Q. $x^{2} - 3|x| + 2 < 0,\$then$\ x$ belongs to
A) $(1,\ 2)$
B) $( - 2, - 1)$
C) $( - 2, - 1) \cup (1,2)$
D) $( - 3,\ 5)$
Explanation
(c)
$x^{2} - 3|x| + 2 < 0$
$\Rightarrow$ $|x|^{2} - 3|x| + 2 < 0$
$\Rightarrow \left( |x| - 1 \right)\left( |x| - 2 \right) < 0$
$\Rightarrow 1 < |x| < 2$
$\Rightarrow - 2 < x < - 1\$or$\ 1 < x < 2$
$\therefore x\ \epsilon( - 2, - 1) \cup (1,2)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.