$$\begin{gathered} \begin{array}{l}LetQbethesetofallrationalnumbersin[0,1]and \\ f:[0,1]\to [0,1]bedefinedby\\ f\left(x\right)=\{\begin{array}{l}xforx\in Q\\ 1-xfor x\notin Q\end{array} \\ Thentheset S=\{x\in [0,1]:\left(fof\right)\left(x\right)\}isequalto\\ \end{array} \end{gathered}$$

$\begin{array}{l}f\left(x\right)=\{\begin{array}{l}xforx\in Q\\ 1-xforx\notin Q\end{array}\\ Ifx\in Qthenf\left(x\right)=x\\ fof\left(x\right)=f\left(f\left(x\right)\right)=f\left(x\right)=x\\ Ifx\notin Qthenf\left(x\right)=1-x\\ fof\left(x\right)=f\left(f\left(x\right)\right)=f(1-x)=1-(1-x)=1-1+x=x\\ fof\left(x\right)=\{\begin{array}{l}xwherex\in Q,x\in [0,1]\\ xwherex\notin Q,x\in [0,1]\end{array}\\ \therefore S=[0,1]\end{array}$