Question

Let $f$ be a real valued function, defined on $R-\{-1,1\}$ and given by $f\left(x\right)=3{\log }_e\left|\frac{x-1}{x+1}\right|-\frac{2}{x-1}.$ Then in which of the following intervals, function $f\left(x\right)$ is increasing ?

• A. $(-\infty ,-1)\cup ([\frac{1}{2},\infty )-\{1\left\}\right)$
• B. $(-1,\frac{1}{2}]$
• C. $(-\infty ,\infty )-\{-1,1\}$
  
• D. $(-\infty ,\frac{1}{2}]-\{-1\}$

Answer 1

The correct option is A $(-\infty ,-1)\cup ([\frac{1}{2},\infty )-\{1\left\}\right)$

$f\left(x\right)=3{\log }_e\left|\frac{x-1}{x+1}\right|-\frac{2}{x-1}$
$f^{\prime }\left(x\right)=3(\frac{x+1}{x-1}\times \frac{x+1-(x-1)}{(x+1{)}^2})+\frac{2}{(x-1{)}^2}$
$=\frac{6}{(x-1)(x+1)}+\frac{2}{(x-1{)}^2}$
$=\frac{4(2x-1)}{(x+1)(x-1{)}^2}$


$\therefore x\in (-\infty ,-1)\cup [\frac{1}{2},\infty )-\left\{1\right\}$