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Q. The true set of values of $x$ satisfying $|x|-|x-3|=3$, is

Complex Numbers and Quadratic Equations

Solution:

$\text { Case-I : } x <0 $
$- x + x -3=3$
$-3=3 \text { (no solution) }$
$\text { Case-II : } 0 \leq x <3 $
$x+x-3=3 $
$x=3 \text { (no solution) } $
$\text { Case-III : } x \geq 3 $
$x-(x-3)=3 $
$3=3 $
$\therefore x \in[3, \infty)$