If $f:[0,1] \to [0,1]$ be defined by $f(x) = \left\{ {\begin{array}{cccccccccccccccccccc}{x,}&{{\rm{ if \quad }}x{\rm{ \quad is \quad rational }}}\\{1 - x,}&{{\rm{ if \quad }}x{\rm{ \quad is \quad irrational }}}\end{array}} \right.$.
then $(fof)x$ is
If $f:[0,1] \to [0,1]$ be defined by $f(x) = \left\{ {\begin{array}{cccccccccccccccccccc}{x,}&{{\rm{ if \quad }}x{\rm{ \quad is \quad rational }}}\\{1 - x,}&{{\rm{ if \quad }}x{\rm{ \quad is \quad irrational }}}\end{array}} \right.$.
then $(fof)x$ is
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It is given that,, $f:[0,1] \to [0,1]$ be defined by
$f(x) = \left\{ {\begin{array}{cccccccccccccccccccc}{x,}&{{\rm{ if \quad }}x{\rm{ \quad is~~ rational }}}\\{1 - x,}&{{\rm{ if \quad }}x{\rm{ \quad is \quad irrational }}}\end{array}} \right.$
$\therefore$ $(fof)x = f(f(x)) = x$
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