Let (21-a+21+a ), f(a), (3a+3-a ) be in A.P. and be the minimum value of f(a). Then the value of the integral e( -1) e( ) dx (e2x-e-2x ) is :
- 12 e ( 43 )
- 14 e ( 43 )
- 12 e ( 85 )
- 14 e ( 85 )
Answer
(2) 14 e ( 43 )
Solution
A.P. 2f(a)= (21-a+21+a )+ (3a+3-a )=2 (2a+2-a )+ (3a+3-a )
By A.M. G.M.: 2a+2-a 2 and 3a+3-a 2 (equality at a=0)
2f(a) 6 f(a) 3 =3
I= e2 e3 dxe2x-e-2x= e2 e3 e2x dxe4x-1
Put e2x=u 2e2xdx=du; limits u=4 to u=9
I= 12 49 duu2-1= 14 [ e u-1u+1 ]49
I= 14 [ e 810- e 35 ]= 14 e ( 45 53 )
I= 14 e ( 43 )
Answer: (2)