming a 3 term quadratic equation and an attempt to solve A1 for 60 y = only www.theallpapers.com

Page 4 Mark Scheme Syllabus Paper GCE O LEVEL – October/November 2013 0606 13 © Cambridge International Examinations 2013 3 θ θ θ θ θ 2 2 2 2 2 sin cos sin sin tan − = − = θ θ θ θ 2 2 2 2 cos cos sin sin − = ( ) θ θ θ 2 2 2 cos cos 1 sin − = θ θ 2 4 cos sin = θ θ 2 4 sec sin Alt solution 1 Using tan2 θ = sin2 θ sec2 θ LHS = sin2 θ sec2 θ – sin2 θ = sin2 θ (sec2 θ – 1) = sin2 θ tan2 θ = sin4 θ sec2 θ Alt solution 2 RHS = sin4 θ sec2 θ = θ θ θ 2 2 2 cos sin sin = ( ) θ θ θ 2 2 2 cos cos 1 sin − = θ θ θ θ 2 2 2 2 cos cos sin sin − = θ θ θ θ θ 2 2 2 2 2 cos cos sin cos sin − = tan2 θ – sin2 θ M1 M1 M1 A1 [4] M1 M1 M1 A1 M1 M1 M1 A1 Marks are awarded only if they can lead to a complete proof for the methods other than those shown below M1 for dealing with tan and a fraction M1 for factorising M1 for use of identity 1 sin cos 2 2 = + θ θ A1 for all correct M1 use of tan2 x = sin2 xsec2 x M1 for factorising M1 for use of identity A1 for all correct M1 for splitting sin4 θ and use of identity M1 for multiplication M1 for writing as two terms and cancelling A1 for all correct www.theallpapers.com

Page 5 Mark Scheme Syllabus Paper GCE O LEVEL – October/November 2013 0606 13 © Cambridge International Examinations 2013 4 (i) ( ) ( ) ( )4 2 2 2 3 3 2 e e 2 3 d d + + − + = x x x x y x x = ( ) ( )3 2 3 2 2 + + x x e x , A = 2 Alt solution x y d d = e2x ( ) ( ) ( ) 2 2 3 3 2 3 2 − − + + + − x e x x = ( ) ( )3 2 3 2 e 2 + + x x x , A = 2 (ii) x = –2, y = e–4 M1 A2, 1, 0 A1 [4] M1 A2,1,0 A1 B1, B1 [2] M1 for attempt at quotient rule –1 for each error Must be convinced of correct simplification e.g. sight of (x + 3 –1) or (x + 2)(x + 3) M1 for attempt at product rule –1 for each error Must be convinced of correct simplification e.g. sight of (x + 3 –1) or (x + 2)(x + 3) Accept 1/e4 5 (i) ( ) ( ) 3 2 2 f f x x = ( ) 3 3 2 2 x = or 3 3 2 1 2 2               = 2-5 Alt method 4 1 2 1 f =       5 2 4 1 f − =       (ii) f ' (x) = g ' (x) 6x2 = 4 – 10x Leading to (3x – 1) (x + 2) = 0 2 , 3 1 − = x M1 A1 [2] M1 A1 B1 B1 M1 A1 [4] M1 for ( ) 3 3 2 2 x = or 3 3 2 1 2 2               For 2-5only M1 for f of their f       2 1 For 2-5only B1 for 6x2 B1 for 4 – 10x M1 for solution of quadratic equation obtained from differentiation of both A1 for both www.theallpapers.com

Page 6 Mark Scheme Syllabus Paper GCE O LEVEL – October/November 2013 0606 13 © Cambridge International Examinations 2013 6 Area under the curve: 2 0 3 2 2 0 3 4 d 4       − = − ∫ x x x x = ( ) 0 3 2 2 2 4 −       − = 3 2 10 Area of trapezium = ( )( ) 2 3 2 2 4 2 1 = + Shaded area = 3 2 10 – 2 3 Shaded area = 3 2 Or: Equation of chord: x y 2 4 − = Shaded area = ∫ + − − 2 0 2 d 2 4 4 x x x 2 0 3 2 3 2 2         −x x = 3 2 M1 A1 DM1 B1 M1 A1 [6] B1 M1 M1 √A1 DM1 A1 [6] M1 for attempt to integrate DM1 for application of limits B1 for area of trapezium, allow unsimplified M1 for subtraction of the two areas Must be in this form B1 for the equation of the chord unsimplified M1 for subtraction M1 for attempt to integrate √A1for       − − 3 2 3 2 x x m or equivalent, where m is the gradient of their chord DM1 for application of limits www.theallpapers.com

