GATE CS 2013 Q8 — Regular and Context-free Languages

To find the language represented by $L_1 L_2^* \cup L_1^*$, we evaluate each part of the expression step-by-step using the given languages $L_1 = \phi$ and $L_2 = \{a\}$:

1. Evaluate $L_1 L_2^*$:
   The Kleene closure of $L_2$ is $L_2^* = \{a\}^* = \{\epsilon, a, aa, aaa, \dots\}$.
   Concatenating the empty language $\phi$ with any language $L$ yields the empty language $\phi$.
   $$L_1 L_2^* = \phi \cdot \{a\}^* = \phi$$
2. Evaluate $L_1^*$:
   By definition, the Kleene star of any language $L$ always contains the empty string $\epsilon$ because $L^* = L^0 \cup L^1 \cup L^2 \cup \dots$, and $L^0 = \{\epsilon\}$.
   For the empty language $L_1 = \phi$:
   $$L_1^* = \phi^* = \{\epsilon\}$$
3. Evaluate the union:
   Combining the two parts using the union operator:
   $$L_1 L_2^* \cup L_1^* = \phi \cup \{\epsilon\} = \{\epsilon\}$$

Therefore, the expression represents the language $\{\epsilon\}$, which corresponds to Option A.