VICTORIA JUNIOR COLLEGE Preliminary Examination. Paper 1 September 2015

Transcription

1 VICTORIA JUNIOR COLLEGE Prelimiary Eamiatio MATHEMATICS (Higher ) 70/0 Paper September 05 Additioal Materials: Aswer Paper Graph Paper List of Formulae (MF5) 3 hours READ THESE INSTRUCTIONS FIRST Write your ame ad CT group o all the work you had i. Write i dark blue or black pe o both sides of the paper. You may use a soft pecil for ay diagrams or graphs. Do ot use staples, paper clips, glue or correctio fluid. Aswer all the questios. Give o-eact umerical aswers correct to 3 sigificat figures, or decimal place i the case of agles i degrees, uless a differet level of accuracy is specified i the questio. You are epected to use a approved graphig calculator. Usupported aswers from a graphig calculator are allowed uless a questio specifically states otherwise. Where usupported aswers from a graphig calculator are ot allowed i a questio, you are required to preset the mathematical steps usig mathematical otatios ad ot calculator commads. You are remided of the eed for clear presetatio i your aswers. At the ed of the eamiatio, faste all your work securely together. The umber of marks is give i brackets [ ] at the ed of each questio or part questio. This documet cosists of 5 prited pages VJC 05 VICTORIA JUNIOR COLLEGE [Tur over

2 A sequece w, w, w 3,... is such that w = + [( ) w +] +, where +, ad = a, where a is a costat. w Use the method of mathematical iductio to prove that w = a + ( ). [5] The fuctio h is give by 3 h: a + b + c + d,, where a, b, c ad d are real costats. The graph of y = h( ) passes through the poit (,). Give that (, ) is a statioary poit, fid three liear equatios ivolvig a, b, c ad d. [3] ab By writig each of a, b ad c i terms of d, fid the eact set of values of d such that 0 c. a + b + d 3 The curve C has equatio y =, where a, b ad d are costats. Give that the lie y = + 3 is a asymptote to C, fid the values of a ad b. [3] [3] Give further that d < 6, fid the coordiates of ay poits of itersectio with the - ad y- aes, leavig your aswer i terms of d. Hece sketch C, statig the equatios of ay asymptotes. [] The fuctio g is defied by where λ is a real costat. g: e 7, >λ, (i) Fid the eact miimum value of λ such that the iverse of g eists. [] Usig the value of λ foud i (i), (ii) fid g ( ), [] (iii) sketch the graphs of y = g ( ), y = g ( ) ad y gg ( ) = o a sigle diagram. [3]

3 5 The diagram shows the curve C with equatio y f ( ) asymptotes to C. C has a miimum ( a,0) ad a maimum poit ( a, a) cuts the y-ais at the poit ( 0, 5a) ( ). y y = f y = a O ( a,0) =. The lies y = a ad = are, where a >. C O separate diagrams, sketch the graphs of (i) y = f, [3] ( ) (ii) y f' ( ) =. [3] 0 Fid the value of ( ) f' d, leavig your aswer i terms of a. [] a ( a, a) ( 0, 5a) = 6 O Jauary 05, Mrs Koh put $000 ito a ivestmet fud which pays compoud iterest at a rate of 8% per aum o the last day of each year. She puts a further $000 ito the fud o the first day of each subsequet year util she retires. (i) If she retires o 3 December 00, show that the total value of her ivestmet o her retiremet day is $8635, correct to the earest dollar. [] O Jauary 05, Mr Woo put $000 ito a savigs pla that pays o iterest. O the first day of each subsequet year, he saves $80 more tha the previous year. Thus, he saves $080 o Jauary 06, $60 o Jauary 07, ad so o. (ii) By formig a suitable iequality, fid the year i which Mr Woo will first have saved over $8635 i total. [] [Tur over 3

4 7 A tak iitially cotais 00 litres of solutio with 00 kg of salt dissolved i it. A solutio cotaiig 0.5 kg of salt per litre flows ito the tak at a rate of litres per miute ad the solutio flows out at the same rate. You should assume that the iflow is istataeously ad thoroughly mied with the cotets of the tak. If the amout of salt i the tak is q kg at the ed of t miutes, show that dq q dt =. [] Fid the time take for the cocetratio of salt i the tak to reach 0.6 kg per litre. [5] (Cocetratio of salt = the amout of salt per uit volume of solutio i the tak.) State what happes to q for large values of t. Sketch a graph of q agaist t. [3] 8 (i) If z = + i y, where, y y, prove that ( ) ( ) z * = z *. [] (ii) Solve the equatio z = Ö i, givig your aswers eactly i the form + iy. [] (iii) Use your aswers i part (ii) to solve the equatio w = + 6Ö i. [] (iv) The roots i part (ii) are represeted by z ad z. Give that arg ( z ) =, fid arg ( zz ), givig your aswer i terms of. [] A curve C has parametric equatios = e si, y = e cos, where <. (i) Sketch C, idicatig clearly the aial itercepts. [] (ii) C cuts the y- ad -aes at poits A ad B respectively. A particle moves alog C from A to B, with its -coordiate icreasig at a costat rate of 0. uits per secod. Fid the 6 eact rate of chage of its y-coordiate whe = e. [3] (iii) The taget at the poit P o C is parallel to the y-ais. Fid the equatio of this taget. [] (iv) The poit Q o C is such that agle POQ =. Fid the area of triagle OPQ. [5]

