Q99 / 155 · WBJEE 2022 · Mathematics
The maximum value of $f(x)=e^{\sin x}+e^{\cos x} ; x \in R$ is
- A$2 \mathrm{e}$
- B$2 \sqrt{\mathrm{e}}$
- Correct answerC$2 \mathrm{e}^{1 / \sqrt{2}}$
- D$2 e^{-1 / \sqrt{2}}$
Solution
$f(x)=e^{\sin x}+e^{\cos x}$
$\begin{aligned}
&f^{\prime}(x)=e^{\sin x} \cdot \cos x-e^{\cos x} \cdot \sin x \\
&\begin{aligned}
f^{\prime \prime}(x) &=e^{\sin x} \cos ^2 x+e^{\cos x} \cdot \sin ^2 x-\sin x \cdot e^{\sin x}-\cos x \cdot e^{\cos x} \\
&=e^{\sin x}\left(1-\sin x-\sin ^2 x\right)+e^{\cos x}\left(1-\cos ^2 x-\cos x\right) \\
f^{\prime}(\pi / 4) &=0 \text { and } f^{\prime \prime}(\pi / 4) < 0 \\
f^{\prime}(x) &=0 \text { at } x=\pi / 4+2 n \pi \text { or } 5 \pi / 4+2 n \pi(n \in z)
\end{aligned}
\end{aligned}$
$f^{\prime}(\pi / 4+2 n \pi)=0$ and $f^{\prime \prime}(\pi / 4+2 n \pi) < 0$
$\therefore f_{\max }=f(\pi / 4+2 n \pi), n \in Z=2 e^{1 / \sqrt{2}}$
Hence, the correct answer is option: 'C'.