Cambridge IGCSE Mathematics - Additional 0606 — 2017 Oct/Nov Paper 2 · Variant 1

* 7 6 9 0 4 0 8 7 2 6 * This document consists of 15 printed pages and 1 blank page. DC (CW) 135370/2 © UCLES 2017 [Turn over ADDITIONAL MATHEMATICS 0606/21 Paper 2 October/November 2017 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/17 © UCLES 2017 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/17 © UCLES 2017 [Turn over 1 Solve the inequality ( )( ) x x 1 5 12 2 - - . [4] 2 Show that sin sin tan sec 1 1 1 1 2 i i i i - - + = . [4]

4 0606/21/O/N/17 © UCLES 2017 3 Solve the equation ( ) ( ) log log x x 10 5 2 7 5 5 + = + - . [4]

5 0606/21/O/N/17 © UCLES 2017 [Turn over 4 Solve the following simultaneous equations for x and y, giving each answer in its simplest surd form. x y 3 4 + = x y 2 5 3 - = [5]

6 0606/21/O/N/17 © UCLES 2017 5 (i) Find x x 3 2 5 d d + ` j. [2] (ii) Use your answer to part (i) to find ( ) x x 3 2 30 d 2 + y . [2] (iii) Hence evaluate ( ) x x 3 2 30 d 2 1 2 + y . [2]

7 0606/21/O/N/17 © UCLES 2017 [Turn over 6 It is given that p q M 2 3 = - c m where p and q are integers. (i) If det M =13, find an equation connecting p and q. [1] (ii) Given also that p q p q M 4 6 3 3 3 12 2 2 = - - - - + e o, find a second equation connecting p and q. [2] (iii) Find the value of p and of q. [4]

8 0606/21/O/N/17 © UCLES 2017 7 Find y in terms of x, given that x y x x 6 2 d d 2 2 3 = + and that when x = 1, y = 3 and x y d d 1 = . [6]

9 0606/21/O/N/17 © UCLES 2017 [Turn over 8 Given that ( ) z a a 3 3 = + + and z b 79 3 2 = + , find the value of each of the integers a and b. [6]

10 0606/21/O/N/17 © UCLES 2017 9 (i) Expand (1 + x)4, simplifying all coefficients. [1] (ii) Expand (6 - x)4, simplifying all coefficients. [2] (iii) Hence express (6 - x)4 - (1+ x)4 = 175 in the form ax3 + bx2 + cx + d = 0, where a, b, c and d are integers. [2]

11 0606/21/O/N/17 © UCLES 2017 [Turn over (iv) Show that x = 2 is a solution of the equation in part (iii) and show that this equation has no other real roots. [5]

12 0606/21/O/N/17 © UCLES 2017 10 In this question i is a unit vector due east and j is a unit vector due north. Units of length and velocity are metres and metres per second respectively. The initial position vectors of particles A and B, relative to a fixed point O, are 2i + 4j and 10i +14j respectively. Particles A and B start moving at the same time. A moves with constant velocity i + j and B moves with constant velocity -2i - 3j. Find (i) the position vector of A after t seconds, [1] (ii) the position vector of B after t seconds. [1] It is given that X is the distance between A and B after t seconds. (iii) Show that X 2 = (8 - 3t)2 + (10 - 4t)2. [3]

13 0606/21/O/N/17 © UCLES 2017 [Turn over (iv) Find the value of t for which (8 - 3t)2 + (10 - 4t)2 has a stationary value and the corresponding value of X. [4]

14 0606/21/O/N/17 © UCLES 2017 11 The line y = kx + 3, where k is a positive constant, is a tangent to the curve x2 - 2x + y2 = 8 at the point P. (i) Find the value of k. [4] (ii) Find the coordinates of P. [3] (iii) Find the equation of the normal to the curve at P. [2]

15 0606/21/O/N/17 © UCLES 2017 12 (i) Differentiate ( ) cosx 1 - with respect to x. [2] (ii) Hence find x y d d given that ( ) tan cos y x x 4 1 = + - . [2] (iii) Using your answer to part (ii) find the values of x in the range 0 G x G 2r such that x y d d = 4. [6]

