Cambridge IGCSE Mathematics - Additional 0606 — 2015 Oct/Nov Paper 1 · Variant 1

This document consists of 16 printed pages. DC (LEG/CGW) 98539/3 © UCLES 2015 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education * 6 1 3 3 6 0 6 8 8 6 * ADDITIONAL MATHEMATICS 0606/11 Paper 1 October/November 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 0606/11/O/N/15 © UCLES 2015 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

3 0606/11/O/N/15 © UCLES 2015 [Turn over 1 Find the range of values of k for which the equation kx k x xk 8 2 2 + = - has 2 real distinct roots. [4]

4 0606/11/O/N/15 © UCLES 2015 2 A curve, showing the relationship between two variables x and y, passes through the point , P 1 3 - ^ h. The curve has a gradient of 2 at P. Given that d d x y 5 2 2 =- , find the equation of the curve. [4]

5 0606/11/O/N/15 © UCLES 2015 [Turn over 3 Show that sec cosec sec cosec 1 1 2 2 i i i i - + - = . [5]

6 0606/11/O/N/15 © UCLES 2015 4 (a) 6 books are to be chosen from 8 different books. (i) Find the number of different selections of 6 books that could be made. [1] A clock is to be displayed on a shelf with 3 of the 8 different books on each side of it. Find the number of ways this can be done if (ii) there are no restrictions on the choice of books, [1] (iii) 3 of the 8 books are music books which have to be kept together. [2] (b) A team of 6 tennis players is to be chosen from 10 tennis players consisting of 7 men and 3 women. Find the number of different teams that could be chosen if the team must include at least 1 woman. [3]

7 0606/11/O/N/15 © UCLES 2015 [Turn over 5 Variables x and y are such that ln y x x 3 2 1 2 = - + ^ ^ h h. (i) Find the value of d d x y when x 2 = . [4] (ii) Hence find the approximate change in y when x changes from 2 to 2.03. [2]

8 0606/11/O/N/15 © UCLES 2015 6 It is given that  = ,where is an integer x x x 1 12 | G G " , and that sets A, B, C and D are such that A = multiples of 3 " ,, B = prime numbers " ,, C = odd integers " ,, D = even integers " ,. Write down the following sets in terms of their elements. (i) A B + [1] (ii) A C , [1] (iii) A C + l [1] (iv) D B , l ^ h [1] (v) Write down a set E such that E D 1 . [1]

9 0606/11/O/N/15 © UCLES 2015 [Turn over 7 Two variables, x and y, are such that y Axb = , where A and b are constants. When lny is plotted against lnx, a straight line graph is obtained which passes through the points . , . 1 4 5 8 ^ h and . , . 2 2 6 0 ^ h. (i) Find the value of A and of b. [4] (ii) Calculate the value of y when x = 5. [2]

10 0606/11/O/N/15 © UCLES 2015 8 Find the equation of the tangent to the curve y x x 5 2 1 2 = + - at the point where x = 2. [7]

11 0606/11/O/N/15 © UCLES 2015 [Turn over 9 You are not allowed to use a calculator in this question. (i) Find d x x 4 + y . [2] (ii) A B y = √4 + x y x 5 O The diagram shows the graph of y x 4 = + , which meets the y-axis at the point A and the line x 5 = at the point B. Using your answer to part (i), find the area of the region enclosed by the curve and the straight line AB. [5]

12 0606/11/O/N/15 © UCLES 2015 10 A E D B C 10 cm The diagram shows two circles, centres A and B, each of radius 10 cm. The point B lies on the circumference of the circle with centre A. The two circles intersect at the points C and D. The point E lies on the circumference of the circle centre B such that ABE is a diameter. (i) Explain why triangle ABC is equilateral. [1] (ii) Write down, in terms of r, angle CBE. [1] (iii) Find the perimeter of the shaded region. [5]

13 0606/11/O/N/15 © UCLES 2015 [Turn over (iv) Find the area of the shaded region. [3]

