Cambridge IGCSE Mathematics - Additional 0606 — 2014 Oct/Nov Paper 2 · Variant 2

This document consists of 15 printed pages and 1 blank page. DC (LK/SLM) 97553 © UCLES 2014 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 4 9 9 3 1 9 7 9 6 9 * ADDITIONAL MATHEMATICS 0606/22 Paper 2 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 0606/22/O/N/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 0606/22/O/N/14 © UCLES 2014 [Turn over 1 (a) On each of the Venn diagrams below shade the region which represents the given set. A C B A C B ( ) A B C + , l ( A B C + , ) l [2] (b) In a year group of 98 pupils, F is the set of pupils who play football and H is the set of pupils who play hockey. There are 60 pupils who play football and 50 pupils who play hockey. The number that play both sports is x and the number that play neither is x 30 2 - . Find the value of x. [3]

4 0606/22/O/N/14 © UCLES 2014 2 Solve the inequality ( ) x x x 9 2 1 1 2 2 1 + - + . [3] 3 Solve the following simultaneous equations. ( ) log log x y 3 2 2 2 + = + ( ) log x y 3 2 + = [5]

5 0606/22/O/N/14 © UCLES 2014 [Turn over 4 The functions f and g are defined for real values of x by ( )x x 1 3 f = - - for , x 1 2 ( )x x x 2 3 2 g = - - for . x 2 2 (i) Find gf(37). [2] (ii) Find an expression for ( )x f 1 - . [2] (iii) Find an expression for ( ). x g 1 - [2]

6 0606/22/O/N/14 © UCLES 2014 5 The number of bacteria B in a culture, t days after the first observation, is given by B 500 400e . t 0 2 = + . (i) Find the initial number present. [1] (ii) Find the number present after 10 days. [1] (iii) Find the rate at which the bacteria are increasing after 10 days. [2] (iv) Find the value of t when B 10000 = . [3]

7 0606/22/O/N/14 © UCLES 2014 [Turn over 6 (i) Calculate the coordinates of the points where the line y x 2 = + cuts the curve . x y 10 2 2 + = [4] (ii) Find the exact values of m for which the line y mx 5 = + is a tangent to the curve . x y 10 2 2 + = [4]

8 0606/22/O/N/14 © UCLES 2014 7 A particle moving in a straight line passes through a fixed point O. The displacement, x metres, of the particle, t seconds after it passes through O, is given by . sin x t t 2 = + (i) Find an expression for the velocity, v ms 1 - , at time t. [2] When the particle is first at instantaneous rest, find (ii) the value of t, [2] (iii) its displacement and acceleration. [3]

9 0606/22/O/N/14 © UCLES 2014 [Turn over 8 (i) Given that y x x 2 2 2 = + , show that ( ) x y x kx 2 d d 2 2 = + , where k is a constant to be found. [3] (ii) Hence find ( ) x x x 2 d 2 2 + c e dd . [2]

10 0606/22/O/N/14 © UCLES 2014 9 Integers a and b are such that ( ) a a b 3 5 5 51 2 + + - = . Find the possible values of a and the corresponding values of b. [6]

11 0606/22/O/N/14 © UCLES 2014 [Turn over 10 (i) Prove that . sec cosec cot tan x x x x - = [4] (ii) Use the result from part (i) to solve the equation sec cosec cot x x x 3 = for 0° x 1 1 360°. [4]

12 0606/22/O/N/14 © UCLES 2014 11 P Q O S R 0.8 rad 5 cm x cm The diagram shows a sector OPQ of a circle with centre O and radius x cm. Angle POQ is 0.8 radians. The point S lies on OQ such that OS = 5 cm. The point R lies on OP such that angle ORS is a right angle. Given that the area of triangle ORS is one-fifth of the area of sector OPQ, find (i) the area of sector OPQ in terms of x and hence show that the value of x is 8.837 correct to 4 significant figures, [5]

13 0606/22/O/N/14 © UCLES 2014 [Turn over (ii) the perimeter of PQSR, [3] (iii) the area of PQSR. [2]

14 0606/22/O/N/14 © UCLES 2014 12 (i) Show that x 2 - is a factor of x x 3 14 32 3 2 - + . [1] (ii) Hence factorise x x 3 14 32 3 2 - + completely. [4]

15 0606/22/O/N/14 © UCLES 2014 The diagram below shows part of the curve y x x 3 14 32 2 = - + cutting the x-axis at the points P and Q. y x P Q O y = 3x – 14 + 32 x2 (iii) State the x-coordinates of P and Q. [1] (iv) Find ( ) x x x 3 14 32 d 2 - + y and hence determine the area of the shaded region. [4]

16 0606/22/O/N/14 © UCLES 2014 BLANK PAGE 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.

® 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 0606/22 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 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 22 © Cambridge International Examinations 2014 1 (a) (b) No.in H only = x − 50 ; No in F only = x − 60 Sum: 98 2 30 60 50 = − + + − + − x x x x x = 14 B1 B1 B1 M1 A1 Both written or on diagram Add at least 3 terms each with x involved and equate to 98 soi 2 ( ) 2 2 2 9 2 1 1 8 2 oe isw x x x x + −< + < 2 1 2 1 < < − x M1 A1 A1 Expand and collect terms 3 ( ) 2 2 log 3 log 2 3 4 x y x y + = + → + = ( ) 2 log 3 8 x y x y + = → + = ( ) x x − = + 8 4 3 5 29 5.8, oe 2.2 oe x x y = → = = B1 B1 M1 A1 A1 Eliminate y or x from two linear three term equations

