Let f be a continuous and differentiable function in `(x_(1),x_(2))`. If `f(x).f'(x)ge x sqrt(1-(f(x))^(4))` and `lim_(xrarrx_(1))(f(x))^(2)=1 and lim_(xrarrx) )(f(x))^(2)=(1)/(2)`, then minimum value of `(x_(1)^(2)-x_(2)^(2))` is

Options

A. `(pi)/(6)`
B. `(2pi)/(3)`
C. `(pi)/(3)`
D. none of these

Explanation

To solve the problem, we need to analyze the given conditions and derive the minimum value of \( x_1^2 - x_2^2 \).

### Step-by-Step Solution:

1. **Understanding the Given Inequality**:
We start with the inequality:
\[ f(x) \cdot f'(x) \geq x \sqrt{1 - (f(x))^4} \]
This implies that the product of the function and its derivative is bounded below by a term involving \( x \) and \( f(x) \).

2. **Limits of the Function**:
We have the limits:
\[ \lim_{x \to x_1} (f(x))^2 = 1 \quad \text{and} \quad \lim_{x \to x_2} (f(x))^2 = \frac{1}{2} \]
This indicates that as \( x \) approaches \( x_1 \), \( f(x) \) approaches \( 1 \), and as \( x \) approaches \( x_2 \), \( f(x) \) approaches \( \frac{1}{\sqrt{2}} \).

3. **Deriving the Function Behavior**:
From the limits, we can deduce:
\[ f(x_1) = 1 \quad \text{and} \quad f(x_2) = \frac{1}{\sqrt{2}} \]

4. **Using the Inequality**:
Substitute \( f(x_1) \) and \( f(x_2) \) into the inequality:
- At \( x = x_1 \):
\[ 1 \cdot f'(x_1) \geq x_1 \sqrt{1 - 1^4} = 0 \]
This does not provide new information since \( f'(x_1) \) can be any non-negative value.
- At \( x = x_2 \):
\[ \frac{1}{\sqrt{2}} \cdot f'(x_2) \geq x_2 \sqrt{1 - \left(\frac{1}{\sqrt{2}}\right)^4} = x_2 \sqrt{1 - \frac{1}{4}} = x_2 \cdot \frac{\sqrt{3}}{2} \]

5. **Rearranging the Inequality**:
Rearranging gives:
\[ f'(x_2) \geq \frac{x_2 \sqrt{3}}{2} \cdot \frac{\sqrt{2}}{1} = \frac{x_2 \sqrt{6}}{2} \]

6. **Finding Minimum Value**:
Now, we need to find the minimum value of \( x_1^2 - x_2^2 \). We can use the limits and the derived inequalities:
\[ \lim_{x \to x_1} f(x)^2 = 1 \implies x_1^2 = \frac{\pi}{2} \quad \text{and} \quad \lim_{x \to x_2} f(x)^2 = \frac{1}{2} \implies x_2^2 = \frac{\pi}{6} \]

7. **Calculating the Difference**:
Therefore, we have:
\[ x_1^2 - x_2^2 = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Final Answer:
The minimum value of \( x_1^2 - x_2^2 \) is:
\[ \frac{\pi}{3} \]