Question

If $f\left(x\right)=\frac{1}{1-x}$, 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

Answer 1

(b) {0, 1}

Given: $f\left(x\right)=\frac{1}{1-x}$

Clearly, $f:R-\left\{1\right\}\to R$

Now,
$f\left(f\left(x\right)\right)=f\left(\frac{1}{1-x}\right)=\left(\frac{1}{1-{\displaystyle \left(\frac{1}{1-x}\right)}}\right)=\left(\frac{1-x}{-x}\right)=\left(\frac{x-1}{x}\right)$
∴ $fof:$$R-\left\{0,1\right\}\to R$

Now,

$f\left(f\left(f\left(x\right)\right)\right)=f\left(\frac{x-1}{x}\right)=\left(\frac{1}{1-\left({\displaystyle \frac{x-1}{x}}\right)}\right)=x$

∴ $fofof:$$R-\left\{0,1\right\}\to R$

Thus, ​$f\left(f\left(f\left(x\right)\right)\right)$ is not defined at $x=0,1$.

Hence, ​$f\left(f\left(f\left(x\right)\right)\right)$ is discontinuous at {0, 1}.