Question

Suppose $f\left(x\right)=3x^3-13x^2+14x-2$, it is assumed that $f\left(x\right)=0$ will have 3 root say $\alpha ,\beta$ and $\gamma$, where $\alpha <\beta <\gamma$.

The value of $ta{n}^{-1}\alpha +ta{n}^{-1}\beta +ta{n}^{-1}\gamma$ is:

• A. $\frac{\pi }{2}$
• B. $\frac{3\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\frac{3\pi }{4}$

Answer 1

Answer: (D)

$\frac{3\pi }{4}$

$f\left(x\right)=3x^3-13x^2+14x-2$
$\alpha ,\beta ,\gamma$ are roots
$\therefore \alpha +\beta +\gamma =-\left(\frac{-13}{3}\right)=\frac{13}{3}$
$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{14}{3}$
$\alpha \beta \gamma =-\left(\frac{-2}{3}\right)=\frac{2}{3}$
${\tan }^{-1}\alpha +{\tan }^{-1}\beta +{\tan }^{-1}\gamma$
$={\tan }^{-1}\left[\frac{\alpha +\beta }{1-\alpha \beta }\right]+{\tan }^{-1}\gamma$
$={\tan }^{-1}\left[\frac{\frac{\alpha +\beta }{1-\alpha \beta }+\gamma }{1-\left(\frac{\alpha +\beta }{1-\alpha \beta }\right)\gamma }\right]$
$={\tan }^{-1}\left[\frac{\alpha +\beta +\gamma -\alpha \beta \gamma }{1-\alpha \beta -\alpha \gamma +\beta \gamma }\right]$
$={\tan }^{-1}\left[\frac{(\alpha +\beta +\gamma )-\left(\alpha \beta \gamma \right)}{1-(\alpha \beta +\beta \gamma +\gamma \alpha )}\right]$
$={\tan }^{-1}\left[\frac{\frac{13}{3}-\frac{2}{3}}{1-\frac{14}{3}}\right]$
$={\tan }^{-1}\left[\frac{11}{-11}\right]$
$={\tan }^{-1}(-1)$
$=\frac{3\pi }{4}$