If $(\cot^{-1} x)^2 -7(\cot^{-1}x)+10>0$ then $x$ lies in the interval?(One or more than one correct)
A)$(\cot 5, \cot 2)$
B)$(-\infty,\cot 5)\cup (\cot2,\infty)$
C)$(-\infty,\cot 5)$
D)$(\cot2,\infty)$
My attempt-
I reached upto this point- $(t-5)(t-2)$ where $t=\cot^{-1}x$.
I dont know what to do after that since $\cot^{-1}$ is a decresing function will it have any impact ?
From your analysis, you should see that $\cot^{-1}(x) \in R \setminus [2,5]$. But taking range of $\cot^{-1}(x)$ to be $(0, \pi)$, we can only have $(0,2)$ as solution
$$\cot^{-1}(x) \in (0,2) \\ \implies x \in (\cot(2),\lim_{x \to 0^{+}} \cot (x))$$
Or equivalently $x \in (\cot(2), \infty)$