Question

lf $f\left(x\right)=\frac{1}{1-x}$ then the points of discontinuity of $\left(fofof\right)\left(x\right)$ is:

• A. $\{0,1\}$
• B. $\{0,\pm 1\}$
• C. {1}
• D. $\{\pm 1\}$

Answer 1

The correct option is A $\{0,1\}$

Given $f\left(x\right)=\frac{1}{1-x}$
$\left(fof\right)\left(x\right)=\frac{1}{1-\frac{1}{1-x}}$
$=\frac{x-1}{x}$
$\left(fofof\right)\left(x\right)=\frac{\frac{1}{1-x}-1}{\frac{1}{1-x}}=x$
Now, $\left(fofof\right)\left(x\right)=x$ is in itself everywhere continuous.
However, $\left(fof\right)\left(x\right)$ is discontinuous at $x=0$ and $f\left(x\right)$ is discontinuous at $x=1$ as their denominator become 0 at these points.