Question

Consider a function $f:R-\left\{1\right\}\to R-\left\{1\right\}$ defined by $f\left(x\right)=\frac{x+a}{x-1},$ where $a$ is a real constant. If $f$ is not constant in its domain, then which of the following is/are CORRECT ?

• A. domain and range of $f$ are the same
• B. $\left(fof\right)\left(\frac{1}{x}\right)=\frac{1}{x}$
• C. $f\left(\frac{1}{f\left(f\right(x\left)\right)}\right)-f\left(f\left(f\left(\frac{1}{x}\right)\right)\right)=0$
• D. $f$ has its own inverse

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A domain and range of $f$ are the same
B $\left(fof\right)\left(\frac{1}{x}\right)=\frac{1}{x}$
C $f\left(\frac{1}{f\left(f\right(x\left)\right)}\right)-f\left(f\left(f\left(\frac{1}{x}\right)\right)\right)=0$
D $f$ has its own inverse

$y=f\left(x\right)=\frac{x+a}{x-1}$
Clearly, domain is $R-\left\{1\right\}$
$f^{\prime }\left(x\right)=\frac{(x-1)-(x+a)}{(x-1{)}^2}=-\frac{(a+1)}{(x-1{)}^2}$
Since, $f$ is not constant, hence $a\ne -1$.
Hence $f^{\prime }\left(x\right)$ is either always increasing or decreasing, thus one-one.

Now, $y=\frac{x+a}{x-1}$
$\Rightarrow yx-y=x+a\Rightarrow x(y-1)=y+a$
$\Rightarrow x=\frac{y+a}{y-1}$
Clearly, $x\in$ Domain$\left(f\right)$ for all $y\in R-\left\{1\right\}$
$\therefore$ Range $=R-\left\{1\right\}$ and thus $f$ is onto.
Therefore $f^{-1}$ is defined.
and $f^{-1}\left(x\right)=\frac{x+a}{x-1}$

We can see $f$ and $f^{-1}$ are same.
Now, $\left(fof\right)\left(\frac{1}{x}\right)=fo{f}^{-1}\left(\frac{1}{x}\right)=\frac{1}{x}$

We can write $f\left(\frac{1}{f\left(f\right(x\left)\right)}\right)-f\left(f\left(f\left(\frac{1}{x}\right)\right)\right)$ as
$f\left(\frac{1}{f\left(f^{-1}\right(x\left)\right)}\right)-f\left(f\left(f^{-1}\left(\frac{1}{x}\right)\right)\right)$
$=f\left(\frac{1}{x}\right)-f\left(\frac{1}{x}\right)=0[\because f(f^{-1}\left(y\right))=y]$