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