Question

The values of $x$ satisfying the inequality $[{\tan }^{-1}x{]}^2-[{\tan }^{-1}x]-2\le 0$, where $[.]$ denotes integer part function, is

• A. $[-\tan 1,\infty )$
• B. $[-\frac{\pi }{4},\tan 2]$
• C. $[-\tan 1,\tan 3]$
• D. $[\tan 1,\tan 3]$

Answer 1

The correct option is A $[-\tan 1,\infty )$

Given : $[{\tan }^{-1}x{]}^2-[{\tan }^{-1}x]-2\le 0$
Assuming $\left[{\tan }^{-1}x\right]=y$
$\Rightarrow y^2-y-2\le 0\Rightarrow (y-2)(y+1)\le 0\Rightarrow y\in [-1,2]\Rightarrow \left[{\tan }^{-1}x\right]\in [-1,2]\Rightarrow {\tan }^{-1}x\in [-1,3)$
But ${\tan }^{-1}x\in (-\frac{\pi }{2},\frac{\pi }{2})$
$\Rightarrow {\tan }^{-1}x\in [-1,\frac{\pi }{2})$
We know that ${\tan }^{-1}x$ is an increasing function
$\therefore x\in [-\tan 1,\infty )$