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. $\left(\frac{1}{2}\right)(1+\sqrt{1+4{\log }_{2}x})$
• C. $\left(\frac{1}{2}\right)(1-\sqrt{1+4{\log }_{2}x})$
• D. Not defined

Answer 1

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

$f\left(x\right)=2^{x(x+1)},$ i.e., $y=2^{x(x+1)},$

$\Rightarrow {\log }_{2}y=x(x-1)=x^2-x$

$\Rightarrow x^2-x-{\log }_{2}y=0$

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

Since $x\in [1,\infty )$ $\therefore$ -ve sign is ruled out.

$\therefore \Rightarrow x=\frac{1}{2}(1+\sqrt{1+4{\log }_{2}y})$

$\Rightarrow f^{-1}\left(x\right)=\frac{1}{2}(1+\sqrt{1+4{\log }_{2}y})$

$\Rightarrow f^{-1}\left(x\right)=\frac{1}{2}(1+\sqrt{1+4{\log }_{2}x})$