Cambridge IGCSE Mathematics - Additional 0606 — 2014 May/June Paper 2 · Variant 1

This document consists of 15 printed pages and 1 blank page. DC (RW/SW) 73390/3 © UCLES 2014 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 8 6 5 2 2 2 4 7 5 1 * ADDITIONAL MATHEMATICS 0606/21 Paper 2 May/June 2014 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 0606/21/M/J/14 © UCLES 2014 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/M/J/14 © UCLES 2014 [Turn over 1 Find the set of values of x for which x x x 2 1 + ^ h . [3] 2 Without using a calculator, express 6 1 3 2 + - ^ h in the form a b 3 + , where a and b are integers to be found. [4]

4 0606/21/M/J/14 © UCLES 2014 3 (i) On the axes below, sketch the graph of y x x 4 2 = - + ^ ^h h showing the coordinates of the points where the curve meets the x-axis. [2] y x O (ii) Find the set of values of k for which x x k 4 2 - + = ^ ^h h has four solutions. [3]

5 0606/21/M/J/14 © UCLES 2014 [Turn over 4 The expression x ax bx 2 12 3 2 + + + has a factor x 4 - and leaves a remainder of 12 - when divided by x 1 - . Find the value of each of the constants a and b. [5]

6 0606/21/M/J/14 © UCLES 2014 5 (i) Express x x 2 6 2 - + in the form p x q r 2 - + ^ h , where p, q and r are constants to be found. [3] (ii) Hence state the least value of x x 2 6 2 - + and the value of x at which this occurs. [2]

7 0606/21/M/J/14 © UCLES 2014 [Turn over 6 (a) Find the coefficient of x5 in the expansion of x 3 2 8 - ^ h . [2] (b) (i) Write down the first three terms in the expansion of x 1 2 6 + ^ h in ascending powers of x. [2] (ii) In the expansion of ax x 1 1 2 6 + + ^ ^h h , the coefficient of x2 is 1.5 times the coefficient of x. Find the value of the constant a. [4]

8 0606/21/M/J/14 © UCLES 2014 7 Given that a curve has equation y x x 1 2 = + , where x 0 2 , find (i) x y d d , [2] (ii) x y d d 2 2 . [2] Hence, or otherwise, find (iii) the coordinates and nature of the stationary point of the curve. [4]

9 0606/21/M/J/14 © UCLES 2014 [Turn over 8 A sector of a circle of radius r cm has an angle of i radians, where r 1 i . The perimeter of the sector is 30 cm. (i) Show that the area, A cm2, of the sector is given by A r r 15 2 = - . [3] (ii) Given that r can vary, find the maximum area of the sector. [3]

10 0606/21/M/J/14 © UCLES 2014 9 Solutions to this question by accurate drawing will not be accepted. The points , A p 1 ^ h, , B 1 6 ^ h, , C q 4 ^ h and , D 5 4 ^ h, where p and q are constants, are the vertices of a kite ABCD. The diagonals of the kite, AC and BD, intersect at the point E. The line AC is the perpendicular bisector of BD. Find (i) the coordinates of E, [2] (ii) the equation of the diagonal AC, [3]

11 0606/21/M/J/14 © UCLES 2014 [Turn over (iii) the area of the kite ABCD. [3]

12 0606/21/M/J/14 © UCLES 2014 10 Find x y d d when (i) cos sin y x x 2 3 = J L KK N P OO, [4] (ii) ln tan y x x l = + . [4]

13 0606/21/M/J/14 © UCLES 2014 [Turn over 11 (a) Solve 2 64 1 x x 5 2 = - . [4] (b) By changing the base of log 4 a 2 , express log log 4 2 l a a 2 + ^ ^h h as a single logarithm to base a. [4]

14 0606/21/M/J/14 © UCLES 2014 12 The functions f and g are defined by x x x 1 2 f = + ^ h for x 0 2 , x x 1 g = + ^ h for x 1 2- . (i) Find 8 fg^ h. [2] (ii) Find an expression for x f 2^ h, giving your answer in the form bx c ax + , where a, b and c are integers to be found. [3] (iii) Find an expression for x g 1 - ^ h, stating its domain and range. [4]

15 0606/21/M/J/14 © UCLES 2014 (iv) On the same axes, sketch the graphs of y x g = ^ h and y x g 1 = - ^ h, indicating the geometrical relationship between the graphs. [3] y x O

16 0606/21/M/J/14 © UCLES 2014 BLANK PAGE 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. 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.

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2014 series 0606 ADDITIONAL MATHEMATICS 0606/21 Paper 2, 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 May/June 2014 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 21 © Cambridge International Examinations 2014 1 [ ] 0 2 > + x x critical values 0 and −1 soi 0 1 < < − x M1 A1 A1 expands and rearranges condone space, comma, “and” but not “or” Mark final answer. 2 ( ) 2 3 1 6 + or ( )( ) 2 3 1 3 6 + + = b a 3 2 4 6 + or ( )( ) 3 2 4 3 6 + + = b a 3 2 4 3 2 4 3 2 4 6 − − × + AND attempting to multiply out 3 3 6 − isw M1 M1 M1 A1 for dealing with the negative index (condone treating 6 as have negative index at this stage) for squaring for rationalising or for obtaining a pair of simultaneous equations 6 6 4 = + b a and 0 4 2 = + b a 3 (i) (ii) O -2 4 x = 1 (only) soi y = ± 9 (only) 9 0 < < k B1 B1 B1 B1 B1 correct shape x intercepts marked or implied by tick marks, for example or seen nearby; condone y intercept omitted can be implied by second B1 or 9 ,9 + ± = k or –9 or both; must be strict inequality in k; condone space, comma, “and”, “or” 4 Attempt to find f(4) or f(1) or division to a remainder 128 + 16a + 4b + 12 = 0 or better (16a + 4b = –140) 2 + a + b + 12 = –12 or better (a + b = –26) Solves linear equations in a and b a = –3, b = –23 M1 A1 A1 M1 A1 condone one error both

