If the solve the problem

Solution:

(d) $3 \pi / 4$

The given function is $f(x)=e^{x} \sin x$.

Differentiating the given function with respect to x, we get 

$f^{\prime}(x)=e^{x} \cos x+\sin x e^{x}$

 

$\Rightarrow f^{\prime}(c)=e^{c} \cos c+\sin c e^{c}$

Now, $e^{x} \cos x$ is continuous and derivable in $R$.

 

Therefore, it is continuous on $[0, \pi]$ and derivable on $(0, \pi)$.

$\therefore f^{\prime}(c)=0$

$\Rightarrow e^{c}(\cos c+\sin c)=0$

$\Rightarrow(\cos c+\sin c)=0 \quad\left(\because e^{c} \neq 0\right)$

$\Rightarrow \tan c=-1$

$\Rightarrow c=\frac{3 \pi}{4}, \frac{7 \pi}{4}, \ldots$

 

$\therefore c=\frac{3 \pi}{4} \in(0, \pi)$

Thus, $c=\frac{3 \pi}{4} \in(0, \pi)$ for which Rolle's theorem holds.

Hence, the required value of c is 3π/4.