Cambridge IGCSE Mathematics - Additional 0606 — 2014 May/June Paper 1 · Variant 2

ADDITIONAL MATHEMATICS 0606/12 Paper 1 May/June 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. This document consists of 15 printed pages and 1 blank page. DC (NF/CGW) 73328/3 © UCLES 2014 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 2 8 0 6 8 4 6 6 6 5 *

2 0606/12/M/J/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 0606/12/M/J/14 © UCLES 2014 [Turn over 1 Show that sin cos cos sin A A A A 1 1 + + + can be written in the form sec p A, where p is an integer to be found. [4]

4 0606/12/M/J/14 © UCLES 2014 2 (a) On the Venn diagrams below, draw sets A and B as indicated. (i) (ii) A B A B = [2] (b) The universal set % and sets P and Q are such that n(%) = 20, ( ) P Q 15 n , = , ( ) P 13 n = and ( ) P Q 4 n + = . Find (i) ( ) Q n , [1] (ii) ( ) P Q n , l ^ h, [1] (iii) ( ) P Q n + l . [1]

5 0606/12/M/J/14 © UCLES 2014 [Turn over 3 (i) Sketch the graph of y x x 2 1 2 = + - ^ ^h h for x 2 3 G G - , showing the coordinates of the points where the curve meets the x- and y-axes. [3] (ii) Find the non-zero values of k for which the equation x k x 2 1 2 + = - ^ ^h h has two solutions only. [2]

6 0606/12/M/J/14 © UCLES 2014 4 The region enclosed by the curve sin y x 2 3 = , the x-axis and the line x = a , where a 0 1 1 1 radian, lies entirely above the x-axis. Given that the area of this region is 3 1 square unit, find the value of a. [6]

7 0606/12/M/J/14 © UCLES 2014 [Turn over 5 (i) Given that 2 4 8 1 x y 5 # = , show that x y 5 2 3 – + = . [3] (ii) Solve the simultaneous equations 2 4 8 1 x y 5 # = and 7 49 1 x y 2 # = . [4]

8 0606/12/M/J/14 © UCLES 2014 6 (a) Matrices X, Y and Z are such that X 2 1 3 2 = c m, Y 1 4 6 3 5 7 = f p and Z 1 2 3 = ^ h. Write down all the matrix products which are possible using any two of these matrices. Do not evaluate these products. [2] (b) Matrices A and B are such that A 5 4 2 1 = - - c m and AB 3 6 9 3 = - - c m. Find the matrix B. [5]

9 0606/12/M/J/14 © UCLES 2014 [Turn over 7 The diagram shows a circle, centre O, radius 8 cm. Points P and Q lie on the circle such that the chord PQ = 12 cm and angle POQ i = radians. P Q O 8 cm 12 cm θ rad (i) Show that .1 696 i = , correct to 3 decimal places. [2] (ii) Find the perimeter of the shaded region. [3] (iii) Find the area of the shaded region. [3]

10 0606/12/M/J/14 © UCLES 2014 8 (a) (i) How many different 5-digit numbers can be formed using the digits 1, 2, 4, 5, 7 and 9 if no digit is repeated? [1] (ii) How many of these numbers are even? [1] (iii) How many of these numbers are less than 60 000 and even? [3] (b) How many different groups of 6 children can be chosen from a class of 18 children if the class contains one set of twins who must not be separated? [3]

11 0606/12/M/J/14 © UCLES 2014 [Turn over 9 A solid circular cylinder has a base radius of r cm and a volume of 4000 cm3. (i) Show that the total surface area, A cm2, of the cylinder is given by A r r 8000 2 2 r = + . [3] (ii) Given that r can vary, find the minimum total surface area of the cylinder, justifying that this area is a minimum. [6]

12 0606/12/M/J/14 © UCLES 2014 10 In this question i is a unit vector due East and j is a unit vector due North. At 12 00 hours, a ship leaves a port P and travels with a speed of 26 kmh–1 in the direction i j 5 12 + . (i) Show that the velocity of the ship is i j 10 24 + ^ h kmh–1. [2] (ii) Write down the position vector of the ship, relative to P, at 16 00 hours. [1] (iii) Find the position vector of the ship, relative to P, t hours after 16 00 hours. [2] At 16 00 hours, a speedboat leaves a lighthouse which has position vector i j 120 81 + ^ h km, relative to P, to intercept the ship. The speedboat has a velocity of i j 22 30 - + ^ h kmh–1. (iv) Find the position vector, relative to P, of the speedboat t hours after 16 00 hours. [1]

13 0606/12/M/J/14 © UCLES 2014 [Turn over (v) Find the time at which the speedboat intercepts the ship and the position vector, relative to P, of the point of interception. [4]

14 0606/12/M/J/14 © UCLES 2014 11 (a) Solve tan tan x x 5 0 2 + = for ° ° x 0 180 G G . [3] (b) Solve cos sin y y 2 1 0 2 - = - for ° ° y 0 360 G G . [4]

15 0606/12/M/J/14 © UCLES 2014 (c) Solve sec z 2 2 6 r - = ` j for z 0 G G r radians. [4]

