Let f be a real valued function, defined onR--1,1 and given by fx=3log_e(x-1)/(x+1)-2/(x-1). Then in which of the following intervals, function fx is increasing?

Correct answer: (A) -∞,-1∪1/2,∞-1. fx=3log_ex-1-3log_ex+1-2/(x-1) f^(')x=3/(x-1)-3/(x+1)+(2x-1^2f^(')x=42x-1x-1^2x+1f^(')x≥0⇒ x∈-∞,-1∪1)/2,1∪1,∞ or -∞,-1∪1/2,∞-1

Q45 / 90 · JEE Main 2021 (16 Mar Shift 2) · Mathematics

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

Correct answer
A
$(- \infty, - 1) \cup ([\frac{1}{2}, \infty) - \{1\})$
B
$(- \infty, \infty) - \{- 1, 1\}$
C
$(- 1, \frac{1}{2}]$
D
$(- \infty, \frac{1}{2}] - \{- 1\}$

Solution

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

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

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

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

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

Hence, the correct answer is option: 'A'.

Let f be a real valued function, defined onR--1,1 and given by fx=3logex-1x+1-2x-1. Then in which of the following intervals, function fx is increasing?

Solution

Relevant Questions

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