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The value of $\mathrm{c}$ in Rolle's Theorem for the function $f(x)=e^{x} \sin x, x \in[0, \pi]$ is
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Verified Answer
The correct answer is:
$\frac{3 \pi}{4}$
Since Rolle's theorem is satisfied
$$
\begin{array}{c}
\therefore \mathrm{f}^{\prime}(\mathrm{c})=0 \Rightarrow \mathrm{e}^{\mathrm{c}} \sin \mathrm{c}+\cos \mathrm{e}^{\mathrm{c}}=0 \\
\Rightarrow \mathrm{e}^{\mathrm{c}}\{\sin \mathrm{c}+\cos \mathrm{c}\}=0 \\
\therefore \operatorname{sinc}+\operatorname{cosc}=0 \quad\left(\because \mathrm{e}^{\mathrm{c}} \neq 0\right) \\
\Rightarrow \operatorname{tanc}=-1 \Rightarrow \mathrm{c}=\tan ^{-1}(-1)=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}
\end{array}
$$
$$
\begin{array}{c}
\therefore \mathrm{f}^{\prime}(\mathrm{c})=0 \Rightarrow \mathrm{e}^{\mathrm{c}} \sin \mathrm{c}+\cos \mathrm{e}^{\mathrm{c}}=0 \\
\Rightarrow \mathrm{e}^{\mathrm{c}}\{\sin \mathrm{c}+\cos \mathrm{c}\}=0 \\
\therefore \operatorname{sinc}+\operatorname{cosc}=0 \quad\left(\because \mathrm{e}^{\mathrm{c}} \neq 0\right) \\
\Rightarrow \operatorname{tanc}=-1 \Rightarrow \mathrm{c}=\tan ^{-1}(-1)=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}
\end{array}
$$
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