16 0606/12/M/J/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. 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. BLANK PAGE

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2014 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 May/June 2014 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 12 © Cambridge International Examinations 2014 1 ( ) ( ) A A A A cos sin 1 sin 1 cos 2 2 + + + ( ) A A A A A cos sin 1 sin sin 2 1 cos 2 2 + + + + = ( ) ( ) A A A cos sin 1 sin 1 2 + + = A A sec 2 cos 2 = Alternative: ( ) ( )( ) A A A A A A cos sin 1 sin 1 sin 1 sin 1 cos + + − + − = ( ) A A A A A cos sin 1 sin 1 sin 1 cos 2 + + − − = ( ) A A A A A cos sin 1 cos sin 1 cos 2 + + − = A A A A cos sin 1 cos sin 1 + + − A A sec 2 cos 2 = = M1 M1 DM1 A1 M1 M1 M1 A1 M1 for obtaining a single fraction, correctly M1 for expansion of ( )2 sin 1 A + and use of identity DM1 for factorisation and cancelling of ( ) A sin 1+ factor A1 for use of A A sec cos 1 = and final answer M1 for multiplying first term by A A sin 1 sin 1 − − M1 for expansion of ( )( ) A A sin 1 sin 1 + − and use of identity M1 for simplification of the 2 terms A1 for use of A A sec cos 1 = and final answer 2 (a) (i) (i) (b) (i) (ii) (iii) 6 5 9 B1 B1 B1 B1 B1

Page 3 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 12 © Cambridge International Examinations 2014 3 (i) (ii) Maximum point occurs when 8 25 = y so 8 25 > k B1 B1 B1 M1 A1 B1 for shape B1 for y = 2 (must have a graph) B1 for x = – 0.5 and 2 (must have a graph) M1 for obtaining the value of y at the maximum point, by either completing the square, differentiation, use of discriminant or symmetry. Must have the correct sign for A1 Ignore any upper limits 4 3 1 d 3 sin 0∫ = a x x 3 1 d = x 3 1 3 cos 3 2 0 =    − a x 3 1 3 2 3 cos 3 2 =      − −      − a 5.0 3 cos = a 3 3 π = a , 9 π = a B1,B1 M1 A1 M1 A1 B1 for cos3 k x only, B1 for x 3 cos 3 2 − only M1 for correct substitution of the correct limits into their result A1 for correct equation M1 for correct method of solution of equation of the form k ma = cos A1 allow 0.349, must be a radian answer 5 (i) (ii) 3 2 5 2 2 2 − = × y x leads to 3 2 5 − = + y x 1 49 7 2 = × y x can be written as 0 4 = + y x Solving 3 2 5 − = + y x and 0 4 = + y x leads to , 3 2 − = x 6 1 = y B1, B1 DB1 B1 B1 M1 A1 B1 for y 2 2 , B1 for 3 2−, B1 for dealing with indices correctly to obtain given answer B1 for either y 4 7 or 0 7 seen B1 for 0 4 = + y x M1 for solution of their simultaneous equations, must both be linear A1 for both, allow equivalent fractions only

Page 4 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 12 © Cambridge International Examinations 2014 6 (a) (b) YX and ZY       − − = − 9 3 3 6 1 A B ,       − −       − = 9 3 3 6 2 5 1 4 3 1       − − − = 3 21 9 18 3 1 or       − − 1 7 3 6 Alternative method:       − − =             − − 9 3 3 6 2 1 5 4 b d a c Leads to 3 2 5 = −c a , 9 2 5 = −d b 6 4 − = + − c a , 3 4 − = + − d b Solutions give matrix       − − − 3 21 9 18 3 1 or       − − 1 7 3 6 B1,B1 M1 B1,B1 DM1 A1 M1 A2,1,0 M1 A1 B1 for each, must be in correct order, M1 for pre-multiplication by A-1 B1 for 3 1 − , B1 for       2 5 1 4 DM1 for attempt at matrix multiplication A1 allow in either form M1 for a complete method to obtain 4 equations –1 for each incorrect equation M1 for solution to find 4 unknowns A1 for a correct, final matrix

Page 5 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 12 © Cambridge International Examinations 2014 7 (i) (ii) (iii) 8 6 2 sin = θ , 8481 .0 2 = θ or better or θ cos 128 8 8 12 2 2 2 − + = 6961 .1 = θ or better or using areas θ sin 8 2 1 7 2 12 2 1 2 = × × oe 9922 .0 sin = θ , 4455 .1 = θ or 6961 .1 Arc length = ( ) 8 696 .1 2 × − π ( ) 36.7 or 697 . 36 Perimeter = ( ) 8 696 .1 2 12 × − + π = 7. 48 Area = ( ) 696 .1 sin 2 8 696 .1 2 2 8 2 2 + − π 179 awrt ,6. 178 ,5. 178 = Alternative: Area = ( )       − − 696 .1 sin 2 8 696 .1 8 2 1 8 2 2 2 π M1 A1 M1 A1 M1 A1 A1 M1,M1 A1 M1 for a complete method to find either θ or 2 θ Answer given. M1 for using the area of the triangle in 2 different forms A1 for choosing the correct angle. M1 for correct attempt at a minor or major arc length A1 for correct major arc length, allow unsimplified A1 for 48.7 or better M1 for correct attempt to find area of major sector M1 for correct attempt to find area of triangle, using any method M1 for attempt at area of circle – area of minor sector M1 for area of triangle

