Cambridge IGCSE Mathematics - Additional 0606 — 2014 Oct/Nov Paper 1 · Variant 3

This document consists of 15 printed pages and 1 blank page. DC (LK/SLM) 81861/3 © UCLES 2014 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 8 5 3 8 8 2 7 9 5 0 * ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 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 0606/13/O/N/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 0606/13/O/N/14 © UCLES 2014 [Turn over 1 The diagram shows the graph of cos y a bx c = + for ° ° x 0 360 G G , where a, b and c are positive integers. O y x 180° 1 7 360° State the value of each of a, b and c. [3] a = b = c =

4 0606/13/O/N/14 © UCLES 2014 2 The line y x 4 8 = + cuts the curve xy x 4 2 = + at the points A and B. Find the exact length of AB. [5]

5 0606/13/O/N/14 © UCLES 2014 [Turn over 3 The universal set  is the set of real numbers. Sets A, B and C are such that A = :x x x 5 6 0 2 + + = " ,, B = :x x x x 3 2 1 0 - + + = ^ ^ ^ h h h " ,, C = :x x x 3 0 2 + + = " ,. (i) State the value of each of n(A), n(B) and n(C). [3] n A = ^ h n B = ^ h n C = ^ h (ii) List the elements in the set A B , . [1] (iii) List the elements in the set A B + . [1] (iv) Describe the set Cl. [1]

6 0606/13/O/N/14 © UCLES 2014 4 (a) Solve sin cos x x 3 5 0 + = for ° ° x 0 360 G G . [3] (b) Solve cosec y 3 4 2 r + = J L KK N P OO for y 0 G G r radians. [5]

7 0606/13/O/N/14 © UCLES 2014 [Turn over 5 (a) A drinks machine sells coffee, tea and cola. Coffee costs $0.50, tea costs $0.40 and cola costs $0.45. The table below shows the numbers of drinks sold over a 4-day period. Coffee Tea Cola Tuesday 12 2 1 Wednesday 9 3 0 Thursday 8 5 1 Friday 11 2 0 (i) Write down 2 matrices whose product will give the amount of money the drinks machine took each day and evaluate this product. [4] (ii) Hence write down the total amount of money taken by the machine for this 4-day period. [1] (b) Matrices X and Y are such that X = 2 5 4 1 - J L KK N P OO and XY = I, where I is the identity matrix. Find the matrix Y. [3]

8 0606/13/O/N/14 © UCLES 2014 6 The diagram shows a sector, AOB, of a circle centre O, radius 12 cm. Angle . AOB 0 9 = radians. The point C lies on OA such that OC = CB. 0.9 rad O C A B 12 cm (i) Show that OC = 9.65 cm correct to 3 significant figures. [2] (ii) Find the perimeter of the shaded region. [3]

9 0606/13/O/N/14 © UCLES 2014 [Turn over (iii) Find the area of the shaded region. [3] 7 Solve the equation log log x x 1 2 18 9 5 5 + = - ^ h. [5]

10 0606/13/O/N/14 © UCLES 2014 8 (i) Given that f ln x x x3 = ^ h , show that f ln x x 3 1 = + l^ ^ h h. [3] (ii) Hence find ln d x x 1 + ^ h y . [2] (iii) Hence find ln d x x 1 2y in the form ln p q + , where p and q are integers. [3]

11 0606/13/O/N/14 © UCLES 2014 [Turn over 9 (a) Given that the first 3 terms in the expansion of qx 5 p - ^ h are x rx 625 1500 2 - + , find the value of each of the integers p, q and r. [5] (b) Find the value of the term that is independent of x in the expansion of x x 2 4 1 3 12 + J L KK N P OO . [3]

12 0606/13/O/N/14 © UCLES 2014 10 (a) Solve the following simultaneous equations. 25 5 1 y x 3 2 = - 27 3 81 y x 1 = - [5]

13 0606/13/O/N/14 © UCLES 2014 [Turn over (b) The diagram shows a triangle ABC such that AB 2 3 = + ^ h cm, BC 1 2 3 = + ^ h cm and AC 2 = cm. B C A 2 cm 1 2 3 cm + ^ h 2 3 cm + ^ h Without using a calculator, find the value of cosA in the form a b 3 + , where a and b are constants to be found. [4]

14 0606/13/O/N/14 © UCLES 2014 11 The diagram shows part of the curve y x x 5 1 2 = + - ^ ^h h . O x y y = (x + 5)(x – 1)2 (i) Find the x-coordinates of the stationary points of the curve. [5]

15 0606/13/O/N/14 © UCLES 2014 (ii) Find d x x x 5 1 2 + - ^ ^h h y . [3] (iii) Hence find the area enclosed by the curve and the x-axis. [2] (iv) Find the set of positive values of k for which the equation x k x 5 1 2 + = - ^ ^h h has only one real solution. [2]

