Inverse Trigonometric Functions Question 69
Question: If $ \sum\limits_{i=1}^{2n}{{{\cos }^{-1}}x_{i}=0} $ then $ \sum\limits_{i=1}^{2n}{x_{i}} $ is
Options:
A) n
B) 2n
C) $ \frac{n( n+1 )}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Since $ 0\le {{\cos }^{-1}}x_{i}\le \pi $, it follows that $ {{\cos }^{-1}}x_{i}=0 $ for all i.
$ \therefore $ $ x_{i}=1 $ for all $ i\therefore \sum\limits_{i=1}^{2n}{x_{i}=2n} $
BETA
