If `f(x)` is a differentiable function satisfying `f^(')(x)lt2` for all `xepsilonR` and `f(1)=2,` then greatest possible integral value of `f(3)` is

Explanation

To solve the problem, we need to find the greatest possible integral value of \( f(3) \) given the conditions on the function \( f(x) \).

### Step-by-Step Solution:

1. **Understand the Given Information:**
We know that \( f(x) \) is a differentiable function, which implies it is continuous. We also have the condition that \( f'(x) < 2 \) for all \( x \in \mathbb{R} \) and that \( f(1) = 2 \).

2. **Apply the Mean Value Theorem:**
The Mean Value Theorem states that if \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one \( c \in (a, b) \) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
In our case, let \( a = 1 \) and \( b = 3 \). Therefore, we can write:
\[ f'(c) = \frac{f(3) - f(1)}{3 - 1} \]

3. **Substitute Known Values:**
We know \( f(1) = 2 \). Substituting this into the equation gives:
\[ f'(c) = \frac{f(3) - 2}{2} \]

4. **Use the Condition on the Derivative:**
Since we are given that \( f'(x) < 2 \) for all \( x \), we can substitute this into our equation:
\[ \frac{f(3) - 2}{2} < 2 \]

5. **Solve the Inequality:**
Multiplying both sides of the inequality by 2 gives:
\[ f(3) - 2 < 4 \]
Adding 2 to both sides results in:
\[ f(3) < 6 \]

6. **Determine the Greatest Possible Integral Value:**
The greatest integer less than 6 is 5. Therefore, the greatest possible integral value of \( f(3) \) is:
\[ \boxed{5} \]