The point in the interval `(0,2pi)` where `f(X) =e^(x)` sinx has maximum slope is

Options

A. `(pi)/4`
B. `(pi)/2`
C. `pi`
D. `(3pi)/2`

Answer

B

Explanation

Given `f(x)=e^(x) sin x`
`:.f'(x)=e^(x)cosx+sin x e^(x)`
and `f''(x)=-e^(x)sinx+(cosx)e^(x)+e^(x)cosx+e^(x)sinx`
`=2e^(x)cosx`
For maximum slope put
`f''(x)=0`
`implies2e^(x)cosx=0`
`impliescosx=0`
`implies x=(pi)/2, (3p)/2, AAx epsilon[0,2pi]`
Now `f'''(x)=2 [-e^(x)sinx+e^(x) cosx]`
At `x=(pi)/2, f'''(x)lt0`
and at `x=(3pi)/2,f'''(x)gt0`
`:.` Slope is maximum at `x=(pi)/2`