Question

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

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

Answer 1

The correct option is C $(0,1]$

$\frac{x^2+e}{x^2+1}>0\forall x\in R$
So, domain of $f$ is $R$

${\log }_e\left(\frac{x^2+e}{x^2+1}\right)$
$={\log }_e\left(\frac{x^2+1+e-1}{x^2+1}\right)$
$={\log }_e(1+\frac{e-1}{x^2+1})$

Now, $0\le x^2<\infty$
$\Rightarrow 1\le x^2+1<\infty$
$\Rightarrow 0<\frac{1}{x^2+1}\le 1$
$\Rightarrow 0<\frac{e-1}{x^2+1}\le e-1$
$\Rightarrow 1<1+\frac{e-1}{x^2+1}\le e$

Since, ${\log }_{e}x$ is strictly increasing function.
$\therefore {\log }_{e}1<{\log }_e(1+\frac{e-1}{x^2+1})\le {\log }_{e}e$
$\Rightarrow 0<{\log }_e(1+\frac{e-1}{x^2+1})\le 1$

Hence, $R\left(f\right)=(0,1]$