Let `f: R rarrR` be a differentiable function satisfying f (1) = 2 . If `alpha ,beta` are real numbers satisfying `alpha^(2) + beta^(2)=1` then `f(alphax).f(betax)=f(x)` for all `x inR` Then the value of `f(0)+f'(0)` is _______.

Answer

To solve the problem, we need to find the value of \( f(0) + f'(0) \) given the conditions of the function \( f \).

### Step 1: Analyze the given functional equation
We are given that \( f(\alpha x) \cdot f(\beta x) = f(x) \) for all \( x \in \mathbb{R} \) where \( \alpha^2 + \beta^2 = 1 \).

### Step 2: Substitute \( x = 0 \)
Substituting \( x = 0 \) into the functional equation:
\[ f(\alpha \cdot 0) \cdot f(\beta \cdot 0) = f(0) \implies f(0) \cdot f(0) = f(0) \implies f(0)^2 = f(0). \]
This implies that \( f(0) \) can be either \( 0 \) or \( 1 \).

### Step 3: Consider the case \( f(0) = 0 \)
If \( f(0) = 0 \), then substituting back into the functional equation gives:
\[ f(\alpha x) \cdot f(\beta x) = f(x). \]
For \( x = 1 \):
\[ f(\alpha) \cdot f(\beta) = f(1) = 2. \]
This suggests that both \( f(\alpha) \) and \( f(\beta) \) would need to be non-zero, which contradicts the assumption that \( f(0) = 0 \). Hence, \( f(0) \) cannot be \( 0 \).

### Step 4: Conclude \( f(0) = 1 \)
Thus, we conclude that \( f(0) = 1 \).

### Step 5: Differentiate the functional equation
Now, we differentiate the functional equation \( f(\alpha x) \cdot f(\beta x) = f(x) \) with respect to \( x \):
Using the product rule:
\[ \frac{d}{dx}[f(\alpha x) \cdot f(\beta x)] = f'(\alpha x) \cdot \alpha \cdot f(\beta x) + f(\alpha x) \cdot f'(\beta x) \cdot \beta = f'(x). \]

### Step 6: Substitute \( x = 0 \) in the differentiated equation
Substituting \( x = 0 \):
\[ f'(\alpha \cdot 0) \cdot \alpha \cdot f(\beta \cdot 0) + f(0) \cdot f'(\beta \cdot 0) \cdot \beta = f'(0). \]
This simplifies to:
\[ f'(0) \cdot \alpha \cdot 1 + 1 \cdot f'(0) \cdot \beta = f'(0). \]
Thus:
\[ f'(0)(\alpha + \beta) = f'(0). \]

### Step 7: Analyze the equation
From \( \alpha^2 + \beta^2 = 1 \) and \( \alpha + \beta = 1 \), we can conclude that \( f'(0) \) must equal \( 0 \) because if \( f'(0) \neq 0 \), we would have \( \alpha + \beta = 1 \) leading to a contradiction.

### Step 8: Final calculation
Thus, we have:
\[ f(0) = 1 \quad \text{and} \quad f'(0) = 0. \]
Therefore:
\[ f(0) + f'(0) = 1 + 0 = 1. \]

### Final Answer
The value of \( f(0) + f'(0) \) is \( \boxed{1} \).