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Question

Let α be the only real root of the equation x2011+x2010++x3+x2+x+a0=0, where a0 is a positive number less than 1, then tan1(α)+tan1(1α) is equal to

A
0
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B
π2
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C
π2
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D
π
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Solution

The correct option is C π2
f(x)=x2011+x2010+x2009++x2+x+a0
f(0)=a0>0
f(1)=a01<0
So, the root of the equation lies between (-1, 0)
root of the equation is negative.
Hence, tan1(α)+tan1(1α)=π2


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