A) \[-\frac{\pi }{3}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{3}{\pi }\]
D) \[-\frac{3}{\pi }\]
Correct Answer: B
Solution :
[b] Let \[{{\cos }^{-1}}x=y\] \[\Rightarrow x=\cos y,\] so that \[\frac{1}{2}\le x\le 1\]or \[0\le y\le \frac{\pi }{3}\] and \[\frac{x}{2}+\frac{1}{2}\sqrt{3-3{{x}^{2}}}=\frac{1}{2}\cos y+\frac{\sqrt{3}}{2}\sin y\] \[=\cos \frac{\pi }{3}\cos y+\sin \frac{\pi }{3}\sin y=\cos \left( \frac{\pi }{3}-y \right)\] \[\Rightarrow {{\cos }^{-1}}\left( \frac{x}{2}+\frac{1}{2}\sqrt{3-3{{x}^{2}}} \right)=\frac{\pi }{3}-y\] \[\therefore \] the given expression is equal to \[y+\frac{\pi }{3}-y,\,\,\,i.e,\frac{\pi }{3}\]
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