Cambridge IGCSE Mathematics - Additional 0606 — 2015 Feb/March Paper 1 · Variant 2

This document consists of 16 printed pages. DC (CW/SG) 99296/4 © UCLES 2015 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 0 9 8 5 1 4 8 7 5 5 * ADDITIONAL MATHEMATICS 0606/12 Paper 1 February/March 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 0606/12/F/M/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 0606/12/F/M/15 © UCLES 2015 [Turn over 1 A sports club has members who play a variety of sports. Sets C, F and T are such that C = { members who play cricket}, F = { members who play football}, T = { members who play tennis}. Describe the following in words. (i) C F , [1] (ii) Tl [1] (iii) F T + Q = [1] (iv) ( ) C T 10 n + = [1]

4 0606/12/F/M/15 © UCLES 2015 2 Find the values of k for which the line y kx 3 = - does not meet the curve y x x k 2 3 2 = - + . [5]

5 0606/12/F/M/15 © UCLES 2015 [Turn over 3 (i) On the axes below sketch the graph of y x 4 5 = - , stating the coordinates of the points where the graph meets the coordinate axes. [3] y x O (ii) Solve x 4 5 9 - = . [3]

6 0606/12/F/M/15 © UCLES 2015 4 (i) Write down, in ascending powers of x, the first 3 terms in the expansion of ( )x 3 2 6 + . Give each term in its simplest form. [3] (ii) Hence find the coefficient of x2 in the expansion of ( )( ) x x 2 3 2 6 - + . [2]

7 0606/12/F/M/15 © UCLES 2015 [Turn over 5 (2, 1) (3, 5) e y O x Variables x and y are such that when e y is plotted against x a straight line graph is obtained. The diagram shows this straight line graph which passes through the points (2, 1) and (3, 5). (i) Express y in terms of x. [4] (ii) State the values of x for which y exists. [1] (iii) Find the value of x when ln y 6 = . [1]

8 0606/12/F/M/15 © UCLES 2015 6 (i) Given that tan y x x 2 = , find x y d d . [3] (ii) Hence find the equation of the normal to the curve tan y x x 2 = at the point where x 8 r = . [3]

9 0606/12/F/M/15 © UCLES 2015 [Turn over 7 The polynomial ( )x ax bx x 3 4 p 3 2 = + - - has a factor of x 2 1 - and leaves a remainder of 10 - when divided by x 2 + . (i) Show that a 10 = and find the value of b. [4] (ii) Given that ( ) ( )( ) x x rx sx t 2 1 p 2 = - + + , find the value of each of the integers r, s and t. [2] (iii) Hence find the exact solutions of ( )x 0 p = . [3]

10 0606/12/F/M/15 © UCLES 2015 8 (a) A function f is such that ( ) sin2 f i i = for 0 2 G G i r . (i) Write down the range of f. [1] (ii) Write down a suitable restricted domain for f such that f 1 - exists. [1] (b) Functions g and h are such that ( ) ln x x 2 4 g = + for x 0 2 , ( )x x 4 h 2 = + for x 0 2 . (i) Find g 1 - , stating its domain and its range. [4] (ii) Solve ( )x 10 gh = . [3]

11 0606/12/F/M/15 © UCLES 2015 [Turn over (iii) Solve ( ) ( ) x x g h = l l . [3]

12 0606/12/F/M/15 © UCLES 2015 9 x cm x cm y cm x cm The diagram shows an empty container in the form of an open triangular prism. The triangular faces are equilateral with a side of x cm and the length of each rectangular face is y cm. The container is made from thin sheet metal. When full, the container holds 200 3 cm3. (i) Show that A cm2 , the total area of the thin sheet metal used, is given by A x x 2 3 1600 2 = + . [5]

13 0606/12/F/M/15 © UCLES 2015 [Turn over (ii) Given that x and y can vary, find the stationary value of A and determine its nature. [6]

14 0606/12/F/M/15 © UCLES 2015 10 Do not use a calculator in this question. A B C θ ( )cm 1 2 5 + ( )cm 2 5 + The diagram shows triangle ABC which is right-angled at the point B. The side AB 1 2 5 = + ^ h cm and the side BC 5 2 = + ^ h cm. Angle BCA i = . (i) Find tani in the form a b 5 + , where a and b are integers to be found. [3] (ii) Hence find sec2i in the form c d 5 + , where c and d are integers to be found. [3]

15 0606/12/F/M/15 © UCLES 2015 [Turn over 11 (a) (i) Show that cot tan cosec cos x x x x + = . [3] (ii) Hence solve . cot tan cosec y y y 3 3 3 0 5 + = for y 0 G G r radians, giving your answers in terms of r. [3] Question 11(b) is printed on the next page.

