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Question

Answer 1

Lets see the strings generated by G and G'.

For G,
S -> bAcAb -> bcAb ->bcb

S -> bbSbb -> bbAcAbb -> bbcbb

S -> bbSbb -> bbAcAbb -> bbcAb -> bbcb

So, G is generating all strings where number of b's to the left of 'c' in the input string is greater than or equal to the number of b's to the right. So, L(G) is context-free but not regular.

For G' we have an extra production bA -> A which can condense any number of b's. So, G' will generate all strings over b and c containing at least one b before and after a c. This language is hence regular. 

So, (B) is correct.