Question

Let $f$ be a real valued function, defined on $R -\{-1,1\}$ 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?

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

Answer 1

Answer: (A)

$f ( x )=3 \ln ( x -1)-3 \ell n ( x +1)-\frac{2}{ x -1}$
$f'( x )=\frac{3}{ x -1}-\frac{3}{ x +1}+\frac{2}{( x -1)^{2}}$
$f '( x )=\frac{4(2 x -1)}{( x -1)^{2}( x +1)}$
$f'( x ) \geq 0$
$\Rightarrow x \in(-\infty,-1) \cup\left[\frac{1}{2}, 1\right) \cup(1, \infty)$