Cambridge IGCSE Mathematics - Additional 0606 — 2014 May/June Paper 2 · Variant 3

This document consists of 16 printed pages. DC (NH/CGW) 73394/4 © UCLES 2014 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 9 1 0 4 5 3 8 9 2 1 * ADDITIONAL MATHEMATICS 0606/23 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 0606/23/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 0606/23/M/J/14 © UCLES 2014 [Turn over 1 r cm 1.6 rad 500 cm2 The diagram shows a sector of a circle of radius r cm. The angle of the sector is 1.6 radians and the area of the sector is 500cm2. (i) Find the value of r. [2] (ii) Hence find the perimeter of the sector. [2]

4 0606/23/M/J/14 © UCLES 2014 2 Using the substitution log u x 3 = , solve, for x, the equation log log x 12 3 4 x 3 - = . [5]

5 0606/23/M/J/14 © UCLES 2014 [Turn over 3 In a motor racing competition, the winning driver in each race scores 5 points, the second and third placed drivers score 3 and 1 points respectively. Each team has two members. The results of the drivers in one team, over a number of races, are shown in the table below. Driver 1st place 2nd place 3rd place Alan 3 1 4 Brian 1 4 0 (i) Write down two matrices whose product under matrix multiplication will give the number of points scored by each of the drivers. Hence calculate the number of points scored by Alan and by Brian. [3] (ii) The points scored by Alan and by Brian are added to give the number of points scored by the team. Using your answer to part (i), write down two matrices whose product would give the number of points scored by the team. [1]

6 0606/23/M/J/14 © UCLES 2014 4 (a) Illustrate the following statements using the Venn diagrams below. (i) A B A , = (ii) A B C + + Q = [2] (b) It is given that  is the set of integers between 1 and 100 inclusive. S and C are subsets of , where S is the set of square numbers and C is the set of cube numbers. Write the following statements using set notation. (i) 50 is not a cube number. [1] (ii) 64 is both a square number and a cube number. [1] (iii) There are 90 integers between 1 and 100 inclusive which are not square numbers. [1]

7 0606/23/M/J/14 © UCLES 2014 [Turn over 5 Do not use a calculator in this question. (i) Show that 2 2 4 8 2 2 3 0 2 + - + = ^ ^ h h . [2] (ii) Solve the equation x x 2 2 2 2 0 3 4 2 2 + - + = + ^ ^ h h , giving your answer in the form a b 2 + where a and b are integers. [3]

8 0606/23/M/J/14 © UCLES 2014 6 Find the coordinates of the points of intersection of the curve x y 8 10 1 - = and the line x y 9 + = . [6]

9 0606/23/M/J/14 © UCLES 2014 [Turn over 7 (a) Prove that sec cosec tan cot sin cos 1 i i i i i i + + = + . [3] (b) Given that tanx 12 5 =- and ° ° x 90 180 1 1 , find the exact value of sinx and of cosx, giving each answer as a fraction. [3] Answer sinx = cosx =

10 0606/23/M/J/14 © UCLES 2014 8 A curve is such that x y x x 6 8 3 d d 2 = - + . (i) Show that the curve has no stationary points. [2] Given that the curve passes through the point P(2, 10), (ii) find the equation of the tangent to the curve at the point P, [2] (iii) find the equation of the curve. [4]

11 0606/23/M/J/14 © UCLES 2014 [Turn over 9 Solutions to this question by accurate drawing will not be accepted. The points ( , ) A 2 11 , ( , ) B 2 3 - and ( , ) C 2 1 - are the vertices of a triangle. (i) Find the equation of the perpendicular bisector of AB. [4] The line through A parallel to BC intersects the perpendicular bisector of AB at the point D. (ii) Find the area of the quadrilateral ABCD. [6]

12 0606/23/M/J/14 © UCLES 2014 10 (i) Given that y x x 21 2 2 = + , show that x y x k 21 d d 2 3 = + ^ h , where k is a constant to be found. [5]

13 0606/23/M/J/14 © UCLES 2014 [Turn over (ii) Hence find x x 21 6 d 2 3 + c e dd ^ h and evaluate x x 21 6 d 2 3 10 + 2 c e dd ^ h . [3]

14 0606/23/M/J/14 © UCLES 2014 11 M N P A B O In the diagram OA a 2 = and OB b 5 = . The point M is the midpoint of OA and the point N lies on OB such that : : ON NB 3 2 = . (i) Find an expression for the vector MB in terms of a and b. [2] The point P lies on AN such that AP AN m = . (ii) Find an expression for the vector AP in terms of m, a and b. [2]

15 0606/23/M/J/14 © UCLES 2014 [Turn over (iii) Find an expression for the vector MP in terms of m, a and b. [2] (iv) Given that M, P and B are collinear, find the value of m. [4] Question 12 is printed on the next page.

