Question

Which of the followings set of intervals of $x$ satisfying the inequality $[{\tan }^{-1}x{]}^2-\frac{3\left[{\tan }^{-1}x\right]}{2}+\frac{1}{2}>0$
(where $[.]$ denotes greatest integer function)

• A. $(-\infty ,-\tan 1]$
• B. $[\tan 1,\infty )$
• C. $(-\infty ,\tan 1)$
• D. $(-\infty ,\tan \frac{1}{7}]$

Answer 1

Answer: (A), (C), (D)

$(-\infty ,\tan \frac{1}{7}]$

Given : $[{\tan }^{-1}x{]}^2-\frac{3\left[{\tan }^{-1}x\right]}{2}+\frac{1}{2}>0$
We know, $-\frac{\pi }{2}<{\tan }^{-1}x<\frac{\pi }{2}$
Let $t=\left[{\tan }^{-1}x\right]$, then inequality becomes
$t^2-\frac{3t}{2}+\frac{1}{2}>0$
$\Rightarrow (t-\frac{1}{2})(t-1)>0$
$\Rightarrow t<\frac{1}{2}\text{or}t>1$
$\Rightarrow \left[{\tan }^{-1}x\right]<\frac{1}{2}\text{or}\left[{\tan }^{-1}x\right]>1$
$\Rightarrow -\frac{\pi }{2}<{\tan }^{-1}x<1\text{or}{\tan }^{-1}x\ge 2$
$\Rightarrow -\infty <x<\tan 1\text{or}x\in \varphi [\because -\frac{\pi }{2}<{\tan }^{-1}x<\frac{\pi }{2}]$
$\therefore$ Set of values of $x$ is $(-\infty ,\tan 1)$

And $(-\infty ,-\tan 1]\subset (-\infty ,\tan 1)$
also $(-\infty ,\tan \frac{1}{7}]\subset (-\infty ,\tan 1)$