Question

The complete solution set of the inequality $\frac{1}{1+\ln x}+\frac{1}{1-\ln x}>2$ is

• A. $(\frac{1}{e},1)$
• B. $(1,e)$
• C. $(\frac{1}{e},e)$
• D. $(\frac{1}{e},1)\cup (1,e)$

Answer 1

The correct option is D $(\frac{1}{e},1)\cup (1,e)$

$\frac{1}{1+\ln x}+\frac{1}{1-\ln x}>2$
The above ineuality is defined when
$\left(i\right)x>0$
$\left(ii\right)\ln x\ne \pm 1\Rightarrow x\ne \frac{1}{e},e$

Now, $\frac{2}{1-(\ln x{)}^2}>2$
$\Rightarrow \frac{2}{1-(\ln x{)}^2}-2>0$
$\Rightarrow \frac{2-2+2(\ln x{)}^2}{1-(\ln x{)}^2}>0$
$\Rightarrow \frac{2(\ln x{)}^2}{1-(\ln x{)}^2}>0$
$\Rightarrow 1-(\ln x{)}^2>0[\because 2(\ln x{)}^2>0\forall x>0\text{and}x\ne 1]$
$\Rightarrow (\ln x{)}^2-1<0$
$\Rightarrow \ln x\in (-1,1)$
$\Rightarrow x\in (\frac{1}{e},e)$
$\therefore x\in (\frac{1}{e},1)\cup (1,e)(\because x\ne 1)$