Cambridge IGCSE Mathematics - Additional 0606 — 2013 Oct/Nov Paper 1 · Variant 2

This document consists of 16 printed pages. DC (SJW/CGW) 80816 © UCLES 2013 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 0 9 8 6 6 1 0 4 7 5 * ADDITIONAL MATHEMATICS 0606/12 Paper 1 October/November 2013 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 a pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, 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. www.theallpapers.com

2 0606/12/O/N/13 © UCLES 2013 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x a b b ac 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 www.theallpapers.com

3 0606/12/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use 1 The diagram shows the graph of ( ) sin y a bx c = + for 0 2 x G G r, where a, b and c are positive integers. 4 3 2 1 O y x –1 –2 –3 › 2› State the value of a, of b and of c. [3] a = b = c = 2 Find the set of values of k for which the curve ( ) ( ) y k x x k 1 3 1 2 = + - + + lies below the x-axis. [4] www.theallpapers.com

4 0606/12/O/N/13 © UCLES 2013 For Examiner’s Use 3 Show that 1 1 2 cos sin sin cos sec i i i i i + + + = . [4] www.theallpapers.com

5 0606/12/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use 4 The sets A and B are such that : , cos A x x x 2 1 0 6 0 2 ° ° G G = = $ ., : , tan B x x x 3 0 6 0 2 ° ° G G = = " ,. (i) Find n(A). [1] (ii) Find n(B). [1] (iii) Find the elements of A , B. [1] (iv) Find the elements of A + B. [1] www.theallpapers.com

6 0606/12/O/N/13 © UCLES 2013 For Examiner’s Use 5 (i) Find (9 3 ) sin x x d + y . [3] (ii) Hence show that r ( ) sin x x a b 9 3 d 9 r + = + ry , where a and b are constants to be found. [3] www.theallpapers.com

7 0606/12/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use 6 The function ( ) 4 2 x ax x bx f 3 2 = + + - , where a and b are constants, is such that x 2 1 - is a factor. Given that the remainder when ( )x f is divided by x 2 - is twice the remainder when ( )x f is divided by x 1 + , find the value of a and of b. [6] www.theallpapers.com

8 0606/12/O/N/13 © UCLES 2013 For Examiner’s Use 7 (a) (i) Find how many different 4-digit numbers can be formed from the digits 1, 3, 5, 6, 8 and 9 if each digit may be used only once. [1] (ii) Find how many of these 4-digit numbers are even. [1] (b) A team of 6 people is to be selected from 8 men and 4 women. Find the number of different teams that can be selected if (i) there are no restrictions, [1] (ii) the team contains all 4 women, [1] (iii) the team contains at least 4 men. [3] www.theallpapers.com

9 0606/12/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use 8 (i) On the grid below, sketch the graph of ( )( ) y x x 2 3 = - + for x 5 4 G G - , and state the coordinates of the points where the curve meets the coordinate axes. [4] y x –5 –4 –3 –2 –1 1 3 2 4 O (ii) Find the coordinates of the stationary point on the curve ( )( ) y x x 2 3 = - + . [2] (iii) Given that k is a positive constant, state the set of values of k for which ( )( ) x x k 2 3 - + = has 2 solutions only. [1] www.theallpapers.com

10 0606/12/O/N/13 © UCLES 2013 For Examiner’s Use 9 (a) Differentiate ( ) ln x x 4 2 1 3 + with respect to x. [3] (b) (i) Given that y x x 2 2 = + , show that x y x x 2 4 d d 3 = + + ^ h . [4] www.theallpapers.com

11 0606/12/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use (ii) Hence find y x x x 2 5 20 d 3 + + ^ h . [2] (iii) Hence evaluate 2 7y x x x 2 5 20 d 3 + + ^ h . [2] www.theallpapers.com

12 0606/12/O/N/13 © UCLES 2013 For Examiner’s Use 10 Solutions to this question by accurate drawing will not be accepted. The points ( , ) A 3 2 - and ( , ) B 1 4 are vertices of an isosceles triangle ABC, where angle ° B 90 = . (i) Find the length of the line AB. [1] (ii) Find the equation of the line BC. [3] www.theallpapers.com

