Question

The maximum value of
$\sin (x+\frac{\pi }{5})+\cos (x+\frac{\pi }{5})$, where $x\in (0,\frac{\pi }{2})$ is attained at

• A. $\frac{\pi }{20}$
• B. $\frac{\pi }{15}$
• C. $\frac{\pi }{10}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

$\frac{\pi }{20}$

$\sin (x+\frac{\pi }{5})+\cos (x+\frac{\pi }{5})$

Let it be ${}^{\prime }{y}^{\prime }$

$y=\sin (x+\frac{\pi }{5})+\cos (\pi +\frac{\pi }{5})$

$\Rightarrow \frac{dy}{dx}=\cos (x+\frac{\pi }{5})-\sin (x+\frac{\pi }{5})[\frac{d}{dx}\sin x=\cos x]$

$\Rightarrow \frac{d}{dx}\cos x=-\sin x$

For maximum value

$\Rightarrow \frac{dy}{dx}=0\to \cos (x+\frac{\pi }{5})=\sin (x+\frac{\pi }{5})$

This is only possible when

$x+\frac{\pi }{5}=\frac{\pi }{4}$

$\Rightarrow x=\frac{\pi }{4}-\frac{\pi }{5}=\frac{\pi }{20}$.