If $f(x)=\left\{\begin{matrix}x\,when\,x\,is\,rational\\1-x\,when\,x\,is\,irrational\end{matrix}\right.$ then fof(x) is given as

1 $x$ $1+x$ none of these

$x$

$fof=\left\{\begin{matrix}f(x)\,when\,f(x)\,is\,rational\\1-f(x)\,when\,f(x)\,is\,irrational\end{matrix}\right.$ $=\left\{\begin{matrix}x\,when\,f(x)\,is\,rational\\1-(1-x)\,when\,f(x)\,is\,irrational\end{matrix}\right.=x$ That is, it fof(x) = x.