Consider the following context-free grammar $G$, where $S$, $A$, and $B$ are the variables (non-terminals), $a$ and $b$ are the terminal symbols, $S$ is the start variable, and the rules of $G$ are described as: $$S \rightarrow aaB \mid Abb$$ $$A \rightarrow a \mid aA$$ $$B \rightarrow b \mid bB$$ Which ONE of the languages $L(G)$ is accepted by $G$?

To determine the language generated by the context-free grammar $G$:

1. Language derived from variable $A$: The productions for $A$ are:

   $$A \rightarrow a \mid aA$$

   This generates one or more copies of terminal $a$, i.e., $L(A) = a^+ = \{a^n \mid n \ge 1\}$.

2. Language derived from variable $B$: The productions for $B$ are:

   $$B \rightarrow b \mid bB$$

   This generates one or more copies of terminal $b$, i.e., $L(B) = b^+ = \{b^n \mid n \ge 1\}$.

3. Language derived from the start variable $S$: The productions for $S$ are:

   $$S \rightarrow aaB \mid Abb$$

   • From $S \rightarrow aaB$, the generated strings are $aa b^+ = \{a^2 b^n \mid n \ge 1\}$.

   • From $S \rightarrow Abb$, the generated strings are $a^+ bb = \{a^n b^2 \mid n \ge 1\}$.

Combining both production paths:

Hence, (A) is the correct option.