Question

If $\sum _{i=1}^{2n}{\cos }^{-1}{x}_i=0$, then $\sum _{i=1}^{2n}{x}_i$ is :

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

Answer 1

The correct option is A $n$

$\sum _{i=1}^{2n}{\cos }^{-1}{x}_1=0$

$\therefore {\cos }^{-1}{x}_1+{\cos }^{-1}{x}_2+{\cos }^{-1}{x}_2+{\cos }^{-1}{x}_3....+{\cos }^{-1}{x}_{2n}=0$

$\therefore$ We know that

$-\pi <{\cos }^{-1}x<\pi$

So to have net sum $=0$

We must have ${\cos }^{-1}{x}_i>0$ and ${\cos }^{-1}{x}_i<0$

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