Question

If $\sum _{i=1}^{2n}{\sin }^{-1}{x}_i=n\pi$, then $\sum _{i=1}^{2n}{x}_i$ is equal to

• A. $n/2$
• B. $2n$
• C. $\frac{n(n+1)}{2}$
• D. none of these

Answer 1

The correct option is B $2n$

As $-\frac{\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}$

Hence from question
${\sin }^{-1}{x}_i=\frac{\pi }{2}$ for all $i$

$\therefore x_i=1$ for all $i$

$\therefore \sum _{i=1}^{2n}{x}_i=2n$