Answer: $\mathbf C$
Explanation:
$2f(x) - 3f(\frac{1}{x}) = x^2\tag{1}\;\;\text{[Given]}$
Substitute, $x = \frac{1}{x}$, we get:
$2f(\frac{1}{x}) - 3f(x) = \frac{1}{x^2}\tag{2}$
Multiplying Equation $(1)$ by $3$, we get:
$6f(x) - 9f(\frac{1}{x}) = 3x^2\tag{3}$
Multiplying Equation $(2)$ by $2$, we get:
$4f(\frac{1}{x}) - 6f(x) = \frac{2}{x^2}\implies -6f(x) + 4f(\frac{1}{x}) = \frac{2}{x^2}\tag{4}$
Adding Equations $(3)$ and $(4)$, we get:
$f(\frac{1}{x}) = -\frac{1}{5}\left(3x^2 + \frac{2}{x^2}\right)$
Now, on putting $x = \frac{1}{2}$ in above equation, we get:
$x = -\frac{7}{4}$
$\therefore \mathbf C$ is the correct option.