First slide

Introduction to ITF

Question

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

Moderate

A

0
B

$n$
C

$\frac{n (n + 1)}{2}$
D

none of these

Solution

We know that
$\begin{array}{l} 0 \leq \cos^{- 1} x_{r} \leq π; r = 1, 2, \dots, n \\ ∴ \sum_{r = 1}^{n} \cos^{- 1} x_{r} = 0 \\ \Rightarrow \cos^{- 1} x_{r} = 0 for r = 1, 2, \dots, n \\ \Rightarrow x_{r} = 1 for r = 1, 2, \dots, n \\ ∴ \sum_{r = 1}^{n} x_{r} = \sum_{r = 1}^{n} 1 = n \end{array}$

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