Let f be a continuous and differentiable function in (a, b), if `f(x)`, `f'(x) ge x sqrt(1-(f(x))^(4))`and `lim_(x to a^(-))(f(x))^(2)=1` and `lim_(x to b^(-))(f(x))^(2)=(1)/(2)` . Then the minimum value of `[a^(2)-b^(2)]+80` is ________. ([.] denotes represent greatest integer function)

Answer

To solve the problem step by step, we will analyze the given conditions and derive the necessary inequalities.

### Step 1: Understand the given conditions
We have a function \( f \) that is continuous and differentiable on the interval \( (a, b) \). The conditions provided are:
1. \( f'(x) \geq x \sqrt{1 - (f(x))^4} \)
2. \( \lim_{x \to a^-} (f(x))^2 = 1 \)
3. \( \lim_{x \to b^-} (f(x))^2 = \frac{1}{2} \)

### Step 2: Analyze the limits
From the limits, we can conclude:
- At \( x = a \), \( f(a)^2 = 1 \) implies \( f(a) = 1 \) or \( f(a) = -1 \).
- At \( x = b \), \( f(b)^2 = \frac{1}{2} \) implies \( f(b) = \frac{1}{\sqrt{2}} \) or \( f(b) = -\frac{1}{\sqrt{2}} \).

### Step 3: Establish the inequality
Using the inequality \( f'(x) \geq x \sqrt{1 - (f(x))^4} \), we can integrate both sides. However, first, we need to express \( f'(x) \) in a more manageable form.

### Step 4: Integrate the inequality
We will consider the function \( g(x) = f(x)^2 \). Then, we have:
\[ g'(x) = 2f(x)f'(x) \]
From the inequality, we can derive:
\[ f'(x) \geq x \sqrt{1 - (f(x))^4} \]
Multiplying both sides by \( 2f(x) \) gives:
\[ 2f(x)f'(x) \geq 2x f(x) \sqrt{1 - (f(x))^4} \]
This implies:
\[ g'(x) \geq 2x f(x) \sqrt{1 - (f(x))^4} \]

### Step 5: Analyze the behavior of \( g(x) \)
Since \( g(a) = 1 \) and \( g(b) = \frac{1}{2} \), we can analyze the change in \( g(x) \) over the interval \( [a, b] \).

### Step 6: Establish the relationship between \( a^2 \) and \( b^2 \)
To find \( a^2 - b^2 \), we note that:
\[ g(a) - g(b) = 1 - \frac{1}{2} = \frac{1}{2} \]
This implies that the difference \( a^2 - b^2 \) must be at least \( \frac{1}{2} \).

### Step 7: Calculate the minimum value
We need to find the minimum value of \( [a^2 - b^2] + 80 \). Since \( a^2 - b^2 \geq \frac{1}{2} \), we can say:
\[ [a^2 - b^2] \geq 0 \]
Thus, the minimum value of \( [a^2 - b^2] + 80 \) is:
\[ 0 + 80 = 80 \]

### Step 8: Consider the greatest integer function
Since \( [a^2 - b^2] \) can take values starting from 0, the greatest integer function will yield:
\[ \lfloor 80 \rfloor = 80 \]

### Final Answer
The minimum value of \( [a^2 - b^2] + 80 \) is **80**.