Search any question & find its solution
Question:
Answered & Verified by Expert
If $y=|\cos x|+|\sin x|$, then $\frac{d y}{d x}$ at $x=\frac{2 \pi}{3}$ is
Options:
Solution:
1018 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}(\sqrt{3}-1)$
We have, $y=|\cos x|+|\sin x|$
At $x=\frac{2 \pi}{3}, \cos x$ is negative and $\sin x$ is positive.
$\therefore \quad y=-\cos x+\sin x$
$\Rightarrow \frac{d y}{d x}=\sin x+\cos x$
$\left.\therefore \frac{d y}{d x}\right|_{\text {at } x=2 \pi / 3}=\sin \left(\frac{2 \pi}{3}\right)+\cos \left(\frac{2 \pi}{3}\right)$
$$
=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}
$$
At $x=\frac{2 \pi}{3}, \cos x$ is negative and $\sin x$ is positive.
$\therefore \quad y=-\cos x+\sin x$
$\Rightarrow \frac{d y}{d x}=\sin x+\cos x$
$\left.\therefore \frac{d y}{d x}\right|_{\text {at } x=2 \pi / 3}=\sin \left(\frac{2 \pi}{3}\right)+\cos \left(\frac{2 \pi}{3}\right)$
$$
=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.