Question

Let $Q$ be the set of all rational numbers in $[0,1]$ and $f:[0,1] \rightarrow[0,1]$ be defined by
$f(x)=\begin{cases}x & \text { for } x \in Q \\ 1-x & \text { for } x \notin Q\end{cases}$
Then, the set $S=\{x \in[0,21]:(f \circ f)(x)\}$ is equal to

• A. [0,1]
• B. -Q
• C. [0,1]-Q
• D. (0,1)

Answer 1

Answer: (A)

Given, $f(x)=\begin{cases}x & \text { for } x \in Q \\ 1-x & \text { for } x \notin Q\end{cases}$ is defined for
$f:[0,1] \rightarrow[0,1]$
If $x$ is rational, then
$ f(x) =x $
$\therefore f(f(x)) =f(x)=x$
If $x$ is irrational, then
$f(x)=1-x$
$\therefore $ fof $ (x)=f(1-x)=1-(1-x)$
$=X$
$\therefore $ fof $(x)=x$ is possible for all values of domain