Minimum value of `f(x)=cos^(2)x+(secx)/(4)`, `x in (-(pi)/(2),(pi)/(2))` is

Options

A. `3//2`
B. `3//4`
C. `3//8`
D. none of these

Explanation

To find the minimum value of the function \( f(x) = \cos^2 x + \frac{\sec x}{4} \) for \( x \) in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), we can use the concept of the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution

1. **Identify the Function**:
We start with the function given:
\[ f(x) = \cos^2 x + \frac{\sec x}{4} \]

2. **Rewrite Secant**:
Recall that \( \sec x = \frac{1}{\cos x} \). Thus, we can rewrite the function as:
\[ f(x) = \cos^2 x + \frac{1}{4 \cos x} \]

3. **Apply AM-GM Inequality**:
To apply the AM-GM inequality, we need to express the function in a suitable form. We can rewrite \( f(x) \) as:
\[ f(x) = \cos^2 x + \frac{\sec x}{4} + \frac{\sec x}{4} \]
This allows us to apply AM-GM to the three terms \( \cos^2 x, \frac{\sec x}{4}, \frac{\sec x}{4} \).

4. **Using AM-GM**:
According to the AM-GM inequality:
\[ \frac{\cos^2 x + \frac{\sec x}{4} + \frac{\sec x}{4}}{3} \geq \sqrt[3]{\cos^2 x \cdot \frac{\sec x}{4} \cdot \frac{\sec x}{4}} \]
Simplifying the right-hand side:
\[ \sqrt[3]{\cos^2 x \cdot \frac{1}{4 \cos x} \cdot \frac{1}{4 \cos x}} = \sqrt[3]{\frac{\cos^2 x}{16 \cos^2 x}} = \sqrt[3]{\frac{1}{16}} = \frac{1}{2} \]

5. **Combine Results**:
Thus, we have:
\[ \frac{\cos^2 x + \frac{\sec x}{4} + \frac{\sec x}{4}}{3} \geq \frac{1}{2} \]
Multiplying both sides by 3 gives:
\[ \cos^2 x + \frac{\sec x}{4} + \frac{\sec x}{4} \geq \frac{3}{2} \]

6. **Find Minimum Value**:
Therefore, we can conclude that:
\[ f(x) \geq \frac{3}{4} \]
The minimum value of \( f(x) \) occurs when \( \cos^2 x = \frac{1}{4} \) and \( \sec x = 1 \), which happens at \( x = \frac{\pi}{3} \).

7. **Final Result**:
Thus, the minimum value of \( f(x) \) is:
\[ \boxed{\frac{3}{4}} \]