SOLUTIONS: Trial Exam 2013 MATHEMATICAL METHODS Written Examination 2

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1 The Mathematical Association of Victoria SOLUTIONS: Trial Exam 0 MATHEMATICAL METHODS Written Examination SECTION : Multiple Choice. D. C. B 4. B 5. A 6. D 7. C 8. E 9. B 0. A. D. B. D 4. E 5. D 6. E 7. C 8. A 9. E 0. A. D. C Question x 5 f( x) = = x x The maximal domain is R \{}. The range is R \{ }. D Question The equations of the asymptotes are x= and x= C Question f : R + R, f( x) = log e( x), and g:, R, g( x) = (x ) f g( x) = log (x ) ( ) e f ( g( x) ) = loge ( x ) Dilation by a factor of from the x-axis, dilation by a factor of a from the y-axis and a translation of a of a unit to the right. B The Mathematical Association of Victoria, 0

2 0 MAV Mathematical Methods (CAS) Trial Exam Solutions Question 4 4 f() x = Ax( x+ ) = Ax + 6Ax + Ax + 8Ax 4 f() x = ax + bx 4x + cx A= 4, A= b= 6A= B Question 5 π π Translation of in the negative direction of the x axis y = tan x +, π Refection in the y axis y = tan x + A Question 6 Amplitude of Translation of c in the negative direction from the x axis Range is [!! c,! c] = " #!( + c), (! c) $ % D Question 7 f ( x + h) f ( x) + hf ʹ ( x) f( x+ h) f( x) hfʹ ( x) h = 0., f( x) = x, fʹ ( x) =, fʹ (8) = = x 8 The Mathematical Association of Victoria, 0

3 0 MAV Mathematical Methods (CAS) Trial Exam Solutions h f!(x) = 0." = 0 C Question 8 g ( x) + c = 0 Sketch a possible graph for g. For one solution f needs to be translated down more than units or translated up more than units. { c: c< } { c: c> } E Question 9 g:[ 8,8] R, where g( x) = x 8 x + The graph is not differentiable at the endpoints or the sharp point. x= 8, x= 0, x= 8 B Question 0 gx () = e x Area = 0.5 ( g(0) + g(0.5) + g() + g(.5) ) The Mathematical Association of Victoria, 0

4 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 4 = e 0.5 e e.5 A Question f :[ a, b] R, where f( x) = x The average value will be zero if a and b are equally spaced either side of x = as the area above the line y = 0 will equal the area below the line y = 0., D [ ] Question ( ) f ( x) dx = 5 ( f ( x) ) dx = ( f ( x) ) [ x] = 0 (6 ) = 6 B Question a= t + t v= + t+ c 0 5= + 0+ c t v= + t 5 4 t t d = + 5t+ c D Option A is acceleration against time Option B is velocity against time Option C is y= t 5 Option E is y = Question 4 SD( X ) = = E ( X ) [ E( X )].44 a E The Mathematical Association of Victoria, 0

5 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 5 Question 5 Let X be the number who play a musical instrument out of 60 X ~Bi(60,0.5) Pr( X < 5) = correct to four decimal places D Question = E OR w w w+ w wʹ w+ wʹ w w+ wʹ wʹ w = = E Question 7 6 f() t dt = f() t dt C 6 Question 8 The Mathematical Association of Victoria, 0

6 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 6 a Gradient is a a = Area = a = a a = a = OR A a " ( x!)dx = Solve on the CAS or " a$ x #! x % ' = & (" 9 a # $! % & '! " # $! % + ) & ', * - = a{ 4! } = a = a = A Question 9 Define f( x) = x x = Var( X ) x f ( x) dx xf ( x) dx 0.06 correct to four decimal places E Question 0 Pr X < a = 0. ( ) 95 The Mathematical Association of Victoria, 0

7 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 7 Pr Z < a.4 = = = a =.79; a 8.0 A 8 = Question! " # Pr( X = 0) + Pr( X =) $ %! # $ (! p)"+ (! p)" p% &! " #(! p) ( + p) $ %! " #! p $ % p D OR The Mathematical Association of Victoria, 0

8 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 8 Question dv dt h = r = 750cm h r = V = πr h h = π 9 πh = 7 dv πh = dh 9 / min The Mathematical Association of Victoria, 0

9 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 9 dh dh dv = dt dv dt 9 = 750 π h 6750 = π h END OF SECTION SOLUTIONS C The Mathematical Association of Victoria, 0

10 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 0 SECTION : Extended Answer Solutions Question a.tsa = πrs + πr Curved surface area = π rs = 00 s = r + h M π r r + h = 00 M r h h 00 r + h = π r h = π r = r π r π r = π r π r h = as required M Show that πr b. V = π r h π r = π r π r 4 4 r π r V = as required M Show that c π r 0 > M 0 0 < r < A π The Mathematical Association of Victoria, 0

