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:

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

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

$y=3x^2+1⇒±\sqrt{\frac{y-1}{3}}=x$

at $y=1,x=0$

$y=2, x=±\sqrt{\frac{2-1}{3}}=±\frac{1}{\sqrt{3}}$

so $f^{-1}\{1,2\}=\left\{-\frac{1}{\sqrt{3}}, 0, \frac{1}{\sqrt{3}}\right\}$