If `f(x)` is continuous on `[0,2]` , differentiable in `(0,2) ,f(0)=2, f(2)=8` and `f'(x) le 3` for all `x in (0,2)`, then find the value of `f(1)`.

Options

A. 4
B. 5
C. 6
D. 7

Explanation

To solve the problem, we will use the Mean Value Theorem (MVT) which states that if a function is continuous on a closed interval and differentiable on an open interval, then there exists at least one point in the open interval where the derivative of the function is equal to the average rate of change of the function over that interval.

### Step-by-Step Solution:

1. **Identify the given information:**
- The function \( f(x) \) is continuous on the interval \([0, 2]\).
- The function is differentiable on the interval \((0, 2)\).
- \( f(0) = 2 \)
- \( f(2) = 8 \)
- \( f'(x) \leq 3 \) for all \( x \in (0, 2) \)

2. **Apply the Mean Value Theorem on the interval \([0, 1]\):**
- According to MVT, there exists some \( c_1 \in (0, 1) \) such that:
\[ f'(c_1) = \frac{f(1) - f(0)}{1 - 0} \]
- Substituting the known values:
\[ f'(c_1) = f(1) - 2 \]
- Since \( f'(c_1) \leq 3 \), we have:
\[ f(1) - 2 \leq 3 \]
- Rearranging gives:
\[ f(1) \leq 5 \]

3. **Apply the Mean Value Theorem on the interval \([1, 2]\):**
- There exists some \( c_2 \in (1, 2) \) such that:
\[ f'(c_2) = \frac{f(2) - f(1)}{2 - 1} \]
- Substituting the known values:
\[ f'(c_2) = 8 - f(1) \]
- Since \( f'(c_2) \leq 3 \), we have:
\[ 8 - f(1) \leq 3 \]
- Rearranging gives:
\[ f(1) \geq 5 \]

4. **Combine the results:**
- From the first MVT application, we found \( f(1) \leq 5 \).
- From the second MVT application, we found \( f(1) \geq 5 \).
- Therefore, the only value that satisfies both inequalities is:
\[ f(1) = 5 \]

### Final Answer:
Thus, the value of \( f(1) \) is \( \boxed{5} \).