If a function $f: R_{+} → [-5, \infty)$ given by $f(x)=9 x^2+6 x-5$ is one-one and onto, then $f^{-1}(y)$ is _______.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) - $\frac{\sqrt{(y+6)}-1}{3}$
$f(x)=9 x^2+6 x-5$
$y=(3x+1)^2-6$
$\sqrt{y+6}=3x+1$
$\frac{\sqrt{y+6}-1}{3}=x⇒f^{-1}(y)=\frac{\sqrt{(y+6)}-1}{3}$