Cambridge IGCSE Mathematics - Additional 0606 — 2016 Oct/Nov Paper 1 · Variant 3

This document consists of 15 printed pages and 1 blank page. DC (RW/AR) 116667/2 © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 8 7 6 7 8 8 5 4 6 1 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2016 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/O/N/16 © UCLES 2016 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

3 0606/13/O/N/16 © UCLES 2016 [Turn over 1 On the axes below, sketch the graph of cos y x 2 3 = for ° x 0 180 G G . [3] y 3 2 1 0 –1 –2 –3 30° 60° 90° 120° 150° 180° x 2 Express m m m m m 2 4 3 9 + - in the form Am B + , where A and B are integers to be found. [3]

4 0606/13/O/N/16 © UCLES 2016 3 (i) Given that x p x 3 1 2 3 2 + - =- ^ h , show that, for x to be real, p p 3 9 0 2 H - - . [3] (ii) Hence find the set of values of p for which x is real, expressing your answer in exact form. [3]

5 0606/13/O/N/16 © UCLES 2016 [Turn over 4 (i) Find, in ascending powers of x, the first 3 terms in the expansion of x 2 4 6 - J L KK N P OO . [3] (ii) Hence find the term independent of x in the expansion of x x 4 2 3 2 2 6 + + - 4 x J L KK J L KK N P OO N P OO . [3]

6 0606/13/O/N/16 © UCLES 2016 5 (i) Given that log xy 2 5 9 = , show that log log x y 5 3 3 + = . [3] (ii) Hence solve the equations log xy 2 5 9 = , log log x y 6 3 3 # =- . [5]

7 0606/13/O/N/16 © UCLES 2016 [Turn over 6 (i) Find ln x x 3 11 d d 2 - ^ ^ hh. [2] (ii) Hence show that ln x x x p x c 3 11 3 11 d 2 2 - = - + ^ h y , where p is a constant to be found, and c is a constant of integration. [1] (iii) Given that a ln x x x 3 11 2 d 2 2 - = y , where a 2 2 , find the value of a. [4]

8 0606/13/O/N/16 © UCLES 2016 7 0 1 2 3 4 5 1 2 3 4 5 ln y 1 x (1.5, 3.5) (4.0, 1.5) The variables x and y are such that when lny is plotted against x 1 the straight line graph shown above is obtained. (i) Given that y Ae x b = , find the value of A and of b. [4]

9 0606/13/O/N/16 © UCLES 2016 [Turn over (ii) Find the value of y when . x 0 32 = . [2] (iii) Find the value of x when y 20 = . [2]

10 0606/13/O/N/16 © UCLES 2016 8 (a) (i) Show that cosec sin cosec sec2 i i i i - = . [3] (ii) Hence solve cosec sin cosec 4 i i i - = for ° ° 0 360 1 1 i . [3]

11 0606/13/O/N/16 © UCLES 2016 [Turn over (b) Solve r tan x 3 4 1 + = J L KK N P OO for r x 0 2 1 1 , giving your answers in terms of r. [3]

12 0606/13/O/N/16 © UCLES 2016 9 (a) A team of 5 students is to be chosen from a class of 10 boys and 8 girls. Find the number of different teams that may be chosen if (i) there are no restrictions, [1] (ii) the team must contain at least one boy and one girl. [4]

13 0606/13/O/N/16 © UCLES 2016 [Turn over (b) A computer password, which must contain 6 characters, is to be chosen from the following 10 characters: Symbols ? ! * Numbers 3 5 7 Letters W X Y Z Each character may be used once only in any password. Find the number of possible passwords that may be chosen if (i) there are no restrictions, [1] (ii) each password must start with a letter and finish with a number, [2] (iii) each password must contain at least one symbol. [3]

