Question

If $f\left(x\right)=2x+{\cot }^{-1}x+\log \left(\sqrt{1+x^2}-x\right)$, then f(x)

• A. increases in $\left(0,\infty \right)$
• B. decreases in $\left(0,\infty \right)$
• C. neither increases nor decreases in $\left(0,\infty \right)$
• D. increases in $\left(-\infty ,\infty \right)$

Answer 1

Answer: (A), (D)

The correct options are
A increases in $\left(0,\infty \right)$
D increases in $\left(-\infty ,\infty \right)$

$f\left(x\right)=2x+{\cot }^{-1}x+\log \left(\sqrt{1+x^2}-x\right)\therefore f\text{'}\left(x\right)=2-\frac{1}{1+x^2}+\frac{1}{\sqrt{1+x^2}-x}\left(\frac{x}{\sqrt{1+x^2}}-1\right)\Rightarrow f\text{'}\left(x\right)=\frac{1+2x^2}{1+x^2}-\frac{1}{\sqrt{1+x^2}}\Rightarrow f\text{'}\left(x\right)=\frac{1+2x^2}{1+x^2}-\frac{\sqrt{1+x^2}}{1+x^2}\Rightarrow f\text{'}\left(x\right)=\frac{x^2+\sqrt{1+x^2}\left(\sqrt{1+x^2}-1\right)}{1+x^2}\mathrm{Hence},f\text{'}\left(x\right)>0\mathrm{for}\mathrm{all}x.\mathrm{Hence},f\left(x\right)\mathrm{is}\mathrm{an}\mathrm{increasing}\mathrm{function}\mathrm{in}\left(-\infty ,\infty \right)\mathrm{and},\mathrm{in}\mathrm{particular}f\left(x\right)\mathrm{is}\mathrm{an}\mathrm{increasing}\mathrm{function}\mathrm{in}\left(0,\infty \right).$