If $\alpha $ is the only real root of the equation $x^3 + bx^2 + cx + 1 = 0 (b < c)$, then the value of $tan^{-1} \alpha + tan^{-1}\left(\frac{1}{\alpha }\right)$ is equal to
Answer & explanation
Correct answer: option 2
Let $f(x)=x^3 + bx^2 + cx + 1$.Then,
$f(0) = 1 > 0 $ and $ f(-1) = b - c< 0 $ [∵ b < c ]
$⇒ \alpha $ lies between -1 and 0.
$⇒ \alpha < 0 $
$⇒ tan^{-1}\left(\frac{1}{\alpha }\right) = - \pi + cot^{-1} \alpha $
$⇒ tan^{-1} \alpha + tan^{-1}\frac{1}{\alpha }= -\pi + tan^{-1}\alpha + cot^{-1} \alpha $
$⇒ tan^{-1} \alpha + tan^{-1}\frac{1}{\alpha } = - \pi + \frac{\pi}{2} = -\frac{\pi}{2}$