14 0606/13/O/N/16 © UCLES 2016 10 A curve y x f = ^ h is such that x x 6 8 f e x 2 = - l^ h . (i) Given that the curve passes through the point P ,0 3 - ^ h, find the equation of the curve. [5] The normal to the curve y x f = ^ h at P meets the line y x 2 3 = - at the point Q. (ii) Find the area of the triangle OPQ, where O is the origin. [5]

15 0606/13/O/N/16 © UCLES 2016 11 A particle moving in a straight line has a velocity of v ms–1 such that, t s after leaving a fixed point, v t t 4 8 3 2 = - + . (i) Find the acceleration of the particle when t 3 = . [2] (ii) Find the values of t for which the particle is momentarily at rest. [2] (iii) Find the total distance the particle has travelled when . t 1 5 = . [5]

16 0606/13/O/N/16 © UCLES 2016 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. This document consists of 8 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2016 MARK SCHEME Maximum Mark: 80 Published 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 2016 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 – October/November 2016 0606 13 © UCLES 2016 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 Question Answer Marks Part Marks 1 B1 B1 B1 for symmetrical shape as in the diagram with curved maxima of equal height and cusps on the x-axis for a complete ‘curve’ with all low points on the x-axis and all high points on 2 = y for a complete ‘curve’ meeting the x-axis at o o o 30 , 90 , 150 = x only. 2 = 2 4 9 2 3 − + m m = ( )( ) 2 3 2 3 2 3 − + + m m m = 2 3 − m Alternative Method 9 (4 ) − m m m = 3 (2 ) + m m (Am + B) Comparing coefficients 2A = 4, 3A + 2B = 0, 3 9 = − B M1 A1 A1 M1 A1 A1 for multiplying each term by m , using a common denominator of m or for multiplying numerator and denominator by 3 2 − m m for a correct expression that will cancel ( )( ) 2 3 2 3 2 3 − + + m m m , ( )( ) 2 2 4 9 2 3 (4 9) − − − m m m ( )( ) 2 3 2 3 (2 3) (2 3)(2 3) − + − + − m m m m m , or equivalents for 2 3 − m or 2, 3 = = − A B for correct expansion for correct comparisons to obtain A and B for 2 3 − m or 2, 3 = = − A B 45 90 135 180 −1 −0.5 0.5 1 1.5 2 2.5 x y

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 13 © UCLES 2016 Question Answer Marks Part Marks 3 (i) ( ) 2 3 2 3 0 − + + = x xp p ( ) 2 ( 2 ) 4 3 3 0 − − × × + p p . oe 2 3( 3) + p p . or 2 4 12 36 0 − − p p . 2 3 9 0 − − p p . M1 DM1 A1 for obtaining a 3-term quadratic in the form 2 ( 0) + + = ax bx c for correct substitution of their a, b and c into ‘ 2 4 − b ac ’and use of discriminant. for full correct working, ⩾ the only sign used, ⩾ used before division by 4 and ⩾ used in answer line and penultimate line. (ii) Correct method of solution 2 3 9 0 − − = p p leading to critical values 3 3 5 2 ± = p 3 3 5 3 3 5 , 2 2 − + p p - . M1 A1 A1 for correct substitution in the quadratic formula or for correct attempt to complete the square. (allow 1 sign error in either method) for both correct critical values for correct range 4 (i) (ii) 2 64 48 15 − + x x ( ) ( ) ( ) 4 '64' 2 ' 48' 3 '15' × + × − + × = 205 cao B3 M1 A1 A1 for each correct term for correctly obtaining three products using their coefficients in (i) for two correct out of three products (unsimplified) cao for 205 selected as final answer 5 (i) 9 9 9 log log log = + xy x y 3 3 3 3 log log log 9 log 9 = + x y = 3 3 log log 5 2 2 2 + = x y 3 3 log log 5 + = x y M1 M1 A1 for use of log AB = logA + logB for correct method for change of base. Division by log39 should be seen and not implied. for dealing with 2 correctly and ‘finishing off’ Alternative method 9 5 log 2 = xy 5 5 2 9 3 = = xy 3 log 5 = xy 3 3 log log 5 + = x y M1 M1 A1 for obtaining xy as a power of 3 for correct use of log3 for using law for logs and arriving at correct answer