16 0606/23/M/J/14 © UCLES 2014 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. University of 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 The function f is such that ( )x x 2 3 f = + - for x 4 28 G G . (i) Find the range of f. [2] (ii) Find ( ) 12 f 2 . [2] (iii) Find an expression for ( )x f 1 - . [2] The function g is defined by ( )x x 120 g = for x 0 H . (iv) Find the value of x for which ( )x 20 gf = . [3]

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2014 series 0606 ADDITIONAL MATHEMATICS 0606/23 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 23 © Cambridge International Examinations 2014 1 (i) (ii) 2 2 1 500 r = (1.6) 25 only their 25 + their 25 + their 25 × 1.6 or better 90 M1 A1 M1 A1 25 ± is A0 their 25 must be positive 2 x x 3 log 1 3 log = oe soi u2 – 4u – 12 = 0 oe solve their 3 term quadratic in u Solve log3 x = 6 or log3 x = –2 oe 729 and 9 1 B1 M1 M1 M1 A1 may be implied by u x 1 3 log = oe condone sign errors 3 (i) (ii)       0 3 1 4 1 3 and           1 3 5 or ( )1 3 5 and           0 4 4 1 1 3 Multiplication of compatible matrices       17 22 or ( ) 17 22 as appropriate ( )1 1 with       17 22 or ( ) 17 22 with       1 1 B1 M1 A1 B1 Must be correct shape from candidates product

Page 3 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 23 © Cambridge International Examinations 2014 4 (a) (i) B1 (ii) or B1 any Venn diagram showing three circles which do not all overlap (b) (i) 50 ∉ C B1 (ii) 64 ∈ S ∩ C B1ft ft only on use of ⊄ and ⊂ instead of ∉ and ∈ (iii) n(S′) = 90 B1 5 (i) 16 2 16 8 4 2 2 2 + + =     + Correct completion B1 B1 (ii) Use a ac b b 2 4 2 − ± − Multiply top and bottom by 3 2 2 − 2 2 − M1 M1 A1               +     + = 3 2 2 2 4 2 2 Or 6 2 4 − 6 Eliminate x or y Rearrange to quadratic in x or y x2 – 27x + 72 = 0 or y2 + 9y – 90 = 0 Factorise or solve 3 term quadratic x = 3, x = 24 or y = 6, y = –15 y = 6, y = –15 or x = 3, x = 24 M1 M1 A1 M1 A1 B1

Page 4 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 23 © Cambridge International Examinations 2014 7 (a) (b) θ θ θ θ θ θ sin 1 cos 1 sin cos cos sin + + Clears the fractions in the numerator and denominator using common denominator θ θ θ θ cos sin cos sin 2 2 + + and completion evidence of 13 13 5 sin = x 13 12 cos − = x B1 M1 A1 B1 B1 B1ft ft on their 13 8 (i) (ii) (iii) Attempt to find b2 – 4ac Completely correct argument m = 6(4) – 8(2) + 3 y – 10 = 11(x – 2) or y = 11x – 12 Integrate to 2x3 – 4x2 + 3x(+c) 10 = 2(2)3 – 4(2)2 + 3(2) + c y = 2x3 – 4x2 + 3x + 4 soi M1 A1 M1 A1 B2,1,0 M1 A1 may be in formula or attempt to complete square dep on c being a genuine constant of integration

Page 5 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 23 © Cambridge International Examinations 2014 9 (i) (ii) (0, 7) mAB = 2 perpendicular gradient 2 1 − = 7 2 1 + − = x y mAB = –1 y = –x + 13 Solve their y = –x + 13 and 7 2 1 + − = x y D(12,1) Complete method for area 84 B1 B1 M1 A1 B1 B1 M1 A1 M1 A1 10 (i) (ii) 21 21 d d 2 2 + =     + x x x x Use of quotient rule ( ) ( ) ( ) 21 21 2 21 2 2 2 2 + + × − + x x x x x Multiply each term by ( ) 21 2 + x ( ) ( )2 3 2 2 2 21 2 21 2 + − + x x x leading to k = 42 21 2 6 2 + × x x k Use limits in 21 2 2 + × x x C 55 8 or 0.145 B1 M1 A1 M1 A1 M1 M1 A1 Alt method using product rule 3 2 2 21 21 1 d d     + − =     + x x x x is B1 then M1 A1 as in quotient k must be a constant

Page 6 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 23 © Cambridge International Examinations 2014 11 (i) (ii) (iii) (iv) a = OM MB = 5b – a b ON 3 = AP = λ (3b – 2a) AP MA MP + = a + λ (3b – 2a) Put MB MP µ = Equate components Solve simultaneous equations 7 5 = λ B1 B1 B1 B1 M1 A1 M1 M1 M1 A1 12 (i) (ii) (iii) (iv) 3 < f < 7 f(12) = 5 (f(5) = ) 2 2 + Clear indication of method f–1(x) = (x – 2)2 + 3 gf (x) = ( ) 2 3 120 + − x Attempt to solve their gf (x) = 20 x = 19 B1,B1 B1 B1 M1 A1 B1 M1 A1 If B0 then SC1 for 3 I f I 7 f 2 (x) ( ) 2 3 2 3 +     − + − x earns B1 condone y = (x – 2)2 + 3