Let `f(x)` be continuous functions `f: RvecR` satisfying `f(0)=1a n df(2x)-f(x)=xdot` Then the value of `f(3)` is `2` b. `3` c. `4` d. 5

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To solve the problem, we need to find the function \( f(x) \) that satisfies the given conditions and then evaluate \( f(3) \).

### Step-by-Step Solution:

1. **Given Conditions**:
- We know that \( f(0) = 1 \).
- The functional equation is given by:
\[ f(2x) - f(x) = x \]

2. **Substituting \( x \) with \( \frac{x}{2} \)**:
- By substituting \( x \) with \( \frac{x}{2} \) in the functional equation, we get:
\[ f(x) - f\left(\frac{x}{2}\right) = \frac{x}{2} \]

3. **Substituting \( x \) with \( \frac{x}{4} \)**:
- Now, substitute \( x \) with \( \frac{x}{4} \):
\[ f\left(\frac{x}{2}\right) - f\left(\frac{x}{4}\right) = \frac{x}{4} \]

4. **Continuing the Pattern**:
- Continuing this process, we can write a series of equations:
- \( f(2x) - f(x) = x \)
- \( f(x) - f\left(\frac{x}{2}\right) = \frac{x}{2} \)
- \( f\left(\frac{x}{2}\right) - f\left(\frac{x}{4}\right) = \frac{x}{4} \)
- And so on...

5. **Summing the Series**:
- If we sum these equations, we notice that all intermediate terms cancel out:
\[ f(2x) - f\left(\frac{x}{2^n}\right) = x + \frac{x}{2} + \frac{x}{4} + \ldots + \frac{x}{2^{n-1}} \]
- The right-hand side is a geometric series which sums to:
\[ \text{Sum} = x \left(1 - \frac{1}{2^n}\right) \div \left(1 - \frac{1}{2}\right) = x \left(1 - \frac{1}{2^n}\right) \cdot 2 \]

6. **Taking the Limit as \( n \to \infty \)**:
- As \( n \) approaches infinity, \( f\left(\frac{x}{2^n}\right) \) approaches \( f(0) \):
\[ f(2x) - f(0) = 2x \]
- Since \( f(0) = 1 \):
\[ f(2x) - 1 = 2x \quad \Rightarrow \quad f(2x) = 2x + 1 \]

7. **Finding \( f(x) \)**:
- Now, we can express \( f(x) \) in terms of \( x \):
- Let \( x = \frac{x}{2} \) (i.e., \( x = 1 \)):
\[ f(x) = x + 1 \]

8. **Finding \( f(3) \)**:
- Finally, we substitute \( x = 3 \):
\[ f(3) = 3 + 1 = 4 \]

### Conclusion:
The value of \( f(3) \) is \( 4 \).