Question

If $\alpha$ is the only real root of the equation $x^3+b{x}^2+cx+1=0(b<c)$ then the value of $ta{n}^{-1}\alpha +ta{n}^{-1}\left(\frac{1}{\alpha }\right)$ is

• A. $\frac{\pi }{2}$
• B. $-\frac{\pi }{2}$
• C. \N
• D. can not be determined

Answer 1

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

$f\left(0\right)=1,f(-1)=b-c<0\Rightarrow \alpha \in (-1,0)\Rightarrow \alpha <0\therefore ta{n}^{-1}\left(\alpha \right)+ta{n}^{-1}\left(\frac{1}{\alpha }\right)=ta{n}^{-1}\left(\alpha \right)-\pi +co{t}^{-1}\alpha =-\pi +\frac{\pi }{2}=-\frac{\pi }{2}$