Question

If ${\left({\cot }^{-1}x\right)}^2-7\left({\cot }^{-1}x\right)+10>0$,then $x$ lies in the interval

• A. $(\cot 5,\cot 2)$
• B. $(-\infty ,\cot 5)\cup (\cot 2,\infty )$
• C. $(-\infty ,\cot 5)$
• D. $(\cot 2,\infty )$

Answer 1

The correct option is D $(\cot 2,\infty )$

${\left({\cot }^{-1}x\right)}^2-7\left({\cot }^{-1}x\right)+10>0$

$\Rightarrow y^2-7y+10>0$, where $y={\cot }^{-1}x$

$\Rightarrow (y-2)(y-5)>0$

$\Rightarrow y\in (-\infty ,2)\cup (5,\infty )$

$\Rightarrow {\cot }^{-1}x\in (-\infty ,2)\cup (5,\infty )$

But we know that $0<{\cot }^{-1}x<\pi$

$\Rightarrow {\cot }^{-1}x\in (0,2)$

$\therefore x\in (\cot 2,\infty )$