Question

Let $S_1$ is the complete solution set of the inequality ${\cos }^{-1}x>{\cos }^{-1}{x}^2$ and $S_2$ is the complete solution set of the inequality ${\cot }^{-1}{x}^2-5{\cot }^{-1}x+6>0$, then $S_1\cap S_2$ is

• A. $-1,0$
• B. $\cot 3,0$
• C. $\cot 2,0$
• D. $(-1,\cot 2)$

Answer 1

Answer: (C)

${\cos }^{-1}x>{\cos }^{-1}{x}^2$
$\Rightarrow x<x^2\&x\in -1,1$
$\{\because {\cos }^{-1}x$ is a decreasing function $\}$
$\Rightarrow xx-1>0\&x\in -1,1$
$\Rightarrow x\in -\infty ,0\cup 1,\infty \&x\in -1,1$
$\Rightarrow x\in -1,0\dots 1$
Now ${\cot }^{-1}{x}^2-5{\cot }^{-1}x+6>0$
$\Rightarrow {\cot }^{-1}x-2{\cot }^{-1}x-3>0$
$\Rightarrow {\cot }^{-1}x<2{\text{ or }}^{-1}{\cot }^{-1}x>3$
$x>\cot 2$ or $x<\cot 3\dots 2$
$\{\because {\cot }^{-1}x$ is a decreasing function $\}$

$1\cap 2\Rightarrow x\in \cot 2,0$