Question

If $0\le x\le 2\pi$ and $\left|\cos x\right|\le \sin x$ then

• A. the set of value of x is $x\in [\frac{\pi }{4},\frac{3\pi }{4}]$
• B. the number of solutions that are integral multiples of $\frac{\pi }{4}$ is one
• C. the sum of the largest and the smallest solution is $\frac{3\pi }{4}$
• D. $xϵ[\frac{\pi }{4},\frac{\pi }{3}]\cup (\frac{\pi }{2},\frac{3\pi }{4})$

Answer 1

The correct option is A the set of value of x is $x\in [\frac{\pi }{4},\frac{3\pi }{4}]$

Given, $\mid \cos x\mid \le \sin x$, $0\le x\le 2\pi$

If $0\le x\le \frac{\pi }{2}\Rightarrow \cos x\ge 0\Rightarrow \mid \cos x\mid =\cos x$

$\Rightarrow \tan x\ge 1\Rightarrow x\in [\frac{\pi }{4},\frac{\pi }{2}]$

If $\frac{\pi }{2}<x\le \pi \Rightarrow \cos x\le 0\Rightarrow \mid \cos x\mid =-\cos x$

$\Rightarrow \tan x\le 1\Rightarrow x\in (\frac{\pi }{2},\frac{3\pi }{4}]$

considering both cases,

$x\in [\frac{\pi }{4},\frac{3\pi }{4}]$

Hence option 'B' is correct.