Question

If $f\left(x\right)=\frac{x}{x-1},x\ne 1$, then ${\left(fofo...of\right)}_{\left(19times\right)}\left(x\right)$ is equal to

• A. $\frac{x}{\left(x-1\right)}$
• B. ${\left[\frac{x}{x-1}\right]}^{19}$
• C. $\frac{19x}{x-1}$
• D. $x$

Answer 1

The correct option is A

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

Explanation for the correct option:

Step 1: Find $fof\left(x\right)$

Given that,

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

Now,

$\begin{array}{rcl}fof\left(x\right)& =& f\left(f\left(x\right)\right)\\ & =& f\left(\frac{x}{x-1}\right)\\ & =& \frac{{\displaystyle \frac{x}{x-1}}}{{\displaystyle \frac{x}{x-1}}-1}\\ & =& \frac{{\displaystyle \frac{x}{x-1}}}{{\displaystyle \frac{x-x+1}{x-1}}}\\ & =& x\end{array}$

Step 2: Find $f\left(f\left(f\left(x\right)\right)\right)$:

$\begin{array}{rcl}f\left(f\left(f\left(x\right)\right)\right)& =& f\left(x\right)\left[\mathrm{from}1\right]\\ & =& \frac{x}{x-1}\end{array}$

Step 3: Find the value of ${\left(fofo...of\right)}_{\left(19times\right)}\left(x\right)$:

$\begin{array}{rcl}{\left(fofo...of\right)}_{\left(19times\right)}\left(x\right)& =& f\left(x\right)\left[\mathrm{from}1\right]\\ & =& \frac{x}{x-1}\end{array}$

Hence, the correct option is A.