Question

Show that $\text{f(x)}=2\text{x}+{\text{cot}}^{-1}\text{x}+\text{log}(\sqrt{1+{\text{x}}^2}-\text{x})$ is increasing in $R$.

Answer 1

Given:
$\text{f(x)}=2\text{x}+{\text{cot}}^{-1}\text{x}+\text{log}(\sqrt{1+{\text{x}}^2}-\text{x})$
Differentiating both sides w.r.t. x, we get
$\Rightarrow \text{f'(x)}=2-\frac{1}{1+{\text{x}}^2}+\frac{1}{\sqrt{1+{\text{x}}^2}-\text{x}}[\frac{1}{\sqrt{1+{\text{x}}^2}}-1]$
$=2-\frac{1}{1+{\text{x}}^2}+\frac{1}{\sqrt{1+{\text{x}}^2}-\text{x}}.\frac{\text{x}-\sqrt{1+{\text{x}}^2}}{\sqrt{1+{\text{x}}^2}}$
$\Rightarrow \text{f'(x)}=2-\frac{1}{1+{\text{x}}^2}-\frac{1}{\sqrt{1+{\text{x}}^2}}$
Since. $1+{\text{x}}^2$ & $\sqrt{1+{\text{x}}^2}\ge 1$ for all real x
$\Rightarrow \frac{1}{1+{\text{x}}^2}$ & $\frac{1}{\sqrt{1+{\text{x}}^2}}\le 1$ for all real x
$\therefore 2-\frac{1}{1+{\text{x}}^2}$ & $\frac{1}{\sqrt{1+{\text{x}}^2}}>0$ for all real x
$\Rightarrow \text{f'(x)}\ge 0$ for all real x
Hence, $\text{f(x)}$ is increasing on $R$
Hence proved.