Question

The domain of the function
$f\left(x\right)=\frac{1}{\sqrt{\left|\sin x\right|+\sin x}}$ is

• A. $\left(-2n\pi ,2n\pi \right)$
• B. $\left(2n\pi ,\left(2n+1\right)\pi \right)$
• C. $\left(\left(4n-1\right)\frac{\pi }{2},\left(4n+1\right)\frac{\pi }{2}\right)$
• D. $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

Answer 1

The correct option is A $\left(-2n\pi ,2n\pi \right)$

$f\left(x\right)=\frac{1}{\sqrt{\left|\sin x\right|+\sin x}}$

$\Rightarrow \left|\sin x\right|+\sin x>0$

If $\sin x>0,\sin x+\sin x>0$

$\Rightarrow 2\sin x>0$

$\Rightarrow \sin x>0$

$\sin x$ is positive from $[0,\pi ]$

$\therefore x\in (2n\pi ,(2n+1)\pi )$

If $\sin x<0,,\sin x+\sin x>0$ is not possible $\because 0>0$ is not possible.

Hence domain is $x\in (2n\pi ,(2n+1)\pi )$