5 0 A sequece u, u, u 3,... is defied by i i+ i+ A A A + ui = +, where A is a costat ad i. i! ( i+ )! ( i+ )! Aother sequece v, v, v 3,... is defied by (i) Show that v v + = u, where. i= i + + A A A = A +. [] ( + )! ( + )! (ii) Hece, fid N + + A A N v = N ( + )! ( + )!. [] Hece eplai why N + + A A N v = N ( + )! ( + )! coverges as N, ad write dow the value of the limit i terms of A. [3] The equatios of the plae, ad the lies l ad l are give by : a y + z = 3, l : = y, z = 5, where a is costat, ad λ is a real parameter. ( ) l ( ) l : r = i + j + 3k + a + a i j k, Give that the shortest distace from the poit P with coordiates (,, ) to is, show that a =. [] (i) Give that A is a poit o l ad B is a poit o l, fid the positio vectors of A ad B such that AB is perpedicular to both l ad l. [] (ii) Show that l is i the plae. [] (iii) Give that l is parallel to plae, fid the vector equatio of the lie of reflectio of l i. [3] (iv) Fid the cartesia equatio of plae p which is perpedicular to plae ad also cotais l. [3] 5

6 05 VJC JC Prelim Paper Solutios Q) Let P be the statemet: w = a + ( ), LHS of P = w = a (give) RHS of P = a() + ( ) = a P is true Assume P k is true for some k i.e. wk = ak + ( k ) We wat to show P k + is true i.e. w = k ak ( + + ) + k LHS of Pk+ = wk+ = [( k+ ) wk +] k = {( k + )[ ak + ( k )]+} k ( k+ )( k ) + = ak ( + ) + k k = ak ( + ) + k = a( k + ) + k = RHS of P k + Pk is true P k + is true Sice we have show that () P is true ad () Pk is true P k + is true. By mathematical iductio, P is true for all positive itegers. Q) Sub (,) ad (, ) ito y = h( ). a+ b+ c+ d = () 8a+ b+ c+ d = -----() Sice (,) is also the statioary poit, h '() = 0. i.e. a+ b+ c= (3) Usig the GC, a= d 3 5 b= + d c= d ab 0 c 3 5 d + d 0 d 6 { d : d or d < 0}

7 Q3) a + b + y = d = + 3+ k By observatio, a = + b + d = k ( )( ) Compare coefficets of : b= 3 b= + d y = Give: d < 6 Asymptotes: y = + 3, = Aial itercepts: whe = 0, Whe y = 0, d y = + d = 0 ( )( d ) ± ± 8d = = d The coordiates are 0,, 8 d + 8 d,0,,0. 8d + 8d Let α = ad β = y y = + 3 d ( 0, ) ( α,0) O ( β,0) = Q(i) dy Let y = e 7. So, = e 7. d d y =0 e 7 = 0 d = l 7 mi l= l 7 y y = e 7 Q(ii) Let = g ( ) g( ) = e 7 = From the GC, y = y y = e = 3.3

8 Q(iii) ( 7 7 l 7,l 7) y 0.6 ( 7 7 l 7,7 7 l 7) y = g ( ) 0.6 ( l 7,7 7 l 7) y = gg y = g ( ) ( ) Q5(i) y = f ( ) y = f ( ) y = 0 (, 0 ) ( ) a, a y ( ) 0, 5a Q5(ii) y = f'( ) ( a,0) y = a ( a,0) y = = ( ) ( ) a = 0 ( ) f( 0) ( a) f( a) =( 5a) + a+ ( a) 0 0 a Q6(i) f' d = f = 3a Amout at ed of year.08(000) = ( ) (.08) + 000(.08) 3 : ( ) 3 = 000( ) 000( ) 3

9 Amout at the ed of year 00 = 000 (.08) ( ) ( ) 6.08 (.08) = = 8635 (to earest dollar) Q6(ii) 6 S = > 8635 terms S = [ (000) + ( )(80) ] > > 0 < 5.87 (N.A.) or > Least umber of years that he still eeds to save = 36 The year at which Mr Woo s savigs i this savigs pla will first eceed $8635 = = 050 Q7 Rate of salt flowig ito tak per miute is ( 0.5) =.5 kg Rate of salt flowig out per miute is q= 0.03q 00 dq Therefore, = q. dt dq = q dt dq= dt q l q = t + C q = Ae 0.03t 0.03t q= Be Whe t = 0, q = 00, (00) = B B = t q =.5e.6 kg per litre = = 6 kg of salt i the tak ( ) Thus =.5e t =. mi (3 s.f) 0.03t

