Let f : (–1, 1) \rightarrow R be such that f (cos 4\theta) = \frac{2}{2 - \left(sec\right)^{2} \theta} for \theta \in \left(\right. 0 , \textrm{ }\textrm{ } \frac{\pi}{4} \left.\right) \textrm{ }\textrm{ } \cup \textrm{ }\textrm{ } \left(\right. \frac{\pi}{4} , \textrm{ }\textrm{ } \frac{\pi}{2} \left.\right) . Then the value(s) of f \textrm{ } \left(\right. \frac{1}{3} \left.\right) is(are)

Engineering

Mathematics

Introduction to Functions

Question

Let f : (–1, 1) $\to$ R be such that f (cos 4 $θ$ ) = $\frac{2}{2 - \sec^{2} θ}$ for $θ \in (0, \frac{π}{4}) \cup (\frac{π}{4}, \frac{π}{2})$ . Then the value(s) of $f (\frac{1}{3})$ is(are)

$1 + \sqrt{\frac{2}{3}}$

$1 + \sqrt{\frac{3}{2}}$

$1 - \sqrt{\frac{3}{2}}$

$1 - \sqrt{\frac{2}{3}}$

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$\cos 4 θ = \frac{1}{3} \Rightarrow 2 \cos^{2} 2 θ - 1 = \frac{1}{3} \Rightarrow \cos^{2} 2 θ = \frac{2}{3} \Rightarrow \cos 2 θ = \pm \sqrt{\frac{2}{3}}$

$\Rightarrow \cos 2 θ = \pm \sqrt{\frac{2}{3}}$

But $θ \in (0, \frac{π}{4}) \cup (\frac{π}{4}, \frac{π}{2}) \Rightarrow 2 θ \in (0, \frac{π}{2}) \cup (\frac{π}{2}, π)$

$f (\cos 4 θ) = \frac{2 \cos^{2} θ}{2 \cos^{2} θ - 1} \Rightarrow \frac{1 + \cos 2 θ}{\cos 2 θ} = 1 + \sec 2 θ$

$f (\cos 4 θ) = 1 + \sqrt{\frac{3}{2}} or 1 - \sqrt{\frac{3}{2}}$