021.08s.final.sol

021.08s.final.sol - MATH021 Concise Calculus Final...

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MATH021 Concise Calculus Final Examination Solution, Spring 08 Part I: Multiple Choice Questions Question 1 2 3 4 5 6 7 8 9 10 Total Answer d a e c d c b a b c Question 11 12 13 14 15 Answer e a d b c 1. Find the domain of the function f ( x ) = radicalbigg ( x + 1)( x - 3) 1 - 2 x . (a) x ≤ - 1 or x 3 (b) - 1 x < 1 2 or x 3 (c) x ≥ - 1 or x 3 or x < 1 2 (d) x ≤ - 1 or 1 2 < x 3 (e) - 1 x 3 or 1 2 < x 3 Solution The answer is ( d ). The domain is determined by the inequality ( x + 1)( x - 3) 1 - 2 x 0. 2. Solve the inequality : | 2 x - 3 | < | x + 2 | . (a) 1 3 < x < 5 (b) x < 3 2 (c) 1 10 < x < 9 4 (d) 1 2 < x < 3 4 (e) x > 2 Solution The answer is ( a ). | 2 x - 3 | < | x + 2 | ⇐⇒ vextendsingle vextendsingle vextendsingle vextendsingle 2 x - 3 x + 2 vextendsingle vextendsingle vextendsingle vextendsingle < 1 ⇐⇒ - 1 < 2 x - 3 x + 2 < 1 0 < 2 x - 3 x + 2 + 1 = 3 x - 1 x + 2 and 0 > 2 x - 3 x + 2 - 1 = x - 5 x + 2 ( - 2 < x or x > 1 3 ) and - 2 < x < 5 1 3 < x < 5 3. If b = 3 + log 2 x + 1 2 x - 1 , express x in terms of b . (a) x = 2 b + 3 2 b +1 - 3 (b) x = 2 b 3 + 1 2 b 3 - 1 (c) x = 2 b 1 + 1 2 b 3 + 2 (d) x = 2 b 1 + 4 2 b - 3 (e) x = 2 b + 8 2 b +1 - 8 Solution The answer is (e). b = 3 + log 2 x + 1 2 x - 1 ⇐⇒ b - 3 = log 2 x + 1 2 x - 1 ⇐⇒ x + 1 2 x - 1 = 2 b 3 x + 1 = 2 b 2 x - 2 b 3 ⇐⇒ x = 2 b 3 + 1 2 b 2 - 1 = 2 3 (2 b + 8) 2 3 (2 b +1 - 8) ⇐⇒ x = 2 b + 8 2 b +1 - 8
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1 4. Find the period of the function y = 3 sin ( x 2 π + π 4 ) . (a) 1 (b) 4 π (c) 4 π 2 (d) 4 (e) 3 π 4 Solution The answer is (c). Period = 2 π 1 2 π = 4 π 2 . 5. Find lim x 4 x - 4 x + 5 - 3 . (a) 0 (b) 2 (c) 4 (d) 6 (e) Does not exist Solution The answer is (d). Either by l’Hˆ opital’s rule, or lim x 4 x - 4 x + 5 - 3 = lim x 4 x - 4 x + 5 - 3 · x + 5 + 3 x + 5 + 3 = lim x 4 ( x + 5 + 3) = 9 + 3 = 6 6. Find the horizontal asymptote of the function y = (4 x 2 + 3)( x - 3) 2 x 3 - 2 x 2 + 3 x + 1 . (a) y = 0 (b) y = 1 2 (c) y = 2 (d) y = 7 2 (e) Does not exist Solution The answer is ( c ). lim x + (4 x 2 + 3)( x - 3) 2 x 3 - 2 x 2 + 3 x + 1 lim x + 4 x 3 (1 + 3 4 x 2 )(1 - 3 x ) 2 x 3 (1 - 2 x + 3 x 2 + 1 x 3 ) = 4 2 = 2. 7. A particle is moving along the x -axis so that its position at time t is given by the function x = t 4 - 2 t 3 + t - 5 . Find the acceleration of the particular at t = 2. (a) 18 (b) 24 (c) 36 (d) 52 (e) 63 Solution The answer is ( b ). Acceleration = d 2 x dt 2 vextendsingle vextendsingle vextendsingle vextendsingle t =2 = (12 t 2 - 12 t ) vextendsingle vextendsingle vextendsingle vextendsingle t =2 = 24 8. Find the slope of the tangent line at the point (6 , 1) to the curve defined by the equation xy - y 3 = 5 . (a) - 1 3 (b) - 1 2 (c) 3 2 (d) 2 (e) 5 3 Solution The answer is ( a ). d dx bracketleftBig xy - y 3 bracketrightBig = d dx (5) ⇐⇒ y + x dy dx - 3 y 2 dy dx = 0 Putting in x = 6, y = 1, we have dy dx vextendsingle vextendsingle vextendsingle vextendsingle x =6 , y =1 = - 1 3 .
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2 9. Find a constant k so that the function y = e kx sin 4 x satisfies the equation y ′′ - 6 y + 25 y = 0 .
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