Question

If $f\left(x\right)={\log }_e(x^2+1)-e^{-x}+1\forall x\in R,$ then which of the following is CORRECT?

• A. $f\left(x\right)$ is an increasing function
• B. $f\left(x\right)$ is strictly increasing function
• C. $f\left(x\right)$ is decreasing function
• D. $f\left(x\right)$ is strictly decreasing function

Answer 1

The correct option is B $f\left(x\right)$ is strictly increasing function

$f\left(x\right)={\log }_e(x^2+1)-e^{-x}+1$
$\Rightarrow f^{\prime }\left(x\right)=\frac{2x}{1+x^2}+e^{-x}\Rightarrow f^{\prime }\left(x\right)=e^{-x}+\frac{2}{x+\left(\frac{1}{x}\right)}$
$f^{\prime }\left(0\right)=1$
For $x>0,f^{\prime }\left(x\right)>0$
$\text{For}x<0,-1<\frac{2}{x+\left(\frac{1}{x}\right)}<0\text{and}{e}^{-x}>1$
Hence, $(\frac{2x}{1+x^2}+e^{-x})>0$
$\Rightarrow f\left(x\right)$ is a strictly increasing function $\forall x\in R$