Question

The solution set of inequality $\left(\cot ^{-1} x\right)\left(\tan ^{-1} x\right)+\left(2-\frac{\pi}{2}\right) \cot ^{-1} x-3 \tan ^{-1} x-3\left(2-\frac{\pi}{2}\right)>0$, is

• A. $x \in(\tan 2, \tan 3)$
• B. $x \in(\cot 3, \cot 2)$
• C. $x \in(-\infty, \tan 2) \cup(\tan 3, \infty)$
• D. $x \in(-\infty, \cot 3) \cup(\cot 2, \infty)$

Answer 1

Answer: (B)

Given $\left(\cot ^{-1} x\right)\left(\tan ^{-1} x\right)+\left(2-\frac{\pi}{2}\right) \cot ^{-1} x-3 \tan ^{-1} x-3\left(2-\frac{\pi}{2}\right)>0$
$\Rightarrow \cot ^{-1} x\left(\tan ^{-1} x+2-\frac{\pi}{2}\right)-3\left(\tan ^{-1} x+2-\frac{\pi}{2}\right)>0$
$\left(\right.$ As $\left.\tan ^{-1} x-\frac{\pi}{2}=-\cot ^{-1} x\right)$
$\Rightarrow\left(\cot ^{-1} x-3\right)\left(2-\cot ^{-1} x\right) >0 $ $\Rightarrow\left(\cot ^{-1} x-3\right)\left(\cot ^{-1} x-2\right)< 0$
$\Rightarrow 2 < \cot ^{-1} x < 3 \Rightarrow \cot 3< x <\cot 2 $ (As $\cot ^{-1} x$ is a decreasing function.)
Hence $x \in(\cot 3, \cot 2)$