14 0606/11/O/N/15 © UCLES 2015 11 (a) A function f is such that f x x x 6 4 2 = + + ^ h for x 0 H . (i) Show that x x 6 4 2 + + can be written in the form x a b 2 + + ^ h , where a and b are integers. [2] (ii) Write down the range of f. [1] (iii) Find f 1 - and state its domain. [3]

15 0606/11/O/N/15 © UCLES 2015 [Turn over (b) Functions g and h are such that, for x R d , g e x x = ^ h and h x x 5 2 = + ^ h . Solve h g x 37 2 = ^ h . [4] Question 12 is printed on the next page.

16 0606/11/O/N/15 © UCLES 2015 12 The line x y 2 1 0 - + = meets the curve x y 3 19 2 + = at the points A and B. The perpendicular bisector of the line AB meets the x-axis at the point C. Find the area of the triangle ABC. [9] 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 October/November 2015 series 0606 ADDITIONAL MATHEMATICS 0606/11 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 2015 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 2015 0606 11 © Cambridge International Examinations 2015 Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied www without wrong working 1 0 ) 8 2 ( 2 = + − + k x k kx 0 4 2 > −ac b so ) 0 ( 4 ) 8 2 ( 2 2 > − − k k ) 0 ( 4 64 32 4 2 2 > − + − k k k leading to 2 < k only M1 DM1 DM1 A1 for attempt to obtain a 3 term quadratic in the form 0 2 = + + c bx ax ,where b contains a term in k and a constant for use of ac b 4 2 − for attempt to simplify and solve for k A1 must have correct sign 2 ) ( 5 d d c x x y + − =       When d 1, 2 d = − = y x x leading to 3 5 d d − − = x x y d x x y + − − = 3 2 5 2 When 3 ,1 = − = y x leading to x x y 3 2 5 2 5 2 − − = Alternative scheme: c bx ax y + + = 2 so b ax x y + = 2 d d When 1 − = x , 2 d d = x y so 2 2 = + − b a a x y 2 d d 2 2 = so 2 5 − = a , 3 − = b , 2 5 = c M1 A1 DM1 A1 M1 A1 DM1 A1 for attempt to integrate, do not penalise omission of arbitrary constant. Must have ... d d = x y for attempt to integrate their x y d d , but penalise omission of arbitrary constant. for use of c bx ax y + + = 2 , differentiation and use of conditions to give an equation in a and b for a correct equation for a second differentiation to obtain a for a, b and c all correct

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 11 © Cambridge International Examinations 2015 3 2 2 (sec 1) (cosec 1) sec cosec θ θ θ θ − + − = LHS θ θ cot tan + = θ θ θ θ sin cos cos sin + = θ θ θ θ cos sin cos sin 2 2 + = θ θ cos sin 1 = sec cosec θ θ = Alternate scheme: LHS θ θ cot tan + = θ θ tan 1 tan + = θ θ tan 1 tan 2 + = θ θ tan sec2 = θ θ θ sec tan sec × = cosec sec θ θ = B1 B1 M1 M1 A1 B1 M1 B1 M1 A1 may be implied by the next line for dealing with θ tan and θ cot in terms of θ sin and θ cos for attempt to obtain as a single fraction for the use of 1 cos sin 2 2 = + θ θ in correct context Must be convinced as AG may be implied by subsequent work for attempt to obtain as a single fraction for use of the correct identity for ‘splitting’ 2 sec θ Must be convinced as AG 4 (a) (i) (ii) (iii) (b) 28 20160 )3 4 5 ( 6 × × × oe to give 360 2 ) 3 4 5 ( 6 × × × × 720 = Either 7 210 6 7 6 10 − = −C C 203 = Or 1W 5M 63 = 2W 4M 105 = 3W 3M 35 = Total 203 = B1 B1 B1 B1 B1, B1 B1 B1 B1 B1 for realising that the music books can be arranged amongst themselves and consideration of the other 5 books for the realisation that the above arrangement can be either side of the clock. B1 for , 6 10C B1 for 6 7C for 1 case correct, must be considering more than 1 different case, allow C notation for the other 2 cases, allow C notation for final result

