Question

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

• A. $(\cot 5,\cot 2)$
• B. $(-\infty ,\cot 5)\cup (\cot 2,\infty )$
• C. $(-\infty ,\cot 5)$
• D. $(cot2,\infty )$

Answer 1

Answer: (D)

$(cot2,\infty )$

Let $co{t}^{-1}\left(x\right)=y$
$\Rightarrow y^2-7y+10>0$
$\Rightarrow (y-2)(y-5)>0$
$\Rightarrow y<2$ or $y>5$
$\Rightarrow co{t}^{-1}x<2$ and $co{t}^{-1}x>5$
The range of inverse co-tangent function is $(0,\pi )$.
Hence, $co{t}^{-1}x$ cannot be greater than $5$.
Hence, $co{t}^{-1}x<2$ .
Hence, the solution set is $(cot2,\infty )$
(Note : cot 2 is negative)