Page 7 Mark Scheme Syllabus Paper GCE O LEVEL – October/November 2013 0606 13 © Cambridge International Examinations 2013 7 (i) 2t2 – 2 (t2 – t + 1) Leading to, 2 3 = t (ii) A =       3 7 2 6 , A–1 =       − − 6 7 2 3 4 1       =             11 10 3 7 2 6 y x             − − =       11 10 6 7 2 3 4 1 y x       − =       1 2 y x , leading to x = 2, y = –1 B1 M1 A1 [3] B1, B1 B1 M1 A1 [5] Correct determinant seen unsimplified M1 for simplification and solution A1 for solution of det A=1only, not 1/det A=1 B1 for 4 1 , B1 for matrix B1 for dealing correctly with the factor of 2 M1 for pre-multiplying their       11 10 by their A–1 to obtain a column matrix Allow       − =       1 2 y x for A1 8 (i) ( ) 5.7 sin 4 2 1 2 = θ 215 .1 , 16 15 sin = = θ θ … (ii) 4 2 1 2 sin CD = θ , (CD = 4.567) Arc length = 6(1.215) Perimeter = 2 + 2 + 6(1.215) + their CD = awrt 15.9 (iii) Area = ( ) 5.7 215 .1 6 2 1 2 − = 14.4 (awrt) M1 A1 [2] M1 B1 M1 A1 [4] B1 M1 A1 [3] M1 for attempt to find the area of the triangle and equate to 7.5 A1 for solution to obtain the given answer Solution must include 1.2153…. or 1.2154 M1 for attempt to find CD B1 for arc length M1 for sum of 4 appropriate lengths B1 for sector area M1 for subtraction of the 2 areas www.theallpapers.com

Page 8 Mark Scheme Syllabus Paper GCE O LEVEL – October/November 2013 0606 13 © Cambridge International Examinations 2013 9 (a) (i) 6 (1– cos2 x) = 5 + cos x 6 cos2 x + cos x – 1=0 (3 cos x – 1) (2 cos x + 1) =0 x = 70.5° x = 120° (ii) cos x = sin y 3 1 sin = y only so y = 19.5°, 160.5° (b) cot z (4 cot z – 3) = 0 cot z = 0, 2 π = z 3 4 tan , 4 3 cot = = z z so z = 0.927 M1 M1 A1, A1 [4] DM1 √A1, √A1 [3] M1 B1 M1 A1 [4] M1 for use of sin2x = (1 – cos2x) correctly M1 for solution of a 3 term quadratic in cos and attempt at solution of a trig equation A1 for each correct solution DM1 for relating cos x and sin y or other correct method of solution M1 for attempt to use a factor B1 for 2 π (1.57) M1 dealing with cot and attempt at solution 10 (i) lg s (ii) lgs 0.3 0.6 0.78 0.9 lgt 1.4 0.8 0.44 0.19 (iii) No marks in this part unless lgt v lgs graph is used Gradient : n = –2 (allow –2.1→ –1.9) Intercept : log k, or other method k = 100 (allow 120 90 → ) Alt method Using simultaneous equations, points used must lie on the plotted line. (iv) When t = 4, lg t = 0.6 so lg s = 0.69 4.9 s = (allow 2.5 8.4 → ) B1 [1] M1 DM1 A1 [3] M1A1 M1, A1 [4] M2 A1, A1 M1 A1 [2] Allow in table or on graph if no contradiction No marks for graph unless lgt against lgs (or lnt against lns) M1 for 3 or more points correct DM1 for a line through 3 or 4 correct points A1 all points correct with a straight line extending at least from first point to last point M1 calculates gradient A1 for n = –2 M1 for use of intercept and dealing with logarithm correctly (can use another point) Must attempt to solve 2 valid equations. k = 100 and n = –2 M1 for valid method using either the correct graph or using lgt = nlgs + lgk or t = ksn using their n and their k www.theallpapers.com

Page 9 Mark Scheme Syllabus Paper GCE O LEVEL – October/November 2013 0606 13 © Cambridge International Examinations 2013 11 (i) k x x 0 2 2 e 4 5 e     + − 3 4 5 1 e 4 5 e 2 2 =      + −       + −k k e2k + 4 5 e–2k 4 12 − = 0 4e4k – 12e2k + 5 = 0 (ii) 4y2 – 12y + 5 =0 leading to e2k = , 2 5 e2k = 2 1 k = 0.458, –0.347 B1, B1 M1 M1 A1 [5] M1 M1 A1, A1 [4] B1 for each term integrated correctly, allow unsimplified M1 for application of limits to an integral of the form Ae2x + Be-2x M1 for equating to 4 3 and attempt to rearrange to obtain a 3 term equation. Must be using an integral of the form Ae2x + Be-2x Answer given, so must be convinced M1 for solution of quadratic equation M1 for solving equations involving exponentials A1 for each www.theallpapers.com