Question

The solution set of the inequality ${\left({\tan }^{-1}x\right)}^2\le \left({\tan }^{-1}x\right)+6$ is

• A. $\left[-tan2,tan3\right]$
• B. $\left[tan2,-tan3\right]$
• C. $\left[tan2,tan3\right]$
• D. $\left(-\in fty,\in fty\right)$

Answer 1

Answer: (D)

Let ${\text{tan}}^{-1}x=u$
So, inequality becomes $u^2-u-6\le 0$
$\Rightarrow -2\le u\le 3$
So, $-2\le {\text{tan}}^{-1}x\le 3$
As $-\frac{\pi }{2}<{\text{tan}}^{-1}x<\frac{\pi }{2}$
$\Rightarrow -2\le {\text{tan}}^{-1}x\le 3,\forall x\in R$
Therefore, the solution set of the inequality ${\left({\tan }^{-1}x\right)}^2\le \left({\tan }^{-1}x\right)+6$ is $(-\infty ,\infty )$