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# Let $f :[0,1]\rightarrow[0,1]$ be defined by$$f(x)= \left \{ \begin{array}{1 1}\;x,if\;x\;is\;rational & \;\\1-x,if\;x\;is\;irrational & \;\end{array} \right.$$ Then $(f o f)x$ is $$\begin{array}{1 1}(A)\;constant & (B)\;1+x\\(C)\;x & (D)\;none\;of\;these\end{array}$$

Can you answer this question?

Toolbox:
• $fof(x)=f(f(x))$
$f:[0,1] \to [0,1]$

$f(x)= \left\{ \begin{array}{1 1} x & \quad if\;x\;is\;rational \\ 1-x & \quad if\;x\;is\;irrational \end{array} \right.$

$fof(x)-f(f(x))$

when x is rational

$=f(x)$

=x

when x is irrational

$fof(x)=f(1-x)$

$=1-(1-x)$

$=x$

Hence $(fof) (x)=x$

'c' option is correct

answered Mar 5, 2013 by