11 0 MAV Mathematical Methods (CAS) Trial Exam Solutions d. Solve Vʹ () r = 0 or find the maximum value 0 r = A 4 π 000 V( r ) = A max 7 4 π The Mathematical Association of Victoria, 0

12 0 MAV Mathematical Methods (CAS) Trial Exam Solutions e , 7 π 4 π 0, 0 π Shape A Coordinates A Drawn to scale ½ A Open circles ½ A Round down f. TSA for 000 cones = 000( πrs + πr ) M Substitute π rs = 00 and 0 r = M 4 π TSA for 000 cones = cm A The Mathematical Association of Victoria, 0

13 0 MAV Mathematical Methods (CAS) Trial Exam Solutions Question a. i. = ( 0, 7.6), B = (, 5.077) A xa ii. g(x) = f (!x) 5 g ( x) = cos x + 5 A 0 5 or g(x) = 5 " cos! 0 x + % $ '+ 5 # 5& iii. domain:! " x " 0 A A b. i. Area = f (x)dx! 0 = = 6.04 m correct to three decimal places A The Mathematical Association of Victoria, 0

14 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 4 ii. By symmetry, Total area = =.607 m correct to three decimal places A iii. a = M dx Area: f ( x) ( ) = Area:.7 m correct to three decimal places A c. i. C = ( 0.500, 6.7) By symmetry, = ( 0.500, 6.7) A F A The Mathematical Association of Victoria, 0

15 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 5 ii. AO = , BD =, FC =, M CE = = ( ) =.87 FE M Total length = =.04 m (nearest cm) A The Mathematical Association of Victoria, 0

16 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 6 Question a. i. % non-defective: 00! ( 4 + 6) = 90% A ii. Let X mm be the diameter of a cylinder. ( ) = 0.96!Pr% Z < z = 5. " µ Pr X <5. # $! & ( = 0.96 ' y X 5. µ = σ eq A 4.7 µ Pr = σ ( X < 4.7) = 0.94 Pr Z < z = The Mathematical Association of Victoria, 0

17 0 MAV Mathematical Methods (CAS) Trial Exam Solutions µ = σ eq A Solving the simultaneous for µ and! : µ =4.95,! = 0.5 M show that 95% of cylinders: 4.95±! 0.5=4.6 mm;5.7 mm A iii. Pr ( 4.8 < X < 5.0) = = 0.48 correct to two decimal places A b. i. Y is number of non-defective cylinders out of 8: Binomial n = 8, p = 0.9 M Pr( Y! 7) = Pr( Y = 7) + Pr( Y = 8) The Mathematical Association of Victoria, 0

18 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 8! = # " 8 7 $! &' ' 0.+ # 8 % " 8 $ &' ' 0. 0 % = 0.8 correct to four decimal places A OR Binomial n = 8, p = 0. M Pr( Y!) = Pr( Y = 0) + Pr( Y =) = 0.8 correct to four decimal places A ii. Let $P be the profit per box. P = { 65,! 45} M E ( P) = 65! 0.8+ "45! = 5.855"8.405 = $44.44 A iii = correct to four decimal places A The Mathematical Association of Victoria, 0

19 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 9 c. i. Transition matrix M! % purchased after two years: # " ! $ # & # % # # " $ &! & = # & " & % $ & % 94% purchase from this company (market share) A ii. Long run: = OR = Overall percentage of cylinders purchased from this company: 95% A The Mathematical Association of Victoria, 0

20 0 MAV Mathematical Methods (CAS) Trial Exam Solutions 0 iii. The long term market share of the competitor is 5% A Question 4 a. Solve = x for x x 5 5, y = g y = f y = f y = x = 0 y = g with 5 5, and (0, 0) A Shape for y = A x Sharp point at (, 0) Asymptotes A A The Mathematical Association of Victoria, 0

21 0 MAV Mathematical Methods (CAS) Trial Exam Solutions b., x < 0 x x f( x) = +, 0 < x < x 4 x ½ = A Round down c. 5 ( xdx ) + dx x x = A = loge + 4 A d. The area will be same when y = x + k crosses the right hand branch of the hyperbola when x >. x-intercept, for y = x + k is x= k. A x coordinate of the point of intersection with y = and x > and k < 0 is x k k k + x = A Solve or Solve k k k+ k k k dx ( x + k) dx = loge + x 4 k for k. A k k k+ k k k+ dx ( x + k) dx = 0.90 x for k. A =.7 correct to one decimal place A k The Mathematical Association of Victoria, 0

22 0 MAV Mathematical Methods (CAS) Trial Exam Solutions e. There will be three solutions when y = x + k crosses the left hand branch of the hyperbola twice. x coordinates of the point of intersection with y = and x < 0 and k > 0 are x ± k k k+ x = A from part d. Solve k k > 0 for k > 0. M k > A The Mathematical Association of Victoria, 0 END OF SECTION SOLUTIONS