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

(-infty,-1 )cup ( [(1)/(2),infty )-\1\ )

Differentiation and its Application: Let f be a real valued function, defined on mathbfR-\-1,1\ and given… | JEE Main 2021

JEE Main 2021 (Online) 16th March Evening ShiftQuestion 1

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

Differentiation and its Application

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

Solution

Concept. Simplify $f'$ to one fraction and read off its sign. Formulas used. (1) $\dfrac{d}{dx}\ln\left|\dfrac{x-1}{x+1}\right|=\dfrac{1}{x-1}-\dfrac{1}{x+1}$Step 1 - $f'$. By (1), $$f'(x)=\frac{3}{x-1}-\frac{3}{x+1}+\frac{2}{(x-1)^2}=\frac{4(2x-1)}{(x-1)^2(x+1)}.$$ Step 2 - Sign. $(x-1)^2>0$, so $f'$ has the sign of $(2x-1)(x+1)$: positive for $x<-1$ and for $x>\frac12$. Step 3 - Answer. $f$ increases on $(-\infty,-1)\cup\left(\left[\frac12,\infty\right)-\{1\}\right)$. Answer: (B)

Differentiation and its Application - JEE Mathematics question (JEE Main 2021 (Online) 16th March Evening Shift)

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