Consider a differentiable `f:R to R` for which `f(1)=2 and f(x+y)=2^(x)f(y)+4^(y)f(x) AA x , y in R.`
The value of f(4) is

Options

A. 160
B. 240
C. 200
D. None of these

Explanation

To solve the problem, we need to find the value of \( f(4) \) given the functional equation and the initial condition. Let's go through the solution step by step.

### Step 1: Write down the functional equation
We are given:
\[ f(x+y) = 2^x f(y) + 4^y f(x) \]
for all \( x, y \in \mathbb{R} \), and we know that \( f(1) = 2 \).

### Step 2: Substitute specific values for \( x \) and \( y \)
Let's first substitute \( x = 1 \) and \( y = 1 \):
\[ f(1+1) = 2^1 f(1) + 4^1 f(1) \]
This simplifies to:
\[ f(2) = 2 \cdot 2 + 4 \cdot 2 = 4 + 8 = 12 \]

### Step 3: Find \( f(0) \)
Next, substitute \( x = 0 \) and \( y = 0 \):
\[ f(0+0) = 2^0 f(0) + 4^0 f(0) \]
This simplifies to:
\[ f(0) = 1 \cdot f(0) + 1 \cdot f(0) \implies f(0) = 2f(0) \]
This implies that \( f(0) = 0 \).

### Step 4: Substitute \( y = 2 \) and \( x = 2 \)
Now, substitute \( x = 2 \) and \( y = 2 \):
\[ f(2+2) = 2^2 f(2) + 4^2 f(2) \]
This simplifies to:
\[ f(4) = 4 \cdot 12 + 16 \cdot 12 = 48 + 192 = 240 \]

### Conclusion
Thus, the value of \( f(4) \) is:
\[ \boxed{240} \]