16 0606/21/O/N/17 © UCLES 2017 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 a registered trademark. This document consists of 7 printed pages. © UCLES 2017 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/21 Paper 2 October/November 2017 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 International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the October/November 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 2 of 7 The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark M Method marks, awarded for a valid method applied to the problem. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. For accuracy marks to be given, the associated Method mark must be earned or implied. B 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 in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’ is used to indicate that a particular M or B mark is dependent on an earlier mark in the scheme. Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working nfww not from wrong working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 3 of 7 Question Answer Marks Guidance 1 2 6 7( 0) x x − − > B1 ( )( ) 7 1 ( 0) x x − + > M1 Critical values 7 and –1 A1 7 or 1 x x > < − A1 2 ( ) ( ) ( )( ) 1 sinθ 1 sin 1 sin 1 sin θ θ θ + − − − + M1 Dealing with fractions ( ) 2 2sin 1 sin θ θ = − A1 Simplification 2 2sin cos θ θ = M1 Use of identity (seen anywhere) 2tan sec θ θ = M1 Use of sin tan cos θ θ θ = and 1 sec cos θ θ = (seen anywhere) 3 5 2 log 25 = B1 ( ) ( ) 5 5 5 log 25 log 7 log 25 7 x x + − = − ( ) 10 5 25 7 x x + = − M1 180 15x = M1 Equate, clear brackets and collect terms. 12 x = A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 4 of 7 Question Answer Marks Guidance 4 ( ) 2 4 3 5 3 x x − − = M1 Eliminate y 5 3 8 2 3 1 x + = + A1 ( )( ) ( )( ) 5 3 8 2 3 1 2 3 1 2 3 1 x + − = + − M1 Multiply by ( ) a b c + as appropriate 2 3 x = + A1 1 2 3 y = − A1 Alternative method ( ) 3 5 3 2 4 y y + + = M1 Eliminate x ( ) 11 2 3 1 y − = + A1 ( ) ( )( ) 11 2 3 1 2 3 1 2 3 1 y − − = + − M1 Multiply by ( ) a b c + as appropriate 1 2 3 y = − A1 2 3 x = + A1 5(i) ( ) 2 d 5 5 3 2 3 d 3 2 x x x −  = − + ×   +   M1 ( ) 2 5 3 2 x − − + A1 3 × 5(ii) ( ) ( ) 2 30 10 d 3 2 3 2 x x x   − =   + +     ∫ M1 ( ) 1 3 2 x + A1 10 × − 5(iii) ( ) 2 1 10 10 10 3 2 8 5 x   − = − +   +     M1 Insert limits and subtract 3 4 = A1 6(i) 2 3 13 q p + = B1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 5 of 7 Question Answer Marks Guidance 6(ii) Multiply matrices correctly M1 2 12 p pq + = A1 6(iii) ( ) 4 13 3 24 p p p + − = M1 Eliminate q 2 3 17 24 0 p p − + = A1 ( )( ) 3 8 3 0 p p − − = M1 Solve 3, 2 p q = = A1 7 ( ) 2 2 d 1 3 d y x C x x = − + B2 B1 for 2 3x B1 for 2 1 x − . d 1, 1 1 d y x C x = = → = − B1 3 1 y x x D x = + − + 1, 3 2 x y D = = → = B2 B1 for two correct terms in x 3 1 2 y x x x = + − + B1 8 ( ) ( ) 2 2 2 3 3 2 3 3 z a a a a = + + + + 79 3 b = + M1 ( ) 2 2 3 3 79 a a + + = and ( ) 2 3 a a b + = A1 FT Equate correctly to obtain both eqns 2 2 + 3 +18 + 27 = 79 a a a 2 4 18 52 0 a a + − = M1 Expand and simplify to obtain 3 term quadratic ( )( ) 2 4 26 0 a a − + = M1 2, a = 20 b = A2 A1 for each 9(i) 2 3 4 1 4 6 4 x x x x + + + + B1 9(ii) 2 3 4 1296 864 216 24 x x x x − + − + B2 Minus 1 each error. 9(iii) 2 3 1295 868 210 28 175 x x x − + − = M1 Subtract and equate to 1 3 2 28 210 868 1120 0 x x x − + − = A1

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 6 of 7 Question Answer Marks Guidance 9(iv) ( ) ( ) ( ) 3 2 28 2 210 2 868 2 1120 − + − M1 Inserts 2 x = 224 840 1736 1120 0 = − + − = ( ) 2 x − is a factor A1 ( )( ) 2 2 28 154 560 x x x − − + M1A1 M1 for 28 and 560 seen oe A1 for –154 2 4 0 b ac − < shown B1 10(i) ( ) ( ) 2 4 t = + + + r i j i j A B1 10(ii) ( ) ( ) 10 14 2 3 t = + + − − r i j i j B B1 10(iii) (8 10 ) ( 3 4 ) t − = + + − − r r i j i j B A M1 ( ) ( ) 2 2 2 8 3 10 4 X t t = − + − M1A1 10(iv) Differentiate M1 ( )( ) ( )( ) 2 d 2 8 3 3 2 10 4 4 d X t t t = − − + − − oe A1 2 d 0 2.56 d 0.4 X t t X = →= → = B2 B1 for value of t B1 for value of X . 11(i) ( ) 2 2 2 3 8 x x kx − + + = M1 Eliminate y ( ) ( ) 2 2 1 6 2 1 0 k x k x + + − + = A1 ( ) ( ) 2 2 2 4 0 6 2 4 1 0 b ac k k − = → − − + = M1 3 4 k = A1 Answer given 11(ii) 2.5 2 2 1.5625 b x x a − − = → = × M1 = – 0.8 A1 0.75 0.8 3 2.4 y = × − + = A1 FT

0606/21 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 7 of 7 Question Answer Marks Guidance 11(iii) Eqn of PQ 2.4 0.8 y x − + = 4 3 − M1 3 4 4 y x → = − A1 12(i) ( ) 1 2 d cos 1 = sin d cos x x x x − × M1 2 1 cos x A1 sinx × 12(ii) 2 2 d 4sin sec d cos y x x x x = + B1 2 sec x B1 2 4sin cos x x 12(iii) 2 1 4 sin 4 cos cos cos x x x x + × = M1 Equate their (i) to 4 and multiply by 2 cos x 2 1 4sin 4cos x x →+ = M1 Use of identity and simplify 2 4sin 4sin 3 0 x x + − = A1 ( )( ) 2sin 1 2sin 3 0 x x − + = M1 Solve 5 , 6 6 x π π = A2 A1 for each