Let `f:R rarr R` be a continuous function such that
`f(x)-2f(x/2)+f(x/4)=x^(2)`.
The equation f(x)-x-f(0)=0 have exactly

A. no solution
B. one solution
C. two solution
D. infinite solution

Options

A. no solution
B. one solution
C. two solution
D. infinite solution

Answer

To solve the given problem, we need to analyze the functional equation provided and find the nature of the solutions to the equation \( f(x) - x - f(0) = 0 \).

### Step 1: Analyze the functional equation

The functional equation is given as:

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

This is a second-order difference equation. We will look for a polynomial solution of the form \( f(x) = ax^2 + bx + c \).

### Step 2: Substitute the polynomial into the functional equation

Substituting \( f(x) = ax^2 + bx + c \) into the functional equation:

1. Calculate \( f\left(\frac{x}{2}\right) \):
\[ f\left(\frac{x}{2}\right) = a\left(\frac{x}{2}\right)^2 + b\left(\frac{x}{2}\right) + c = \frac{ax^2}{4} + \frac{bx}{2} + c \]

2. Calculate \( f\left(\frac{x}{4}\right) \):
\[ f\left(\frac{x}{4}\right) = a\left(\frac{x}{4}\right)^2 + b\left(\frac{x}{4}\right) + c = \frac{ax^2}{16} + \frac{bx}{4} + c \]

Now substitute these into the functional equation:

\[ f(x) - 2f\left(\frac{x}{2}\right) + f\left(\frac{x}{4}\right) = ax^2 + bx + c - 2\left(\frac{ax^2}{4} + \frac{bx}{2} + c\right) + \left(\frac{ax^2}{16} + \frac{bx}{4} + c\right) \]

### Step 3: Simplify the equation

Now simplify the left-hand side:

\[ = ax^2 + bx + c - \left(\frac{ax^2}{2} + bx + 2c\right) + \left(\frac{ax^2}{16} + \frac{bx}{4} + c\right) \]

Combine like terms:

\[ = \left(a - \frac{a}{2} + \frac{a}{16}\right)x^2 + \left(b - b + \frac{b}{4}\right)x + (c - 2c + c) \]

This simplifies to:

\[ \left(a - \frac{a}{2} + \frac{a}{16}\right)x^2 + \frac{b}{4}x + 0 \]

### Step 4: Set coefficients equal to \( x^2 \)

We need this to equal \( x^2 \):

1. Coefficient of \( x^2 \):
\[ a - \frac{a}{2} + \frac{a}{16} = 1 \]

2. Coefficient of \( x \):
\[ \frac{b}{4} = 0 \implies b = 0 \]

3. Constant term:
\[ 0 = 0 \text{ (which is always true)} \]

### Step 5: Solve for \( a \)

Now, solve the equation for \( a \):

\[ a - \frac{a}{2} + \frac{a}{16} = 1 \]

Multiply through by 16 to eliminate the fraction:

\[ 16a - 8a + a = 16 \implies 9a = 16 \implies a = \frac{16}{9} \]

### Step 6: Write the function

Thus, we have:

\[ f(x) = \frac{16}{9}x^2 + c \quad \text{(where \( c \) is a constant)} \]

### Step 7: Analyze the equation \( f(x) - x - f(0) = 0 \)

Now, substitute \( f(0) \):

\[ f(0) = c \]

The equation becomes:

\[ \frac{16}{9}x^2 + c - x - c = 0 \implies \frac{16}{9}x^2 - x = 0 \]

Factoring out \( x \):

\[ x\left(\frac{16}{9}x - 1\right) = 0 \]

### Step 8: Find the solutions

This gives us:

1. \( x = 0 \)
2. \( \frac{16}{9}x - 1 = 0 \implies x = \frac{9}{16} \)

Thus, there are exactly **two solutions**.

### Conclusion

The answer to the question is **C. two solutions**.

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