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

Answer: (A)

$x^2+5|x|-6=0$
$|x{|}^2+5|x|-6=0$
$|x{|}^2+6|x|-|x|-6=0$
$|x|(|x|+6)-1(|x|+6)=0$
$(|x|-1)(|x|+6)=0$
$\Rightarrow |x|=1$
or $|x|=6$
But $|x|\ne -6$
(since modulus can not give negative values)
$\therefore |x|=1$
$\therefore x=\pm 1$
$\alpha =1,\beta =-1$
$\therefore \left|{\tan }^{-1}\alpha -{\tan }^1\beta \right|=\left|{\tan }^{-1}1-{\tan }^{-1}(1)\right|$
$=\left|\frac{\pi }{4}-\left(-\frac{\pi }{4}\right)\right|$
$=\left|\frac{\pi }{2}\right|$