16 0606/13/O/N/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. 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 the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME for the October/November 2014 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 2014 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 2014 0606 13 © Cambridge International Examinations 2014 1 4 2 3 = = = c b a B1 B1 B1 2 16 2 = x or 0 3 4 2 = + −y y 4 ± = x 3 ,1 = y Points (–4, 1) and (4, 3) Line AB = 2 2 2 8 + 68 = or 17 2 M1 A1 A1 M1 A1 for correct elimination of one variable and attempt to form a quadratic equation in x or y. for use of Pythagoras theorem allow either form 3 (i) (ii) (iii) (iv) ( ) ( ) ( ) 0 n 3 n 2 n = = = C B A { } 3 ,3 ,2 ,1 − − − = ∪B A { } 2 − = ∩B A ξ, ’the universal set’, R, ‘real numbers’, } { : x x∈ B1 B1 B1 B1 B1 B1 B0 for n(2), {2},{0}, ∅ , {} etc. 4 (a) (b) 3 5 tan − = x ° ° = 0. 301 , 0. 121 x 2 1 4 3 sin =       + π y 6 4 3 π π = + y , 6 5π , 6 13π , 6 17π 3 12 y π = − , 12 7π , 12 23π , 12 31π 36 7π = y , 36 23π , 36 31π (0.611, 2.01 and 2.71) M1 A1 A1ft M1 A1 DM1 A1, A1 Correct statement or 67 .1 tan − = x A1 for either correct solution ft from their first solution for dealing correctly with cosec and attempt to solve subsequent equation for 6 π , 6 5π , or 6 13π , or 6 17π for correct order of operations A1 for one correct solution A1 for both the other correct solutions and no others in range.

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 13 © Cambridge International Examinations 2014 5 (a) (i) (ii) (b)             =                       30 .6 45 .6 70 .5 25 .7 45 .0 4.0 5.0 0 2 11 1 5 8 0 3 9 1 2 12 or ( )           0 1 0 1 2 5 3 2 11 8 9 12 45 .0 4.0 5.0 ( ) 30 .6 45 .6 70 .5 25 .7 = 25.70 I X Y X Y 1 1 − − = = or       − = 2 5 4 1 22 1 Y or           − 22 2 22 5 22 4 22 1 Alternative method:       =             − 1 0 0 1 1 5 4 2 d c b a 0 4 2 ,1 4 2 = + = + d b c a 1 5 ,0 5 = + − = + − d b c a leading to       − = 2 5 4 1 22 1 oe M1 DM1 A2,1,0 B1 M1 A1 A1 M1 A1 A1 for correct compatible matrices in the correct order. Allow 1 error in each matrix. Allow if done in cents for a correct method for multiplying their matrices to obtain an appropriate 4 by 1 or 1 by 4 matrix. A2 all correct or A1 3 correct elements. Allow 25.7 for matrix algebra for       22 1 for       − 2 5 4 1 k for a complete method using simultaneous equations 22 1 = a and 22 5 = c or 22 4 − = b and 22 2 = d for correct matrix

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 13 © Cambridge International Examinations 2014 6 (i) (ii) (iii) OC 6 9.0 cos = or ( ) 8.1 sin 12 9.0 sin − = π OC ... 652 .9 9.0 cos 6 = = OC or ( ) ... 652 .9 8.1 sin 9.0 sin 12 = − = π OC Perimeter = ( ) ( ) 652 .9 12 652 .9 12 9.0 − + + × = 22.8 Area = ( )      − × −       × × 8.1 sin 652 .9 2 1 9.0 12 2 1 2 2 π 64.8 − 45.36 = 19.4 to 19.5 Alternative Method: ( ) 8.1 sin 652 .9 652 .9 12 2 1 × × − ( ) 9.0 sin 9.0 12 2 1 2 − 11.04 + 8.40 Area =19.4 to 19.5 M1 A1 B1 M1 A1 B1 B1 B1 B1 B1 B1 for correct use of cosine, sine rule, cosine rule or any other valid method for manipulating correctly to OC = 9.652(35…) Must have 4th figure (or more) for rounding for arc length for attempt to add the correct lengths for area of sector, allow unsimplified for area of isosceles triangle ( ) ( ) ( ) 8.1 sin ... 2 65 .9 2 1 2 − π or ( ) 9.0 tan 6 12 2 1 × or ( ) ( ) 9.0 sin ... 2 65 .9 12 2 1 × × , allow unsimplified. for answer in range 19.4 to 19.5 for area of triangle ACB, unsimplified for area of segment, unsimplified answer in range 19.4 to 19.5 7 ( ) 9 18 log log 2 1 5 5 − = + x x ( ) 9 18 log log 5 log 5 2 5 5 − = + x x 9 18 5 2 − = x x ( )( ) 0 3 3 5 = − − x x 5 3 = x , 3 B1, B1 M1 DM1 A1 B1 for dealing with ‘1’, B1 for dealing with ‘2’ for a correct use of addition or subtraction of logarithms for elimination of logarithms to form a 3 term quadratic and for solution of quadratic for both x values

