If $\sum\limits^{2n}_{r=1}sin^{-1}x_r = n\, \, \pi, $ then $\sum\limits^{2n}_{r=1} x_r $ is equal to
Answer & explanation
Correct answer: option 2
We have ,
$-\frac{\pi}{2} ≤ sin^{-1} x_i ≤ \frac{\pi}{2}$ for i = 1, 2...., 2n
$∴ \sum\limits^{2n}_{r=1}sin^{-1}x_r = n\, \, \pi, $
$ ⇒ sin^{-1} x_r =\frac{\pi}{2}; r = 1, 2, ...., 2n$
$⇒ x_r = 1; r = 1, 2, ..., 2n$
$⇒ \sum\limits^{2n}_{r=1} x_r = \sum\limits^{2n}_{r=1} 1 = 2n $