Question

Let $f\left(x\right)={\tan }^{-1}x-\frac{In\left|x\right|}{2},x\ne 0.$. Then $f\left(x\right)$ is increasing in

• A. $\left(0,\infty \right)$
• B. $\left(-\infty ,0\right)$
• C. $\left(1,\infty \right)$
• D. none of these

Answer 1

Answer: (B)

$\left(-\infty ,0\right)$

$f\left(x\right)={\tan }^{-1}x-\frac{\ln |x|}{2},x\ne 0$

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

$\Rightarrow f^{\prime }\left(x\right)=0$

$\Rightarrow \frac{1}{1+x^2}-\frac{1}{2x}=0$

$\Rightarrow \frac{1}{1+x^2}=\frac{1}{2x}$

$\Rightarrow x^2+1=2x$

$\Rightarrow x^2-2x+1=0$

$\Rightarrow (x-1{)}^2=0$

$\therefore x=1$

Hence

$f\left(x\right)$ is increasing in $(-\infty ,0)$