Let f be a real valued function, defined on R - -1,1 and given by… - Vidyadip

MCQ

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?

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

✓

Answer

Correct option: A.

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

a
$f(x)=3 \ell n(x-1)-3 \ell n(x+1)-\frac{2}{x-1}$

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

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

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

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

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