Question

If ${\left[{\tan }^{-1}x\right]}^2-2\left[{\tan }^{-1}x\right]+1\le 0$, where $[.]$ denotes greatest integer $\le x,x$ belongs to

• A. $(\tan 1,\tan 2)$
• B. $[\frac{\pi }{4},\frac{\pi }{2}]$
• C. $[\tan 1,\infty )$
• D. None of these

Answer 1

The correct option is A $(\tan 1,\tan 2)$

Using $(a-b{)}^2=a^2-2ab+b^2$

The given inequation is ${\{\left[{\tan }^{-1}x\right]-1\}}^2\le 0$

$\Rightarrow \left[{\tan }^{-1}x\right]=1$ as ${\{\left[{\tan }^{-1}x\right]-1\}}^2\nless 0$

$\Rightarrow 1\le {\tan }^{-1}x<2$ ( using property of greatest integer function)

$\Rightarrow x\in (\tan 1,\tan 2)$