Let f be a real valued function, defined on R - \ - 1, 1\ and given by f(x) = 3 _ e |(x - 1)/(x + 1)| - (2)/(x - 1) . Then in which of the following intervals, function f(x) is increasing?

Correct answer: A. (- ∞ , - 1) ([(1)/(2), ∞) - \ 1\ ) f(x) = 3 _ e (x - 1) - 3 _ e (x + 1) - (2)/(x - 1) f ^ ' (x) = (3)/(x - 1) - (3)/(x + 1) + 2 (x - 1) ^ 2 f ^ ' (x) = 4(2x - 1) (x - 1) ^ 2 (x + 1) f ^ ' (x) ≥ 0 ⇒ x ∈ (- ∞ , - 1) [(1)/(2), 1) (1, ∞) or (- ∞ , - 1) ([(1)/(2), ∞) - \ 1\ )

JEE Main 202116 MarchEvening Shift

Exam: JEE Main
Year: 2021
Date: 16 March
Shift: Evening Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2021 · 16 March · Evening Shift · Mathematics · Application of Derivatives

Let f be a real valued function, defined on R - \ - 1, 1\ and given...

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

Options

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

Correct answer

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

Step-by-step solution

f\left(x\right) = 3\log_{e}\left(x - 1\right) - 3\log_{e}\left(x + 1\right) - \dfrac{2}{x - 1} $$ {f}^{'}\left(x\right) = \dfrac{3}{x - 1} - \dfrac{3}{x + 1} + \dfrac{2}{{\left(x - 1\right)}^{2}} $$ {f}^{'}\left(x\right) = \dfrac{4\left(2x - 1\right)}{{\left(x - 1\right)}^{2}\left(x + 1\right)} $$ {f}^{'}\left(x\right) \ge 0 $$ ⇒ x \in \left(- \infty , - 1\right) \cup \left[\dfrac{1}{2}, 1\right) \cup \left(1, \infty\right) $or \left(- \infty , - 1\right) \cup \left(\left[\dfrac{1}{2}, \infty\right) - \left\{1\right\}\right)$

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