Question

If $f\left(x\right)=\frac{1}{1-x}$ , then the points of discontinuity of the function $f^{30}\left(x\right)$ where $f^n\left(x\right)=fof......of$ ( $n$ times) are

• A. $x=2,1$
• B. $x=0,1$
• C. $x=1,2$
• D. no points of discontinuity

Answer 1

Answer: (B)

Clearly, $x=1$ is a point of discontinuity of the function $f\left(x\right)=\frac{1}{1-x}$ .
if $x\ne 1$ , then
$\left(fof\right)\left(x\right)=f\left[f\left(x\right)\right]=f\left(\frac{1}{1-x}\right)=\frac{x-1}{x},$ which is discontinuous at $x=0$ .
If $x\ne 0$ and $x\ne 1$ , then
$\left(fofof\right)\left(x\right)=f\left[\left(fof\right)\left(x\right)\right]=f\left(\frac{x-1}{x}\right)=x$
Which is continuous everywhere.
Hence, $f^{30}\left(x\right)=x,$ which is continuous everywhere.
So, the only points of discontinuity are $x=0$ and $x=1$