Cambridge IGCSE Mathematics - Additional 0606 — 2002 Oct/Nov Paper 2 · Variant 1

TIME 2 hours INSTRUCTIONS TO CANDIDATES Write your name, Centre number and candidate number in the spaces provided on the answer paper/answer booklet. Answer all the questions. Write your answers on the separate answer paper provided. If you use more than one sheet of paper, fasten the sheets together. 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. INFORMATION FOR CANDIDATES 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. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL EXAMINATIONS ADDITIONAL MATHEMATICS 0606/2 PAPER 2 OCTOBER/NOVEMBER SESSION 2002 2 hours Additional materials: Answer paper Electronic calculator Graph paper Mathematical tables This question paper consists of 5 printed pages and 3 blank pages. SP (NH/TC) S40047/2 © CIE 2002 [Turn over www.XtremePapers.com

2 0606/2/O/N/02 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, . Binomial Theorem (a + b)n = an + an–1 b + an–2 b2 + … + an–r br + … + bn, where n is a positive integer and = . 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1. sec2 A = 1 + tan2 A. cosec2 A = 1 + cot2 A. Formulae for ∆ABC = = . a2 = b2 + c2 – 2bc cos A. ∆= bc sin A. 1 2 c sin C b sin B a sin A n! (n – r)!r! ) n r( ) n r( ) n 2( ) n 1( x b b ac a = ± – – 2 4 2

3 0606/2/O/N/02 [Turn over 1 Write down the inverse of the matrix and use this to solve the simultaneous equations 4x + 3y + 7 = 0, 7x + 6y + 16 = 0. [4] 2 Find the first three terms in the expansion, in ascending powers of x, of (2 + x)6 and hence obtain the coefficient of x2 in the expansion of (2 + x – x2)6. [4] 3 Given that k = and that p = , express in its simplest surd form (i) p, (ii) p – . [5] 4 Given that
= {x: – 5 < x < 5}, A = {x: 8 > 2x + 1}, B = {x: x2 > x + 2}, find the values of x which define the set A
B. [6] 5 (a) The producer of a play requires a total cast of 5, of which 3 are actors and 2 are actresses. He auditions 5 actors and 4 actresses for the cast. Find the total number of ways in which the cast can be obtained. [3] (b) Find how many different odd 4-digit numbers less than 4000 can be made from the digits 1, 2, 3, 4, 5, 6, 7 if no digit may be repeated. [3] 6 The cubic polynomial f(x) is such that the coefficient of x3 is –1 and the roots of the equation f(x) = 0 are 1, 2 and k. Given that f(x) has a remainder of 8 when divided by x – 3, find (i) the value of k, (ii) the remainder when f(x) is divided by x + 3. [6] 7 (i) Differentiate xsin x with respect to x. [2] (ii) Hence evaluate ∫0 π–2 xcos x dx. [4] 1 p 1 + k 1 – k 1 3 ) 4 3 7 6 (

4 0606/2/O/N/02 8 (i) Sketch the graph of y = ln x. [2] (ii) Determine the equation of the straight line which would need to be drawn on the graph of y = ln x in order to obtain a graphical solution of the equation x2ex–2 = 1. [4] 9 (a) Find, in its simplest form, the product of a 1 3 + b 2 3 and a 2 3 – a 1 3b 2 3 + b 4 3. [3] (b) Given that 22x+2 × 5x–1 = 8x × 52x, evaluate 10x. [4] 10 In the diagram, OA →= a, OB →= b, AM →= MB →and OP →= 1 3OB →. (i) Express AP →and OM →in terms of a and b. [3] (ii) Given that OQ →=
OM →, express OQ →in terms of
, a and b. [1] (iii) Given that AQ →= AP →, express OQ →in terms of , a and b. [2] (iv) Hence find the value of
and of . [3] 11 A car moves on a straight road. As the driver passes a point A on the road with a speed of 20ms–1, he notices an accident ahead at a point B. He immediately applies the brakes and the car moves with an acceleration of ams–2, where a = – 6 and t s is the time after passing A. When t = 4, the car passes the accident at B. The car then moves with a constant acceleration of 2ms–2 until the original speed of 20ms–1 is regained at a point C. Find (i) the speed of the car at B, [4] (ii) the distance AB, [3] (iii) the time taken for the car to travel from B to C. [2] Sketch the velocity-time graph for the journey from A to C. [2] 3t 2 A O B P Q M b a

5 0606/2/O/N/02 12 Answer only one of the following two alternatives. EITHER The diagram shows a greenhouse standing on a horizontal rectangular base. The vertical semicircular ends and the curved roof are made from polythene sheeting. The radius of each semicircle is r m and the length of the greenhouse is l m. Given that 120m2 of polythene sheeting is used for the greenhouse, express l in terms of r and show that the volume, V m3, of the greenhouse is given by V = 60r – . [4] Given that r can vary, find, to 2 decimal places, the value of r for which V has a stationary value. [3] Find this value of V and determine whether it is a maximum or a minimum. [3] OR The diagram shows part of the curve y = x2 ln x, crossing the x-axis at Q and having a minimum point at P. (i) Find the value of at Q. [4] (ii) Show that the x-coordinate of P is . [3] (iii) Find the value of at P. [3] d2y dx2 1 e dy dx Q O x y P y = x2 ln x πr3 2 l m r m

