A) n
B) 2n
C) \[\frac{n\left( n+1 \right)}{2}\]
D) None of these
Correct Answer: B
Solution :
[b] Since \[0\le {{\cos }^{-1}}{{x}_{i}}\le \pi ,\therefore {{\cos }^{-1}}{{x}_{i}}=0\] for all i. \[\therefore \]\[{{x}_{i}}=1\] for all \[i\therefore \sum\limits_{i=1}^{2n}{{{x}_{i}}=2n}\]
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