QuesSol2006 - FACULTY OF ARTS AND SCIENCE University of...

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FACULTY OF ARTS AND SCIENCE University of Toronto FINAL EXAMINATIONS, APRIL/MAY 2006 MAT 133Y1Y Calculus and Linear Algebra for Commerce PART A. MULTIPLE CHOICE 1. [3 marks] The derivative of x 3 x 2 + 1 is: ± A 3 x 2 2 x x 2 + 1 ± B 3 x 2 2 x ± C 3 x 2 x 2 + 1 + x 3 1 2 x 2 + 1 ± D 3 x 2 x 2 + 1 + x 3 2 x ± E 3 x 2 x 2 + 1 + x 4 x 2 + 1 2. [3 marks] lim x 1 2 x - x - 1 ( x - 1) 2 = ± A - 1 4 ± B 1 ± C 1 2 ± D - 1 2 ± E 1 4 1
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3. [3 marks] The function f ( x ) = x ( x - 1) 2 on the interval ( -∞ , 2] has ± A an absolute minimum value of - 1 4 and no absolute maximum value ± B an absolute minimum value of - 1 and no absolute maximum value ± C an absolute minimum value of - 1 4 and an sbsolute maximum value of 2 ± D an absolute minimum value of - 1 and an absolute maximum value of 2 ± E no absolute maximum or minimum values 4. [3 marks] Z e 1 3 ln( x 2 ) x dx = ± A 8 / ln 3 ± B (9 - e ) /e ± C 4 ± D 4 / ln 3 ± E 4 ln 3 5. [3 marks] If the demand equation for a product is p = 300 - q 2 and the supply equation is p = 10 q +100 then the Producers’ Surplus at market equilibrium is ± A 500 ± B 100 ± C 2500 ± D 0 ± E 2000 / 3 2
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6. [3 marks] Given that: f (1) = 1 , f (2) = - 2 , f (3) = - 1 , f (4) = 3 , f (5) = 4 ; if we use Simpson’s Rule with n = 4 , then the value of Z 5 1 f ( x ) dx is approximately: ± A 7 ± B 5 / 2 ± C 7 / 3 ± D 5 / 3 ± E 7 / 2 7. [3 marks] Z 3 1 x 2 ln x dx is equal to ± A 9 ln 3 - 26 9 ± B 9 ln 3 ± C 9 ln 3 - 4 ± D 9 ln 3 + 26 9 ± E 9 ln 3 + 4 8. [3 marks] Z x 2 + x + 1 x 3 + x 2 dx is equal to: ± A ln | x 3 + x 2 | + C ± B ln | x + 1 |- 1 x + C ± C ln | x ( x + 1) | + C ± D ln | x | + ln | x 3 + x 2 | + C ± E 2 x + 1 3 x 2 + 2 x + C 3
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9. [3 marks] Z 2 x - 1 x 3 dx is ± A 1 4 ± B 1 2 ± C 7 16 ± D 3 8 ± E undefned; the integral diverges 10. [3 marks] IF products A and B have joint demand Functions q A ( p A , p B ) = 128 - p A + 8 p B - p A p B q B ( p A , p B ) = 288 - 6 p A - 18 p B + 2 p A p B then they are complementary products provided ± A p A < 8 and p B > 3 ± B p A > 8 and p B < 3 ± C p A < 9 ± D p A < 8 and p B < 3 ± E p A > 8 and p B > 3 4
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11. [3 marks] If f ( x, y ) = xy where x = x ( t ) and y = y ( t ) satisfy x (1) = 3 y (1) = - 2 dx dt (1) = 4 dy dt (1) = 5 , then df dt (1) = ± A 2 ± B - 8 ± C - 10 ± D 7 ± E 12 12. [3 marks] If xyz = 1 , then ∂z ∂x = ± A - z y ± B - y x ± C - x z ± D - x y ± E - z x 5
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13. [3 marks] If g ( u, v, w ) = ( u 2 + 3 v ) 2 4 w - 5 , then at u = - 1 , v = 1 , w = 2 , 2 g ∂u∂w = ± A 4 9 ± B 16 3 ± C 8 9 ± D 8 3 ± E 64 9 14. [3 marks] The function f ( x, y ) = 1 3 x 3 + 1 2 y 2 +
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QuesSol2006 - FACULTY OF ARTS AND SCIENCE University of...

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