Question

Consider a function $f:x\to \frac{x+a}{x-1}:x\in R-\left\{1\right\}$where $a$ is a real constant, If $f$ is not a constant function answer the following $f\left(\frac{1}{f\left(f\right(x\left)\right)}\right)-f\left(f\left(f\left(\frac{1}{x}\right)\right)\right)$

• A. $x$
• B. $x+1$
• C. $0$
• D. $2x$

Answer 1

The correct option is C $0$

$f\left(f\right(x\left)\right)$ $=\frac{\frac{x+a}{x-1}+a}{\frac{x+a}{x-1}-1}$
$=\frac{x+a+ax-a}{x+a-x+1}$
$=\frac{x(a+1)}{a+1}$ $=x$. ...(i)
$f\left(\frac{1}{x}\right)$ $=\frac{1+ax}{1-x}$
Hence, $f\left(f\right(\frac{1}{x}\left)\right)$
$=\frac{\frac{1+ax}{1-x}+a}{\frac{1+ax}{1-x}-1}$
$=\frac{1+ax+a-ax}{1+ax-1+x}$
$=\frac{1+a}{x(1+a)}$ $=\frac{1}{x}$.
Hence, $f\left(\frac{1}{f\left(f\right(x\left)\right)}\right)-f\left(f\right(f\left(\frac{1}{x}\right)\left)\right)$
$=f\left(\frac{1}{x}\right)-f\left(\frac{1}{x}\right)$
$=0$.