If $F:[1,∞) → [2, ∞)$ is given by $f(x) = x +\frac{1}{x}$, then $f^{-1}(x)$ equals.

$\frac{x+\sqrt{x^2-4}}{2}$

The correct answer is Option (2) → $\frac{x+\sqrt{x^2-4}}{2}$

Clearly, $f: [1, ∞) → [2, ∞)$ is a bijection.

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

$⇒x +\frac{1}{x}=y$

$⇒x^2-xy+1=0$

$⇒x=\frac{y±\sqrt{y^2-4}}{2}$

$⇒x=\frac{y+\sqrt{y^2-4}}{2}$    $[∵x≥1]$

$⇒f^{-1}(y)=\frac{y+\sqrt{y^2-4}}{2}$

Hence, $f^{-1}(x)=\frac{x+\sqrt{x^2-4}}{2}$ for all $x∈[1, ∞)$