Question

If $\alpha ,\beta ,$ and $\gamma$ are the roots of ${\tan }^{-1}x+{\tan }^{-1}(x+1)={\tan }^{-1}3x$, then

• A. $\alpha +\beta +\gamma =0$
• B. $\alpha \beta +\beta \gamma +\gamma \alpha =-1/3$
• C. $\alpha \beta \gamma =1$
• D. ${|\alpha -\beta |}_{max}=1$

Answer 1

Answer: (B)

$\alpha \beta +\beta \gamma +\gamma \alpha =-1/3$

$ta{n}^{-1}\left(x\right)+ta{n}^{-1}(x+1)=ta{n}^{-1}\left(3x\right)$
$ta{n}^{-1}\left(\frac{2x+1}{1-x^2-x}\right)=ta{n}^{-1}3x$
Therefore after taking $tan$ on both the sides, we get
$2x+1=3x(1-x^2-x)$
$2x+1=3x-3x^3-3x^2$
$3x^3+3x^2-x+1=0$

$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =-\frac{1}{3}$