Cambridge IGCSE Mathematics - Additional 0606 — 2014 Oct/Nov Paper 1 · Variant 1

This document consists of 15 printed pages and 1 blank page. DC (LK/SLM) 81862/3 © UCLES 2014 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 8 0 2 3 6 5 2 7 1 3 * ADDITIONAL MATHEMATICS 0606/11 Paper 1 October/November 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.

2 0606/11/O/N/14 © UCLES 2014 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/14 © UCLES 2014 [Turn over 1 Find the coordinates of the stationary point on the curve y x x 16 2 = + . [4] 2 (a) On the axes below, sketch the curve cos y x 3 2 1 = - for x 0 180 ° ° G G . [3] O 2 –2 –4 –6 4 6 45° 90° 135° 180° y x (b) (i) State the amplitude of sin x 1 4 2 - . [1] (ii) State the period of tan x 5 3 1 + . [1]

4 0606/11/O/N/14 © UCLES 2014 3 A curve is such that y x x 3 2 d d = + for x 3 2- . The curve passes through the point ( , ) 6 10 . (i) Find the equation of the curve. [4] (ii) Find the x-coordinate of the point on the curve where y = 6. [1]

5 0606/11/O/N/14 © UCLES 2014 [Turn over 4 (i) Using the substitution y 5x = , show that the equation ( ) 5 5 2 2 5 x x x 2 1 1 - + = + + can be written in the form ay by 2 0 2 + + = , where a and b are constants to be found. [2] (ii) Hence solve the equation ( ) 5 5 2 2 5 x x x 2 1 1 - + = + + . [4]

6 0606/11/O/N/14 © UCLES 2014 5 (i) Find the equation of the tangent to the curve nl y x x 3 = - at the point on the curve where x 1 = . [4] (ii) Show that this tangent bisects the line joining the points ( , ) 2 16 - and ( , ) 12 2 . [2]

7 0606/11/O/N/14 © UCLES 2014 [Turn over 6 (i) Given that the coefficient of x2 in the expansion of ( ) px 2 6 + is 60, find the value of the positive constant p. [3] (ii) Using your value of p, find the coefficient of x2 in the expansion of ( )( ) x px 3 2 6 - + . [3]

8 0606/11/O/N/14 © UCLES 2014 7 Matrices A and B are such that a a b b 3 2 A = - J L KK N P OO and B a b b a 2 2 = - J L KK N P OO , where a and b are non-zero constants. (i) Find A 1 – . [2] (ii) Using your answer to part (i), find the matrix X such that XA = B. [4]

9 0606/11/O/N/14 © UCLES 2014 [Turn over 8 The point P lies on the line joining ( , ) A 2 3 - and ( , ) B 10 19 such that : : AP PB 1 3 = . (i) Show that the x-coordinate of P is 1 and find the y-coordinate of P. [2] (ii) Find the equation of the line through P which is perpendicular to AB. [3] The line through P which is perpendicular to AB meets the y-axis at the point Q. (iii) Find the area of the triangle AQB. [3]

10 0606/11/O/N/14 © UCLES 2014 9 The table shows experimental values of variables x and y. x 2 2.5 3 3.5 4 y 18.8 29.6 46.9 74.1 117.2 (i) By plotting a suitable straight line graph on the grid below, show that x and y are related by the equation y abx = , where a and b are constants. [4]

11 0606/11/O/N/14 © UCLES 2014 [Turn over (ii) Use your graph to find the value of a and of b. [4]

12 0606/11/O/N/14 © UCLES 2014 10 (a) (i) Find how many different 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 and 6 if no digit is repeated. [1] (ii) How many of the 4-digit numbers found in part (i) are greater than 6000? [1] (iii) How many of the 4-digit numbers found in part (i) are greater than 6000 and are odd? [1] (b) A quiz team of 10 players is to be chosen from a class of 8 boys and 12 girls. (i) Find the number of different teams that can be chosen if the team has to have equal numbers of girls and boys. [3] (ii) Find the number of different teams that can be chosen if the team has to include the youngest and oldest boy and the youngest and oldest girl. [2]

13 0606/11/O/N/14 © UCLES 2014 [Turn over 11 (a) Solve cos cot x x 2 3 3 = for ° ° x 0 90 G G . [5] (b) Solve sec y 2 2 r + =- J L KK N P OO for y 0 G G r radians. [4]

