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

The correct option is B $(-\infty ,\tan 1]$

Given, ${\left({\tan }^{-1}x\right)}^2-3{\tan }^{-1}x+2\ge 0$

Replacing ${\tan }^{-1}\left(x\right)$ by $t$, we get

$\therefore t^2-3t+2\ge 0$

$(t-1)(t-2)\ge 0$

$t\ge 2$ and $t\le 1$

Now ${\tan }^{-1}\left(x\right)ϵ(\frac{-\pi }{2},\frac{\pi }{2})$
And
$\tan \left(2\right)=-2.1$

Hence, $x\in (-\infty ,\tan 1)$