Question

If $f\left(x\right)=\frac{1}{1-x}$ then the points of discontinuity of $\left(fff\right)\left(x\right)$ is

• A. $(0,1)$
• B. $\left(0,\pm 1\right)$
• C. $\left(1\right)$
• D. $\left(\pm 1\right)$

Answer 1

Answer: (A)

$(0,1)$

$\Rightarrow$$f\left(x\right)=\frac{1}{1-x}$

$\Rightarrow$$f\left(f\left(x\right)\right)=\frac{1}{1-f\left(x\right)}=\frac{1}{1-\frac{1}{1-x}}=\frac{1-x}{1-x-1}$

$\Rightarrow$$f\left(f\left(x\right)\right)=\frac{1-x}{-x}=\frac{x-1}{x}$

$\Rightarrow$$f\left(f\left(f\left(x\right)\right)\right)=\frac{1}{1-f\left(f\left(x\right)\right)}$

$=\frac{1}{1-\left(\frac{x-1}{x}\right)}=\frac{1.x}{x-x+1}$

$=\frac{x}{1}$

$f\left(f\left(f\left(x\right)\right)\right)=x$

Is continuous for all $x$.

$x=\left\{0,1\right\}$ is point of discontinuity.