Year 2009 VCE Mathematical Methods CAS Solutions Trial Examination 2

Transcription

1 Yer 9 VCE Mthemticl Methods CAS Solutions Tril Emintion KILBAHA MULTIMEDIA PUBLISHING PO BOX 7 KEW VIC AUSTRALIA TEL: () FAX: () 987 kilbh@gmil.com KILBAHA PTY LTD 9

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3 Mthemticl Methods CAS Tril Emintion 9 Solutions Section Pge SECTION ANSWERS A B C D E A B C D E A B C D E A B C D E 5 A B C D E A B C D E 7 A B C D E 8 A B C D E 9 A B C D E A B C D E A B C D E A B C D E A B C D E A B C D E 5 A B C D E A B C D E 7 A B C D E 8 A B C D E 9 A B C D E A B C D E A B C D E A B C D E KILBAHA PTY LTD 9

4 Mthemticl Methods CAS Tril Emintion 9 Solutions Section Pge SECTION Question Answer D log ( ) f = g differentiting using the product rule e g f ( ) = g ( ) loge ( ) + e e e f = g loge ( e) + g e e e f = + = e Question Answer A p+ + ( p+ ) y = nd p+ y = q, in mtri form s p = y q The determinnt p + = p( p + ) = p p = ( p + p ) = ( p )( p + ) p so when p= ndp= there is no unique solution. If p = the equtions become () nd () + y = + y = q + y = nd + y = q when q = the two equtions re the sme eqution, so when p= ndq= there is n infinite number of solutions. Question Answer B f ( ) e Question y 7+ e + e d= = + verge vlue is Answer E = into y = ( + ) or y+ = ( + ) y = y' + nd = ' + become ' = nd y' = y in mtri form T y = + y KILBAHA PTY LTD 9

5 Mthemticl Methods CAS Tril Emintion 9 Solutions Section Pge 5 Question 5 Answer C Let y = m + c nd y = +, the tngent to the grph t dy dy the point P, where =. = + = = m so B. is true. d d = = y = 9+ 9 = P, is on the tngent, At y = m+ c = 9+ c c= so A. is true, lso D. is true. b The re A= ( y y) d = b=, so tht = + ( ) + ( + ) E. is true, C. is flse. Question Answer A f ( + h) f ( ) + hf ( ) with f ( ) = = h=. so tht f.f Question 7.99 Answer B A m c d The required re is below the -is, so tking the bsolute vlue, mkes the re positive. A. is true ( ) d this is lso equl to D. which is true ( ) by symmetry C. is true ( ) t, now the inverse function is d. The grph of re bounded by the curve nd the y-is is Question 8 Answer C d, y =, this crosses the y-is = y y = + y = +, the + d, so tht E. is true, B. is flse. f : y = + dom f = R rn f =, f = y + trnsposing y = y =± but rn f = R dom f =, so we must tke the negtive, :(, ), f R f = KILBAHA PTY LTD 9

6 Mthemticl Methods CAS Tril Emintion 9 Solutions Section Pge Question 9 Answer A ( f ( ) ) d f ( ) d [ ] f = f, f is n even function, nd f d=, then f d= 5 = = 5 = Question Answer B The function is not defined when =, ll of A, C, D. nd E. re flse, The function is n even function, symmetricl bout the y-is. y = log e ( ) dy loge( ) ( loge( ) ) d = + = + dy for turning points, d =, since loge ( ) = = e =, e the grph hs minimums t =± e y Question Answer E f : y = b + b f = b+ b= y b= y b y b b f ( ) = y = b+ so f = f b The domin nd rnge of both f nd f re R \{ b }. Since nd b, the grph of y = f ( ) psses through, b b nd the grph of y = f ( ) psses through b, b. All of A. B. C. D. re true, however E. is flse y = f nd y = f lwys intersects on the line y = t the points The grph of ( b, b ) ± ± only if >. KILBAHA PTY LTD 9

7 Mthemticl Methods CAS Tril Emintion 9 Solutions Section Pge 7 Question Answer C dy cos cos y d sin = = = + c to find c, use d 5π = sin + c= + c= c= y = sin now when = y = sin = 5π y = Question Answer D b b b + b b y = = = b+ hs y = b s horizontl symptote nd = s verticl symptote. Question Let f [ π ] R f ( ) Answer D π :,, = cos. The period is T = = π The grph of f is trnsformed by reflection in the -is, the rule is g( ) = cos, we only hve one-qurter of cycle now diltion of fctor from the y-is, replce with g:, [ π ] R, g( ) = cos since we must hve one-qurter of cycle, the new domin is [ ], π then diltion by fctor of from the -is, multiply y by :,, = cos Answer E the eqution becomes g [ π ] R g( ) Question 5 ( A B ) + b p= ( A B ) + p= b ( A B ) = + p ( + b) Pr or Pr Pr A A B p b p b B p? b KILBAHA PTY LTD 9

8 Mthemticl Methods CAS Tril Emintion 9 Solutions Section Pge 8 Question f = f = 8 f = turning points t =, =± for the function to be one-one, we require < Answer D y Question 7 Answer B d (?,.) n ( X = ) = e e loge (.) log (.) X = Bi n= p= Betty winning gme. Pr.. n log. log. n = 5. so n= e Question 8 Answer B d X = N μx = μ, σ = 9 X μ μ μ Pr ( X > μ ) = Pr Z > Pr Z.5 = > = μ Question 9 Answer C Let g ( ) = f ( t) dt then g ( ) = f ( ) g = g = f = 9 g = f = KILBAHA PTY LTD 9

