The function $f$ satisfies $f(x)+3f(1-x)=x^2$ for every real $x$. If $S=\{x \mid f(x)=0 \}$, which one is true?

$\textbf{(A)}$ $S$ is an infinite set.

$\textbf{(B)}$ $\{0,1\} \subset S$

$\textbf{(C)}$ $S=\phi$

$\textbf{(D)}$ $S = \{(3+\sqrt 3)/2, (3-\sqrt 3)/2\}$

$\textbf{(E)}$ None of above
1. Given the functional equation:
   \[
   f(x) + 3f(1-x) = x^2
   \]
   We need to find the function \( f(x) \).

2. To solve for \( f(x) \), we substitute \( x \) with \( 1-x \) in the given equation:
   \[
   f(1-x) + 3f(x) = (1-x)^2
   \]
   Simplifying the right-hand side:
   \[
   f(1-x) + 3f(x) = 1 - 2x + x^2
   \]

3. Now we have a system of linear equations:
   \[
   \begin{cases}
   f(x) + 3f(1-x) = x^2 \\
   f(1-x) + 3f(x) = 1 - 2x + x^2
   \end{cases}
   \]

4. Let's solve this system. Multiply the first equation by 3:
   \[
   3f(x) + 9f(1-x) = 3x^2
   \]

5. Subtract the second equation from the modified first equation:
   \[
   (3f(x) + 9f(1-x)) - (f(1-x) + 3f(x)) = 3x^2 - (1 - 2x + x^2)
   \]
   Simplify:
   \[
   8f(1-x) = 2x^2 + 2x - 1
   \]
   \[
   f(1-x) = \frac{1}{8}(2x^2 + 2x - 1)
   \]

6. Substitute \( f(1-x) \) back into the first equation:
   \[
   f(x) + 3\left(\frac{1}{8}(2x^2 + 2x - 1)\right) = x^2
   \]
   Simplify:
   \[
   f(x) + \frac{3}{8}(2x^2 + 2x - 1) = x^2
   \]
   \[
   f(x) + \frac{3}{4}x^2 + \frac{3}{4}x - \frac{3}{8} = x^2
   \]
   \[
   f(x) = x^2 - \frac{3}{4}x^2 - \frac{3}{4}x + \frac{3}{8}
   \]
   \[
   f(x) = \frac{1}{4}x^2 - \frac{3}{4}x + \frac{3}{8}
   \]

7. To find the set \( S \) where \( f(x) = 0 \):
   \[
   \frac{1}{4}x^2 - \frac{3}{4}x + \frac{3}{8} = 0
   \]
   Multiply through by 4 to clear the fraction:
   \[
   x^2 - 3x + 3 = 0
   \]

8. Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \frac{3 \pm \sqrt{9 - 12}}{2}
   \]
   \[
   x = \frac{3 \pm \sqrt{-3}}{2}
   \]
   Since the discriminant is negative, there are no real roots.

9. Therefore, \( f(x) = 0 \) has no real solutions, implying \( S = \emptyset \).

The final answer is \( \boxed{S = \emptyset} \)