A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[-\frac{1}{2}\]
D) \[-\frac{3}{4}\]
Correct Answer: A
Solution :
\[\frac{d}{dx}\left\{ {{\cos }^{2}}\left( {{\cot }^{-1}}\sqrt{\frac{2+x}{2-x}} \right) \right\}\] put \[x=2\cos \theta \] ??(i) \[\Rightarrow \] \[\frac{d}{dx}\left\{ {{\cos }^{2}}\left( {{\cot }^{2}}\sqrt{\frac{2+\cos \theta .2}{2-\cos \theta .2}} \right) \right\}\] \[=\frac{d}{dx}\left\{ {{\cos }^{2}}\left( {{\cot }^{-1}}\sqrt{\frac{1+\cos \theta }{1-\cos \theta }} \right) \right\}\] \[=\frac{d}{dx}\left\{ {{\cos }^{2}}\left( {{\cot }^{-1}}\sqrt{\frac{2{{\cos }^{2}}\frac{\theta }{2}}{2{{\sin }^{2}}\frac{\theta }{2}}} \right) \right\}\] \[=\frac{d}{dx}\{{{\cos }^{2}}({{\cot }^{-1}}(\cot \theta /2))\}\] \[=\frac{d}{dx}\left\{ {{\cos }^{2}}\frac{\theta }{2} \right\}=\frac{1}{2}.\frac{d}{dx}(1+\cos \theta )\] \[=\frac{d}{dx}\left( \frac{1}{2}+\frac{1}{2}.\frac{x}{2} \right)\] \[=\frac{d}{dx}\left( \frac{1}{2}+\frac{x}{4} \right)\] [from Eq. (i)] \[=(0+1/4)=1/4\]
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