Page 6 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 12 © Cambridge International Examinations 2014 8 (a) (i) (ii) (iii) (b) 720 240 Starts with either a 2 or a 4: 48 ways Does not start with either a 2 or a 4: 96 ways (i.e. starts with 1 or 5) Total = 144 Alternative 1: Ends with a 2, starts with a 1,4 or 5 : 72 ways Ends with a 4, starts with a 1,2 or 5 : 72 ways Total =144 Alternative 2: ( ) 3 4 2 2 240 P × × − or ( ) ( ) 3 4 3 4 2 2 4 P P − × × 144 = Alternative 3: 1 2 3 4 1 3 P P P × × or 2 4 3 × × 144 = With twins : ( ) 1820 4 16 = C Without twins: ( ) 8008 6 16 = C Total: 9828 Alternative: ( ) 5 16 6 18 2 C C × − 9828 = B1 B1 B1 B1 B1 B1 B1 B1 B2 B1 B2 B1 B1 B1 B1 B1,B1 B1 allow unevaluated allow unevaluated must be evaluated B2 for correct expression seen, allow P notation Allow P notation here, for B2 B1 for …, 6 18 − C , B1 for 5 16 2 C ×

Page 7 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 12 © Cambridge International Examinations 2014 9 (i) (ii) 2 4000 r h π = or 4000 2 = h r π 2 2 2 r rh A π π + = 2 2 2 4000 2 r r r A π π π + = r r r A π 4 8000 d d 2 + − = When 0 d d = r A , π 4 8000 3 = r leading to 1390 , 1395 = A π 4 16000 d d 3 2 2 + = r r A , which, is positive so a minimum. B1 M1 A1 B1, B1 M1 M1 A1 √B1 M1 for substitution of h or rh π into their equation for A A1 Answer given B1 for each term correct M1 for equating to zero and attempt to find 3 r M1 for substitution of their r to obtain A. A1 for 1390 or awrt 1395 √B1 for a complete correct method and conclusion.

Page 8 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 12 © Cambridge International Examinations 2014 10 (i) (ii) (iii) (iv) (v) Velocity = ( )j i 12 5 13 1 26 + × = j i 24 10 + Alternative 1: 2 2 24 10 24 10 + = + j i 26 = Showing that one vector is a multiple of the other, hence same direction Alternative 2: 13 12 5 2 2 = + , 26 13 = k , so 2 = k Velocity ( )j i 12 5 2 + = , Velocity j i 24 10 + = Alternative 3: Use of trig: 5 12 tan = α , ° = 4. 67 α Velocity j i 4. 67 sin 26 4. 67 cos 26 + ° Velocity = j i 24 10 + Position vector = ( )j i 24 10 4 + or j i 96 40 + ( ) ( )t j i j i 24 10 96 40 + + + oe ( ) ( )t j i j i 30 22 81 120 + − + + oe t t 22 120 10 40 − = + or t t 30 81 24 96 + = + 30 18 or 5.2 : t = Position vector j i 156 65 + = M1 A1 M1 A1 M1 A1 M1 A1 B1 M1 A1 B1 M1 A1 DM1 A1 M1 for ( )j i 12 5 13 1 + M1 for working from given answer to obtain the given speed A1 for a completely correct method M1 for attempt to obtain the ‘multiple’ and apply to the direction vector A1 for a completely correct method M1 for reaching this stage A1 for a completely correct method Allow either form for B1 M1 for their ( ) ( )t j i ii 24 10 + + or ( ) ( ) 4 24 10 + × + t j i A1 correct answer only M1 for equating like vectors A1 Allow for t = 2.5 DM1 for use of t to obtain position vector A1 cao

Page 9 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0606 12 © Cambridge International Examinations 2014 11 (a) (b) (c) ( ) 0 5 tan tan = + x x 0 tan = x , ° ° = 180 , 0 x 5 tan − = x , ° = 3. 101 x 2(1–sin2 y) – sin y – 1 = 0 2 sin2 y + sin y – 1 = 0 (2 sin y – 1) (sin y + 1) = 0 2 1 sin = y , y = 30°, 150° sin y = –1, y = 270° 2 1 6 2 cos =       −π z 3 6 2 π π =       − z 4 π z = or 0.785 or better 3 5 6 2 π π z =       − 12 11π z = or 2.88 or better B1,B1 B1 M1 A1,A1 A1 M1 A1 M1 A1 B1 for each , must be from correct work M1 for use of correct identity and attempt to solve resulting 3 term quadratic equation. M1 for dealing with sec correctly and obtaining 3 π or 1.05 M1 for obtaining a second equation 3 2 6 2 π π π their z − =       − oe