Answer:
Option B
Explanation:
Given , f(x)=ex(sin x-cos x)
On differentiating both sides w.r.t x, we get
$ f'(x)= e^{x}\frac{d}{dx}(\sin x-\cos x)+(\sin x-\cos x )\frac{d}{dx}(e^{x})$
[ by using product rule of derivative]
$= e^{x}(\cos x+\sin x)+(\sin x-\cos x )e^{x}= 2e^{x} \sin x$
We know that, if Rolle's theorm is verified,
then their exist $c\in \left(\frac{\pi}{4},\frac{5 \pi}{4}\right)$ , such that f'(c)=0
$\therefore$ $2e^{c} \sin c =0$ $\Rightarrow$ $\sin c=0$
$\Rightarrow$ $c=\frac{\pi}{2}$ $\in \left(\frac{\pi}{4},\frac{5 \pi}{4}\right)$
