A) 0
B) 1
C) 2
D) 4
Correct Answer: A
Solution :
Here, \[x+{{x}^{2}}\,+{{x}^{3}}+{{x}^{4}}+.....\infty =\frac{x}{1-x};\,\,|x|<1\] and \[-6+6x-6{{x}^{2}}+....\infty =\frac{-6}{1+x};\,|x|<1\] So, that \[{{\tan }^{-1}}\,\left( \frac{x}{1-x} \right)+{{\cot }^{-1}}\,\left( \frac{-6}{1+x} \right)=\frac{\pi }{2}\] \[\Rightarrow \,\,\frac{x}{1-x}\,=\frac{-6}{1+x}\]\[\Rightarrow \,{{x}^{2}}+x=-6+6x\] \[\Rightarrow \,{{x}^{2}}-5x+6=0\Rightarrow \,(x-2)\,(x-3)\,=0\Rightarrow \,x=2,3\]But \[-1<x<1\] (Given)You need to login to perform this action.
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