A) \[\Rightarrow \]
B) \[I=\frac{E}{R'}\]
C) \[E=-\frac{nd\phi }{dt},\]
D) 1
Correct Answer: B
Solution :
Let\[y={{\tan }^{-1}}\left( \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right)\] Putting \[x=\tan \theta ,\]we get \[y={{\tan }^{-1}}\left( \frac{\sec \theta -1}{\tan \theta } \right)={{\tan }^{-1}}\left( \tan \frac{\theta }{2} \right)\]\[=\frac{1}{2}{{\tan }^{-1}}x\] On differentiating both sides w.r.t. x, we get \[\frac{dy}{dx}=\frac{1}{2(1+{{x}^{2}})}\]and\[z={{\tan }^{-1}}\left( \frac{2x\sqrt{1-{{x}^{2}}}}{1-2{{x}^{2}}} \right)\] Putting \[x=\sin \theta ,\] we get \[z={{\tan }^{-1}}\left( \frac{2\sin \theta \cos \theta }{\cos 2\theta } \right)={{\tan }^{-1}}(\tan 2\theta )\] \[\Rightarrow \]\[z=2\theta =2{{\sin }^{-1}}x\] On differentiating both sides w.r.t. x, we get \[\frac{dz}{dx}=\frac{2}{\sqrt{1-{{x}^{2}}}}\] Thus,\[\frac{dy}{dz}=\frac{dy/dx}{dz/dx}=\frac{1}{4(1+{{x}^{2}})}\sqrt{1-{{x}^{2}}}\] \[\therefore \]\[{{\left[ \frac{dy}{dz} \right]}_{z=0}}=\frac{1}{4}\]
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