CAMBRIDGE INTERNATIONAL EXAMINATIONS NOVEMBER 2002 INTERNATIONAL GCSE UNIVERSITY of CAMBRIDGE Local Examinations Syndicate

Page 1 | Mark Scheme [ IGCSE Examinations — November 2002 Sree Paper | 0606 2 | 4[4) Inverse = {© ~9 ft Bi Bt -7 4 3 x)_1(6 -3/-7)_(2 MIAt y} 3\-7 4 )l-16) (5 2 [4] 2 +x 2x x + 8X8 opty? B2, 1,0 1x2 Fl casi ince erry OF wi Telit, Chet = 64 + 192x + 240x” Replace x by x-x* => coefficient of x? = -192+240 = 48 M1 A Sc. + —— tae toe VE41 VE+1 3 [5] (i) Either 3 V3 or * + M1 tt V3-1" W341 v3 v3 Simplify > 2+V3 Ai 2 1 2- v3 _2-V3 MtAt “o4J32-J3 4-3 p-La24 V5-@-V5)= 25 At Or 5. an? 6+4V3, By p +857 57 243 Multiply by 2-43 = aye M1 At 2-3 {—_—__—__— 4 [6] Solving inequalities: A x<3.5 B1 B X-x-2=0 > (x-2)(x+1)=0 => x=-1,2 M1 At xX -x-2>0 => x<-1, x>2 Required values -5<x<-1 Al M1 2<x<35 Al

Page 2 Mark Scheme Syllabus | Paper IGCSE Examinations — November 2002 0606 2 5 [6] (a) Either ,C; = 5x4*3 or 4c, = 4x3 B1 1x2x3 1x2 Product = 10x6 = 60 M1 A1 (b) Either, ending in 1 (or3). => 2*5%*4 or, endingin5(or7) = 3x54] B1 Adding all4 cases => 40+40+60+60 = 200 M1A1 _| __ 6 [6] (i) f(x) = -(x=-1)(x-2) (x-4) M1A1 #(3) = -2x1x(3-k) =8 > k=7 M1A1 (ii) f(-3) = -(-4)(-5)(-10) = 200 7 (6) (i) xu sin x) = sin x + x cos x M1A4 (ii) [x cos x dx = x sinx - [sin x dx M1 fain x dx = =cos x IDM1 L x sin x + cos x Z-1= 0871 Al Atles.ec. 8 [6] (i) y [- -2 asx -— 0; thro’ (1,0); +7 asx") | B2,1,0 1 x 2 “2h (ii) Taxe Legs nx?s lane stat me =F Riay aur-Bs O Ai LL Make Inx the subject = Inx =-%(x-2) => _ lineis y=1-x2 M1 A1 9 [7] (a) Correct combination of indices M1 Either (a - a'@b”* + b*) x a! = a—a?b™ + aps? Or (a? — ap +b) xp? = a8 apt + p? Al Sum = a+b? Al (b) 2? = 4 x 2% ots 6F+5 er=2™ B2,1,0 © 10 = 4/5 M1 At

Page 3 Mark Scheme Syllabus Paper IGCSE Examinations — November 2002 0606 2 | 10 [9] | (i) AP =b/3-a OM = a/2 + bi2 B1 MIA (ii) OQ = A(al2 + b/2) Biv ° B (ili) OQ = OA+pAP = a+ (b/3—a) M1 AI (iv) Comparing coefficients 4/2 =1-p and 24/2 = p/3 M1 Solving ALEK ne% M1 A1 —— _ 2 44 [11] (i) v= fet- ear = Sot (+e) M1At 2 4 [Vi-0 = 20 => c= 20 [Ves = 12-244+20 = 8 Al Al - 3t? e 2 qv (i) [(S-- 6t + 20)d t= —-- 3t? +20¢ M1A1 4 4 AB =[ | = 16 - 48 + 80 = 48 Al (ii) ve=8, vce=20 => tac = (20-8)/2=6 miarl (iv) v 20 curve B1 8 straight line Bit t ne ‘ “ | 12 [10] Az=nPetnrl > 1 = (120-27)/ar Bint Either awe prestnest ¢ Ve war (eS = 60r—-%nP (Ray M1A1 rT dV/dr = 60-32 7/2 = 0 when ? = 40/y/= 3.57 B1 M1A1 Stationary value of V = 143. (142.73) @vid? = —3xr <Oforr>0 => maximum [or any valid method ] (i) dy/dx = x? x 4x + 2xInx AtQ, y=O0 => Inx=0 >x=1 [dy/ dx] ,21 = 1 csc (ii) AtP, dy/dx =O => x(1+2Inx)=0 => Inx= -% M1 At => x=e*=1/ Ve (= 0.6%5) (AG) At (ili) d?y/ dx? = d(x + 2x Inx)/dx = 1+2Inx + (2x *1/x) M1 AL = 3+2Inx Al [ey de] x= = 34+2(-%) = 2eso