Suppose `f: RvecR^+` be a differentiable function such that `3f(x+y)=f(x)f(y)AAx ,y in R` with `f(1)=6.` Then the value of `f(2)` is `6` b. `9` c. `12` d. `15`

Options

A. 6
B. 9
C. 12
D. 15

Explanation

To solve the problem, we need to find the value of \( f(2) \) given the functional equation \( 3f(x+y) = f(x)f(y) \) for all \( x, y \in \mathbb{R} \) and the condition \( f(1) = 6 \).

### Step-by-Step Solution:

1. **Substituting Values**:
Let's first substitute \( x = 0 \) and \( y = 1 \) into the functional equation:
\[ 3f(0 + 1) = f(0)f(1) \]
This simplifies to:
\[ 3f(1) = f(0)f(1) \]

2. **Using the Given Value**:
We know that \( f(1) = 6 \):
\[ 3 \cdot 6 = f(0) \cdot 6 \]
This simplifies to:
\[ 18 = 6f(0) \]
Dividing both sides by 6 gives:
\[ f(0) = 3 \]

3. **Finding \( f(2) \)**:
Next, we substitute \( x = 1 \) and \( y = 1 \) into the functional equation:
\[ 3f(1 + 1) = f(1)f(1) \]
This simplifies to:
\[ 3f(2) = f(1) \cdot f(1) = 6 \cdot 6 = 36 \]

4. **Solving for \( f(2) \)**:
Now we can solve for \( f(2) \):
\[ f(2) = \frac{36}{3} = 12 \]

Thus, the value of \( f(2) \) is \( 12 \).

### Final Answer:
The value of \( f(2) \) is \( 12 \).