A) 0
B) \[\frac{1}{2}\]
C) \[-\frac{1}{2}\]
D) 1
Correct Answer: D
Solution :
\[{{[\sqrt{1+{{x}^{2}}}-x]}^{-1}}=\frac{1}{\sqrt{1+{{x}^{2}}}-x}\times \frac{(\sqrt{1+{{x}^{2}}}+x)}{(\sqrt{1+{{x}^{2}}}+x)}\] \[=\frac{\sqrt{1+{{x}^{2}}}+x}{1+{{x}^{2}}-{{x}^{2}}}=x+\sqrt{1+{{x}^{2}}}=x+{{(1+{{x}^{2}})}^{1/2}}\] \[=x+1+\frac{1}{2}{{x}^{2}}+\frac{1}{2}\left( -\frac{1}{2} \right)\frac{{{x}^{4}}}{2!}+...\]. Obviously, coefficient of x is 1.
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