Cambridge IGCSE Mathematics - Additional 0606 — 2016 Oct/Nov Paper 2 · Variant 1

This document consists of 16 printed pages. DC (NH/JG) 116665/2 © UCLES 2016 [Turn over * 1 0 7 8 7 6 6 1 6 2 * ADDITIONAL MATHEMATICS 0606/21 Paper 2 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. Cambridge International Examinations Cambridge International General Certificate of Secondary Education

2 0606/21/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 0606/21/O/N/16 © UCLES 2016 [Turn over 1 Solve the equation 4 3 x x - = . [3] 2 Without using a calculator, find the integers a and b such that 3 1 3 1 3 3 a b + + - = - . [5]

4 0606/21/O/N/16 © UCLES 2016 3 Solve the equation 2 2 10 1 lg lg x x - + = c m . [5] 4 The number of bacteria, N, present in a culture can be modelled by the equation 7000 2000 N e 0.05t = + - , where t is measured in days. Find (i) the number of bacteria when t = 10, [1]

5 0606/21/O/N/16 © UCLES 2016 [Turn over (ii) the value of t when the number of bacteria reaches 7500, [3] (iii) the rate at which the number of bacteria is decreasing after 8 days. [3]

6 0606/21/O/N/16 © UCLES 2016 5 The curve with equation 2 7 2 y x x x 3 2 = + - + passes through the point A (−2, 16). Find (i) the equation of the tangent to the curve at the point A, [3] (ii) the coordinates of the point where this tangent meets the curve again. [5]

7 0606/21/O/N/16 © UCLES 2016 [Turn over 6 (i) Prove that 1 1 tan cos cot sin cos sin x x x x x x + - + = - . [4] (ii) Hence solve the equation 1 1 3 4 180° 180° tan cos cot sin sin cos x x x x x x x for 1 1 + - + = - - . [4]

8 0606/21/O/N/16 © UCLES 2016 7 14 cm x cm P S R Q x cm 3 cm (i) Show that the area, A cm2, of the trapezium PQRS is given by 7 9 x A x2 = + - ^ h . [2]

9 0606/21/O/N/16 © UCLES 2016 [Turn over (ii) Given that x can vary, find the stationary value of A. [7]

10 0606/21/O/N/16 © UCLES 2016 8 The function x f^ h is given by 1 3 1 x x x f 3 3 = + - ^ h for 0 G x G 3. (i) Show that 1 x x kx f 3 2 2 = + l^ ^ h h , where k is a constant to be determined. [3] (ii) Find 1 x x x d 3 2 2 + ^ h y and hence evaluate 1 x x x d 3 2 1 2 2 + ^ h y . [4]

11 0606/21/O/N/16 © UCLES 2016 [Turn over (iii) Find f −1(x), stating its domain. [4]

12 0606/21/O/N/16 © UCLES 2016 9 The line 4 y kx = - , where k is a positive constant, passes through the point P (0, −4) and is a tangent to the curve 2 8 x y y 2 2 + - = at the point T. Find (i) the value of k, [5]

13 0606/21/O/N/16 © UCLES 2016 [Turn over (ii) the coordinates of T, [3] (iii) the length of TP. [2]

14 0606/21/O/N/16 © UCLES 2016 10 The town of Cambley is 5 km east and p km north of Edwintown so that the position vector of Cambley from Edwintown is 1000 5000 pm c metres. Manjit sets out from Edwintown at the same time as Raj sets out from Cambley. Manjit sets out from Edwintown on a bearing of 020° at a speed of 2.5 ms–1 so that her position vector relative to Edwintown after t seconds is given by 2.5 20° 2.5 70° cos cos t t c m metres. Raj sets out from Cambley on a bearing of 310° at 2 ms–1. (i) Find the position vector of Raj relative to Edwintown after t seconds. [2]

15 0606/21/O/N/16 © UCLES 2016 [Turn over Manjit and Raj meet after T seconds. (ii) Find the value of T and of p. [5] Question 11 is printed on the next page.

