Question

Let $A=\left\{x\in R:x\ge 1\right\}$. The inverse of the function, $f:A\to A$ given by $f\left(x\right)=2^{x\left(x-1\right)},\mathrm{is}$

(a) ${\left(\frac{1}{2}\right)}^{x\left(x-1\right)}$

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

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

(d) not defined

Answer 1

$$\begin{gathered} \text{Let}{f}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow f\left(y\right)=x \\ \Rightarrow 2^{y\left(y-1\right)}=x \\ \Rightarrow 2^{y^2-y}=x \\ \Rightarrow y^2-y={\log }_{2}x \\ \Rightarrow y^2-y+\frac{1}{4}={\log }_{2}x+\frac{1}{4} \\ \Rightarrow {\left(y-\frac{1}{2}\right)}^2=\frac{4{\log }_{2}x+1}{4} \\ \Rightarrow y-\frac{1}{2}=\pm \frac{\sqrt{4{\log }_{2}x+1}}{2} \\ \Rightarrow y=\frac{1}{2}\pm \frac{\sqrt{4{\log }_{2}x+1}}{2} \\ \Rightarrow y=\frac{1}{2}+\frac{\sqrt{4{\log }_{2}x+1}}{2}\left(\text{∵ y ≥1}\right) \\ \text{So,}{f}^{-1}\left(x\right)=\frac{1}{2}(1+\sqrt{1+4{\log }_{2}x})\text{[from}\left(1\right)\text{]} \end{gathered}$$
So, the answer is (b).