If for all real values of a one root of the equation `x^2-3ax + f(a)=0` is double of the other, then f(x) is equal to

Options

A. `2x`
B. `x^(2)`
C. `2x^(2)`
D. `2sqrt(x)`

Explanation

To solve the problem, we need to analyze the given quadratic equation and the conditions provided.

### Step-by-Step Solution:

1. **Understanding the Roots**:
We have the quadratic equation \( x^2 - 3ax + f(a) = 0 \). We are told that one root is double the other. Let’s denote the roots as \( t \) and \( 2t \).

2. **Sum of the Roots**:
According to Vieta's formulas, the sum of the roots \( t + 2t \) should equal \( -\frac{b}{a} \). Here, \( b = -3a \) and \( a = 1 \).
\[ t + 2t = 3t = -\frac{-3a}{1} = 3a \]
Thus, we have:
\[ 3t = 3a \implies t = a \]

3. **Product of the Roots**:
The product of the roots \( t \cdot 2t \) should equal \( \frac{c}{a} \). Here, \( c = f(a) \) and \( a = 1 \).
\[ t \cdot 2t = 2t^2 = f(a) \]
Substituting \( t = a \):
\[ 2(a^2) = f(a) \]
Therefore, we can express \( f(a) \) as:
\[ f(a) = 2a^2 \]

4. **Generalizing for \( f(x) \)**:
Since the problem asks for \( f(x) \), we can replace \( a \) with \( x \) in our derived expression:
\[ f(x) = 2x^2 \]

### Final Answer:
Thus, the function \( f(x) \) is:
\[ f(x) = 2x^2 \]