16 0606/21/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. 11 Mr and Mrs Coldicott have 5 sons and 4 daughters. All 11 members of the family play tennis. Six members of the family enter a tennis competition where teams consist of 4 males and 2 females. Find the number of different teams of 4 males and 2 females that could be selected if (i) there are no further restrictions, [2] (ii) Mr and Mrs Coldicott must both be in the team, [2] (iii) either Mr or Mrs Coldicott is in the team but not both. [3]

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 6 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/21 Paper 2 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 21 © 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 4 3 1 x x x − = → = 4 3 x x − = − 0.6 x = OR ( ) 2 2 4 3 x x − = 2 15 24 9 0 x x − + = ( )( ) 3 1 5 3 0 x x − − = 1 and 0.6 x x = = B1 M1 A1 B1 M1 A1 www use of x − or ( ) 4 3 x − − but not both. solve correct 3 term quadratic www 2 ( ) ( ) ( )( )( ) 3 1 3 1 3 3 3 1 3 1 a b − + + = − − + ( ) 2 3 3 = − oe 2 a b + = 6 a b −+ = − 2 and 4 b a = − = M1 DM1 A1 DM1 A1 Common denominator or ( )( ) 3 1 3 1 × − + equate constant terms and 3 terms. both correct solve two linear equations to obtain a = or b = both correct 3 2 2lg lg x x = 1 lg10 = 2 2 10 2 lg lg lg 2 10 x x x x   +   − =     +     oe ( )( ) 2 2 10 100 0 2 5 10 0 x x x x − − = → + − = 10 only x = B1 B1 B1 M1 A1 soi anywhere soi anywhere soi division; logs may be removed obtain correct 3 term quadratic equation and attempt to solve 5 x = − must not remain.

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 21 © UCLES 2016 Question Answer Marks Part Marks 4 (i) 0.5 10 7000 2000e 8213or 8210 t N − = → = + = B1 Do not accept non integer responses. (ii) 0.05 0.05 7500 7500 7000 2000e 500 e 2000 t t N − − = → = + = ln0.25 0.05 ln0.25 0.05 27.7 (days) t t − = →= − = M1 M1 A1 insert and make -0.05t e subject take logs and make t the subject awrt 27.7 (iii) 0.05 d 100e d t N t − = − d 8 67 d N t t = → = ± (.0) M1 A1 A1 0.05 e t k − where k is a constant k = −100 or 0.05 2000 − × awrt 67 ± mark final answer 5 (i) 2 d 3 4 7 d y x x x = + − d 2 12 8 7 3 d y x x = −→ = −− = − Equation of tangent : 16 3 3 10 2 y y x x − = −→ = − + + B1 M1 A1 insert 2 x = − into their gradient and use ( ) 2, 16 − and their gradient of tangent in equation of line. (ii) Tangent cuts curve again 3 2 2 7 2 3 10 x x x x + − + = − + 3 2 2 4 8 0 x x x + − − = ( )( )( ) 2 2 2 0 x x x + + − = 2, 4 x y = = M1 A1 M1 A1A1 equate curve and their linear answer from (i). factorise: ( ) 2 x ± and a two or three term quadratic is sufficient. Allow long division withhold final A1 if (2, 4) not clearly identified as their sole answer. 6 (i) cos sin cos sin sin cos 1 tan 1 cot 1 1 cos sin x x x x x x x x x x − = − + + + + 2 2 cos sin cos +sin cos +sin x x x x x x = − ( )( ) ( ) cos sin cos +sin cos +sin x x x x x x − = M1 M1 A1 A1 sin cos tan and cot cos sin x x x x x x = = Attempt to multiply by cos and sin x x AG (ii) sin cos 3sin 4cos 5cos 4sin x x x x x x − + = − = 5 tan 4 x = 51.3 , 128.7 x = ° − ° M1 A1 A1A1 equate and collect sin and cos x x oe FT from tan x k =

