For every twice differentiable function f(x) the value of `|f(x)| le 3 AA x in R` and for some `alpha`
`f(alpha) + (f'(alpha))^(2) =80.` Number of integral values that `(f'(x))^(2)` can take between (0,77) are equal to `"________"`

Answer

To solve the problem, we need to analyze the given conditions and derive the possible values for \((f'(x))^2\).

### Step 1: Understand the given conditions
We are given that for every twice differentiable function \(f(x)\), the absolute value of \(f(x)\) is bounded:
\[ |f(x)| \leq 3 \quad \text{for all } x \in \mathbb{R} \]
This implies:
\[ -3 \leq f(x) \leq 3 \]

### Step 2: Analyze the equation at \(x = \alpha\)
We also have the condition:
\[ f(\alpha) + (f'(\alpha))^2 = 80 \]
From this, we can express \((f'(\alpha))^2\) in terms of \(f(\alpha)\):
\[ (f'(\alpha))^2 = 80 - f(\alpha) \]

### Step 3: Determine the bounds for \((f'(\alpha))^2\)
Since \(-3 \leq f(\alpha) \leq 3\), we can substitute these bounds into the equation for \((f'(\alpha))^2\):
1. If \(f(\alpha) = 3\):
\[ (f'(\alpha))^2 = 80 - 3 = 77 \]
2. If \(f(\alpha) = -3\):
\[ (f'(\alpha))^2 = 80 - (-3) = 80 + 3 = 83 \]

Thus, we have:
\[ 77 \leq (f'(\alpha))^2 < 83 \]

### Step 4: Identify the range of \((f'(x))^2\)
The values of \((f'(x))^2\) can take values in the interval:
\[ (0, 77) \quad \text{and} \quad (83, \infty) \]
However, we are only interested in the values between \(0\) and \(77\).

### Step 5: Count the integral values in the interval
The integral values in the interval \((0, 77)\) are:
\[ 1, 2, 3, \ldots, 76 \]
This gives us a total of \(76\) integral values.

### Final Answer
The number of integral values that \((f'(x))^2\) can take between \(0\) and \(77\) is:
\[ \boxed{76} \]