solutions_M212_final_Fall_2004

solutions_M212_final_Fall_2004 - Solutions for M212 final...

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Unformatted text preview: Solutions for M212 final exam Fall 2004 PART A 1 1. f- 1 ( x ) = 1+ x 2+ x 2. (sec- 1 ) ( x ) = 1 sec (sec- 1 ( x )) = 1 sec(sec- 1 ( x )) sec 2 (sec- 1 ( x ))- 1 = 1 x x 2- 1 , hence d dx [sec- 1 ( e x )] = (sec- 1 ) ( e x ) d dx e x = e x e x e 2 x- 1 = 1 e 2 x- 1 3. tan( x ) = sin( x ) cos( x ) = sin( x ) 1- sin 2 ( x ) , thus tan 2 ( x )(1- sin 2 ( x )) = sin 2 ( x ), hence sin 2 ( x ) = tan 2 ( x ) 1+tan 2 ( x ) , and so sin( x ) = tan( x ) 1+tan 2 ( x ) . Therefore: sin(tan- 1 (2)) = tan(tan- 1 (2)) 1+tan 2 (tan- 1 (2)) = 2 5 4. Write 3 x +2 ( x +2) 2 ( x +3) = A x +2 + B ( x +2) 2 + C x +3 and get: R 3 x +2 ( x +2) 2 ( x +3) dx = A ln | x + 2 | - B x +2 + C ln | x + 3 | 9. e x = n =0 x n n ! , and thus e- 1 = n =0 (- 1) n n ! = s 4 + R 4 with s 4 = (- 1) 0! + (- 1) 1 1! + (- 1) 2 2! + (- 1) 3 3! + (- 1) 4 4! = 1- 1 + 1 2- 1 6 + 1 24 = 9 24 = 3 8 and | R 4 | 1 5! = 1 120 . 10. i) The series is absolutely convergent (AC), because for all x we have | cos( x ) | 1, and thus n =0 | cos(4 n ) 4 n | n =0 1 4 n , and the latter series is convergent as it is a geometric series with x = 1 4 . ii) The series is conditionally convergent (CC): { ln( n ) } n 2 is a monotonic increasing sequence, and { 1 ln( n ) } n 2...
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solutions_M212_final_Fall_2004 - Solutions for M212 final...

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