Question

Given the function $f\left(x\right)=\frac{1}{1-x},$ the number of point(s) of discontinuity of the composite function $y=f^{3n}\left(x\right),$ where, $f^n\left(x\right)=fof\cdots of\left(n\text{times}\right)\left(x\right),(n\in N)$ is

• A. $2$
• B. $2n$
• C. $3n$
• D. $1$

Answer 1

The correct option is A $2$

​​​​​$f\left(x\right)=\frac{1}{1-x}\Rightarrow f^2\left(x\right)=fof\left(x\right)=\frac{1}{1-f\left(x\right)}\Rightarrow f^2\left(x\right)=\frac{1}{1-\left(\frac{1}{1-x}\right)}=\frac{x-1}{x}\Rightarrow f^3\left(x\right)=fofof\left(x\right)=\frac{1}{1-f^2\left(x\right)}\Rightarrow f^3\left(x\right)=\frac{1}{1-\left(\frac{x-1}{x}\right)}=x\Rightarrow f^4\left(x\right)=\frac{1}{1-f^3\left(x\right)}\Rightarrow f^4\left(x\right)=\frac{1}{1-x}=f\left(x\right)$
So, it can be observed that
$f^{3n}\left(x\right)=x$
For $f^3\left(x\right)$ is discontinuous for $x=0,1$
Similarly, for $f^{3n}\left(x\right)=x$
$x=0,1$ are points of discontinuity.