Question

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

Answer 1

Let ${\text{cosec}}^{-1}x=t,x\in (-\infty ,-1]\cup [1,\infty ),t\in [-\frac{\pi }{2},\frac{\pi }{2}]-\left\{0\right\}$

$t^2-2t\ge \frac{\pi }{6}(t-2)$
$(t-2)(t-\frac{\pi }{6})\ge 0$

as $t\le \frac{\pi }{2}\Rightarrow t-2<0$
$\Rightarrow t-\frac{\pi }{6}\le 0\Rightarrow t\in [-\frac{\pi }{2},0)\cup (0,\frac{\pi }{6}]$
$\Rightarrow x\in (-\infty ,-1]\cup [2,\infty )$