Question 47

1. For $L_1 = \{a^n w a^n \mid w \in \{a, b\}^*\}$:
   Setting $n = 0$, $a^0 w a^0 = w$. Since $w$ is any arbitrary string in $\{a, b\}^*$, setting $n = 0$ allows $L_1$ to generate every possible string in $\{a, b\}^*$. Thus $L_1 = \{a, b\}^*$, which is a regular language.
   (Even if $n \ge 1$ were required, any string starting and ending with 'a' belongs to $L_1$ by setting $n=1$, giving $a\{a, b\}^*a$, which is also regular).

2. For $L_2 = \{w x w^R \mid w, x \in \{a, b\}^*, |w|, |x| > 0\}$:
   Since $|w| > 0$, $|w| \ge 1$. Let $|w| = 1$:

   • If $w = a$, then $w^R = a$, producing strings of the form $a x a$ where $|x| > 0$.
   • If $w = b$, then $w^R = b$, producing strings of the form $b x b$ where $|x| > 0$.
     Any string that starts and ends with the same character and has length at least 3 can be factored into $w x w^R$ with $|w|=1$. Conversely, any string in $L_2$ has length $\ge 3$ and starts and ends with the same character. Thus $L_2 = a(a+b)^+a + b(a+b)^+b$, which is regular.

Since $L_1$ and $L_2$ are both regular, they are also context-free (every regular language is context-free). Therefore, Statement (A), Statement (B), and Statement (C) are all TRUE.