Let R be the set of all real numbers and f : R → Range f be given by $f(x)=3 x^2+1$. Then $f^{-1}\{1,2\}$ is:

$\left\{-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\}$ $\left\{-\frac{1}{\sqrt{3}}, 0, \frac{1}{\sqrt{3}}\right\}$ $\left[-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right]$ $\left[0, \frac{1}{\sqrt{3}}\right]$

$\left\{-\frac{1}{\sqrt{3}}, 0, \frac{1}{\sqrt{3}}\right\}$

The correct answer is Option (2) → $\left\{-\frac{1}{\sqrt{3}}, 0, \frac{1}{\sqrt{3}}\right\}$