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 21 © UCLES 2016 Question Answer Marks Part Marks 7 (i) 2 9 h x = − ( ) ( ) 2 2 9 14 9 7 2 x A x x x x − = + + = − + B2/1/0 Must be clear that 2 9 x − is the height of the trapezium. 14 2x + oe must be seen AG (ii) ( ) ( ) 0.5 2 2 d 1 9 7 9 2 d 2 A x x x x x − = − + + − × − 2 2 d 0 9 7 d A x x x x = → − = + 2 2 7 9 0 x x + − = 1 x = 1 6 2 A = or 8 8 or 512 or 22.6 M1 A2/1/0 M1 A1 A1 A1 product rule on correct function minus 1 each error , allow unsimplified. equate to 0 and simplify to a linear or quadratic equation. correct three term quadratic obtained Extra positive answer loses penultimate A1. ignore negative solution. 8 (i) ( ) ( ) ( ) ( ) 3 2 3 2 2 3 1 9 3 1 3 f ' 1 x x x x x x + − − = + ( ) 2 2 3 12 1 x x = + M1 A1 A1 quotient rule or product rule all correct www beware 6 6 9 9 x x − gets A0 (ii) ( ) 2 2 3 2 2 3 1 3 1 1 3 1 d 12 1 1 x x x x x   − =   +   + ∫ 1 23 2 12 9 2 7 54   = −     = M1 A1 DM1 A1 3 3 3 1 1 x c x − × + FT 1 1 2 c their = top limit – bottom limit in their integral. or 0.130 or 0.1296 or 0.12 (iii) 3 3 3 1 1 y x y − = + 3 1 3 x y x + = − ( ) 1 f x − = 3 1 3 x x + − Domain : 6 1 2 7 x −- - B1 B1 B1 B1 make 3 y or 3x the subject FT take cube root (as long as 3 y or 3x equals a fraction with terms in or x y only) oe FT change and x y – can be done at any time Allow upper limit of 2.86 . Do not isw

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 21 © UCLES 2016 Question Answer Marks Part Marks 9 (i) tangent touches circle ( ) ( ) 2 2 4 2 4 8 x kx kx + − − − = 2 2 2 8 2 16 0 k x x kx kx + − − + = or better Equal roots as tangent touches circle : 2 4 b ac = ( ) ( ) 2 2 10 4 1 16 k k − = + × 2 36 64 k = 4 3 k = + only M1 A1 DM1 A1 A1 eliminate y or x allow unsimplified use of discriminant on 3 term quadratic soi oe any inequality loses last A1 (ii) 2 b x a − = 4 10 3 25 9 x × → = 12 4 5 5 x y = = − OR tangent 4 4 3 y x = − cuts radius 3 1 4 y x = − + at 12 5 x = 4 5 y = − OR Obtain 2 25 120 144 0 x x − + = oe ( )( ) 5 12 5 12 0 12 4 5 5 x x x y − − = = → = − M1 A1A1 M1 A1 A1 M1 A1A1 use 2 b x a − = find equation of radius and attempt to solve with tangent obtain any 3 term quadratic using their non zero k and reach x = … (iii) ( ) ( ) 2 2 0 2.4 4 0.8 4 TP = − + −+ = M1A1 M1 for using their T and ( ) 0 , 4 − . Signs must be correct.

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 21 © UCLES 2016 Question Answer Marks Part Marks 10 (i) 5000 2cos40 1000 2cos50 jr t p −     = +         B1 B1 xcoordinate oe y coordinate oe (ii) 2.5 cos70 5000 2 cos40 t t = − 5000 2.5cos70 2 cos40 t = + 2095 = awrt or 2090 or 2100 ( ) 2.5cos20 2cos50 2095 1000p − × = 2.23 p = awrt M1 DM1 A1 M1 A1 equate their x values (must be 3 terms) make t the subject allow one sign error equate their y values(must be 3 terms) and insert their or t t . 11 (i) Free choice : no. of ways 6 5 4 2 15 10 C C × = × 150 = B1 B1 6 4 C × another n r C term only 5 2 C × and answer or vice versa (ii) Both Mr and Mrs Coldicott 5 4 3 1 10 4 C C × = × 40 = B1 B1 5 3 C × another n r C term only 4 1 C × and answer or vice versa (iii) Mr C and not Mrs C ( ) 5 4 3 2 60 C C × = Not Mr C and Mrs C ( ) 5 4 4 1 20 C C × = Total = 80 OR Total = (i) – (ii) – neither Neither = 5 4 4 2 30 C C × = Total = 150 – 40 – 30 = 80 B1 B1 B1 M1 A1 A1 An incorrect final answer does not affect the awarding of the first two B1 marks. www