Question:
Let $\mathrm{f}$ be a real valued function, defined on $\mathrm{R}-\{-1,1\}$ and given by
$f(x)=3 \log _{e}\left|\frac{x-1}{x+1}\right|-\frac{2}{x-1}$
Then in which of the following intervals, function $f(x)$ is increasing?
Correct Option: 1
Solution:
$f(x)=3 \ell n(x-1)-3 \ell n(x+1)-\frac{2}{x-1}$
$f^{\prime}(x)=\frac{3}{x-1}-\frac{3}{x+1}+\frac{2}{(x-1)^{2}}$
$f^{\prime}(x)=\frac{4(2 x-1)}{(x-1)^{2}(x+1)}$
$f^{\prime}(x) \geq 0$
$\Rightarrow \quad x \in(-\infty,-1) \cup\left[\frac{1}{2}, 1\right) \cup(1, \infty)$