16 0606/12/F/M/15 © UCLES 2015 (b) Solve sin cos z z 2 8 5 2 + = for z 0 360 ° ° 1 1 . [4] 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.

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME for the March 2015 series 0606 ADDITIONAL MATHEMATICS 0606/12 Paper 12, 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 March 2015 series for most Cambridge IGCSE® components.

Page 2 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2015 0606 12 © Cambridge International Examinations 2015 1 (i) Members who play football or cricket , or both B1 (ii) Members who do not play tennis B1 (iii) There are no members who play both football and tennis B1 (iv) There are 10 members who play both cricket and tennis. B1 2 k x x kx + − = − 3 2 3 2 ( ) ( ) 0 3 3 2 2 = + + + − k k x x Using ac b 4 2 − , ( ) ( ) ( ) ( ) 0 3 2 4 3 2 < + × × − + k k ( )( ) ( ) 0 5 3 < − + k k Critical values 5 ,3 − = k so 5 3 < < − k M1 DM1 DM1 A1 A1 for attempt to obtain a 3 term quadratic equation in terms of x for use of ac b 4 2 − for attempt to solve quadratic equation, dependent on both previous M marks for both critical values for correct range 3 (i) B1 B1 B1 for shape, must touch the x-axis in the correct quadrant for y intercept for x intercept (ii) 9 5 4 ± = −x or ( ) 81 5 4 2 = −x leading to 13 1, 5 = − = x x M1 A1, A1 for attempt to obtain 2 solutions, must be a complete method A1 for each 4 (i) 2 4860 2916 729 x x + + B1,B1 B1 B1 for each correct term (ii) 2 × their 4860 – their 2916 = 6804 M1 A1 for attempt at 2 terms, must be as shown −2 −1 1 2 3 4 −1 1 2 3 4 5 6 7 8 x y

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2015 0606 12 © Cambridge International Examinations 2015 5 (i) gradient = 4 Using either (2, 1) or (3, 5), 7 − = c e 4 = + y x c so ( ) 7 4 ln − = x y Alternative method: 1 2 5 1 3 2 − − = − − y x or equivalent e 4 7 = − y x so ( ) 7 4 ln − = x y B1 M1 M1,A1 M1 A1 M1 A1 for gradient, seen or implied for attempt at straight line equation to obtain a value for c for correct method to deal with e y for attempt at straight line equation using both points allow correct unsimplified for correct method to deal with e y (ii) 4 7 > x B1ft ft on their 4 7 − x (iii) ( ) 7 4 ln 6 ln − = x so 4 13 = x B1ft ft on their 4 7 − x 6 (i) ( ) 2 2 2 tan 2 sec 2 d d x x x x x y − = Or ( ) ( ) 1 2 2 d 2sec 2 tan2 d − − = + − y x x x x x M1 A2,1,0 for attempt to differentiate a quotient (or product) –1 each error (ii) When ( ) 2.546 8 , 8 π π = = y x When 64 1 2 d d , 8 2 π π π − = = x y x 2 64 32 π π − = (3.701) Equation of the normal: ( )      − − − = − 8 2 32 8 2 π π π π x y 65 .2 27 .0 + − = x y (allow 2.66) B1 M1 A1 for y-coordinate (allow 2.55) for an attempt at the normal, must be working with a perpendicular gradient allow in unsimplified form in terms of π or simplified decimal form

