If the function f: R→R be defined by f(x)=x^2-1, then f^-1(8) is :

The correct answer is Option (3) → {-3, 3}

$y=x^2-1$ so $x=±\sqrt{y+1}$

at $y=8$

$f^{-1}(8)=x=±\sqrt{8+1}$

$f^{-1}(8)=±3$

$f^{-1}(8)∈\{-3, 3\}$