A) 4
B) 1
C) ¼
D) ?1/4
Correct Answer: C
Solution :
\[u={{\tan }^{-1}}\left\{ \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right\}\]and \[v=2{{\tan }^{-1}}x\] Put \[x=\tan \theta \]in u and v; \[u={{\tan }^{-1}}\left\{ \frac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta } \right\}\]and \[v=2\theta \] \[u={{\tan }^{-1}}\left\{ \frac{\sec \theta -1}{\tan \theta } \right\}\]and \[v=2\theta \] \[\frac{3\,({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)}{(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)}\]and \[v=2\theta \] \[u=\theta /2\]and \[v=2\theta \]; \[\therefore \frac{du}{dv}=\frac{du/d\theta }{dv/d\theta }=\frac{1/2}{2}=\frac{1}{4}\].
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