13 0606/12/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use (iii) Find the coordinates of each of the two possible positions of C. [6] www.theallpapers.com

14 0606/12/O/N/13 © UCLES 2013 For Examiner’s Use 11 (a) It is given that the matrix 2 4 3 1 A = J L KK N P OO. (i) Find A + 2I. [1] (ii) Find A2. [2] (iii) Using your answer to part (ii) find the matrix B such that A2B = I. [2] www.theallpapers.com

15 0606/12/O/N/13 © UCLES 2013 [Turn over For Examiner’s Use (b) Given that the matrix C x x x 1 1 2 = - + - x 1 - J L KK N P OO, show that det 0 C ! . [4] 12 (a) A function f is such that ( ) 3 1 x x f 2 = - for x 10 8 G G - . (i) Find the range of f. [3] (ii) Write down a suitable domain for f for which f 1 - exists. [1] Question 12(b) is printed on the next page. www.theallpapers.com

16 0606/12/O/N/13 © UCLES 2013 For Examiner’s Use (b) Functions g and h are defined by ( ) 4 2 x g ex = - for x R d , ( ) 5 ln x x h = for x 0 2 . (i) Find ( )x g 1 - . [2] (ii) Solve ( ) 18 x gh = . [3] 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. www.theallpapers.com

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2013 series 0606 ADDITIONAL MATHEMATICS 0606/12 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 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. www.theallpapers.com

Page 2 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 12 © Cambridge International Examinations 2013 Mark Scheme Notes Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated method mark is earned (or implied). B Accuracy 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 generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. • The symbol implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously ‘correct’ answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2, 1, 0 means that the candidate can earn anything from 0 to 2. www.theallpapers.com

Page 3 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 12 © Cambridge International Examinations 2013 1 3, 2, 1 a b c = = = B1, B1, B1 [3] B1 for each 2 Using b2 – 4ac, 9 = 4 (k + 1)2 4k2 + 8k – 5 = 0       − = 2 1 , 2 5 k To be below the x-axis 2 5 − < k Or: ( ) 3 1 2 d d − + = x k x y when ( )1 2 3 ,0 d d + = = k x x y ( ) ( ) ( ) ( )1 1 2 9 1 4 9 1 2 + + + − + + = ∴ k k k k y To lie under the x-axis, y < 0 ( ) ( ) ( ) ( ) 0 1 1 2 9 1 4 9 1 2 < + + + − + + ∴ k k k k leading to 9 = 4(k + 1)2 or equivalent then as for previous method M1 DM1 A1 A1 [4] M1 M1 for any use of b2 – 4ac DM1 for solution of their quadratic in k A1 for critical value(s), 2 1 not necessary A1 for 2 5 − < k only M1 for a complete method to this point. www.theallpapers.com

Page 4 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 12 © Cambridge International Examinations 2013 3 ( ) ( ) θ θ θ θ θ θ θ θ sin 1 cos cos sin 1 sin 1 cos cos sin 1 2 2 + + + + + + + ( ) ( ) ( ) ( ) θ θ θ θ θ θ θ θ θ θ θ θ sec 2 sin 1 cos sin 1 2 sin 1 cos sin 2 2 sin 1 cos cos sin sin 2 1 2 2 = + + = + + = + + + + = Alternative solution: ( ) ( ) θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ sec 2 tan sec tan sec sec 2 tan sec tan sec 2 sec 2 tan sec 1 tan tan sec 2 sec tan sec 1 tan sec tan sec 1 tan sec 2 2 2 2 = + + = + + = + + + + = + + + = + + + M1 DM1 DM1 A1 [4] M1 DM1 DM1 A1 M1 for dealing with the fractions, denominator must be correct, be generous with numerator M1 for expansion and use of 1 sin cos 2 2 = + θ θ M1 for attempt to factorise A1 for obtaining final answer correctly M1 for dealing with the fractions M1 for expansion and use of θ θ 2 2 sec 1 tan = + DM1 for attempt to factorise A1 for obtaining final answer correctly 4 (i) n (A) = 3 (ii) n (B) = 4 (iii) = ∪B A {60°, 240°, 300, 420°, 600°} (iv) = ∩B A {60°, 420°} B1 [1] B1 [1] √B1 [1] √B1 [1] If elements listed for (i), then they must be correct elements to get B1 leading to n (A) = 3. If they are not listed and correct answer given then B1. If elements listed for (ii), then they must be correct elements leading to n (B) = 4 to get B1. If they are not listed and correct answer given then B1. Follow through on any sets listed in (i) and (ii). Do not allow any repetitions. Follow through on any sets listed in (i) and (ii). www.theallpapers.com

