If $f(x)=\frac{1}{1-x}$, then the set of points discontinuity of the function $f(f(f(x)))$ is
(a) $\{1\}$
(b) $\{0,1\}$
(c) $\{-1,1\}$
(d) none of these
(b) $\{0,1\}$
Given: $f(x)=\frac{1}{1-x}$
Clearly, $f: R-\{1\} \rightarrow R$
Now,
$f(f(x))=f\left(\frac{1}{1-x}\right)=\left(\frac{1}{1-\left(\frac{1}{1-x}\right)}\right)=\left(\frac{1-x}{-x}\right)=\left(\frac{x-1}{x}\right)$
$\therefore$ fof $: R-\{0,1\} \rightarrow R$
Now,
$f(f(f(x)))=f\left(\frac{x-1}{x}\right)=\left(\frac{1}{1-\left(\frac{x-1}{x}\right)}\right)=x$
$\therefore$ fofof $: R-\{0,1\} \rightarrow R$
Thus, $f(f(f(x)))$ is not defined at $x=0,1$.
Hence, $f(f(f(x)))$ is discontinuous at $\{0,1\}$.
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