CUET Preparation Today
If the function $f: R→R$ be defined by $f(x)=x^2-1,$ then $f^{-1}(8)$ is : |
{0, 3} {-1, 1} {-3, 3} {-2, 2} |
{-3, 3} |
The correct answer is Option (3) → {-3, 3} $y=x^2-1$ so $x=±\sqrt{y+1}$ at $y=8$ $f^{-1}(8)=x=±\sqrt{8+1}$ $f^{-1}(8)=±3$ $f^{-1}(8)∈\{-3, 3\}$ |