Page 3 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 21 © Cambridge International Examinations 2014 5 (i) (ii) 8 47 4 1 2 2 +       − x (5.875) isw 8 47 is min value when 4 1 = x B3,2,1,0 B1ft + B1ft one mark for each of p, q, r correct; allow correct equivalent values. If B0, then SC2 for 8 47 4 1 2 +       − x , or SC1 for correct values but incorrect format strict ft their 8 47 and their 4 1 ; each value must be correctly attributed; condone 8 47 = y for B1, or       8 47 , 4 1 for B1B1 6 (a) (b) (i) (ii) ( )5 3 3 8 2 3 ± × × C or              ± 5 3 8 8 3 2 3 C –48384 1 + 12x + 60x2 Coefficient of x correct or correct ft (12+a) soi Coefficient of x2 correct or correct ft (60+12a) soi ( ) ( ) a their a their 12 60 12 5.1 + = + × − 4 M1 A1 B2,1,0 B1ft B1ft M1 A1 condone 5 5 8 2 , x C − can be in expansion ignore additional terms. If B0, allow M1 for 3 correct unsimplified terms ft 2 60 12 1 x x their + + ft 2 60 12 1 x x their + + no x or x2 7 (i) (ii) (iii) 2 1 1 1 2 x x + − 2 3 2 1 2 3 x x − Attempting to solve their 0 d d = x y x = 1 y = 3 Substitute their x = 1 into their 2 2 d d x y ; or examines x y d d or y on both sides of their x = 1 Complete and correct determination of nature. If correct, minimum. B1 + B1 B1ft + B1ft M1 A1 M1 A1 or equivalent with negative indices or equivalent with negative indices. Strict ft must achieve x =… (allow slips) SC2 for (1, 3) stated, nfww for using their value from 0 d d = x y must be from correct work

Page 4 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 21 © Cambridge International Examinations 2014 8 (i) (ii) 30 2 = + θ r r giving r r 2 30 − = θ Substitute their expression for θ into θ 2 2 1 r A = Correct simplification to 2 15 r r A − = AG 0 2 15 = −r 5.7 = r 25 . 56 M1 M1 A1 M1 A1 A1 correct arc formula + (2)r rearranged their 0 d d = r A 56.3 is A0 unless 56.25 seen; if M0, then SC2 for A = 56.25 with no working; or SC1 for r = 7.5 with no working 9 (i) (ii) (iii) (3, 5) 2 1 5 1 4 6 − =       − − = BD m       − ÷ − = 2 1 1 AC m seen or used ( ) 3 2 5 − = − x y or 1 , 2 − = + = c c x y or better p = 1 q = 7 [A(1, 1) C(4, 7)] Method for finding area numerically 15 B1B1 M1 M1 A1 M1 M1 A1 column vector B0B1 can be implied by second M1 could be in (ii) e.g.       × + × × + × × − 4 2 1 3 1 2 1 3 1 2 1 24 or shoelace method SC2 for 15 with no working 10 (i) (ii) x 2 sin 2 − and       3 cos 3 1 x Attempt at product rule       −       3 sin 2 sin 2 3 cos 2 cos 3 1 x x x x isw x 2 sec and x 1 Attempt at quotient rule (with given quotient) ( )( ) ( ) ( )2 2 ln l tan 1 ln 1 sec x x x x x + − + isw B1+B1 M1 A1ft B1 + B1 M1 A1 each trig function correctly differentiated ft x k 2 sin 1 and       3 cos 2 x k provided 2 ,1 k k are non-zero or rearrangement to correct product and attempt at product rule penalise poor bracketing if not recovered

Page 5 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 21 © Cambridge International Examinations 2014 11 (a) (b) 6 5 2 2 2 − − = x x x2 – 5x + 6 = 0 Correct method of solution of their 3 term quadratic x = 2 or x = 3 Correct change of base to a a a 2 log 4 log a a a a log 2 log 4 log + 1 log = a a used soi simplification to 4 log a M1 M1 M1 A1 B1 M1 M1 A1 Or ( ) =       = − 64 1 ln 2 ln 5 2 x x −6ln2 their “6” base a only at this stage but can recover at end for a a log 2 log 2 log + =

Page 6 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 21 © Cambridge International Examinations 2014 12 (i) (ii) (iii) (iv) f(3) 4 6 oe 1 1 2 1 2 2 + +       + x x x x A correct and valid step in simplification Correctly simplified to 1 3 4 + x x Putting y = g(x), changing subject to x and swopping x and y or vice versa g–1(x) = x2 –1 (Domain) x > 0 (Range) g–1(x) > –1 x y 0 -1 -1 M1 A1 M1 dM1 A1 M1 A1 B1 B1 B1 + B1 B1 or ( ) ( ) ( ) 1 1 1 2 fg + + + = x x x allow omission of 2(…..) in numerator or (…..) + 1 in denominator, but not both. e.g. multiplying numerator and denominator ,1 by + x , or simplifying 1 1 2 + + x x to 1 1 2 + + + x x x condone x = y2 –1; reasonable attempt at correct method condone y =….. , f−1 =…… condone y > −1 f−1 > −1 correct graphs; −1 need not be labelled but could be implied by ‘one square’ idea of reflection or symmetry in line y = x must be stated.