Question

Show that $f\left(x\right)=2x+co{s}^{-1}x+log(\sqrt{1+x^2}-x)$ is increasing in $R$.

Answer 1

we have $f\left(x\right)=2x+co{s}^{-1}x+log(\sqrt{1+x^2}-x)$

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

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

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

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

For increasing function $f^{\prime }\left(x\right)\ge 0$

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

$\Rightarrow 1+2x^2\ge \sqrt{1+x^2}$

$\Rightarrow (1+2x^2{)}^2\le 1+x^2$

$\Rightarrow 1+4x^4+4x^2\ge 1+x^2$

$4x^4+3x^2\ge 0$

$\Rightarrow x^2(4x^2+3)\ge 0$

Which is true for any real values of $x$

Hence $f\left(x\right)$ is increasing in $R$