CUET Preparation Today
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 _______. |
$\frac{\sqrt{y+1}+6}{3}$ $\frac{\sqrt{y+5}+1}{3}$ $\frac{\sqrt{(y+6)}+1}{3}$ $\frac{\sqrt{(y+6)}-1}{3}$ |
$\frac{\sqrt{(y+6)}-1}{3}$ |
The correct answer is Option (4) - $\frac{\sqrt{(y+6)}-1}{3}$ $f(x)=9 x^2+6 x-5$ $y=(3x+1)^2-6$ $\sqrt{y+6}=3x+1$ $\frac{\sqrt{y+6}-1}{3}=x⇒f^{-1}(y)=\frac{\sqrt{(y+6)}-1}{3}$ |
