Question

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

$(-\infty ,\tan 1]$

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