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 11 © Cambridge International Examinations 2015 5 (i) (ii) )1 2 ln( 1 2 4 ) 3 ( d d 2 2 + + + − = x x x x x y when d 8 2, ln9 d 9 = = − + y x x oe or 1.31or better ≈ ∂y (answer to (i)) × 0.03 , 0393 .0 = allow awrt 0.039 B1 M1 A1 A1 M1 A1FT for correct differentiation of ln function for attempt to differentiate a product for correct product, terms must be bracketed where appropriate for correct final answer for attempt to use small changes follow through on their numerical answer to (i) allow to 2 sf or better 6 (i) (ii) (iii) (iv) (v) {3} = ∩B A } 12 , 11 ,9 ,7 ,6 ,5 ,3 ,1 { = ∪C A 11} 7, 5, {1, = ∩ ′ C A 9} {1, ) ( = ′ ∪B D Any set containing up to 5 positive even numbers ⩽ 12 B1 B1 B1 B1 B1 7 (i) (ii) 25 .0 8.0 2.0 Gradient = = 25 .0 = b Either c + = ) 2.2 ( 25 .0 6 Or c + = ) 4.1( 25 .0 8.5 leading to 233 = A or 5.45 e Alternative schemes: Either Or c b + = ) 2.2 ( 6 ( ) b A 2.2 6 e e = c b + = ) 4.1( 8.5 ( ) b A 4.1 8.5 e e = Leading to 233 = A or 45 .5 e and 25 .0 = b Either 25 .0 5 233× = y Or 233 ln 5 ln 25 .0 ln + = y leading to 348 = y M1 A1 M1 A1 M1 DM1 A1, A1 M1 A1 for attempt to find the gradient for a correct substitution of values from either point and attempt to obtain c or solution by simultaneous equations dealing with ln c A = for 2 simultaneous equations as shown for attempt to solve to get at least one solution for one unknown A1 for each for correct use of either equation in attempt to obtain y using their value of A and of b found in (i)

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 11 © Cambridge International Examinations 2015 8 5 )1 2 ( ) 5 )( 2 ( 2 1 ) 5 ( 2 d d 2 2 1 2 2 1 2 + − + − + = − x x x x x x y or )1 2 ( ) 5 )( 2 ( 2 1 ) 5 ( 2 d d 2 3 2 2 1 2 − + − + = − − x x x x x y When 1 ,2 = = y x and 9 4 d d = x y (allow 0.444 or 0.44) Equation of tangent: ) 2 ( 9 4 1 − = − x y )1 4 9 ( + = x y B1 M1 A1 B1, B1 M1 A1 for ( ) 2 1 2 5 ) 2 ( 2 1 − + x x for a quotient or ( ) 3 2 2 1 (2 ) 5 2 − − + x x for a product allow if either seen in separate working for attempt to differentiate a quotient or a correct product for all correct, allow unsimplified B1 for each for attempt at straight line, must be tangent using their gradient and y allow unsimplified. 9 (i) (ii) ( ) ) ( 4 3 2 2 3 c x + + Area of trapezium       × × = 5 5 2 1 5. 12 = Area ( )       × × −     + = 5 5 2 1 4 3 2 5 0 2 3 x 2 25 3 16 27 3 2 − −       × = 6 1 = or awrt 0.17 Alternative scheme: Equation of AB 2 5 1 + = x y Area ∫       + − + = 5 0 d 2 5 1 4 x x x ( ) 5 0 2 2 3 2 10 4 3 2       − − + = x x x 2 25 3 16 27 3 2 − −       × = 6 1 = or awrt 0.17 B1,B1 M1 A1 M1 A1 A1 M1 M1 A1 A1 A1 B1 for ( )2 3 4 x k + only, B1 for ( )2 3 4 3 2 x + only Condone omission of c for attempt to find the area of the trapezium for correct use of limits using ( )2 3 4 x k + only (must be using 5 and 0) for 3 16 18 − or equivalent for a correct attempt to find the equation of AB for correct use of limits using ( )2 3 4 x k + only (must be using 5 and 0) for 3 16 18 − or equivalent for 12.5 or equivalent

