Question

If $|2x+1|+|x-2|+|x+3|<8$, then $x\in$

• A. $(-2,1)-\{-\frac{1}{2}\}$
• B. $[-\frac{1}{2},1)$
• C. $(-2,-\frac{1}{2})$
• D. $(-2,1)$

Answer 1

The correct option is D $(-2,1)$

If $x<-3$,
$-2x-1-x+2-x-3<8$
$\Rightarrow -x<2.5$
$\Rightarrow x>-2.5$ No solution

If $-3\le x<-\frac{1}{2}$,
$-2x-1-x+2+x+3<8$
$\Rightarrow x>-2\Rightarrow x\in (-2,-\frac{1}{2})$

If $-\frac{1}{2}\le x<2$,
$2x+1-x+2+x+3<8\Rightarrow x<1\Rightarrow x\in [-\frac{1}{2},1)$

If $x\ge 2$,
$2x+1+x-2+x+3<8$
$\Rightarrow x<1.5$ No solution

Hence, $x\in (-2,1)$