Question

The function $f\left(x\right)=x-ln|2x+1|,xϵ(-100,\frac{-1}{2})\cup (\frac{-1}{2},\frac{1}{2})$ is decreasing in interval

• A. $(\frac{-1}{2},\frac{1}{2})$
• B. $(-100,\frac{-1}{2})$
• C. $(\frac{-1}{2},0)$ only
• D. $(0,\frac{1}{2})$ only

Answer 1

The correct option is A $(\frac{-1}{2},\frac{1}{2})$

Given $f\left(x\right)=x-log|2x+1|$

Differentiate function w.r.t. x, we get,

$f^{\prime }\left(x\right)=\frac{d}{dx}[x-log|2x+1|]$

$\therefore f^{\prime }\left(x\right)=1-\frac{1}{2x+1}\times \frac{d}{dx}(2x+1)$

$\therefore f^{\prime }\left(x\right)=1-\frac{1}{2x+1}\times 2$

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

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

For function to be decreasing, $f^{\prime }\left(x\right)<0$

$\therefore \frac{2x-1}{2x+1}<0$

Case 1) $2x-1<0$ and $2x+1>0$

$\therefore 2x<1$ and $2x>-1$

$\therefore x<\frac{1}{2}$ and $x>-\frac{1}{2}$

Thus, function is decreasing in the interval $(\frac{-1}{2},\frac{1}{2})$

Case 2) $2x-1>0$ and $2x+1<0$

$\therefore 2x>1$ and $2x<-1$

$\therefore x>\frac{1}{2}$ and $x<-\frac{1}{2}$

Satisfying both the conditions simultaneously is not possible.