If $f : [0, 1] \to [0, 1]$ be defined by $f(x) = \begin{cases} x, & \text{if } x \text{ is rational} \\ 1 - x, & \text{if } x \text{ is irrational} \end{cases}$

$x$

The correct answer is Option (3) → $x$ ##

Given that, $f : [0, 1] \to [0, 1]$ be defined by

$f(x) = \begin{cases} x, & \text{if } x \text{ is rational} \\ 1 - x, & \text{if } x \text{ is irrational} \end{cases}$

$∴(f of)x = f(f(x)) = x$