Let `f:R->R` be a function such that `f((x+y)/3)=(f(x)+f(y))/3 ,f(0) = 0 and f'(0)=3` ,then

Options

A. a quadratic function
B. continuous but not differerntiable
C. differerntiable in R
D. bounded in R

Explanation

To solve the problem, we need to analyze the given functional equation and the conditions provided.

### Step-by-Step Solution:

1. **Given Functional Equation**:
We start with the functional equation:
\[ f\left(\frac{x+y}{3}\right) = \frac{f(x) + f(y)}{3} \]

2. **Substituting Values**:
Let's substitute \( x = 0 \) and \( y = 0 \) into the functional equation:
\[ f\left(\frac{0+0}{3}\right) = \frac{f(0) + f(0)}{3} \]
This simplifies to:
\[ f(0) = \frac{2f(0)}{3} \]
Since \( f(0) = 0 \) (given), this equation holds true.

3. **Using the Derivative Condition**:
We know that \( f'(0) = 3 \). This means that the function is differentiable at \( x = 0 \) and the slope of the tangent line at this point is 3.

4. **Assuming a Linear Form**:
Given the nature of the functional equation, we can assume that \( f(x) \) is a linear function of the form:
\[ f(x) = kx \]
where \( k \) is a constant.

5. **Finding the Constant \( k \)**:
We know from the derivative condition that:
\[ f'(x) = k \]
Therefore, from \( f'(0) = 3 \), we have:
\[ k = 3 \]
Thus, we can express the function as:
\[ f(x) = 3x \]

6. **Verifying the Functional Equation**:
We need to verify if \( f(x) = 3x \) satisfies the original functional equation:
\[ f\left(\frac{x+y}{3}\right) = 3\left(\frac{x+y}{3}\right) = x + y \]
And,
\[ \frac{f(x) + f(y)}{3} = \frac{3x + 3y}{3} = x + y \]
Both sides are equal, confirming that \( f(x) = 3x \) is indeed a solution.

7. **Analyzing the Options**:
- **Option 1**: Quadratic function - **False** (since \( f(x) = 3x \) is linear).
- **Option 2**: Continuous but not differentiable - **False** (since \( f(x) = 3x \) is differentiable).
- **Option 3**: Differentiable in \( \mathbb{R} \) - **True** (as \( f(x) = 3x \) is differentiable everywhere).
- **Option 4**: Bounded in \( \mathbb{R} \) - **False** (since \( f(x) = 3x \) is unbounded).

Thus, the correct answer is that the function \( f(x) \) is differentiable in \( \mathbb{R} \).