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 13 © Cambridge International Examinations 2014 8 (i) (ii) (iii) ( ) ( ) 3 3 2 ln 3 f x x x x x +      × = ′ x ln 3 3+ = , ( ) x ln 1 3 + = or ( ) x x x ln 3 f = ( ) x x x x ln 3 1 3 f +       × = ′ , ( ) x ln 1 3 + = ( ) 3 ln d n 1 1 3 x x x x = + ∫ or x xln 3 ∫ = + 3 ln 3 1 d ln 1 x x x x or x xln ∫ − x x x d 1 ln or ∫ −     x x x d 1 ln 3 1 3 [ ]2 1 ln x x x − or 2 1 3 ln 3 1     −x x x 1 2 2 ln 2 + − = 4 ln 1+ − = M1 B1 A1 B1 M1 A1 M1 A1 DM1 DM1 A1 for differentiation of a product for differentiation of 3 ln x for simplification to gain given answer for use of x x ln 3 ln 3 = for differentiation of a product for simplification to gain given answer for realising that differentiation is the reverse of integration and using (i) for using answer to (ii) and subtracting ∫ x d 1 dependent on M mark in (ii) for correct application of limits from correct working 9 (a) (b) 625 5 = p , so 4 = p ( ) 1500 5 1 1 4 − = − − q C p ( ) 1500 125 4 − = − × q 3 = q r q C p = − 2 2 2 4 5 1350 = r ( ) 3 3 9 3 12 4 1 2       x x C Term is 1760 B1 M1 A1 M1 A1 M1 DM1 A1 their p substituted in ( ) q C p p − −1 15 or in ( ) qx C p p − −1 15 unsimplified their p and q substituted in ( )2 2 25 q C p p − − or ( )2 2 25 qx C p p − − unsimplified for identifying correct term for attempt to evaluate correct expression must be evaluated

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 13 © Cambridge International Examinations 2014 10 (a) (b) ( ) 1 5 5 2 3 2 = − y x or ( ) 4 1 3 3 3 3 = − y x oe 4 6 − = y x 1 3 + = y x Leads to 3 5 ,6 = = y x Using the cosine rule: ( ) ( ) ( ) A cos 3 2 4 2 3 2 3 2 1 2 2 2 + − + + = + ( ) ( ) ( ) 3 2 4 4 3 4 7 3 4 13 cos + − − + − + = A oe ( ) 3 2 3 2 3 2 2 1 cos − − × + − = A 2 3 1 cos + − = A M1 A1 A1 M1 A1 M1 DM1 DM1 A1 for obtaining one correct equation in powers of 5, 3, 25, 27 or 81 for 4 6 − = y x oe linear equation for 1 3 + = y x oe linear equation for attempt to solve linear simultaneous equations which have been obtained correctly for both. for correct substitution in cosine rule, may use in form of cos ... A = for attempt to make cosA subject and simplify for rationalisation.

Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 13 © Cambridge International Examinations 2014 11 (i) (ii) (iii) (iv) ( ) ( ) ( )2 1 1 2 5 d d − + − + = x x x x y ( )( ) 9 3 1 d d + − = x x x y When 0 d d = x y 3 1 − = = x x Alternative method: 5 9 3 2 3 + − + = x x x y 9 6 3 d d 2 − + = x x x y When 0 d d = x y 3 1 − = = x x ∫ + − + x x x x d 5 9 3 2 3 ) ( 5 2 9 4 2 3 4 c x x x x + + − + = 1 5 2 3 4 5 2 9 4 −       + − + x x x x 1 9 625 225 1 5 125 25 4 2 4 2     = + − + − − − −         = 108 When 32 ,3 = − = y x 32 > k M1 A1 DM1 A1 A1 M1 A1 DM1 A1 A1 M1 A2,1,0 M1 A1 M1 A1 for differentiation of a product, allow unsimplified correct for equating to zero and solution of quadratic for expansion of brackets and differentiation of each term of a 4 term cubic for equating to zero and solution of 3 term quadratic from correct quadratic equation from correct quadratic equation for correct attempt to obtain and integrate a 4 term cubic A2 for 4 correct terms or A1 for 3 correct terms for correct substitution of limits 1 and –5 for their (ii) for realising that the y-coordinate of the maximum point is needed.