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 $x\neq 1$ , then
$\left(f o f\right)\left(x\right)=f\left[f \left(x\right)\right]=f\left(\frac{1}{1 - x}\right)=\frac{x - 1}{x},$ which is discontinuous at $x=0$ .
If $x\neq 0$ and $x\neq 1$ , then
$\left(f o f o f\right)\left(x\right)=f\left[\left(f o f\right) \left(x\right)\right]=f\left(\frac{x - 1}{x}\right)=x$
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 $x=1$