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If ∑_(i=1)²ⁿ cos⁻¹(x_i)=0, then the value of ∑_(i=1)²ⁿ x_i is

1. n
2. 2n
3. (n(n+1))/(2)
4. None of these

Correct answer: 2n

Solution

Each cos^-1(x_i) lies in [0, pi] and is >= 0. A sum of 2n non-negative terms equals 0 only if every term is 0, so each x_i = 1. Therefore sum of x_i = 2n.

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