For every twice differentiable function f(x) the value of `|f(x)| ge 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 given problem, we will follow these steps:

### Step 1: Analyze the given conditions
We know that for every twice differentiable function \( f(x) \), the condition \( |f(x)| \geq 3 \) holds for all \( x \in \mathbb{R} \). This implies that:
- \( f(x) \geq 3 \) or \( f(x) \leq -3 \).

### Step 2: Use the provided equation
We are given that for some \( \alpha \):
\[ f(\alpha) + (f'(\alpha))^2 = 80 \]
From this, we can rearrange to express \( (f'(\alpha))^2 \):
\[ (f'(\alpha))^2 = 80 - f(\alpha) \]

### Step 3: Consider the implications of the modulus condition
Since \( |f(\alpha)| \geq 3 \), we can analyze two cases based on the sign of \( f(\alpha) \):

1. **Case 1:** \( f(\alpha) \geq 3 \)
- In this case, \( f(\alpha) \) can take values starting from 3 upwards. Thus:
\[ (f'(\alpha))^2 = 80 - f(\alpha) \leq 80 - 3 = 77 \]

2. **Case 2:** \( f(\alpha) \leq -3 \)
- Here, \( f(\alpha) \) can take values starting from \(-3\) downwards. Thus:
\[ (f'(\alpha))^2 = 80 - f(\alpha) \geq 80 + 3 = 83 \]

### Step 4: Combine the results
From the two cases, we have:
- From Case 1: \( (f'(\alpha))^2 \leq 77 \)
- From Case 2: \( (f'(\alpha))^2 \geq 83 \)

This means \( (f'(x))^2 \) can take values in the intervals:
- \( (0, 77] \) (from Case 1)
- \( [83, \infty) \) (from Case 2)

### Step 5: Count the integral values in the specified range
We need to find the integral values that \( (f'(x))^2 \) can take between \( 0 \) and \( 77 \):
- The integral values from \( 0 \) to \( 77 \) are \( 1, 2, 3, \ldots, 76 \).
- The total number of integral values is \( 76 \) (as \( 1 \) to \( 76 \) inclusive).

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