Cambridge IGCSE Mathematics - Additional 0606 — 2015 Oct/Nov Paper 1 · Variant 3

This document consists of 16 printed pages. DC (AC/CGW) 98537/3 © UCLES 2015 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 1 8 0 7 0 4 5 4 9 1 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 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/O/N/15 © UCLES 2015 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/15 © UCLES 2015 [Turn over 1 On the Venn diagrams below, shade the regions indicated. (i) A B C A B C [1] (ii) A B C A B C [1] (iii) A B C A B ' C [1]

4 0606/13/O/N/15 © UCLES 2015 2 Solve cos x 2 3 4 1 2 r - = J L KK N P OO for x 0 3 G G r . [4]

5 0606/13/O/N/15 © UCLES 2015 [Turn over 3 (a) Matrices A and B are such that 2 4 1 3 A = J L KK N P OO and 3 6 8 0 1 2 B = J L KK N P OO. Find AB. [2] (b) Given that matrix 4 2 6 8 X = - J L KK N P OO, find the integer value of m and of n such that m n X X I 2 = + , where I is the identity matrix. [5] (c) Given that matrix a a 3 2 Y = J L KK N P OO, find the values of a for which det Y = 0. [2]

6 0606/13/O/N/15 © UCLES 2015 4 You are not allowed to use a calculator in this question. A B C (4√3+1) cm The diagram shows triangle ABC with side ( ) AB 4 3 1 = + cm. Angle B is a right angle. It is given that the area of this triangle is 2 47 cm2. (i) Find the length of the side BC in the form ( ) a b 3 + cm, where a and b are integers. [3] (ii) Hence find the length of the side AC in the form p 2 cm, where p is an integer. [2]

7 0606/13/O/N/15 © UCLES 2015 [Turn over 5 Find the equation of the normal to the curve tan y x 5 3 = - at the point where x 4 r = . [5]

8 0606/13/O/N/15 © UCLES 2015 6 (i) On the axes below, sketch the graph of y x x 4 12 2 = - - showing the coordinates of the points where the graph meets the axes. [3] y x O (ii) Find the coordinates of the stationary point on the curve y x x 4 12 2 = - - . [2] (iii) Find the values of k such that the equation x x k 4 12 2 - - = has only 2 solutions. [2]

9 0606/13/O/N/15 © UCLES 2015 [Turn over 7 A curve, showing the relationship between two variables x and y, is such that d d cos x y x 6 3 2 2 = . Given that the curve has a gradient of 4 3 at the point , 9 3 1 r - c m, find the equation of the curve. [6]

10 0606/13/O/N/15 © UCLES 2015 8 (a) Given that the first 4 terms in the expansion of ( ) kx 2 8 + are x px qx 256 256 2 3 + + + , find the value of k, of p and of q. [3] (b) Find the term that is independent of x in the expansion of x x 2 2 9 - c m . [3]

11 0606/13/O/N/15 © UCLES 2015 [Turn over 9 (a) Five different books are to be arranged on a shelf. There are 2 Mathematics books and 3 History books. Find the number of different arrangements of books if (i) the Mathematics books are next to each other, [2] (ii) the Mathematics books are not next to each other. [2] (b) To compete in a quiz, a team of 5 is to be chosen from a group of 9 men and 6 women. Find the number of different teams that can be chosen if (i) there are no restrictions, [1] (ii) at least two men must be on the team. [3]

12 0606/13/O/N/15 © UCLES 2015 10 6 cm C A B Y X 6 cm 10 cm The diagram shows an isosceles triangle ABC such that AC = 10 cm and AB = BC = 6 cm. BX is an arc of a circle, centre C, and BY is an arc of a circle, centre A. (i) Show that angle ABC = 1.970 radians, correct to 3 decimal places. [2] (ii) Find the perimeter of the shaded region. [4]

13 0606/13/O/N/15 © UCLES 2015 [Turn over (iii) Find the area of the shaded region. [3]

14 0606/13/O/N/15 © UCLES 2015 11 The line x y 2 0 - + = intersects the curve x y x 2 2 1 0 2 2 - + + = at the points A and B. The perpendicular bisector of the line AB intersects the curve at the points C and D. Find the length of the line CD in the form a 5, where a is an integer. [10]

15 0606/13/O/N/15 © UCLES 2015 [Turn over Question 12 is printed on the next page.

