Given set A = x | 1 <= x2 < 9 and set B = 0, 1, 2, 3, 4, find the intersection of A and B, assuming x represents integers.

Set $A$ is defined as $\{x \mid 1 \leq x^2 < 9\}$. This inequality can be broken down into two parts:

1. $1 \leq x^2$
2. $x^2 < 9$

For $1 \leq x^2$:
Taking the square root of both sides, we get $\sqrt{1} \leq \sqrt{x^2}$, which simplifies to $1 \leq |x|$.
This means $x \leq -1$ or $x \geq 1$.

For $x^2 < 9$:
Taking the square root of both sides, we get $\sqrt{x^2} < \sqrt{9}$, which simplifies to $|x| < 3$.
This means $-3 < x < 3$.

Combining these conditions, we need $x$ such that ($x \leq -1$ or $x \geq 1$) AND ($-3 < x < 3$).
The values of $x$ that satisfy both conditions are:
$-3 < x \leq -1$ or $1 \leq x < 3$.

Since set $B$ consists of integers and the options are sets of integers, we assume $x$ must be an integer.
For the interval $-3 < x \leq -1$, the integers are $\{-2, -1\}$.
For the interval $1 \leq x < 3$, the integers are $\{1, 2\}$.
So, the integer elements of set $A$ are $A = \{-2, -1, 1, 2\}$.

Set $A = \{-2, -1, 1, 2\}$
Set $B = \{0, 1, 2, 3, 4\}$
The intersection $A \cap B$ consists of elements that are present in both sets.
Comparing the elements, we find that 1 and 2 are common to both sets.
Therefore, $A \cap B = \{1, 2\}$.

The calculated intersection is $\{1, 2\}$, which corresponds to option B.

The final answer is $\text{B. } \{1, 2\}$.