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Unformatted text preview: MATH1804/Solution to AS4/ZJH/2009 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1804: University Mathematics A Assignment 4 1. Solution : (a) Taking derivative, f ( x ) = 1- ln x x 2 Let f ( x ) = 0 , we obtain a critical point at x = e . When x < e, f is increasing. When x > e, f is decreasing. (b) According to part (a), f ( x ) is increasing for x < e , achieve its maximum at x = e , and is decreasing for x > e . Since π > 3 > e , therefore f ( π ) < f (3) < f ( e ) which implies ln π π < ln 3 3 which is the same as π 3 < 3 π 2. Solution: Write f ( x ) = R x e t sin tdt , g ( x ) = ln x , then R ln x e t sin tdt = f ( g ( x )). By using the Fundamental Theorem of Calculus and the chain rule, we have d dx Z ln x e t sin tdt = f ( g ( x )) · g ( x ) = e t sin t t = g ( x ) · 1 x = x sin (ln x ) 1 x = sin (ln x ) 3. (a) By using the Fundamental Theorem of Calculus, we have F ( x ) = e- x 2 > 0 for any x ∈ R , and hence F ( x ) is strictly increasing on R ....
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