Question

If $y=|\cos x|+|\sin x|$, then $\frac{dy}{dx}$ at $x=\frac{2\pi }{3}$ is

• A. $\frac{1-\sqrt{3}}{2}$
• B. $0$
• C. $\frac{1}{2}(\sqrt{3}-1)$
• D. $Noneofthese$

Answer 1

Answer: (C)

We have, $y=|\cos x|+|\sin x|$
At $x=\frac{2\pi }{3},\cos x$ is -ve and $\sin x$ is +ve
$\therefore y=-\cos x+\sin x$
$\Rightarrow \frac{dy}{dx}=\sin x+\cos x$
$\therefore \frac{dy}{dx}{|}_{atx=2\pi /3}=\sin \left(\frac{2\pi }{3}\right)+\cos \left(\frac{2\pi }{3}\right)$
$=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}$