16 0606/13/O/N/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. 12 (a) Given that 2 4 128 x x y 2 1 # = - + and 27 9 1 y y x 4 2 = - - , find the value of each of the integers x and y. [4] (b) Solve ( ) 2 5 5 1 0 z z 2 + - = . [4]

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME for the October/November 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 October/November 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 – October/November 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) (ii) (iii) A C B A B C B C A B1 B1 B1 2 cos ( ) 2 1 4 3 ± =       −π x oe 3 3 , , 4 4 4 4 x π π π π − = − 3 3, 3, 3 4 4 4 4 4 4 x π π π π π π       = − + ÷ + ÷ + ÷             oe 0 = x and 6 π (or 0 and 0.524) 3 π = x (or 1.05) M1 DM1 A2/1/0 division by 2 and square root correct order of operations in order to obtain a solution A2 for 3 solutions and no extras in the range A1 for 2 solutions A0 for one solution or no solutions

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 13 © Cambridge International Examinations 2015 3 (a) (b) (c)       10 32 30 4 16 12       +       − =       − − 1 0 0 1 8 2 6 4 76 8 24 28 n m m 6 24 = − or m 2 8 = − giving 4 − = m n m + = 4 28 or n m + − = 8 76 44 = n 0 6 2 = − a so 6 ± = a B2,1,0 B2,1,0 B1 M1 A1 B2,1,0 B2 for 6 elements correct, B1for 5 elements correct B2 for 4 correct elements in X2 B1 for 3 correct elements in X2 For 4 m = −using correct I complete method to obtain n B2 for 6 ± = a or 45 .2 ± = a , with no incorrect statements seen or B1 for 6 ± = a or 45 .2 ± = a seen or B1 for 6 = a and no incorrect working 4 (i) (ii) ( ) 1 47 4 3 1 2 2 BC + × = ( ) ( ) ( )1 3 4 1 3 4 1 3 4 47 − − × + = BC 1 3 4 − = BC Alternative method ( ) 1 47 4 3 1 2 2 BC + × = ( )( ) 4 3 1 3 47 a b + + = Leading to 47 12 = + b a and 0 4 = + b a Solution of simultaneous equations 1 3 4 − = BC ( ) ( ) 2 2 4 3 1 4 3 1 + + − ( ) ( ) 48 8 3 1 48 8 3 1 = + + + − + 98 2 = AC 2 7 = AC or 7 = p B1 M1 A1 B1 M1 A1 B1FT B1cao correct use of the area correct rationalisation Dependent on all method being seen Dependent on all method seen including solution of simultaneous equations 6 correct FT terms seen 98 and 7 2 or 98 and p = 7

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 13 © Cambridge International Examinations 2015 5 When , 2 4 x y π = = 2 d 5sec d y x x = When d , 10 4 d y x x π = = Equation of normal       − − = − 4 10 1 2 π x y 10 20 0 or 10 20.8 0 4 y x y x π + − − = + − = oe B1 B1 B1 M1 A1 2 = y x 2 sec 5 10 from differentiation       − − = − 4 10 1 2 π x their their y allow unsimplified 6 (i) (ii) (iii) ( ) 2,16 0 = k 16 > k B1 B1 B1 M1 A1 B1 B1 shape intercepts on x-axis intercept on y-axis for a curve with a maximum and two arms (2, ±16) seen or( ) 2, k where 0 > k (2, 16) or x = 2 and y = 16 only

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 13 © Cambridge International Examinations 2015 7 ( ) d 2sin3 d y x c x = + c + = 2 3 2 3 4 d 2sin3 3 3 d y x x = + ( ) 2 cos3 3 3 3 y x x d = − + + d +       + − = − 9 3 3 3 cos 3 2 3 1 π π 2 3 cos3 3 3 3 3 y x x π = − + − B1 M1 A1 B1FT M1 A1 x 3 sin 2 finding constant using d sin3 d y k x c x = + making use of d 4 3 d y x = and 9 π = x Allow with 20 .5 = c or 27 FT integration of their x k 3 sin finding constant d for d cx x k + + 3 cos Allow π 577 .0 20 .5 3 cos 667 .0 − + − = x x y or better 8 (a) (b) ( ) 3 3 2 2 8 1792 1792 1024 256 2 x k x k kx kx + + + = + 4 1 = k 112 p = 28 q = 3 2 6 3 9 2      − x x C      − 6 6 8 84 x x leading to 672 − B1 B1FT B1FT M1 DM1 A1 FT 1792 multiplied by their k2 FT 1792 multiplied by their k3 correct term seen Term selected and 23 and 9 3 C correctly evaluated

