Let `f:R rarr R` be a continuous function such that
`f(x)-2f(x/2)+f(x/4)=x^(2)`.
f'(0) is equal to

Options

A. 0
B. 1
C. f(0)
D. `-f(0)`

Answer

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

\[ f(x) - 2f\left(\frac{x}{2}\right) + f\left(\frac{x}{4}\right) = x^2. \]

### Step 1: Substitute \( x \) with \( \frac{x}{2} \)

We will replace \( x \) with \( \frac{x}{2} \) in the original equation:

\[ f\left(\frac{x}{2}\right) - 2f\left(\frac{x}{4}\right) + f\left(\frac{x}{8}\right) = \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}. \]

### Step 2: Substitute \( x \) with \( \frac{x}{4} \)

Next, we replace \( x \) with \( \frac{x}{4} \):

\[ f\left(\frac{x}{4}\right) - 2f\left(\frac{x}{8}\right) + f\left(\frac{x}{16}\right) = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}. \]

### Step 3: Set up a system of equations

Now we have three equations:

1. \( f(x) - 2f\left(\frac{x}{2}\right) + f\left(\frac{x}{4}\right) = x^2 \) (Equation 1)
2. \( f\left(\frac{x}{2}\right) - 2f\left(\frac{x}{4}\right) + f\left(\frac{x}{8}\right) = \frac{x^2}{4} \) (Equation 2)
3. \( f\left(\frac{x}{4}\right) - 2f\left(\frac{x}{8}\right) + f\left(\frac{x}{16}\right) = \frac{x^2}{16} \) (Equation 3)

### Step 4: Analyze the pattern

From the equations, we can observe that they form a pattern. If we continue this process, we can derive a general form:

\[ f\left(\frac{x}{2^n}\right) - 2f\left(\frac{x}{2^{n+1}}\right) + f\left(\frac{x}{2^{n+2}}\right) = \frac{x^2}{4^n}. \]

### Step 5: Summing the equations

If we sum these equations as \( n \) approaches infinity, we can find a relationship between \( f(x) \) and \( x^2 \). The left-hand side will converge to \( f(x) \) as the terms will cancel out, leading us to:

\[ f(x) = \frac{4}{3} x^2 + C, \]

where \( C \) is a constant.

### Step 6: Find \( f'(x) \)

Now, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}\left(\frac{4}{3} x^2 + C\right) = \frac{8}{3} x. \]

### Step 7: Evaluate \( f'(0) \)

To find \( f'(0) \):

\[ f'(0) = \frac{8}{3} \cdot 0 = 0. \]

Thus, the value of \( f'(0) \) is:

\[ \boxed{0}. \]