Q. If $f:R \rightarrow R$ is defined by $f(x) = x^{2} - 6x - 14$, then $f^{- 1}(2)$ equals to
A) {2, 8}
B) {-2, 8}
C) {-2, -8}
D) $\{\phi\}$
Explanation
(d)
Let $y = x^{2} - 6x - 14\ \ \Rightarrow \ \ y = (x - 3)^{2} - 23$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \pm \sqrt{y + 23} + 3$
$\Rightarrow \ \ \ \ \ \ \ f^{- 1}(x) = \pm \sqrt{x + 23} + 3$
$\therefore\ \ \ \ \ \ \ f^{- 1}(2) = \pm \sqrt{25} + 3 = - 2,\ 8$
It means we do not define a inverse function
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ f^{- 1}(2) = \phi$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.