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 13 © Cambridge International Examinations 2015 9 (a) (i) (ii) (b) (i) (ii) Number of arrangements with Maths books as one item = 4! or 4 × 3! or Maths books can be arranged 2! ways and History 3! ways = 2! × 3! 2 × 4! or 2 × 4 × 3! or 4 × 2 × 3! = 48 5! – 48 or !3 2 6 × × 72 3003 135 6 3003 − − 2862 or 2M 3W = 720 3M 2W = 1260 4M 1W = 756 5M = 126 2862 M1 A1 M1 A1 B1 M1 B1 A1 M1 B1 A1 ( ) 4! 2 × or ( ) 2 !3 4 × × oe ( ) 2! 3! 4 × × or ( ) 4 !3 2 × × oe A1 for 48 5! – their answer to (i) or for 3 2 6 × × their answer to (i) – 6 9 4 6 × −C 135 subtracted complete correct method using 4 cases, may be implied by working. Must have at least one correct any 3 correct

Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 13 © Cambridge International Examinations 2015 10 (i) (ii) (iii) ABC cos 6 6 2 6 6 10 2 2 2 × × × − + = or 6 5 2 sin =      ABC or 36 11 10 sin 1 − − = π ABC 9702 .1 = ABC 2 = XY Arc length       − 2 970 .1 6 π oe Perimeter =             − + 2 970 .1 6 2 2 π = 9.03 2 11 5 2 1 2 970 .1 6 2 1 2 ×       × × −       − × π = 4.50 or 4.51 or better M1 A1 B1 B1 M1 A1 M1 M1 A1 correct cosine rule statement or correct statement for 2 sin ABC or equating areas oe 1.9702 or better for XY ( may be implied by later work, allow on diagram) correct arc length (unsimplified) their 2 + 2 × 6 × their angle C sector area using their C area of ∆ ABM where M is the midpoint of AC, or (∆s ABY and BXY) or ∆ABC Answers to 3sf or better

Page 8 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 13 © Cambridge International Examinations 2015 11 0 3 2 2 = − − x x or 0 5 6 2 = + − y y leading to (3, 5) and (–1, 1) Midpoint (1, 3) (Gradient – 1) Perpendicular bisector x y − = 4 Meets the curve again if 0 15 10 2 = − + x x or 0 41 18 2 = + − y y leading to 5 2 10, 9 2 10 x y = −± = m ( ) ( ) 2 2 2 10 4 10 4 + = CD 5 8 = CD M1 A1,A1 B1cao M1 M1 A1,A1 M1 A1 substitution and simplification to obtain a three term quadratic equation in one variable A1 for each ‘pair’ from a correct quadratic equation, correctly obtained. midpoint perpendicular bisector, must be using their perpendicular gradient and their midpoint substitution and simplification to obtain a three term quadratic equation in one variable. A1 for each ‘pair’ Pythagoras using their coordinates from solution of second quadratic. ( ) ( )2 2 1 2 2 1 y y x x − + − must be seen if not using correct coordinates. A1 for 5 8 from 320 and all correct so far.

Page 9 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 13 © Cambridge International Examinations 2015 12 (a) (b) ( ) 7 2 1 2 2 2 2 = × + − y x x and ( ) ( ) 1 3 3 4 3 2 2 = − − y x y ( ) 7 2 1 2 = + + − y x x oe ( ) ( ) 4 3 2 2 − = − y x y oe leading to 4, 4 x y = = − Example of Alternative method Method mark as above ( ) 7 2 1 2 = + + − y x x leading to ( ) 2 4 8 x y − = Correctly substituted in ( ) ( ) 1 3 3 4 3 2 2 = − − y x y Leading to ( ) ( )       − − =       − − 4 2 4 8 3 2 4 8 2 2 x x x Leading to 4 = x and 4 − = y ( ) ( )( ) 0 1 5 1 5 2 = + − z z leading to 1 5.2 = z ( )1 5 − = z 5.0 5 = z z = log0.5 log5 or 431 .0 − = z or better M1 A1 A1 A1 M1 A1 A1 A1 M1 A1 DM1 A1 expressing 4x+y , 128 as powers of 2 and 92y–x , 27y–4 as powers of 3 Correct equation from correct working Correct equation from correct working for both As before One of the correct equations in x and y Correct, unsimplified, equation in x or y only Both answers solution of quadratic correct solution correct attempt to solve k z = 5.2 , where k is positive must have one solution only