A) Increases in [0 ,\[\infty \])
B) Decreases in [0 ,\[\infty \])
C) Neither increases nor decreases in (0, \[\infty \])
D) Increases in (?\[\infty \],\[\infty \])
Correct Answer: A
Solution :
We have \[f(x)=2x+{{\cot }^{-1}}x+\log (\sqrt{1+{{x}^{2}}}-x)\] \[\therefore f'(x)=2-\frac{1}{1-{{x}^{2}}}+\frac{1}{\sqrt{1+{{x}^{2}}}-x}\left( \frac{x}{\sqrt{1-{{x}^{2}}}}-1 \right)\] \[=\frac{1+2{{x}^{2}}}{1+{{x}^{2}}}-\frac{1}{\sqrt{1+{{x}^{2}}}}=\frac{1+2{{x}^{2}}}{1+{{x}^{2}}}-\frac{\sqrt{(1+{{x}^{2}})}}{1+{{x}^{2}}}\] \[=\frac{{{x}^{2}}+\sqrt{1+{{x}^{2}}}(\sqrt{1+{{x}^{2}}}-1)}{1+{{x}^{2}}}\ge 0\] for all x Hence f(x) is an increasing function on \[(-\infty ,\,\infty )\] and in particular on \[[0,\ \infty )\].
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