Question
Domain of the function $$f\left( x \right) = \sqrt {\frac{1}{{\sin x}} - 1} ,{\text{ is}}$$
A.
$$\mathop \cup \limits_{n \in I} \left( {2n\pi ,2n\pi + \frac{\pi }{2}} \right)$$
B.
$$\mathop \cup \limits_{n \in I} \left[ {2n\pi ,\left( {2n + 1} \right)\pi } \right]$$
C.
$$\mathop \cup \limits_{n \in I} \left[ {\left( {2n - 1} \right)\pi , 2n\pi} \right ]$$
D.
None of these
Answer :
$$\mathop \cup \limits_{n \in I} \left[ {2n\pi ,\left( {2n + 1} \right)\pi } \right]$$
Solution :
$$\eqalign{
& \frac{1}{{\sin x}} - 1 \geqslant 0;\,\,\frac{{1 - \sin x}}{{\sin x}} \geqslant 0 \cr
& \frac{{\sin x - 1}}{{\sin x}} \leqslant 0 \cr
& \Rightarrow 0 < \sin x \leqslant 1 \cr
& \Rightarrow x \in \mathop \cup \limits_{n \in I} \left[ {2n\pi ,\left( {2n + 1} \right)\pi } \right] \cr} $$