Question

Let $f:R\to R$ and $g:R\to R$ be two functions defined by $f\left(x\right)={\log }_e(x^2+1)-e^{-x}+1$ and $g\left(x\right)=\frac{1-2e^{2x}}{e^x}$. Then, for which of the following range of $\alpha$, the inequality $f\left(g\left(\frac{(\alpha -1{)}^2}{3}\right)\right)>f\left(g(\alpha -\frac{5}{3})\right)$ holds?

• A. $(-2,-1)$
• B. $(2,3)$
• C. $(-1,1)$
• D. $(1,2)$

Answer 1

The correct option is B $(2,3)$

$f\left(x\right)={\log }_e(x^2+1)-e^{-x}+1$
$f^{\prime }\left(x\right)=\frac{2x}{x^2+1}+e^{-x}$
$=\frac{2}{x+\frac{1}{x}}+e^{-x}>0\forall x\in R$
$g\left(x\right)=e^{-x}-2e^x$
$g^{\prime }\left(x\right)=-e^{-x}-2e^x<0\forall x\in R$
$\Rightarrow f\left(x\right)$ is increasing and $g\left(x\right)$ is decreasing function.
$f\left(g\left(\frac{(\alpha -1{)}^2}{3}\right)\right)>f\left(g(\alpha -\frac{5}{3})\right)$
$\Rightarrow \frac{{(\alpha -1)}^2}{3}<\alpha -\frac{5}{3}$
$\Rightarrow {\alpha }^2-5\alpha +6<0$
$\Rightarrow \alpha -2\left)\right(\alpha -3)<0$
$\Rightarrow \alpha \in (2,3)$