A) increases in \[(-\infty ,\infty )\]
B) decreases in \[(0,\infty )\]
C) neither increases nor decreases in \[(0,\infty )\]
D) sometimes increases and sometimes decreases
Correct Answer: A
Solution :
We have, \[f(x)=2x+{{\cot }^{-1}}x+\log (\sqrt{1+{{x}^{2}}}-x)\] On differentiating w.r.t. \[x,\]we get \[f(x)=2-\frac{1}{1+{{x}^{2}}}+\frac{1}{(\sqrt{1+{{x}^{2}}}-x)}\] \[\times \,\,\left( \frac{x}{\sqrt{1+{{x}^{2}}}}-1 \right)\] \[=2-\frac{1}{1+{{x}^{2}}}-\frac{1}{\sqrt{1+{{x}^{2}}}}\] \[=\frac{2+2{{x}^{2}}-1-\sqrt{1+{{x}^{2}}}}{1+{{x}^{2}}}\] \[=\frac{{{x}^{2}}+\sqrt{1+{{x}^{2}}}(\sqrt{1+{{x}^{2}}}-1)}{1+{{x}^{2}}}>0\] \[\Rightarrow \]\[f(x)\]is increasing \[(-\infty ,\infty ).\]
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