Page 5 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 12 © Cambridge International Examinations 2013 5 (i) x x 3 cos 3 1 9 − (+c) (ii) π π 9 3 cos 3 1 9     − x x       − −       − = 3 cos 3 1 3 cos 3 1 9 π π π π 2 1 8 + = π B1, B1, B1 [3] M1 A1, A1 [3] B1 for 9x, B1 for 3 1 or cos3x B1 for – 3 1 cos3x Condone omission of + c M1 for correct use of limits in their answer to (i) A1 for each term 6 2 2 1 8 2 1 f − + + =       b a leading to a + 4b – 8 = 0 f(2) = 2f(–1) 8a + 16 + 2b – 2 = 2(–a + 4 – b –2) leading to 10a + 4b + 10 = 0 or equivalent 2 5 2, = − = ∴ b a M1 A1 M1 A1 DM1 A1 [6] M1 for substitution of 2 1 = x into f (x) A1 for correct equation in any form M1 for attempt to substitute x = 2 or x = –1 into f(x) and use f(2) = ±2f(–1) or 2f(2) = ±f(–1) A1 for a correct equation in any form DM1 (on both previous M marks) for attempt to solve simultaneous equations to obtain either a or b A1 for both correct www.theallpapers.com

Page 6 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 12 © Cambridge International Examinations 2013 7 (a) (i) 360 (ii) 120 (b) (i) 924 (ii) 28 (iii) ( ) ( ) 4 4 2 8 3 4 3 8 924 C C C C × − × − (i.e. 924 – 3M 3W – 2M 4W) (i.e. 924 – 224 – 28 = 672 Or: 4M 2W 2 4 4 8 C C × = 420 Or: 5M 1W 1 4 5 8 C C × = 224 Or: 6M 6 8C = 28 Total = 672 B1 [1] B1 [1] B1 [1] B1 [1] M1 A1 A1 [3] M1 A1 A1 M1 for 3 terms, at least 2 of which must be correct in terms of C notation or evaluated. A1 for any pair (must be evaluated) A1 for final answer M1 for 3 terms, at least 2 of which must be correct in terms of C notation or evaluated. A1 for any pair (must be evaluated) A1 for final answer 8 (i) (ii)      − 4 25 , 2 1 (iii) ) 14 ( 4 25 or 4 25 ≤ < > k k B1 B1 B1 B1 [4] B1, B1 [2] B1 [1] B1 for correct shape B1 for (–3, 0) or –3 seen on graph B1 for (2, 0) or 2 seen on graph B1 for (0, 6) or 6 seen on graph or in a table B1 for each −5 −4 −3 −2 −1 1 2 3 4 −1 1 2 3 4 5 6 7 x y www.theallpapers.com

Page 7 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 12 © Cambridge International Examinations 2013 9 (a) ( )       + + + 1 2 2 4 1 2 n 1 12 3 2 x x x x (b) (i) ( ) ( ) 2 2 1 2 2 2 2 d d 2 1 2 1 + + − + = − x x x x x y ( ) ( ) ( ) ( ) x x x x − + + + = − 2 2 2 2 2 1 ( )2 3 2 4 + + = x x Or: ( ) ( ) ( ) 2 2 2 2 1 2 d d 2 1 2 3 − − + + +      − = x x x x y ( ) ( ) ( ) ( )2 3 2 3 2 4 2 2 2 + + = − + + = − x x x x x (ii) ( ) c x x + + 2 10 (iii) 2 20 3 70 2 10 7 2 − =     + x x 3 40 = M1 A2, 1, 0 [3] M1, A1 DM1 A1 [4] M1, A1 DM1 A1 M1,A1 [2] M1 A1 [2] M1 for differentiation of a correct product –1 for each error M1 for differentiation of a quotient involving ( )2 1 2 + x A1 all correct unsimplified DM1 for attempt to simplify A1 for correct simplification to obtain the given answer M1 for differentiation of a product involving ( ) 2 1 2 − + x A1 all correct unsimplified DM1 for attempt to simplify A1 for correct simplification to obtain the given answer M1 for 2 2 5 1 + × x x or 2 2 5 + × x x A1 correct only, allow unsimplified. Condone omission of + c M1 for correct application of limits in their answer to (b)(ii) www.theallpapers.com

