Let $mathrm{f}$ be a real valued function, defined on

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 $\mathrm{f}(\mathrm{x})$ is increasing?

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

  2. (2) $(-\infty, \infty)-\{-1,1\}$

  3. (3) $\left(-1, \frac{1}{2}\right]$

  4. (4) $\left(-\infty, \frac{1}{2}\right]-\{-1\}$


Correct Option: 1,

Solution:

$f(x)=3 \ln (x-1)-3 \ln (x+1)-\frac{2}{x-1}$

$f^{\prime}(x)=\frac{3}{x-1}-\frac{3}{x+1}+\frac{2}{(x-1)^{2}}$

$f^{\prime}(x)=\frac{4(2 x-1)}{(x-1)^{2}(x+1)}$

$\mathrm{f}^{\prime}(\mathrm{x}) \geq 0$

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

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