A) 0
B) \[\frac{1}{2}\]
C) 1
D) x
Correct Answer: B
Solution :
[b] Let \[y={{\tan }^{-1}}\left[ \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right]\] and \[u={{\tan }^{-1}}x\] Put \[x=\tan \theta \Rightarrow \theta ={{\tan }^{-1}}x\] Then, \[y={{\tan }^{-1}}\left[ \frac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta } \right]\] \[={{\tan }^{-1}}\left[ \frac{\sqrt{{{\sec }^{2}}\theta }-1}{\tan \theta } \right]\] \[={{\tan }^{-1}}\left[ \frac{\sec \theta -1}{\tan \theta } \right]={{\tan }^{-1}}\left[ \frac{\frac{1}{\cos \theta }-1}{\frac{\sin \theta }{\cos \theta }} \right]\] \[={{\tan }^{-1}}\left[ \frac{1-\cos \theta }{\sin \theta } \right]={{\tan }^{-1}}\left[ \frac{2{{\sin }^{2}}\frac{\theta }{2}}{2\sin \frac{\theta }{2},\cos \frac{\theta }{2}} \right]\] \[\left( \begin{align} & \because 1-\cos \theta =2{{\sin }^{2}}\frac{\theta }{2}and \\ & \sin x=2\sin \frac{x}{2}\cdot \cos \frac{x}{2} \\ \end{align} \right)\] \[={{\tan }^{-1}}\left[ \tan \frac{\theta }{2} \right]\] \[\Rightarrow y=\frac{\theta }{2}\Rightarrow y=\frac{{{\tan }^{-1}}x}{2}[\because \theta ={{\tan }^{-1}}x]\] \[\Rightarrow y=\frac{u}{2};\frac{dy}{du}=\frac{1}{2}\]
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