Question

The number of values of $x$ in $[0,2\pi ]$ satisfying the equation $|\cos x-\sin x|\ge \sqrt{2}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

Given equation is $|\cos x-\sin x|\ge \sqrt{2}$

Since $|\cos x-\sin x|\le \sqrt{1+1}=\sqrt{2}$

$\therefore$ we must have $|\cos x-\sin x|=\sqrt{2}$

$\Rightarrow \left|\cos (x+\frac{\pi }{4})\right|=1\Rightarrow \cos (x+\frac{\pi }{4})=1,-1$

$\therefore x+\frac{\pi }{4}=0,2\pi ,4\pi ,6\pi ,...$ or $\pi ,3\pi ,5\pi ,...$

$\therefore x=\frac{3\pi }{4},\frac{7\pi }{4}$