Question

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