Solve this
Question:

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

Solution:

(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 $\{0,1\}$.