Cambridge IGCSE Mathematics - Additional 0606 — 2015 Oct/Nov Paper 2 · Variant 1

This document consists of 16 printed pages. DC (AC/CGW) 98388/4 © UCLES 2015 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 4 0 3 3 8 5 4 1 1 6 * ADDITIONAL MATHEMATICS 0606/21 Paper 2 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 0606/21/O/N/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 0606/21/O/N/15 © UCLES 2015 [Turn over 1 It is given that ( )x x x x 4 4 15 18 f 3 2 = - - + . (i) Show that x 2 + is a factor of ( )x f . [1] (ii) Hence factorise ( )x f completely and solve the equation ( )x 0 f = . [4]

4 0606/21/O/N/15 © UCLES 2015 2 (i) Find, in the simplest form, the first 3 terms of the expansion of ( )x 2 3 6 - , in ascending powers of x. [3] (ii) Find the coefficient of x2 in the expansion of ( )( ) x x 1 2 2 3 6 + - . [2]

5 0606/21/O/N/15 © UCLES 2015 [Turn over 3 Relative to an origin O, points A, B and C have position vectors 5 4 c m, 10 12 - c m and 6 18 - c m respectively. All distances are measured in kilometres. A man drives at a constant speed directly from A to B in 20 minutes. (i) Calculate the speed in kmh–1 at which the man drives from A to B. [3] He now drives directly from B to C at the same speed. (ii) Find how long it takes him to drive from B to C. [3]

6 0606/21/O/N/15 © UCLES 2015 4 (a) Given that 2 3 7 5 4 1 2 2 3 4 0 1 and A B = - = - - f c p m, calculate 2BA. [3] (b) The matrices C and D are given by 1 1 2 6 3 1 2 4 and C D = - = - c c m m. (i) Find C–1. [2] (ii) Hence find the matrix X such that CX + D = I, where I is the identity matrix. [3]

7 0606/21/O/N/15 © UCLES 2015 [Turn over 5 (a) Solve the following equations to find p and q. 8 2 4 q p p q 1 2 1 7 4 # = - + - 81 = 9 3 # [4] (b) Solve the equation ( ) ( ) x x 3 2 1 2 2 g lg lg l - + + = - . [5]

8 0606/21/O/N/15 © UCLES 2015 6 y x 4 2 3 5 1 00 π π 2 3π 2π 2 The figure shows part of the graph of sin y a b cx = + . (i) Find the value of each of the integers a, b and c. [3] Using your values of a, b and c find (ii) d d x y, [2]

9 0606/21/O/N/15 © UCLES 2015 [Turn over (iii) the equation of the normal to the curve at , 2 3 r ` j. [3]

10 0606/21/O/N/15 © UCLES 2015 7 8 cm h cm 6 cm r cm A cone, of height 8 cm and base radius 6 cm, is placed over a cylinder of radius r cm and height h cm and is in contact with the cylinder along the cylinder’s upper rim. The arrangement is symmetrical and the diagram shows a vertical cross-section through the vertex of the cone. (i) Use similar triangles to express h in terms of r. [2] (ii) Hence show that the volume, V cm3, of the cylinder is given by V r r 8 2 3 4 3 r r = - . [1]

11 0606/21/O/N/15 © UCLES 2015 [Turn over (iii) Given that r can vary, find the value of r which gives a stationary value of V. Find this stationary value of V in terms of π and determine its nature. [6]

12 0606/21/O/N/15 © UCLES 2015 8 Solutions to this question by accurate drawing will not be accepted. Two points A and B have coordinates (–3, 2) and (9, 8) respectively. (i) Find the coordinates of C, the point where the line AB cuts the y-axis. [3] (ii) Find the coordinates of D, the mid-point of AB. [1]

13 0606/21/O/N/15 © UCLES 2015 [Turn over (iii) Find the equation of the perpendicular bisector of AB. [2] The perpendicular bisector of AB cuts the y-axis at the point E. (iv) Find the coordinates of E. [1] (v) Show that the area of triangle ABE is four times the area of triangle ECD. [3]

14 0606/21/O/N/15 © UCLES 2015 9 Solve the following equations. (i) sin cos x x 4 2 5 2 0 + = for x 0 180 c c G G [3] (ii) cot cosec y y 3 3 2 + = for y 0 360 c c G G [5]

15 0606/21/O/N/15 © UCLES 2015 [Turn over (iii) cos z 4 2 1 r + =- ` j for z 0 2 G G r radians, giving each answer as a multiple of π [4] Question 10 is printed on the next page.

