A) \[1/2\]
B) \[0\]
C) \[1\]
D) does not exist
Correct Answer: A
Solution :
Given,\[y={{\tan }^{-1}}\left( \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right)\] Putting\[x=\tan \theta \], we get \[\frac{\sqrt{1+{{x}^{2}}}-1}{x}=\frac{\sec \theta -1}{\tan \theta }=\frac{1-\cos \theta }{\sin \theta }=\tan \frac{\theta }{2}\] \[\therefore \] \[y=\frac{1}{2}{{\tan }^{-1}}x\] \[\Rightarrow \] \[y'=\frac{1}{2(1+{{x}^{2}})}\] \[\Rightarrow \] \[y'(0)=\frac{1}{2(1+{{0}^{2}})}=\frac{1}{2}\]
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