Question

If $'\alpha'$ is the only real root of the equation $x^3+bx^2+cx+1=0(b+c)$ , then the value of $\tan^{-1}(\alpha)+\tan^{-1}(\alpha^{-1})$ is equal to

• A. $-\frac{\pi}{2}$
• B. $\frac{\pi}{2}$
• C. 0
• D. cannot be determined

Answer 1

Answer: (A)

Let $ f\left(x\right) = x^{3}+bx^{2} +cx+1$
$\therefore f\left(0\right) = 1>, f\left(-1\right)= b-c <0 $
$ \therefore -1 < \alpha < 0$ and
$ \therefore tan^{-1}\left(\alpha\right) +tan^{-1} \frac{1}{\alpha} = -\frac{\pi}{2} $