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

Answer

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