Question

If $\alpha ,\beta$ are the roots of the equation $\left(ta{n}^{-1}\right(x/5){)}^2+(\sqrt{3}-1\left)ta{n}^{-1}\right(x/5)-\sqrt{3}=0,|\alpha |>|\beta |$ then

• A. $\alpha +\beta =-5\pi /12$
• B. $|\alpha -\beta |=35\pi /12$
• C. $\alpha \beta =-25{\pi }^2/12$
• D. $3\alpha +4\beta =0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $3\alpha +4\beta =0$
B $|\alpha -\beta |=35\pi /12$
C $\alpha \beta =-25{\pi }^2/12$
D $\alpha +\beta =-5\pi /12$

$\left(ta{n}^{-1}\right(x/5)+\sqrt{3})\left(ta{n}^{-1}\right(x/5)-1)=0$
$\Rightarrow ta{n}^{-1}\left(x/5\right)=-\sqrt{3}\Rightarrow \frac{x}{5}=-\frac{\pi }{3}\Rightarrow x=-\frac{5\pi }{3}$
and $ta{n}^{-1}\left(x/5\right)=1\Rightarrow \frac{x}{5}=\frac{\pi }{4}\Rightarrow x=\frac{5\pi }{4}$
Let $\alpha =-\frac{5\pi }{3},\beta =\frac{5\pi }{4}$
$\Rightarrow \alpha +\beta =-5\pi /12,|\alpha -\beta |=35\pi /12$
$\alpha \beta =-25{\pi }^2/12$ and $3\alpha +4\beta =0$