Consider the DFA given below: Which of the following are FALSE? 1. Complement…

Summary of the automaton:

From the start state, reading a 0 goes immediately to the accepting state, and reading any number of 1s stays in a non-accepting region until a 0 is seen. Once a 0 has been seen the machine stays in an accepting sink. Therefore the language accepted is the set of all binary strings that contain at least one 0.

1. Statement 1: "Complement of L(A) is context-free." Verdict: TRUE.

   Reason: The complement is the set of strings with no 0, i.e. all-1 strings (including the empty string). That set is regular (hence context-free).

2. Statement 2: "L(A) = L((11*0 + 0)(0+1)*0*1*)" Verdict: TRUE.

   Reason: The language is all strings containing at least one 0. The given regular expression forces at least one 0 (the left factor ensures a 0 appears) and otherwise allows arbitrary 0/1 content. Thus it denotes the same set Σ*0Σ*.

3. Statement 3: "For the language accepted by A, A is the minimal DFA." Verdict: FALSE.

   Reason: The automaton has two distinct non-accepting states that are equivalent: from either of those states the same set of suffixes leads to acceptance (any suffix containing a 0). They can be merged to obtain a 2-state DFA (start: no 0 seen; accepting sink: 0 seen). Thus the given DFA is not minimal.

4. Statement 4: "A accepts all strings over {0,1} of length at least 2." Verdict: FALSE.

   Reason: Length alone does not guarantee acceptance. For example the string "11" has length 2 but contains no 0, so it is rejected by the automaton.

Conclusion: The false statements are the ones about minimality and about accepting all strings of length at least 2. In other words, statements 3 and 4 are false.