Question

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

Answer 1

We have
$f\left(x\right)=2x+co{t}^{-1}x+log(\sqrt{1+x^2}-x)\therefore f^{\prime }\left(x\right)=2+\left(\frac{-1}{1+x^2}\right)+\frac{1}{(\sqrt{1+x^2}-x)}(\frac{1}{2\sqrt{1+x^2}}.2x-1)=2-\frac{1}{1+x^2}+\frac{1}{\left(\sqrt{1+x^2-x}\right)}.\frac{(x-\sqrt{1+x^2})}{\sqrt{1+x^2}}=2-\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}}=\frac{2+2x^{-}1-\sqrt{1+x^2}}{1+x^2}=\frac{1+2x^2-\sqrt{1+x^2}}{1+x^2}$
For increasing function, $f^{\prime }\left(x\right)\ge 0$
$\Rightarrow \frac{1+2x^2\sqrt{1+x^2}}{1+x^2}\le 0\Rightarrow 1+2x^2\ge \sqrt{1+x^2}\Rightarrow (1+2x^2{)}^2\ge 1+x^2\Rightarrow 1+4x^4+4x^2\ge 1+x^2\Rightarrow 4x^4+3x^2\ge 0\Rightarrow x^2(4x^2+3)\ge 0$
Which is true for any real value of x.
Hence, f(x) is increasing in R.