16 0606/21/O/N/15 © UCLES 2015 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. 10 A particle is moving in a straight line such that its velocity, v ms–1, t seconds after passing a fixed point O is v 6 1 e e t t 2 2 = - - - . (i) Find an expression for the displacement, s m, from O of the particle after t seconds. [3] (ii) Using the substitution u e t2 = , or otherwise, find the time when the particle is at rest. [3] (iii) Find the acceleration at this time. [2]

® 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 0606/21 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 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 21 © 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 nfww not from wrong working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied www without wrong working 1 (i) ( ) f 2 32 16 30 18 0 − = − − + + = B1 All four evaluated terms must be seen. Allow if correct long division used (ii) 2 f( ) ( 2)(4 12 9) x x x x = + − + )3 2 )( 3 2 )( 2 ( − − + = x x x f ( ) 0 2, 1.5 x x = → = − nfww M1 A1 A1 A1 Coefficients 4 and 9 Coefficient –12 All three factors together Allow 1.5 mentioned just once 2 (i) 2 6 2160 576 64 ) 3 2 ( x x x + − = − isw B1B1B1 (ii) 1008 576 2 2160 = × − M1 A1 their final 2160 + 2 × their final –576 3 (i)      − = 8 15 AB 2 2 15 8 ( 17) AB = + = Speed 17 3 51km/hr = × = B1 M1 A1 Allow BA uuur May be implied by later work. Use of Pythagoras on their AB Must be exact (ii)       − = 30 16 BC 2 2 16 30 ( 34) BC = + = Time taken 40 60 51 34 = × = mins (or 3 2 hrs) B1 M1 A1 AllowCB uuur Use of Pythagoras on their BC Allow answers which round to 40 to 2sf. Accept 0.66 or 0.67 hrs. Mark final answer.

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 21 © Cambridge International Examinations 2015 4 (a) 2 1 1 2 4 2 2 3 5 2 3 0 7 4 24 5 48 10 2 5 17 10 34 −   −    =    −          = =         BA B3,2,1,0 –1 each error in 2 × 2 result. Failure to multiply by 2 is one error (b) (i) 1 6 2 1 1 1 8 − −   =     C isw B1 B1 8 1 Matrix (ii) 2 2 1 3 −   − =   − −   I D ( ) 1 6 2 2 2 1 1 1 1 3 8 − − −    = − =    − −    X C I D       − − − = 1 3 18 10 8 1 isw B1 M1 A1 Pre multiply their I – D with their C–1 5 (a) ( ) 14 1 2 1 3 2 2 2 = × + − p q ( ) 4 4 2 3 3 3 = × − q p Solve 16 2 3 = + p q 12 2 = + p q 5, 2 p q = = B1 B1 M1 A1 Correct powers of 2 allow unsimplified isw Correct powers of 3 allow unsimplified isw Attempt to solve their linear equations by eliminating one variable Both correct (b) (3 2)( 1) 50 x x − + = ) 4 )( 13 3 ( 0 52 3 2 − + → = − + x x x x 4 = x 3 13 − = x discarded M1 A1 M1 A1 A1 LHS oe isw 50 from correct processing of 2 lg2 − Solution of their three term quadratic Roots must be obtained from correct quadratic

