Question

Fot the function $f\left(x\right)=\frac{e^x+1}{e^x-1}$, If $n\left(d\right)$ denotes the number of integers which are not in its domain and $n\left(r\right)$ denotes the number of integers which are not in its range,then $n\left(d\right)+n\left(r\right)$ is equal to

Answer 1

$f\left(x\right)=\frac{e^x+1}{e^x-1}$
For domain, $e^x-1\ne 0\Rightarrow x\ne 0$
$\therefore n\left(d\right)=\text{number of integers not in domain}=1$
Let $y=\frac{e^x+1}{e^x-1}$
$\Rightarrow e^{x}y-y=e^x+1$
$\Rightarrow e^x=\frac{y+1}{y-1}$

We know that $e^x>0$
$\Rightarrow \frac{y+1}{y-1}>0\Rightarrow y\in (-\infty ,-1)\cup (1,\infty )$

Hence the range is $(-\infty ,-1)\cup (1,\infty )$

So, $n\left(r\right)=$Number of integers which are not in range$=3\text{i.e,}\{-1,0,1\}$

$\therefore n\left(d\right)+n\left(r\right)=4$