Evaluate Function Value Given Trigonometric Identity Relation

To find the possible value(s) of $f\left(\frac{1}{3}\right)$, we start by simplifying the given expression for $f(\cos 4\theta)$:

Given: $f(\cos 4\theta) = \frac{2}{2 - \sec^2 \theta}, \quad \text{for } \theta \in \left(0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$

Expressing the right-hand side in terms of $\cos \theta$: $f(\cos 4\theta) = \frac{2}{2 - \frac{1}{\cos^2 \theta}} = \frac{2\cos^2 \theta}{2\cos^2 \theta - 1}$

Using the double-angle formulas $2\cos^2 \theta = 1 + \cos 2\theta$ and $2\cos^2 \theta - 1 = \cos 2\theta$, we get: $f(\cos 4\theta) = \frac{1 + \cos 2\theta}{\cos 2\theta} = 1 + \frac{1}{\cos 2\theta}$

We are required to evaluate the function at $\cos 4\theta = \frac{1}{3}$.

Using the identity $\cos 4\theta = 2\cos^2 2\theta - 1$: $2\cos^2 2\theta - 1 = \frac{1}{3}$ $2\cos^2 2\theta = \frac{4}{3} \implies \cos^2 2\theta = \frac{2}{3}$ $\cos 2\theta = \pm \sqrt{\frac{2}{3}}$

Now, we consider the two given intervals for $\theta$:

1. For $\theta \in \left(0, \frac{\pi}{4}\right)$: $2\theta \in \left(0, \frac{\pi}{2}\right) \implies \cos 2\theta > 0$ $\cos 2\theta = \sqrt{\frac{2}{3}}$ Substituting this into the expression for $f$: $f\left(\frac{1}{3}\right) = 1 + \frac{1}{\sqrt{\frac{2}{3}}} = 1 + \sqrt{\frac{3}{2}}$

2. For $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$: $2\theta \in \left(\frac{\pi}{2}, \pi\right) \implies \cos 2\theta < 0$ $\cos 2\theta = -\sqrt{\frac{2}{3}}$ Substituting this into the expression for $f$: $f\left(\frac{1}{3}\right) = 1 - \frac{1}{\sqrt{\frac{2}{3}}} = 1 - \sqrt{\frac{3}{2}}$

Thus, the values obtained from the given relation correspond to options (A) and (B).

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Note on Exam Evaluation: By definition, a function $f: (-1, 1) \to \mathbb{R}$ must assign a unique single value to each input in its domain. Since the relation assigns two different outputs, $1 + \sqrt{\frac{3}{2}}$ and $1 - \sqrt{\frac{3}{2}}$, to the single input $x = \frac{1}{3}$, no such function $f$ exists. Consequently, this question was declared ambiguous/flawed, and Marks to All (Zero Marks to all) was awarded by IIT JEE.

Formally, the intended answers based on the algebraic derivation are (A) and (B).