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Question

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


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