Page 8 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 12 © Cambridge International Examinations 2013 10 (i) 20 or 4.47 (ii) Grad AB = 2 1 , ⊥ grad = –2 ⊥line y – 4 = –2(x – 1) (y = –2x + 6) (iii) Coords of C (x, y) and BC 2 = 20 (x – 1)2 + (y – 4)2 = 20 or Coords of C (x, y) and AC 2 = 40 (x + 3)2 + (y – 2)2 = 40 Need intersection with y = –2x + 6, leads to 5x2 – 10x – 15 = 0 or leads to 5y2 – 40y – = 0 giving x = 3, –1 and y = 0, 8 Or, using vector approach:                             = − + = − = − + = = 0 3 4 2 4 1 OC 8 1 4 2 4 1 OC 2 4 AB B1 [1] M1 M1, A1 [3] M1 A1 DM1 DM1 A1, A1 [6] B1 M1 A1, A1 A1,A1 M1 for attempt at a perp gradient M1 for attempt at straight line equation, must be perpendicular and passing through B. A1 allow unsimplified M1 for attempt to obtain relationship using an appropriate length and the point (1, 4) or (–3, 2) A1 for a correct equation DM1 for attempt to solve with y = –2x + 6 and obtain a quadratic equation in terms of one variable only M1 for attempt to solve quadratic A1 for each ‘pair’ May be implied M1 for correct approach A1 for each element correct A1 for each element correct www.theallpapers.com

Page 9 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 12 © Cambridge International Examinations 2013 11 (a) (i)       3 4 3 4 (ii)       = 13 12 9 16 2 A (iii) B is the inverse matrix of A2       − − = 16 12 9 13 100 1 (b) det C = x(x – 1) – (–1)(x2 – x + 1) det C = 2x2 – 2x + 1 b2 – 4ac < 0, 4 – 8 < 0 No real solutions (so det C ≠ 0) B1 [1] B1, B1 [2] B1, B1 [2] M1 A1 DM1 A1 [4] B1 for any 2 correct elements B1 for all correct Follow through on their A2 M1 for attempt to obtain det C A1 for this correct quadratic expression from a correct det C DM1 for use of discriminant or attempt to solve using the formula, or attempt to complete the square in order to show there are no real roots. A1 for correct reasoning or statement that there are no real roots. www.theallpapers.com

Page 10 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0606 12 © Cambridge International Examinations 2013 12 (a) (i) f(–10) = 299, f(8) = 191 Min point at (0, –1) or when y = –1 ∴range –1 ≤ y ≤ 299 (ii) x ≥ 0 or equivalent (b) (i) ( )       + = − 4 2 n 1 g 1 x x or e g 1 4 2 g 1       + x (ii) gh(x) = g(1n5x) gh(x) = 4e1n5x – 2 20x – 2 = 18, x = 1 Or h(x) = g–1(18) 1n5x = 1n5 leading to x = 1 M1 B1 A1 [3] B1 [1] M1 A1 [2] M1 A1 A1 [3] M1 A1 A1 M1 for substitution of either x = –10 or x = 8, may be seen on diagram B1 May be implied from final answer, may be seen on diagram Must have ≤ for A1, do not allow x Allow any domain which will make f a one-one function Assume upper and lower bound when necessary. M1 for complete method to find the form inverse function, must involve ln or lg if appropriate. May still be in terms of y. A1 must be in terms of x M1 for correct order A1 for correct expression 4e1n5x – 2 A1 for correct solution from correct working M1 for correct order A1 for correct equation A1 for correct solution from correct working www.theallpapers.com