Q. If the function $f:\lbrack 1,\ \infty) \rightarrow \lbrack 1,\ \infty)$ is defined by $f(x) = 2^{x(x - 1)},\$then $f^{- 1}(x)$ is
A) $\left( \frac{1}{2} \right)^{x(x - 1)}$
B) $\frac{1}{2}(1 + \sqrt{1 + 4\log_{2}y})$
C) $\frac{1}{2}(1 - \sqrt{1 + 4\log_{2}y})$
D) $\infty$
Explanation
(b)
Let $y = f(x) = 2^{x(x - 1)}$
$\Rightarrow \ \ \ \ \ \ \ \log_{2}{y = x^{2} - x \Rightarrow x^{2} - x - \log_{2}{y = 0}}$
$\Rightarrow \ \ \ \ \ \ x = \frac{1 \pm \sqrt{1 + 4\log_{2}y}}{2} = \frac{1 + \sqrt{1 + 4\log_{2}y}}{2}$
$\ \ \ \ \ \ \ \ \ \ \ \left\lbrack \begin{array}{r}
\because x = \frac{1 - \sqrt{1 + 4\left( x^{2} - x \right)}}{2} = \frac{1 - (2x - 1)}{2} \\
< 0\ domain\ is\ not\ defined
\end{array} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.