14 0606/11/O/N/14 © UCLES 2014 12 A B R O P Q The position vectors of points A and B relative to an origin O are a and b respectively. The point P is such that OP OA n = . The point Q is such that OQ OB m = . The lines AQ and BP intersect at the point R. (i) Express AQ in terms of m, a and b. [1] (ii) Express BP in terms of n, a and b. [1] It is given that AR AQ 3 = and BR BP 8 7 = . (iii) Express OR in terms of m, a and b. [2]

15 0606/11/O/N/14 © UCLES 2014 (iv) Express OR in terms of n, a and b. [2] (v) Hence find the value of n and of m. [3]

16 0606/11/O/N/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. 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

® 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 2014 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 2014 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 2014 0606 11 © Cambridge International Examinations 2014 1 2 16 2 d d x x x y − = When d 0, d y x = x = 2, y = 12 M1 A1 DM1 A1 for attempt to differentiate all correct for equating x y d d to zero and an attempt to solve for x. A1 for both, but no extra solutions 2 (a) (b) (i) (ii) 4 o 0 6 or 3 π or 1.05 rad B1 B1 B1 B1 B1 for correct shape for max value of 2, starting at (0, 2) and finishing at ( o 180 , 2) for min value of –4 must be positive 3 (i) (ii) ( ) ( ) c x y + + = 2 1 3 4 c +         = 2 1 9 4 10 2 − = c ( ) 2 3 4 2 1 − + = x y ( ) 2 3 4 6 2 1 − + = x 1 = x M1, A1 M1 A1 A1 ft M1 for ( )2 1 3 + x , A1 for ( )2 1 3 4 + x for a correct attempt to find c, but must be from an attempt to integrate Allow A1 for c = –2 ft for substitution into their equation to obtain x; must have the first M1 1 2 3 −4 −3 −2 −1 1 2 3 4 x y 2 -4

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 11 © Cambridge International Examinations 2014 4 (i) (ii) 0 2 7 5 2 = + −y y ( )( ) 0 1 2 5 = − − y y 5 ln 4.0 ln , 5 2 = = x y 569 .0 − = x y = 1, x = 0 B1, B1 M1 M1 A1 B1 B1 for 5, B1 for –7 for solution of quadratic equation from (i) for use of logarithms to solve equation of the type 5 = x k must be evaluated to 3sf or better 5 (i) (ii) x x x y 1 3 d d 2 − = When x = 1, y = 1 and 2 d d = x y Tangent: ( )1 2 1 − = − x y ( )1 2 − = x y Mid-point (5, 9) ( ) 1 5 2 9 − = Alternative Method: Tangent equation 1 2 − = x y Equation of line joining (–2, 16) and (12, 2) 14 + − = x y Solve simultaneously x = 5, y = 9 Mid-point (5, 9) M1 B1 DM1 A1 B1 B1 B1 B1 for attempt to differentiate for y = 1 for attempt to find equation of tangent allow equation unsimplified for midpoint from given coordinates for checking the mid-point lies on tangent for a complete method to find the coordinates of the point of intersection for midpoint from given coordinates 6 (i) (ii) ( ) 2 2 6 240 192 64 2 x p px px + + = + … 2 240 60 p = 1 2 p = ( )( ) ... 240 192 64 3 2 2x p px x + + − Coefficient of x2 is p 192 180 − = 84 B1 M1 A1 B1 ft M1 A1 for 240p2 or 240p2x2 or ( )2 4 2 6 2 px C × × or 2 4 2 6 2 p C × × or 2 2 4 2 6 2 x p C × × for equating their term in x2 to 60 and attempt to solve ft for 192p, 96 or 192 × their p for 180 – 192p

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 11 © Cambridge International Examinations 2014 7 (i) (ii) 1 2 1 3 5 − −   =     b b a a ab A 1 − = X BA = 1 2 5 5 2 2 1 3 5 5   −   −            a b a a a b b b = 0 1 4 2 5 5           B1, B1 M1 DM1 A1 A1 B1 for ab 5 1 , B1 for       − a a b b 3 2 for post-multiplication by inverse matrix for correct attempt at matrix multiplication, needs at least one term correct for their 1 BA− (allow unsimplified) for each correct pair of elements, must be simplified 8 (i) (ii) (iii) 12 16   =     uuur AB , at P, ) 12 ( 4 1 2 + − = x so at P , x = 1 ( ) 16 4 1 3 + = y , y = 7 Gradient of AB = 12 16 , so perp gradient = 4 3 − Perp line: ( )1 4 3 7 − − = − x y ( ) 31 4 3 = + y x 31 0, 4       Q Area AQB = 12.5 B1 B1 M1 M1 A1 B1 ft M1 A1 for convincing argument for x = 1 for y = 7 for finding gradient of perpendicular for equation of perpendicular through their P Allow unsimplified ft on their perpendicular line, may be implied for any valid method of finding the area of the correct triangle, allow use of their Q; must be in the form ( ) 0,q .

