Question

If the solution set of inequality ${\left({\text{cosec}}^{-1}x\right)}^2-2$ ${\text{cosec}}^{-1}x\ge \frac{\pi }{6}\left({\text{cosec}}^{-1}x-2\right)$ is $(-\infty ,m]\cup [n,\infty )$ then $(m+n)$ equals

Answer 1

Answer: 1

${\lambda }^2-2\lambda \ge \frac{\pi }{6}(\lambda -2)$, where $\lambda ={\text{cosec}}^{-1}x$
$(\lambda -2)\cdot \left(\lambda -\frac{\pi }{6}\right)\ge 0$
$\therefore x\in (-\infty ,-1]\cup [2,\infty )$
$\Rightarrow m=-1,n=2$
So, $(m+n)=1$