If $ \sum\limits^{n}_{r=1}cos^{-1}x_r = 0 $, then $\sum\limits^{n}_{r=1} x_r $ equals
Answer & explanation
Correct answer: option 2
We know that
$0 ≤ cos^{-1} x_r ≤\pi ; r = 1, 2, ....., n $
$ ∴\sum\limits^{n}_{r=1}cos^{-1}x_r = 0 $
$⇒ cos^{-1} x_r = 0 $ for r = 1, 2, .....,n
$⇒ x_r = 1 $ for r =1, 2, ....n
$∴\sum\limits^{n}_{r=1} x_r = \sum\limits^{n}_{r=1} 1 = n $