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 11 © Cambridge International Examinations 2015 10 (i) (ii) (iii) (iv) All sides are equal to the radii of the circles which are also equal Angle 3 2π = CBE 3 10 = DE Arc 3 2 10 π × = CE Perimeter 3 20 3 10 20 π + + = 2. 58 or 3. 58 = Area of sector: 3 100 3 2 10 2 1 2 π π = × × Area of triangle: 3 25 3 2 sin 10 2 1 2 = × × π Area 3 25 3 100 + = π or awrt 148 B1 B1 M1 A1 M1 M1 A1 M1 M1 A1 for a convincing argument must be in terms of , 667 .0 allow , π π or better for correct attempt to find DE using their angle CBE for correct DE, allow 17.3 or better for attempt to find arc length with their angle CBE (20.94) for 10 + 10 + DE + an arc length allow unsimplified for sector area using their angle CBE allow unsimplified, may be implied for triangle area using their angle DBE which must be the same as their angle CBE, allow unsimplified, may be implied allow in either form

Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 11 © Cambridge International Examinations 2015 11 (a) (i) (ii) (iii) (b) ( ) 5 3 2 − + x y ⩾ 4 or f ⩾ 4 3 5 − + = x y Domain x ⩾ 4 ) (e h ) g( h 2 2 x x = ) 2 e 5 h( + = x 12 e 25 + = x , 37 12 e 25 = + x leading to 0 x = Alternative scheme 1: (37) h ) hg( 1 − = x 7 (37) h 1 = − ,7 2 e 5 = + x leading to 0 = x Alternative scheme 2: (37) h ) g( 2 − = x 1 (37) h 2 = − ,1 e = x leading to 0 x = B1, B1 B1 M1 A1 B1FT M1 M1 DM1 A1 M1 M1 DM1 A1 M1 M1 DM1 A1 B1 for 3, B1 for 5 − Correct notation or statement must be used for a correct attempt to find the inverse function must be in the correct form and positive root only Follow through on their answer to (ii), must be using x for correct order for dealing with 2 h for solution of equation (dependent on both previous M marks) for correct order for dealing with (37) h -1 for solution of equation (dependent on both previous M marks) for correct order for dealing with (37) h 2 − for solution of equation (dependent on both previous M marks)

Page 8 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 11 © Cambridge International Examinations 2015 12 0 16 6 2 = − + x x or 0 75 10 2 = − + y y leading to 0 ) 2 )( 8 ( = − + x x or 0 ) 15 )( 5 ( = + − y y so 5 ,2 = = y x and 15 ,8 − = − = y x Midpoint ) 5 ,3 ( − − Gradient ,2 = so perpendicular gradient 2 1 − = Perpendicular bisector: )3 ( 2 1 5 + − = + x y ) 0 13 2 ( = + + x y Point C ) 0 , 13 (− Area 0 15 5 0 13 8 2 13 2 1 − − − − = 125 = Alternative method for area: , 125 2 = CM 500 2 = AB Area 500 125 2 1 × × = = 125 M1 DM1 A1, A1 B1 M1 M1 M1 A1 M1 A1 for attempt to obtain a 3 term quadratic in terms of one variable only for attempt to solve quadratic equation A1 for each ‘pair’ of values. for attempt at straight line equation, must be using midpoint and perpendicular gradient for use of 0 = y in their line equation (but not 0 1 2 = + −y x ) for correct attempt to find area, may be using their values for A, B and C (C must lie on the x-axis) for correct attempt to find area may be using their values for A, B and C