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# Question and Answer

## QuestionMaths

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) $$\left[\frac{1}{4},\frac{1}{2}\right]$$
(b) $$[1,3]$$
(c) $$\left[\frac{1}{6},\frac{1}{2}\right]$$
(d) $$[2,4]$$

$$1+{sin}^{2}x\in [1,2]$$
$$\frac{1}{1+{sin}^{2}x}\in \left[\frac{1}{2},1\right]$$
$${sin}^{-1}\left(\frac{1}{1+{sin}^{2}x}\right)\in \left[\frac{\pi }{6},\frac{\pi }{2}\right]$$
$$\frac{K\pi }{6}\in \left[\frac{\pi }{6},\frac{\pi }{2}\right] K\in [1,3]$$
Hence, the correct answer is an option (b).