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
Since the three terms are in 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), so
2f(a) 6 f(a) 3 = 3
I = e2 e3 dxe2x - e-2x = e2 e3 e2x dxe4x - 1
Put e2x = u, so 2e2xdx = du and the limits become 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)