Question

The set of values of $x$ in $[0,2\pi ]$ for which $f\left(x\right)=\sqrt{\sin x-\cos x}$ is defined is

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

Answer 1

The correct option is C $[\frac{\pi }{4},\frac{5\pi }{4}]$

$f\left(x\right)=\sqrt{\sin x-\cos x}$ is defined iff quantity inside square root is non negative.
$\Rightarrow \sin x-\cos x>0$
$\Rightarrow \sin x>\cos x$
$\Rightarrow$ Set of values of $x$ for which graph of $y=\sin x$ is above the graph of $y=\cos x$ in $[0,2\pi ]$.
As $\sin x,\cos x$ are fundamental functions whose graphs are given by
and

now we need to find the set of $x$ for which $\sin x>\cos x$ from the graph

From the above graph, we can conclude that $x\in [\frac{\pi }{4},\frac{5\pi }{4}]$