We are given:
1. $\(x>x^2\)$ (which implies $\(0<x<1\)$ )
2. $\(y=3-|x-2|\)$

We need to compare:
- Quantity A: $y$
- Quantity B: 2

Step 1: Solve $\(x>x^2\)$
This inequality simplifies to:

$$
\(\begin{gathered}
x^2-x<0 \\
x(x-1)<0
\end{gathered}\)
$$


The solution is $\(0<x<1\)$.
Step 2: Analyze $\(y=3-|x-2|\)$
Given $\(0<x<1\)$, let's evaluate $\(|x-2|\)$ :
- Since $\(x<1, x-2<-1\)$, so $\(|x-2|=2-x\)$.

Thus:

$$
\(y=3-(2-x)=1+x\)
$$


Step 3: Determine the Range of $y$
Since $\(0<x<1\)$ :

$$
\(\begin{gathered}
1+0<y<1+1 \\
1<y<2
\end{gathered}\)
$$


Step 4: Compare $y$ and 2
From the range $\(1<y<2\)$ :
- $y$ is always less than 2 (since $y$ approaches but never reaches 2 ).

Verification with Examples
1. For $\(x=0.5\)$ :

$$
\(y=3-|0.5-2|=3-1.5=1.5\)
$$


$$
\(1.5<2 \rightarrow A<B .\)
$$

2. For $\(x \rightarrow 1^{-}\)$:

$$
\(y \rightarrow 3-|1-2|=2^{-}\)
$$


Conclusion
For all valid $\(x(0<x<1)\)$, Quantity A ( $\(y\)$ ) is less than Quantity B (2).