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 11 © Cambridge International Examinations 2014 1 2 3 4 1 2 3 x y 9 (i) (ii) b x a y log log log + = Gradient = log b 4.0 lg = b or 92 .0 ln = b 5.2 = b (allow 2.4 to 2.6) Intercept = log a 47 .0 lg = a or 10 .1 ln = a 3 = a (allow 2.8 to 3.2) Alternative method: Simultaneous equations may be used provided points that are on the plotted straight line are used. 3 = a (allow 2.8 to 3.2) 5.2 = b (allow 2.4 to 2.6) x 2 2.5 3 3.5 4 lg y 1.27 1.47 1.67 1.87 2.07 2 2.5 3 3.5 4 lny 2.93 3.39 3.84 4.31 4.76 B1 M1 A2,1,0 DM1 A1 DM1 A1 DM1 DM1 A1 A1 for the statement, may be seen or implied in later work, for attempt to draw graph of x against log y –1 each error in points plotted for attempt to find gradient and equate it to log b, dependent on M1 in (i) for attempt to equate y-intercept to log a or use their equation with their gradient and a point on the line, dependent on M1 in (i) for a pair of equations using points on the line, dependent on M1 in (i) for solution of these equations, dependent on M1 in (i) A1 for each x logy

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 11 © Cambridge International Examinations 2014 10 (a) (i) (ii) (iii) (b) (i) (ii) 360 60 36 5 12 5 8 C C × 56 792 44352 × = 4 places are accounted for Gender no longer ‘important’ Need 8008 6 16 = C Alternative Method ( ) ( ) ( ) 6 6 10 6 10 6 10 ... 5 0 1 0 6 × + × × C C C C C C 1 60 675 2400 3150 1512 210 8008 + + + + + + = B1 B1 B1 B1, B1 B1 M1 A1 M1 A1 B1 for each, allow unevaluated with no extra terms Final answer must be evaluated and from multiplication for realising that 4 places are accounted or that gender is no longer important for 8008 for at least 5 of the 7 cases, allow unsimplified 11 (a) (b) 0 3 sin 3 cos 3 cos 2 = − x x x 0 3 sin 1 2 3 cos =      − x x Leading to cos3 0, 3 90 , 270 = = o o x x 30 , 90 = o o x and 1 sin3 , 3 30 , 150 2 = = o o x x 10 , 50 = o o x 2 1 2 cos − =       + π y 3 4 , 3 2 2 π π π = + y so 6 5 , 6 π π = y (0.524, 2.62) M1 DM1 A1 DM1 A1 M1 DM1 A1, A1 for use of cos3 cot3 sin3 = x x x , may be implied for attempt to solve 0 3 cos = x correctly from correct factorisation to obtain x A1 for both, no excess solutions in the range for attempt to solve 2 1 3 sin = x correctly to obtain x A1 for both, condone excess solutions for dealing with sec 2 π   +     y correctly for correct order of operations, must not mix degrees and radians

Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 11 © Cambridge International Examinations 2014 12 (i) (ii) (iii) (iv) (v) λ = − uuur AQ b a µ = − uuur BP a b ( ) 1 3 λ = + − uuur OR a b a or ( ) 2 3 λ λ − − b b a 2 1 3 3λ = + a b ( ) 7 8 µ = + − uuur OR b a b or ( ) 1 8 µ µ − − a a b 1 7 8 8 µ = + b a 2 1 1 7 3 3 8 8 λ µ + = + a b b a 21 16 , 8 7 3 2 = = µ µ Allow 0.762 8 3 , 8 1 3 1 = = λ λ Allow 0.375 B1 B1 M1 A1 M1 A1 M1 A1 A1 for a + 1 3 their (i) Allow unsimplified for 7 8 + b their (ii) Allow unsimplified for equating (iii) and (iv) and then equating like vectors