Consider the DFA A given below. Which of the following are FALSE? 1. Complement of $L(A)$ is context-free. 2. $L(A) = L((11^*0+0)(0 + 1)^*0^*1^*)$ 3. For the language accepted by $A$, $A$ is the minimal DFA. 4. $A$ accepts all strings over $\{0, 1\}$ of length at least 2.

Let the states of DFA $A$ be $q_0$ (initial state, top-left), $q_1$ (top-right), and $q_2$ (accepting state, bottom).

• From $q_0$, on input 0 it transitions directly to $q_2$.

• From $q_0$, on input 1 it transitions to $q_1$. From $q_1$, any sequence of 1's loops at $q_1$, and on the first 0, it transitions to $q_2$.

• Once in $q_2$, it loops on both 0 and 1 (trap/absorbing accept state).

Thus, $L(A)$ accepts all strings that contain at least one 0:

Now evaluate each statement:

1. Statement 1: $L(A)$ is regular, so its complement is also regular and therefore context-free. (TRUE)

2. Statement 2: $(11^*0 + 0)(0+1)^*0^*1^* = (1^+0 + 0)(0+1)^* = 1^*0(0+1)^* = (0+1)^*0(0+1)^* = L(A)$. (TRUE)

3. Statement 3: The language $L(A) = (0+1)^*0(0+1)^*$ can be accepted by a 2-state minimal DFA (state $p_0$ loops on 1 and goes to $p_1$ on 0; $p_1$ is accepting and loops on 0, 1). Thus, $A$ (which has 3 states) is not minimal. (FALSE)

4. Statement 4: String 11 has length 2 and only reaches $q_1$ (non-accepting). So $A$ does not accept all strings of length at least 2. (FALSE)

Therefore, statements 3 and 4 are FALSE, which corresponds to option (D).