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The maximum value of $\quad$ $\sin \left(x+\frac{\pi}{5}\right)+\cos \left(x+\frac{\pi}{5}\right)$, where $x \in\left(0, \frac{\pi}{2}\right)$, is attained at
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The correct answer is:
$\frac{\pi}{20}$
Let $\mathrm{y}=\sin \left(\mathrm{x}+\frac{\pi}{5}\right)+\cos \left(\mathrm{x}+\frac{\pi}{5}\right)$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\cos \left(\mathrm{x}+\frac{\pi}{5}\right)-\sin \left(\mathrm{x}+\frac{\pi}{5}\right)$
for maximum value, $\frac{\mathrm{dy}}{\mathrm{dx}}=0$. $\therefore \cos \left(x+\frac{\pi}{5}\right)-\sin \left(x+\frac{\pi}{5}\right)=0$
$\Rightarrow \frac{\cos \left(x+\frac{\pi}{5}\right)}{\sin \left(x+\frac{\pi}{5}\right)}=1$
$\Rightarrow \cot \left(x+\frac{\pi}{5}\right)=\cot \frac{\pi}{4}$
$\Rightarrow x+\frac{\pi}{5}=\frac{\pi}{4}$
$\Rightarrow x=\frac{\pi}{4}-\frac{\pi}{5}=\frac{5 \pi-4 \pi}{20}=\frac{\pi}{20}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\cos \left(\mathrm{x}+\frac{\pi}{5}\right)-\sin \left(\mathrm{x}+\frac{\pi}{5}\right)$
for maximum value, $\frac{\mathrm{dy}}{\mathrm{dx}}=0$. $\therefore \cos \left(x+\frac{\pi}{5}\right)-\sin \left(x+\frac{\pi}{5}\right)=0$
$\Rightarrow \frac{\cos \left(x+\frac{\pi}{5}\right)}{\sin \left(x+\frac{\pi}{5}\right)}=1$
$\Rightarrow \cot \left(x+\frac{\pi}{5}\right)=\cot \frac{\pi}{4}$
$\Rightarrow x+\frac{\pi}{5}=\frac{\pi}{4}$
$\Rightarrow x=\frac{\pi}{4}-\frac{\pi}{5}=\frac{5 \pi-4 \pi}{20}=\frac{\pi}{20}$
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