Question

Given the function $f\left(x\right)=\frac{1}{1-x},$ the number of points of discontinuity of the composite function $y=f^{3n}\left(x\right),where{f}^n\left(x\right)=fofof...\left(ntimes\right)\left(nϵN\right)are0,1.$

• A. $x$
• B. $\frac{1}{x}$
• C. $\frac{-1}{x}$
• D. $\frac{1}{1-x}$

Answer 1

The correct option is A $x$

Clearly, $x=1$ is a point of discontinuity of the function $f\left(x\right)=\frac{1}{1-x},$
Now $f\left[f\left(x\right)\right]=f\left[\frac{1}{1-x}\right]=\frac{1}{1-\frac{1}{1-x}}=\frac{1-x}{-x}=\frac{x-1}{x}$
Hence, $x=0$ is a point of discontinuity of function $\left(fof\right)x$.
Again $\left(fofof\right)x=fo\left[fof\right]x=f\left(\frac{x-1}{x}\right)$ $=\frac{1}{1-\frac{x-1}{x}}=x$
$\therefore f^3\left(x\right)=x.$
Above is continuous every where $\therefore y=f^{3n}\left(x\right)={\left\{f^3\left(x\right)\right\}}^n=x$ which is continuous everywhere.
Ans: A