Question

If the function $f:[1,\infty )\to [1,\infty )$ 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. $\frac{1}{2}(1+\sqrt{1+4{\log }_{2}x})$
• C. $\frac{1}{2}(1-\sqrt{1+4{\log }_{2}x})$
• D. both b and c

Answer 1

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

Let $f\left(x\right)=y$
$y=2^{x(x-1)}\Rightarrow x(x-1)={\log }_{2}y\Rightarrow x^2-x-{\log }_{2}y=0\Rightarrow x=\frac{1\pm \sqrt{1+4{\log }_{2}y}}{2}$
Only $x=\frac{1+\sqrt{1+4{\log }_{2}y}}{2}$ lies in the domain.
$\therefore f^{-1}\left(x\right)=\frac{1}{2}(1+\sqrt{1+4{\log }_{2}x})$