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 21 © Cambridge International Examinations 2015 6 (i) 4 ,2 ,3 = = = c b a B1B1B1 (ii) d 8cos4 d y x x = isw M1 A1FT cx k cos ± and no other term in x 1 ≠ c cx bc cos × and no other term (iii) 8 2 cos 8 d d 2 = = → = π π x y x Eqn: 3 1 1 3.20 8 8 2 y y x x π −   = − → = − +     − DM1 M1 A1 Find their correct numerical x y d d Find equation with their numerical normal gradient ie x y d d 1 − and point , 3 2 π       All correct isw 7 (i) ( ) r h r h − = → − = 6 3 4 6 6 8 M1 A1 Uses correct ratio. Cannot be implied (ii) ( ) r r h r V − × = = 6 3 4 2 2 π π 3 2 3 4 8 r r π π − = B1 AG all steps must be seen Penalise missing brackets at any point in working (iii) 2 4 16 d d r r r V π π − = 4 0 d d = → = r r V ( ) 128 42.7 3 V π π = = 0 8 16 d d 2 2 < − = r r V π π when → = 4 r max M1 A1 M1 A1 A1 B1 Differentiate at least one power reduced by one Attempt to solve – must get r = … Correct value of r. Ignore 0 = r Correct value of V. Condone 134. 2 2 d d V r must be correct and some indication of a negative value seen plus maximum stated

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 21 © Cambridge International Examinations 2015 8 (i) Gradient 8 2 1 9 3 2 AB −   = =   +   isw Equation AB and 2 1 1 0 3.5 0 3 2 2 y x y x −   = → = → = +   +   5.3 = →y B1 M1 A1 Find equation with their gradient and set
= 0 (ii) D is (3, 5) B1 (iii) Gradient perpendicular = –2 Equation perpendicular 2 3 5 − = − − x y ( ) 11 2 + − = → x y M1 A1 Use of 1 2 1 − = × m m on gradient used for their line in (i) (iv) E is (0, 11) A1FT (v) Area of ABE = 3 9 0 3 1 2 8 11 2 2 − − 1 24 99 18 33 45 2 = − + − + = Area of EDC = 3 0 0 3 1 5 3.5 11 5 2 1 10.5 33 11.25 2 = − + = M1 A1 A1 For area of ABE or ECD. 2 1 and their correct 8 elements must be seen. 45 condone from E (0, –4) 11.25 condone from E(0, –4)

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 21 © Cambridge International Examinations 2015 9 (i) 5 tan2 4 x = − (2 128.7, 308.7) x = 3. 64 = x awrt 154.3 awrt M1 A1 A1FT For obtaining and using 5 4 tan2 or 4 5 x = ± ± resulting in = x 2 tanx = … gets M0 their ° + ° 90 3. 64 (ii) 2 cosec 3cosec 4 0 y y + − = or 2 4sin 3sin 1 0 y y − −= ( )( ) cosec 4 cosec 1 0 y y + − = or ( )( ) 4sin 1 sin 1 0 y y + − = 1 sin 4 y = − or sin 1 y = y = 194.5, 345.5, 90 B1 M1 A1A1A1 In any form as a three term quadratic. Solve three term quadratic in cosec y or sin y Answers must be obtained from the correct quadratic (iii) 3 4 π π π − = + z or 3 4 π π π + = + z 12 13 , 12 5 π π = z B1 B1 B1B1 Accept 2.09, 2.10, 1.05, 1.04 π π − − on RHS. Could be implied by final answer Accept 4.19, 4.18, 1.05, 1.04 π π + + on RHS. Could be implied by final answer Answers must be correct multiples of π. 10 (i) 2 2 1 e 3e ( ) 2 t t s t c − = + −+ 0, 0 3.5 t s c = = → = − 2 2 1 e 3e 3.5 2 t t s t −   = + −−     M1 A1 A1 Integrate : coefficient of 2 1 or 3 seen with no change in powers of e. Ignore –t All correct and simplified (ii) 0 6 0 2 = − − → = u u v oe ( )( ) 0 2 3 = + − u u 1 3 ln3 2 u t → = →= or 0.549 M1 DM1 A1 Obtain three term quadratic in u or e2t Condone sign errors. Solve three term quadratic Accept 0.55 No second answer (iii) 2 2 1 ln3 2e 12e 2 t t t a − = → = + 10 4 6 = + = B1 B1 Correct differentiation Allow awrt 10.0 or 9.99. No second answer.