Question

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

• A. ${\left(\frac{1}{2}\right)}^{x(x-1)}$
• B. $\frac{1}{2}\left(1-\sqrt{1+4{\log }_2\left(x\right)}\right)$
• C. $\frac{1}{2}\left(\sqrt{1+4{\log }_2\left(x\right)}\right)$
• D. $\frac{1}{2}\left(1+\sqrt{1+4{\log }_2\left(x\right)}\right)$
• E. Not defined

Answer 1

The correct option is D

$\frac{1}{2}\left(1+\sqrt{1+4{\log }_2\left(x\right)}\right)$

Explanation for the correct option:

Finding the value of $f^{-1}\left(x\right)$:

Given that $f\left(x\right)=2^{\left(x\right(x-1\left)\right)}$

Say $y=2^{\left(x\right(x-1\left)\right)}$

Applying ${\log }_2$ on both sides, we get,

$\begin{array}{crcll}& {\log }_2\left(y\right)& =& {\log }_2{\left(2\right)}^{x\left(x-1\right)}& \\ & & =& x\left(x-1\right){\log }_2\left(2\right)& \left[\because \log \left(m^n\right)=n\log \left(m\right)\right]\\ & & =& x\left(x-1\right)& \\ \Rightarrow & {\log }_2\left(y\right)& =& x^2-x& \\ \Rightarrow & x^2-x-{\log }_2\left(y\right)& =& 0& \\ \Rightarrow & x& =& \frac{1\pm \sqrt{1-4\left(1\right)\left(-{\log }_2\left(y\right)\right)}}{2}& \left[\because \forall a{x}^2+bx+c=0;x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right]\\ \Rightarrow & f^{-1}\left(y\right)& =& \frac{1\pm \sqrt{1+4{\log }_2\left(y\right)}}{2}& \left[\because f\left(x\right)=y\right]\\ \Rightarrow & f^{-1}\left(x\right)& =& \frac{1\pm \sqrt{1+4{\log }_2\left(x\right)}}{2}& \\ \Rightarrow & f^{-1}\left(x\right)& =& \frac{1+\sqrt{1+4{\log }_2\left(x\right)}}{2}& \left[\because f:[1,\infty )\to [1,\infty )\right]\end{array}$

Hence, the correct option is option(D) i.e. $\frac{1}{2}\left(1+\sqrt{1+4{\log }_2\left(x\right)}\right)$