Question

The true set of values of $k$ for which $\sin ^{-1}\left(\frac{1}{1+\sin ^2 x}\right)=\frac{k \pi}{6}$ may have a solution, is

• A. $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$
• B. $[1,3]$
• C. $\left[\frac{1}{6}, \frac{1}{2}\right]$
• D. $[2,4]$

Answer 1

Answer: (B)

As, $ 1 \leq 1+\sin ^2 x \leq 2 \Rightarrow \frac{1}{2} \leq \frac{1}{1+\sin ^2 x} \leq 1 $
$ \Rightarrow \frac{\pi}{6} \leq \sin ^{-1}\left(\frac{1}{1+\sin ^2 x}\right) \leq \frac{\pi}{2}$
$\Rightarrow \frac{\pi}{6} \leq \frac{ k \pi}{6} \leq \frac{\pi}{2} \Rightarrow 1 \leq k \leq 3$