Question

Let $\alpha$ be the only real root of the equation $x^{2011}+x^{2010}+\cdots +x^3+x^2+x+a_0=0$, where $a_0$ is a positive number less than 1, then $ta{n}^{-1}\left(\alpha \right)+ta{n}^{-1}\left(\frac{1}{\alpha }\right)$ is equal to

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{-\pi }{2}$
• D. $\pi$

Answer 1

The correct option is C $\frac{-\pi }{2}$

$f\left(x\right)=x^{2011}+x^{2010}+x^{2009}+\cdots +x^2+x+a_0$
$f\left(0\right)=a_0>0$
$f(-1)=a_0-1<0$
So, the root of the equation lies between (-1, 0)
$\Rightarrow$ root of the equation is negative.
Hence, $ta{n}^{-1}\left(\alpha \right)+ta{n}^{-1}\left(\frac{1}{\alpha }\right)=-\frac{\pi }{2}$