Question

If $f\left(x\right)=\frac{1}{1-x}$, then the points of discontinuity of the function $f^{3n}\left(x\right)$, where $f^n\left(x\right)=fofof...of(n$ times $)$, are

• A. $x=2$
• B. $x=0$
• C. $x=1$
• D. continuous everywhere

Answer 1

Answer: (B), (C)

The correct options are
B $x=0$
C $x=1$

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

$\left(fofof\right)\left(x\right)=f\left[fof\right]=f\left(\frac{x-1}{x}\right)x,$

which is continuous everywhere.

Hence, $f^{3n}\left(x\right)={\left(fofof\right)}^n\left(x\right)=x,$ which is continuous everywhere.

So, the only points of discontinuity are $x=0$ and $x=1.$