Question

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

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

Answer 1

Answer: (A)

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