If the function $\mathrm{f}:[1, \mathrm{\infty })\to [1, \mathrm{\infty })$ is defined by f(x) = 2^{x(x – 1)}, then

(1) one-one:

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=2^{{\mathrm{x}}^2-\mathrm{x}}\\ \Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)=2^{{\mathrm{x}}^2-\mathrm{x}}(2\mathrm{x}-1)\cdot {\log }_2\end{array}$

For f(x) to be one-one, it should be strictly increasing or strictly decreasing.

So, f ′(x) > 0

$\Rightarrow 2^{{\mathrm{x}}^2-\mathrm{x}}(2\mathrm{x}-1)>0$, where 2x^{2 – x} > 0 for all x

$\Rightarrow 2\mathrm{x}-1>0\text{ or }\mathrm{x}>\frac{1}{2}$

Thus, for given domain $[1, \mathrm{\infty })$, f(x) is always increasing. Hence, f is one-one

(2) onto: As f(x) is strictly increasing

$\begin{array}{l}\Rightarrow \text{ Range }\mathrm{f}\left(\mathrm{x}\right)\in \left[\mathrm{f}\right(1),\mathrm{f}(\mathrm{\infty }\left)\right)\\ \Rightarrow \text{ Range }\mathrm{f}\left(\mathrm{x}\right)\in [1,\mathrm{\infty })\\ \Rightarrow \text{ Range of }\mathrm{f}\left(\mathrm{x}\right)=\text{ Co-domain of }\mathrm{f}\left(\mathrm{x}\right),\text{ thus, }\mathrm{f}\text{ is onto. }\end{array}$

(3) Inverse:

As f is one-one and onto, f^–1 can be obtained.

Let y = f(x)

$\begin{array}{l}\Rightarrow \mathrm{y}=2{\mathrm{x}}^{2-\mathrm{x}}\\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}={\log }_2\mathrm{y}\\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}-{\log }_2\mathrm{y}=0\end{array}$

$\Rightarrow \mathrm{x}=\frac{1\pm \sqrt{1+4{\log }_2\mathrm{y}}}{2}$

$\Rightarrow {\mathrm{f}}^{-1}\left(\mathrm{y}\right)=\frac{1+\sqrt{1+4{\log }_2\mathrm{y}}}{2}$                    $[\mathrm{as}\mathrm{y}>0, \mathrm{\forall } \mathrm{x}\in \mathrm{D}]$