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 number of solutions of `f(x)=2` is

Options

A. 0
B. 1
C. 2
D. infinite

Explanation

To solve the problem, we start with the functional equation given:

1. **Given Information**:
- We have a function \( f : \mathbb{R} \to \mathbb{R} \) such that \( f(1) = 2 \).
- The functional equation is:
\[ f(x+y) = 2^x f(y) + 4^y f(x) \quad \forall x, y \in \mathbb{R} \]

2. **Substituting Values**:
- Let's first substitute \( y = 0 \) into the functional equation:
\[ f(x+0) = 2^x f(0) + 4^0 f(x) \]
This simplifies to:
\[ f(x) = 2^x f(0) + f(x) \]
From this, we can deduce:
\[ 0 = 2^x f(0) \quad \forall x \in \mathbb{R} \]
This implies \( f(0) = 0 \).

3. **Rearranging the Functional Equation**:
- Now, we can rewrite the original equation:
\[ f(x+y) - 4^y f(x) = 2^x f(y) \]
- We can also substitute \( x \) with \( 1 \) and \( y \) with \( 1 \):
\[ f(1+1) = 2^1 f(1) + 4^1 f(1) \]
This gives:
\[ f(2) = 2 \cdot 2 + 4 \cdot 2 = 4 + 8 = 12 \]

4. **Finding a General Form for \( f(x) \)**:
- Let's assume a general form for \( f(x) \):
\[ f(x) = k(4^x - 2^x) \]
- We need to find \( k \) using the condition \( f(1) = 2 \):
\[ f(1) = k(4^1 - 2^1) = k(4 - 2) = 2k \]
Setting this equal to 2 gives:
\[ 2k = 2 \implies k = 1 \]
- Therefore, the function simplifies to:
\[ f(x) = 4^x - 2^x \]

5. **Finding the Number of Solutions for \( f(x) = 2 \)**:
- We need to solve:
\[ 4^x - 2^x = 2 \]
- Rewriting \( 4^x \) as \( (2^x)^2 \):
\[ (2^x)^2 - 2^x - 2 = 0 \]
- Let \( u = 2^x \). The equation becomes:
\[ u^2 - u - 2 = 0 \]
- Factoring gives:
\[ (u - 2)(u + 1) = 0 \]
- Thus, \( u = 2 \) or \( u = -1 \). Since \( u = 2^x \) must be positive, we discard \( u = -1 \).
- Therefore:
\[ 2^x = 2 \implies x = 1 \]

6. **Conclusion**:
- The only solution to \( f(x) = 2 \) is \( x = 1 \).
- Thus, the number of solutions of \( f(x) = 2 \) is **1**.