If $f : [0, 1] \to [0, 1]$ be defined by $f(x) = \begin{cases} x, & \text{if } x \text{ is rational} \\ 1 - x, & \text{if } x \text{ is irrational} \end{cases}$, then $f(f(x))$ is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $x$ ##
Given that, $f : [0, 1] \to [0, 1]$ be defined by
$f(x) = \begin{cases} x, & \text{if } x \text{ is rational} \\ 1 - x, & \text{if } x \text{ is irrational} \end{cases}$
$∴(f of)x = f(f(x)) = x$