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If $\sum_{r = 1}^{n} \cos^{- 1} x_{r} = 0,$ then $\sum_{r = 1}^{n} x_{r}$ equals

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n(n+1)2

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Correct option is B

We know that0≤cos−1⁡xr≤π;r=1,2,…,n∴∑r=1n cos−1⁡xr=0⇒cos−1⁡xr=0 for r=1,2,…,n⇒xr=1 for r=1,2,…,n∴ ∑r=1n xr=∑r=1n 1=n

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If ∑r=1n cos−1⁡xr=0, then ∑r=1n xr equals

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If $\sum_{r = 1}^{n} \cos^{- 1} x_{r} = 0,$ then $\sum_{r = 1}^{n} x_{r}$ equals