Question

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

A
(cot5,cot2)
B
(,cot5)(cot2,)
C
(,cot5)
D
(cot2,)

Solution

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

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