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 13 © UCLES 2016 Question Answer Marks Part Marks (ii) ( ) 3 3 log 5 log 6 − = − x x ( ) 2 3 3 log 5log 6 − + = − x x ( ) 2 3 3 log 5log 6 0 − − = x x leading to 3 3 log 6, log 1 = = − x x 1 729, 3 = = x x 1, 729 3 = = y y M1 A1 A1 DM1 A1 for substitution, correct expansion of brackets and manipulation to get a 3 term quadratic for a correct quadratic equation in the form 2 0 + + = ax bx c for both solutions for method of solution of 3 log = x k or 3 log = y k for all x and y correct 6 (i) 2 6 3 11 − x x M1 A1 M1 for 2 3 11 − mx x (ii) 1 6 = p B1 FT for 1 = p m (iii) ( ) 2 1 1 ln 3 11 ln1 ln 2 6 6 − − = a ( ) 2 6 ln 3 11 ln 2 − = a 2 3 11 64 − = a 5 = a only M1 DM1 DM1 A1 for correct use of limits in ( ) 2 ln 3 11 − p x May be implied by following equation for dealing with logs correctly for solution of 2 3 11 − = a k for 5 obtained from an exact method

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 13 © UCLES 2016 Question Answer Marks Part Marks 7 (i) ln ln = + b y A x Gradient: 0.8 = − b Intercept or use of equation: ln 4.7 = A 110 = A Alternative Method 3.5 ln 1.5 = + A b and 1.5 ln 4 = + A b leading to 0.8 = − b ln 4.7 = A and 110 = A Alternative Method e1.5 = Ae4b e3.5 = Ae1.5b leading to 0.8 = − b and 110 = A B1 B1 B1 B1 B1 B1 B1 B1 B1 B1 B1 B1 for equation, may be implied, must be using ln unless recovered for 0.8 = − b oe for ln A = 4.7 oe, allow 4.65 to 4.75 for A = 110, allow 105 to 116 Allow A in terms of e for one equation for 0.8 = − b for ln 4.7 = A for 110 = A or e4.7 for e1.5 = Ae4b or 4.48 = Ae4b for e3.5 =Ae1.5b or 33.1 = Ae1.5b for 0.8 = − b for 110 = A or e4.7 (ii) When 1 0.32, 3.125, ln 2.2 x y x = = = 9 = y ( allow 8.5 to 9.5) or e2.2 M1 A1 for a complete method to obtain y, using either the graph, using their values in the equation for lny or using their values in the equation for y. (iii) When 20, ln 3 = = y y , 1 2.125 = x so 0.47 = x ( allow 0.45 to 0.49) M1 A1 for a complete method to obtain x, using either the graph, using their values in the equation for lny or using their values in the equation for y.

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 13 © UCLES 2016 Question Answer Marks Part Marks 8 (a) (i) 1 cosec sin 1 cosec sin sin sin θ θ θ θ θ θ = − − 2 1 1 sin θ = − or 2 1 sin (1 sin ) sin θ θ θ = − 2 1 cos θ = 2 sec θ = Alternative Method using cosec cosec cosec 1 cosec sin cosec cosec θ θ θ θ θ θ = − − 2 2 cosec cosec 1 θ θ = − 2 2 1 cot cot θ θ + = 2 2 tan 1 sec θ θ = + = M1 DM1 A1 M1 DM1 A1 for using 1 cosec sin θ θ = and either attempt to multiply top and bottom by sin θ or an attempt to combine terms in denominator. for correct use of 1-sin2θ = cos2θ for completing the proof for using 1 sin cosec θ θ = and an attempt to combine terms in denominator. for use of 1+ cot2θ = cosec2θ for completing the proof (ii) 2 1 1 cos , cos 4 2 θ θ = = ± or 2 tan 3, tan 3 θ θ = = ± or 2 3 3 sin , sin 4 2 θ θ = = ± o o o o 60 ,120 , 240 , 300 θ = M1 A1 A1 for using (i) to obtain a value for cos2θ, tan2θ or sin2θ and then taking the square root. for two correct values for two further correct values and no extras in range. (b) π 1 tan 4 3   + =     x π π 7π π 13π π , , 6 4 6 4 6 4 = − − − x π 11π 23π , , 12 12 12   = −     x M1 A1,A1 for correct order of operations, can be implied by π 12 = − x A1 for 11π 12 = x A1 for 23π 12 = x If there are extra solutions in range in addition to the two correct ones then A1A0

Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 13 © UCLES 2016 Question Answer Marks Part Marks 9 (a) (i) (ii) 18 5 8568 = C mmm Either 10 8 4 1 1680 × = C C 10 8 3 2 3360 × = C C 10 8 2 3 2520 × = C C 10 8 1 4 700 × = C C Total = 8260 Or their ( ) 18 10 8 5 5 5 − + C C C ( ) 8568 252 56 − + Total =8260 B1 B1 B2,1,0 B1 B1 B1 B1 B1 for a correct plan B2 4 correct numbers with no extras B1 3 correct numbers (out of 3 or 4) for correct total for correct plan for 252 subtracted for 56 subtracted for correct total (b) (i) (ii) (iii) 10 6 151200 = P 8 4 4 3 × × P = 20160 Answer to (i) - 7 6P =146160 Alternative: 1 symbol: 45360 2 symbols: 75600 3 symbols: 25200 Total: 146160 B1 M1 A1 M1 A1 A1 B2,1,0 B1 for correct unsimplified for correct numerical answer for correct plan for correct unsimplified for correct numerical answer B2 for all 3 correct B1 for 2 correct (out of 2 or 3) for correct sum

Page 8 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 13 © UCLES 2016 Question Answer Marks Part Marks 10 (i) ( ) ( ) 2 2 f 3 4e = − + x x x c passing through ( ) 0, 3 − 0 3 3 0 4e −= × − + c ( ) 2 2 f 3 4e 1 = − + x x x M1 A1 A1 DM1 A1 for one correct term for one correct term 2 3 x or 2 4e − x for a second correct term with no extras for correct method to find c. for correct equation (ii) ( ) f 0 8 ′ = − Normal: 1 3 8 + = y x 8 24 + = y x 2 3 = − y x leads to 8 5 = x oe Area = 1 8 3 2 5 = × × = 2.4 oe B1 M1 DM1 A1 B1 for 1 8 = m for equation of normal using 1 8 = m for solving normal equation simultaneously with y = 2 – 3x to get a value of x for 8 5 = x , 1.6 oe FT for a numerical answer equal to 1 3 their 2 × × x 11 (i) (ii) (iii) 8 8 = − a t When 3 = t , 16 = a 0.5, 1.5 3 2 4 4 3 3 = − + s t t t when 1 2 , 2 3 = = t s when 3, 0 2 = = t s total distance travelled = 4 3 Alternative method B1 B1 B1,B1 M1 A1 DM1 DM1 A1 M1A1 DM1 DM1 A1 for 8 8 t − for 16 B1 for each for at least two terms correct all correct for calculating displacement when either 1 2 = t or 3 2 = t for calculating displacement at 1 2 = t and doubling. for 4 3 oe allow 1.33 As before DM1for calculating displacement when t = 0.5 or for calculating distance travelled between t = 0.5 and t = 1.5 DM1 for doubling distance travelled between t = 0.5 and t = 1.5 or for adding that distance to displacement at t = 0.5 A1 for 4 3 oe allow 1.33