Question

If $\alpha$ and $\beta$ are roots of the equation $x^2+5|x|-6=0$ then the value of $|{\tan }^{-1}\alpha -{\tan }^{-1}\beta |$ is

• A. $\frac{\pi }{2}$
• B. $0$
• C. $\pi$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A $\frac{\pi }{2}$

$x^2+5|x|-6=0$

$|x{|}^2+5|x|-6=0$

$|x{|}^2+6|x|-|x|-6=0$

$|x|(|x|+6)-1(|x|+6)=0$

$(|x|-1)(|x|+6)=0$

$⟹|x|=1$ or $|x|=6$

But $|x|\ne -6$ (since modulus can not give negative values)

$\therefore |x|=1$

$\therefore x=\pm 1$

$\alpha =1,\beta =-1$

$\therefore |{\tan }^{-1}\alpha -{\tan }^1\beta |=|{\tan }^{-1}1-{\tan }^{-1}\left(1\right)|$

$=|\frac{\pi }{4}-(-\frac{\pi }{4})|$

$=\left|\frac{\pi }{2}\right|$