A) \[\sqrt{x-1}\]
B) \[\sqrt{x+1}\]
C) \[\sqrt{{{x}^{2}}+1}\]
D) \[\frac{x}{\sqrt{1+{{x}^{2}}}}\]
Correct Answer: D
Solution :
[d] Area \[=\int_{1}^{b}{f(x)dx=\sqrt{{{b}^{2}}+1}}-\sqrt{2}\] \[=\sqrt{{{b}^{2}}+1}-\sqrt{1+1}\] \[=\left| \sqrt{{{x}^{2}}+1} \right|_{1}^{b}\] \[\therefore f(x)=\frac{d}{dx}\left( \sqrt{{{x}^{2}}+1} \right)=\frac{1}{2}\frac{2x}{\sqrt{{{x}^{2}}+1}}=\frac{x}{\sqrt{{{x}^{2}}+1}}\]
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