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 22 © Cambridge International Examinations 2014 4 (i) (ii) (iii) ( ) f 37 3 = or ( ) ( ) 1 3 2 gf 2 1 3 3 x x x −−− = −− − ( ) 3 2 1 gf 37 6 3 3 − = = − ( ) 2 1 3 3 1 y x y x = −− → + = − ( ) ( ) 2 1 3 1 f x x − + + = oe isw 2 2 3 2 3 2 2 3 2 x y x xy y x xy x y − = − − = − → − = − ( ) 1 3 2 g 2 1 x x x − − = − oe B1 B1 M1 A1 M1 A1 Rearrange and square in any order Interchange x and y and complete Multiply and collect like terms Interchange and complete Mark final answer 5 (i) (ii) (iii) (iv) B = 900 B = 500 + 400e2 = 3455 or 3456 or 3460 0.2 d 80e d t B t   =     ( ) 2 d 10 80e 591 /day d B t t = → = = ( ) 0.2 0.2 10000 500 400e e 23.75 t t = + → = ( ) 0.2 ln23.75 15.8 days t t = = B1 B1 B1 B1 M1 DM1 A1 3455.6 scores B0 awrt e0.2t = k take logs: 0.2t = ln k awrt

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 22 © Cambridge International Examinations 2014 6 (i) (ii) ( ) 10 2 2 2 = + + x x ( )( ) 2 2 3 0 3 1 0 x x x x + − = → + − = Points (1, 3), (–3, –1) isw or elimination of x leads to y2 – 2y – 3 = 0, then as above 10 25 10 2 2 2 = + + + x mx x m ( ) 0 15 10 1 2 2 = + + + mx x m ( ) ( ) 2 2 2 4 0 100 60 1 0 b ac m m − = → − + = 2 3 ± = m oe isw Alternative solution: 2 d d 10 y x x x − = − or d d y x x y = − Result: y x y 5 2 2 + = after inserted in 5 + = mx y Attempt to solve with 10 2 2 = + y x y = 2, x = 6 ± 6 3 ± = m oe B1 M1 A1 A1 B1 M1 A1 A1 B1 M1 A1 A1 3 term quadratic with attempt to solve both x or a pair both y or second pair attempt to use discriminant on three term quadratic. Allow unsimplified cao ± is required allow unsimplified Eliminate x or y both 7 (i) (ii) (iii) v = 2cost + 1 2cost + 1 = 0 2 3 t π = or 2.09 2 2 2 2sin 3.83m 3 3 3 t x π π π   = → = + =     t a sin 2 − = 2 2 1.73 3 ms 3 4 t a π − = = − = − B1 M1 A1 B1 B1ft DB1ft mark final answer equate their v to zero (must be a differential) and attempt to solve to find an angle awrt awrt ft their v (2nd differential) ft using their angle t in correct a awrt 8 (i) (ii) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 d 4 d 2 2 4 x x x x y x x x x k + × − × = = + + = ( ) ( ) 2 2 2 2 1 d 4 2 2 x x x c x x = × + + + ∫ isw M1 A1 A1 B1 B1 apply quotient or product rule unsimplified k=4 does not need to be specifically identified 1 their k × original function

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 22 © Cambridge International Examinations 2014 9 ( ) 45 5 3 5 3 5 3 2 2 + + + = + a a a a oe Equate: 51 45 2 = + + a a and 0 6 = −b a ( )( ) 3 2 0 a a + − = a = –3, 2 b = –18, 12 B1 B1 B1 M1 A1 A1 anywhere Attempt to solve three term quadratic with integer coefficients obtained by equating coeffs Both as correct or one correct pair Both bs correct 10 (i) (ii) secxcosecx = x xsin cos 1 x x x sin cos cot = LHS = x x x sin cos cos 1 2 − oe x x x x tan sin cos sin 2 = = AG 3cot cot tan 2cot tan x x x x x − = → = 2 tan2 = x oe x = 54.7, 125.3, 234.7, 305.3 B1 B1 B1ft B1 M1 A1 A1 A1 anywhere anywhere correct addition of their terms use of identity and cancel equate and collect like terms, allow sign errors 2 values only 2 more values. awrt 11 (i) (ii) (iii) Area of sector = ( ) 2 2 2 1 0.8 0.4 cm 2 x x × × = ( ) 5sin0.8 3.59 SR = = or ( ) 5cos0.8 3.48 OR = = Area of triangle = 2 1 5cos0.8 5sin0.8 6.247cm 2 × = 2 0.08 6.247 8.837cm AG x x = = ( ) 8.84 5 3.84cm SQ = − = ( ) 8.84 5cos0.8 5.35 or 5.36cm PR = − = ( ) 8.84 0.8 7.07cm PQ = × = Perimeter = 19.84 to 19.86 cm or rounded to 19.8 or 19.9 2 Area 4 6.247 25cm PQSR = × = B1 B1 M1 A1 A1 B1 B1 B1 M1 A1 anywhere SR may be seen in stated C absin 2 1 insert correct terms into correct area formulae two lengths from SQ, PR, PQ awrt third length awrt sum 24.95 to 25

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0606 22 © Cambridge International Examinations 2014 12 (i) (ii) (iii) (iv) ( ) ( ) ( ) 3 2 f 2 3 2 14 2 32 0 = − + = Or complete long division ( ) ( )( ) 2 f 2 3 8 16 x x x x = − − − ( ) ( )( )( ) f 2 4 3 4 x x x x = − − + x = 2, 4 ( ) c 32 14 5.1 d 32 14 3 2 2 + − − = + − ∫ x x x x x x Area = 4 2 2 32 14 5.1     − − x x x = (–) 2 B1 M1 A1 M1 A1 B1 B1 B1 M1 A1 3x2 and 16 8x and correct signs Factorise three term quadratic first 2 terms third term correct unsimplified Limits of 2 and 4 and subtract