Question

$\left\{x\in R:\left|\cos x\right|\ge \sin x\right\}\cap \left[0,\frac{3\pi }{2}\right]=$

• A. $\left[0,\frac{\pi }{4}\right]\cup \left[\frac{3\pi }{4},\frac{3\pi }{2}\right]$
• B. $\left[0,\frac{\pi }{4}\right]\cup \left[\frac{\pi }{2},\frac{3\pi }{2}\right]$
• C. $\left[0,\frac{\pi }{4}\right]\cup \left[\frac{5\pi }{4},\frac{3\pi }{2}\right]$
• D. $\left[0,\frac{3\pi }{2}\right]$

Answer 1

Answer: (A)

Given, $\{x\in R:|\cos x|\ge \sin x\}\cap \left[0,\frac{3\pi }{2}\right]$
If we draw the graphs of $|\cos x|$ and $\sin x$, clearly
$|\cos x|\ge \sin x$ when

$x\in \left[0,\frac{\pi }{4}\right]\cup \left[\frac{3\pi }{4},\frac{3\pi }{2}\right]$
$\therefore x\in \left[0,\frac{\pi }{4}\right]\cup \left[\frac{3\pi }{4},\frac{3\pi }{2}\right]\cap \left[0,\frac{3\pi }{2}\right]$
$\Rightarrow x\in \left[0,\frac{\pi }{4}\right]\cup \left[\frac{3\pi }{4},\frac{3\pi }{2}\right]$