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If $ \sum\limits^{n}_{r=1}cos^{-1}x_r = 0 $, then $\sum\limits^{n}_{r=1} x_r $ equals |
0 n $\frac{n(n+1)}{2}$ none of these |
n |
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 $ |