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2015 0606 12 © Cambridge International Examinations 2015 7 (i) 0 4 2 3 4 8 : 2 1 p = − − +       b a Simplifies to 44 2 = + b a ( ) 10 4 6 4 8 : 2 p − = − + + − − b a Simplifies to 3 2 = −b a oe Leads to 17 , 10 = = b a M1 M1 DM1 A1 for correct use of 2 1 = x for correct use of 2 − = x for solution of equations for both, be careful as AG for a, allow verification (ii) ( ) 4 3 17 10 p 2 3 − − + = x x x x ( )( ) 2 2 1 5 11 4 = − + + x x x B2,1,0 –1 each error (iii) 2 1 = x 10 41 11± − = x B1 B1, B1 8 (a) (i) Range 0 < y < 1 B1 (ii) Any suitable domain to give a one-one function B1 e.g. 0 < x < 4 π (b) (i) x y ln 4 2 + = oe 4 2 ln − = y x oe ( ) 4 2 e g 1 − = − x x Domain ∈x Range y > 0 M1 A1 B1 B1 for a complete method to find the inverse must be in the correct form (ii) ( ) 10 4 g 2 = + x ( ) 10 4 ln 4 2 2 = + + x leading to 1.84 x = only Alternative method: ( ) ( ) 10 g 4 h 1 2 − = + = x x ( ) 2 1 e 10 g = − , so 2 2 e 4 = + x leading to 84 .1 = x only M1 DM1 A1 M1 DM1 A1 for correct order for attempt to solve for one solution only for correct order for attempt to solve for one solution only (iii) x x 2 4 = 2 2 = x 2 = x B1 M1 A1 for given equation, allow in this form for attempt to solve, must be using derivatives for one solution only, allow 1.41 or better.

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2015 0606 12 © Cambridge International Examinations 2015 9 (i) Area of triangular face = 4 3 2 3 2 1 2 2 x x = Volume of prism = y x × 4 3 2 3 200 4 3 2 = × y x so 800 2 = y x xy x A 2 4 3 2 2 + × = leading to x x A 1600 2 3 2 + = B1 M1 A1 M1 A1 for area of triangular face for attempt at volume area × their y for correct relationship between x and y for a correct attempt to obtain surface area using their area of triangular face for eliminating y correctly to obtain given answer (ii) 2 1600 3 d d x x x A − = When 3 1600 ,0 d d 3 = = x x A 74 .9 = x so A = 246 3 2 3200 3 d d x x A + = which is positive for 74 .9 = x so the value is a minimum M1 M1 A1 A1 M1 A1ft for attempt to differentiate for equating x A d d to 0 and attempt to solve for correct x for correct A for attempt at second derivative and conclusion, or alternate methods ft for a correct conclusion from completely correct work, follow through on their positive x value. 10 (i) 5 3 6 5 3 6 5 3 6 5 2 1 tan − − × + + = θ 45 36 30 5 12 5 3 6 − − + − = 5 3 8 − = M1 A1, A1 for attempt at θ cot together with rationalisation Must be convinced that a calculator is not being used. A1 for each term (ii) θ θ 2 2 sec 1 tan = + θ 2 cosec 1 5 3 5 16 9 64 = + + − so 3 5 16 9 118 cosec2 − = θ Alternate solutions are acceptable M1 A1, A1 for attempt to use the correct identity or correct use of Pythagoras’ theorem together with their answer to (i) Must be convinced that a calculator is not being used. A1 for each term

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2015 0606 12 © Cambridge International Examinations 2015 11 (a) (i) LHS = y y y y y cos sin sin cos sin 1 + y y y y y cos sin sin cos sin 1 2 2 + = y y y cos sin sin 1 × = y cos = M1 M1 A1 for dealing with cosec, cot and tan in terms of sin and cos for use of 1 cos sin 2 2 = + y y for correct simplification to get the required result. (ii) 5.0 3 cos = z 3 7 , 3 5 , 3 3 π π π = z 5 7 , , 9 9 9 π π π = z M1 A1, A1 for use of (i) and correct attempt to deal with multiple angle A1 for each ‘pair’ of solutions (b) ( ) 5 sin 1 8 sin 2 2 = − + x x 0 3 sin 2 sin 8 2 = − − x x ( )( ) 0 1 sin 2 3 sin 4 = + − x x 2 1 sin , 4 3 sin − = = x x ° ° = 4. 131 , 6. 48 x ° ° 330 , 210 M1 M1 A1, A1 for use of correct identity for attempt to solve quadratic equation A1 for each pair of solutions