1. Given Constraint


\(x<0\)



Testing Case \(1(x=-1 , i.e., -2<x<0 )\):

Quantity A: \(|-1|+|2|=1+2=3 \)
Quantity B: \(|-1+2|=|1|=1 \)
Result: Quantity A > Quantity B (3 > 1).


Testing Case $2(x=-5$, i.e., $x<-2)$ :

Quantity A: \(|-5|+|2|=5+2=7 \)
Quantity B: \(|-5+2|=|-3|=3 \)
Result: Quantity A > Quantity B (7 > 3).


\section*{Algebraic Proof via Triangle Inequality:}

By the Triangle Inequality:


\(|a+b| \leq|a|+|b|\)


Equality ( \(|a+b|=|a|+|b| \) ) holds if and only if $a$ and $b$ have the same sign (or at least one is 0).

Since x<0 and 2>0, x and 2 strictly have opposite signs. Thus, the strict inequality must hold for all x<0 :


\(|x+2|<|x|+|2|\)


Therefore, Quantity \(\mathrm{A}> \) Quantity B for every possible value of x<0 .

Official Answer:
(A) - Quantity A is greater.