Question

The range of $f\left(x\right)$ $=\ln \left(\frac{x^2+e}{x^2+1}\right)$ is

• A. $(0,1)$
• B. $(1,e]$
• C. $(0,1]$
• D. $(1,e)$

Answer 1

Answer: (C)

$(0,1]$

Given,

$f\left(x\right)=\ln \left(\frac{x^2+e}{x^2+1}\right)$

Now,

$\frac{x^2+e}{x^2+1}=\frac{x^2+1-1+e}{x^2+1}=(\frac{x^2+1}{x^2+1}+\frac{e-1}{x^2+1})$

$=1+\frac{e-1}{x^2+1}$

$\therefore \frac{x^2+e}{x^2+1}=1+\frac{e-1}{x^2+1}$
and $1\le x^2+1<\infty \Rightarrow 1\ge \frac{1}{x^2+1}>0\Rightarrow e-1\ge \frac{e-1}{x^2+1}>0$
$\Rightarrow e\ge 1+\frac{e-1}{x^2+1}>1\Rightarrow 1\ge \ln \left(\frac{x^2+e}{x^2+1}\right)>0$

Hence, answer is option $C$