Question

The solution of the inequality ${\left({\tan }^{-1}x\right)}^2-3{\tan }^{-1}x+2\ge 0$ is -

• A. $(-\infty ,\tan 1]\cup [\tan 2,\infty )$
• B. $(-\infty ,\tan 1]$
• C. $(-\infty ,-\tan 1]\cup [\tan 2,\infty )$
• D. $[\tan 2,\infty )$

Answer 1

Answer: (B)

${\left({\tan }^{-1}x\right)}^2-3{\tan }^{-1}x+2\ge 0$
$\left({\tan }^{-1}x-1\right)\left({\tan }^{-1}x-2\right)\ge 0$
we know that ${\tan }^{-1}x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
so ${\tan }^{-1}x\ge 2$ (not possible) or ${\tan }^{-1}x\le 1$
$\Rightarrow x\in (-\infty ,\tan 1]$