9 Mthemticl Methods CAS Tril Emintion 9 Solutions Section Pge 9 Question Answer D Option D. hs nd ( ) Which is the grph required. f = e g = f g = e Question Answer A da dl A = L = L given cm/s dl dt = da da dl =. = L = 9L dt dl dt da = 9 cm /s dt L= Question Answer E b b Pr = = = + = + = A. is true Since ( X ) b ( b) b E ( X) = Pr( X = ) = + b+ = + b= ( b ) B. is true b E( X ) = Pr( X = ) = ( ) + ( ) + b+ = + + b+ b= ( + b) = C. is true, since A. is true. ( ) vr X = E X E X = b = b + 8b D. is true b 5 = = = + + = X E. is flse, E Pr( X ) b ( b ) END OF SECTION SUGGESTED ANSWERS KILBAHA PTY LTD 9

10 Mthemtics Methods CAS Tril Emintion 9 Solutions Section Pge SECTION Question.i f = + c+ d f = + c = is turning point so f ( ) = ( + )( 9) = ( + )( ) but Epnding gives c = 9, lso u =, f = 5= c+ d = + 9+ d so tht d = nd f = v= 7 7 7= 7 v = 7 ii. The grph of y = 9 hs mimum vlue of 5, nd minimum vlue of 7, nd crosses the -is t three distinct points. The grph of y = 9+ d will therefore cross the -is t three distinct points, provided tht d 5, 7 or 5 < d < 7 A b. f = + c+ d f = + c, for two distinct turning points, we require Δ= c > M c< nd d R c. f + p = + p + p + c + p + d f + p = + p + p p+ c + p p + cp+ d = therefore p = p= nd p p+ c= since p= c= = c = nd p p + cp+ d = so d = lterntive method, if so tht p= c= nd d = y = = + y = KILBAHA PTY LTD 9

11 Mthemtics Methods CAS Tril Emintion 9 Solutions Section Pge b d. A = f ( ) d = b= h= n= f = + c+ d L h f f f f = = () () M R= = h f + f () + f + f f f f f = () () = f + f () + f + f 8= f f = d 8 + c+ d = c c = c = Now subtrcting gives + d = d 8= 8 d = 8 M Question. the mplitude is.5, so tht =.5 π π one-hlf cycle is 8, so tht T = = n= n 8 b. y ( k e ) ( k e ) = psses through the origin O(,) nd B (,8) 8= = e k k e = e = k = log e k k = log e ( ) KILBAHA PTY LTD 9

12 Mthemtics Methods CAS Tril Emintion 9 Solutions Section Pge c.i reflect in the y-is trnslte 8 units, to the right, wy from the y-is or trnslte 8 units, to the right prllel to the -is. ii. [ ] k( 8) f :,8 R, f = e A must give domin. k d.i π A = e sin d 8 ii. k π A= + e + cos k π 8 ech term k π k A= + e + cos + + cos k π but e = M k π 8 A = + k k π A = 8 k π p = 8 q= nd r = A Question. the function is continuous the totl re under the curve is one. 8 b t dt + c 8 t dt = f = b= c c= b b + 8c= nd c= b solving gives b= nd c= KILBAHA PTY LTD 9

13 Mthemtics Methods CAS Tril Emintion 9 Solutions Section Pge b. must show point t y.5,. nd zero for t 8ndt G t c. Pr ( T > ) = ( 8 ) 8 t dt or the re of tringle s M Pr ( T > ) = c= Pr ( T > ) = 8 E T t dt t t dt d. = + ( 8 ) M E( T ) =. +. =. minutes KILBAHA PTY LTD 9

14 Mthemtics Methods CAS Tril Emintion 9 Solutions Section Pge e. Since m t dt =. the medin time m is given by ( 8 t) dt =. M ( m m ) + 8 =. 8 solving for m with < m< 8 m =.5 minutes f. X is the running time of the movie in minutes ( 9, ) d X N μ σ = = = Pr ( X > 9) 9 9 = Pr Z > = Pr Z >.5 =.8 d g. Y = Bi( n=, p=.8) ( Y ) = Pr( Y = ) + Pr( Y = ) ( Y ) = + C Pr Pr Pr ( Y ) =. h. Pr ( comedies) = ACC + CAC + CCA M = =.9 M i =.89 or lterntively A.5.5 C =.. A C in the long run, the percentge of movies which re ctions re 8.9% KILBAHA PTY LTD 9

15 Mthemtics Methods CAS Tril Emintion 9 Solutions Section Pge 5 Question. P, O(,) s = d( OP) = ( ) + + s = + = since > s = + M b.i. ds = = for minimum distnce d + 5 = = 5 ds if > consider =.8 =.9> d 5 ds nd if < consider =.7 =.5< d by the sign test it is minimum. M ii. S min =. 8 P, f 8 m = T = c.i t the point m N = 8 = or 8 norml y ( ) y = ii. norml psses through origin (, ) then + = 8 = 5 = = KILBAHA PTY LTD 9

16 Mthemtics Methods CAS Tril Emintion 9 Solutions Section Pge Question 5. ( ) ( ) ( k ) π sin + cos = = where k Z b. since the period of both re π, it follows tht g ( + π ) = g ( ) T = π c. f :, [ π ] R, f ( ) = sin ( ) + cos ( ) sin( ) + cos( ) = sin( ) = cos( ) tn ( ) = 5π π 7π π =,,, = cos sin = d.i. f ( ) ( ) ( ) cos( ) = sin( ) tn ( ) = π π 7π 5π =,,, ii. m π 7π, nd,, min π 5π, nd, e. grph on correct domin, correct -intercepts G nd correct m nd min. G KILBAHA PTY LTD 9

17 Mthemtics Methods CAS Tril Emintion 9 Solutions Section Pge 7 π π f = sin + cos = sin + = sin + π trnslte sin( ), to the left prllel to the -is A = π α = f. END OF SECTION SUGGESTED ANSWERS KILBAHA PTY LTD 9