Question

For 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 a $n\left(r\right)$ denotes the number of integers which are not in its range, then $n\left(d\right)+n\left(r\right)$ is equal to

• A. $2$
• B. $3$
• C. $4$
• D. Infinite

Answer 1

The correct option is A $2$

$f\left(x\right)=\frac{e^x+1}{e^x-1}$

$\therefore n\left(d\right)=f$

As $e^x-1=0$

$\therefore e^x=1$

$\therefore x=0$

$\therefore$ for $x=0$ function is not variable

Also, Let $f\left(4\right)=1$

$\frac{e^x+1}{e^x-1}=1$

$e^x+1=e^x-1$

$\therefore 1=-1$

Not possible

$\therefore n\left(d\right)=1$

$n\left(r\right)=1$

$\therefore n\left(d\right)+n\left(r\right)=2$