Let A = x∈R:x ≥1. The inverse of the function f:A→A given by f(x) = 2…

The correct answer is Option (2) → $\frac{1}{2}\left\{1+\sqrt{1+4\log_2x}\right\}$

It can be easily verified that f: A → A is a bijection.

Let $f(x) = y$. Then,

$f(x)=y$

$⇒2^{x (x − 1)} = y$

$⇒x (x-1) = \log_2 y$

$⇒x^2-x-\log_2 y = 0$

$x=\frac{1±\sqrt{1 + 4 \log_2 y}}{2}$

$⇒x=\frac{1}{2}\{1+\sqrt{1 + 4 \log_2 y}\}$   $[∵x>1]$

$⇒f^{-1}(y)=\frac{1}{2}\left\{1+\sqrt{1 + 4 \log_2 y}\right\}$

Hence $f^{-1}: A → A$ is given by

$f^{-1}(x)=\frac{1}{2}\left\{1+\sqrt{1 + 4 \log_2 y}\right\}$