10 0.03t = ( + ) 0.03t ( ) 0.03q.5 e q = 50 + e 0.03t Whe t is large, e 0 Thus, the amout of salt i the tak decreases to 50kg. q t ( ) q = 50 + e q = 50 t Q8(i) ( LHS = ( + i y) ) * = ( + y + y ) = ( y + y i* ) (i ) i * = i y y ( y ) ( i y) ( ) RHS = ( + i )* = = + iy y i = y y i Q8(ii) Let z = + i y, where y, y ( + i y) = Öi = y () y = Ö +(0) Ö () y = () = = 0 ( )( + 3) = 0 y 0 = =±, y = Ö z = Öi or + Öi Q8(iii) 6Öi ( Ö i) w * = ( Ö i) w = + = + * both sides: ( ) usig (i) : ( w *) = ( Ö i) usig (ii) : w* = Ö(Ö3i ) orö( + Ö3i) = Ö3i or + Ö3i w = + Ö3i or Ö3i 5

11 Q8(iv) z = Öi ad z = +Ö3 i Give: arg ( z ) =. ( zz ) = ( + )( ) = arg ( + 3i + 3) = arg ( + 3i) arg arg 3i 3i Alterative = arg ( z ) ( ) ( z ) = arg + arg = + Give: arg ( z ) = (where < 0 ) ( zz ) = ( z) + ( z ) arg arg arg = + + = + Q(i) y = e si y = e cos e O e Q(ii) d 0. dt =, d y? dt = at e 6 = 6 e si = e = 6 dy e cos e si cos si = = d e cos + e si cos + si d y d d = y dt d dt cos si dy 6 6 = ( 0.) dt cos + si = ( 0. ) = ( ) 6

12 dy cos si Q(iii) = d cos + si At poit P, taget // y-ais dy is udefied d cos + si = 0 ta = = < = e si = e Equatio of taget at poit P is e = Q(iv) Coordiates of P: e, e OP = e + e = e Note that P lies o the lie y = ad OP OQ, the Q lies o the lie of y =. e si = e cos ta = = = e si = e y = e cos = e OQ = e + e = e Area of POQ = ( OP)( OQ) = e e = uits 7

13 Q0(i) v = u+ u + u u 3 A A A = +!! 3! 3 A A A + +! 3!! 3 5 A A A + + 3!! 5! +.. A A A + + ( )! ( )!! + A A A + + ( )!! ( + )! + + A A A + +! ( + )! ( + )! A A A A A = A + + +!! ( + )! ( + )! ( + )! + + A A A = A + ( + )! ( + )! (show) N + + A A Q0(ii) v 7 = N + + ( + )! ( + )! N A N = A 7 N + = N A = A ( N ) 7 N N + 7 = N N A 7 ( 7 ) = A + N N 7 7 N A 7..7 N = A + N ( ) N 7 6 N A = A + N N N = N N N A 7 A A A N v 7 N + + ( + )! ( + )! A 7 A +. 6 N As N, 0 N, so. Also, Hece as N, Series N = Limit = 0 7. N coverges. 8

14 Q) Let M be a poit o plae. 0 OM = 0, MP =, 3 a. = a + 5 a + 5 = a ( 5) 0 5 a + = a a+ a = Qii) l : r = + l 3, λ 3 l : r = 0 + µ, µ λ + µ OA = + 3 λ, OB = µ 3 λ 5 + µ l ggg AB = µ 3l l + ggg ggg AB. = 0, AB. 3 = 0 0 = µ + 7 l, = 7µ + l Solvig, 5 µ =, l = ggg OA ggg =, OB = Q(ii) 0 + µ. = 8+ µ µ + 5= Hece l is i plae.

15 Alterative l : r = 0 + µ 5 0. = = 0 0 l is perpedicular to. l is parallel to. 0. = = 3 5 a poit i l is also i. Hece l is i plae. A Q(ii) l OA + OA' OB = l B reflectio 30 OA' = A ' of l = Lie of reflectio of l i : r = 38 + β 3, β 53 Alterative 0,0,3 o Let D be the poit ( ) ad C be the poit (,, 3 ) o l 0 D CD = 0 = CF = = OF = CF + OC = + = F A ' C l reflectio of l 0

16 OA + OA' OF = OA' = = Lie of reflectio of l i : r = 3 β 3, β + 3 Q(iii) = = 0 r. = 0. 5 Equatio of plae p is + y+ z = 6