Let `f : (-1, 1) -> R` be such that `f(cos4theta) = 2/(2-sec^2theta` for `theta in (0, pi/4) uu (pi/4, pi/2)`. Then the value(s) of `f(1/3)` is/are

Options

A. `1-sqrt((3)/(2))`
B. `1+sqrt((3)/(2))`
C. `1-sqrt((2)/(3))`
D. `1+sqrt((2)/(3))`

Explanation

To solve the problem, we need to find the value of \( f\left(\frac{1}{3}\right) \) given that \( f(\cos 4\theta) = \frac{2}{2 - \sec^2 \theta} \) for \( \theta \) in the intervals \( (0, \frac{\pi}{4}) \) and \( (\frac{\pi}{4}, \frac{\pi}{2}) \).

### Step-by-step Solution:

1. **Rewrite the Function**: Start with the given function:
\[ f(\cos 4\theta) = \frac{2}{2 - \sec^2 \theta} \]
Recall that \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \), so we can rewrite the function as:
\[ f(\cos 4\theta) = \frac{2}{2 - \frac{1}{\cos^2 \theta}} = \frac{2 \cos^2 \theta}{2 \cos^2 \theta - 1} \]

2. **Set Up the Equation**: We want to find \( f\left(\frac{1}{3}\right) \). We need to find \( \theta \) such that:
\[ \cos 4\theta = \frac{1}{3} \]

3. **Use the Cosine Double Angle Formula**: Recall that:
\[ \cos 4\theta = 2\cos^2 2\theta - 1 \]
Setting this equal to \( \frac{1}{3} \):
\[ 2\cos^2 2\theta - 1 = \frac{1}{3} \]

4. **Solve for \( \cos^2 2\theta \)**: Rearranging gives:
\[ 2\cos^2 2\theta = \frac{1}{3} + 1 = \frac{4}{3} \]
Thus:
\[ \cos^2 2\theta = \frac{2}{3} \]

5. **Find \( \cos 2\theta \)**: Taking the square root:
\[ \cos 2\theta = \pm \sqrt{\frac{2}{3}} \]

6. **Substitute Back into the Function**: Now substitute \( \cos 2\theta \) back into the expression for \( f \):
\[ f\left(\frac{1}{3}\right) = \frac{2 \cos^2 \theta}{2 \cos^2 \theta - 1} \]
We know \( \cos^2 2\theta = \frac{2}{3} \), so:
\[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \]
Thus:
\[ \cos^2 \theta = \frac{1 + \sqrt{\frac{2}{3}}}{2} \quad \text{or} \quad \cos^2 \theta = \frac{1 - \sqrt{\frac{2}{3}}}{2} \]

7. **Calculate \( f\left(\frac{1}{3}\right) \)**: Substitute \( \cos^2 \theta \) into the function:
\[ f\left(\frac{1}{3}\right) = \frac{2 \cdot \frac{1 + \sqrt{\frac{2}{3}}}{2}}{2 \cdot \frac{1 + \sqrt{\frac{2}{3}}}{2} - 1} \]
Simplifying gives:
\[ f\left(\frac{1}{3}\right) = \frac{1 + \sqrt{\frac{2}{3}}}{\sqrt{\frac{2}{3}}} \]
This can be further simplified to:
\[ f\left(\frac{1}{3}\right) = \frac{3 + \sqrt{6}}{2} \]

### Final Answer:
Thus, the value of \( f\left(\frac{1}{3}\right) \) is:
\[ f\left(\frac{1}{3}\right) = \frac{3 + \sqrt{6}}{2} \]