question_answer

Show that    \[f(x)=2x+{{\cot }^{-1}}x+\log \,\,(\sqrt{1+{{x}^{2}}}-x)\] is   Increasing in R.

Answer:

Given, \[f(x)=2x+{{\cot }^{-1}}x+\log \,\,(\sqrt{1+{{x}^{2}}}-x)\] On differentiating both sides w.r.t. x, we get \[f'(x)=2+\left( \frac{-1}{1+{{x}^{2}}} \right)+\frac{1}{\sqrt{1+{{x}^{2}}}-x}\]                                     \[\left\{ \left( \frac{1}{2\sqrt{1+{{x}^{2}}}}\cdot 2x \right)-1 \right\}\] \[=2-\frac{1}{1+{{x}^{2}}}+\frac{1}{(\sqrt{1+{{x}^{2}}}-x)}\cdot \frac{(x-\sqrt{1+{{x}^{2}}})}{\sqrt{1+{{x}^{2}}}}\] \[=2-\frac{1}{1+{{x}^{2}}}+\frac{1}{(\sqrt{1+{{x}^{2}}}-x)}\cdot \frac{-(\sqrt{1+{{x}^{2}}}-x)}{\sqrt{1+{{x}^{2}}}}\] \[=2-\frac{1}{1+{{x}^{2}}}-\frac{1}{\sqrt{1+{{x}^{2}}}}=\frac{2+2{{x}^{2}}-1-\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}\] \[=\frac{1+2{{x}^{2}}-\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}\] For increasing function, \[f'(x)\ge 0\] \[\Rightarrow \]  \[\frac{1+2{{x}^{2}}-\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}\ge 0\] \[\Rightarrow \]   \[1+2{{x}^{2}}\ge \sqrt{1+{{x}^{2}}}\] \[[\because \,\,\,1+{{x}^{2}}\,\text{is}\,\,\text{always}\,\,\text{postive}]\] \[\Rightarrow \]   \[{{(1+2{{x}^{2}})}^{2}}\ge 1+{{x}^{2}}\] [On squaring both sides] \[\Rightarrow \]   \[1+4{{x}^{4}}+4{{x}^{4}}\ge 1+{{x}^{2}}\]             \[[\because \,\,{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab]\] \[\Rightarrow \]   \[4{{x}^{4}}+3{{x}^{2}}\ge 0\]         \[\Rightarrow \]   \[{{x}^{2}}(4{{x}^{2}}+3)\ge 0\] Which is true for any real value of x. hence, f(x) is increasing in R.                            Hence proved.