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 given problem step by step, we will analyze the conditions provided and derive the minimum value of \(x_1^2 - x_2^2\).

### Step 1: Understand the given inequalities and limits
We have the inequality:
\[ f(x) f'(x) \geq x \sqrt{1 - (f(x))^4} \]
and the limits:
\[ \lim_{x \to x_1} (f(x))^2 = 1 \quad \text{and} \quad \lim_{x \to x_2} (f(x))^2 = \frac{1}{2}. \]

### Step 2: Rewrite the inequality
We can rewrite the inequality as:
\[ 2f(x) f'(x) \geq 2x \sqrt{1 - (f(x))^4}. \]
This suggests that we can analyze the function \(g(x) = \sin^{-1}(f(x)^2) - x^2\).

### Step 3: Differentiate \(g(x)\)
Differentiating \(g(x)\):
\[ g'(x) = \frac{2f(x)f'(x)}{\sqrt{1 - (f(x))^4}} - 2x. \]
From the given inequality, we know that \(g'(x) \geq 0\), which implies that \(g(x)\) is a non-decreasing function.

### Step 4: Evaluate limits at \(x_1\) and \(x_2\)
Using the limits provided:
\[ \lim_{x \to x_1} g(x) = \sin^{-1}(1) - x_1^2 = \frac{\pi}{2} - x_1^2, \]
\[ \lim_{x \to x_2} g(x) = \sin^{-1}\left(\frac{1}{2}\right) - x_2^2 = \frac{\pi}{6} - x_2^2. \]

### Step 5: Establish the relationship between \(x_1\) and \(x_2\)
Since \(g(x)\) is non-decreasing:
\[ \frac{\pi}{2} - x_1^2 \leq \frac{\pi}{6} - x_2^2. \]
Rearranging gives:
\[ x_1^2 - x_2^2 \geq \frac{\pi}{2} - \frac{\pi}{6}. \]
Calculating the right-hand side:
\[ \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}. \]

### Step 6: Conclusion
Thus, the minimum value of \(x_1^2 - x_2^2\) is:
\[ \boxed{\frac{\pi}{3}}. \]