Question

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

• A. ${\left(\frac{1}{2}\right)}^{x(x-1)}$
• B. $\frac{1}{2}(1+\sqrt{1+4lo{g}_{2}x})$
• C. $\frac{1}{2}(1-\sqrt{1+4lo{g}_{2}x})$
• D. not defined

Answer 1

The correct option is B $\frac{1}{2}(1+\sqrt{1+4lo{g}_{2}x})$

Let $y=2^{x(x-1)}$, where $y\ge 1$ as $x\ge 1$
Taking $lo{g}_2$ on both sides, we get
$lo{g}_{2}y=lo{g}_2{2}^{x(x-1)}$
$\Rightarrow lo{g}_{2}y=x(x-1)$
$\Rightarrow x^2-x-lo{g}_{2}y=0$
$\Rightarrow x=\frac{1\pm \sqrt{1+4lo{g}_{2}y}}{2}$
For $y\ge 1,lo{g}_{2}y\ge 0\Rightarrow 4lo{g}_{2}y\ge 0$ $\Rightarrow 1+4lo{g}_{2}y\ge 1$
$\Rightarrow \sqrt{1+4lo{g}_{2}y}\ge 1$
$\Rightarrow -\sqrt{1+4lo{g}_{2}y}\le -1$
$\Rightarrow 1-\sqrt{1+4lo{g}_{2}y}\le 0$
But $x\ge 1$
So, $x=1-\sqrt{1+4lo{g}_{2}y}$ is not possible.
Therefore, we take $x=\frac{1}{2}(1+\sqrt{(1+4lo{g}_{2}y)})$
$\Rightarrow f^{-1}\left(y\right)=\frac{1}{2}(1+\sqrt{1+4lo{g}_{2}y})$
$\Rightarrow f^{-1}\left(x\right)=\frac{1}{2}(1+\sqrt{1+4lo{g}_{2}y})$