Correct : C

Given:
Context-free grammars:
G1: S → aS | B, B → b | bB
G2: S → aA | bB, A → aA | B | ε, B → bB | ε
To find:
The languages generated by G1 and G2.
Solution:
First, let me analyze grammar G1.
Looking at the non-terminal B:
B → b | bB
This means B produces at least one 'b', so B generates b⁺, or bⁿ where n ≥ 1 (that is, n > 0).
Now looking at the start symbol S:
S → aS | B
S can produce zero or more 'a's before switching to B.
So S generates a*B, which gives a^mbⁿ where m ≥ 0 and n > 0.
Therefore, L(G1) = {a^mbⁿ | m ≥ 0 and n > 0}.
Now let me analyze grammar G2.
First, look at B:
B → bB | ε, which means B generates b* (zero or more 'b's).
Next, look at A:
A → aA | B | ε, which generates a*B = a*b*.
Now, look at the productions for start symbol S:
S → aA | bB
Case 1: S → aA
Since A generates a*b*, aA produces strings with at least one 'a' followed by any number of 'b's (including zero 'b's).
So this case gives a^mbⁿ where m ≥ 1 and n ≥ 0.
Case 2: S → bB
Since B generates b*, bB produces strings with zero 'a's and at least one 'b'.
So this case gives a^mbⁿ where m = 0 and n ≥ 1.
Combining both cases from S, the grammar generates all strings of the form a^mbⁿ except the empty string (m = 0, n = 0).
So at least one of m or n must be strictly positive.
This gives the condition: m > 0 or n > 0.
Therefore, L(G2) = {a^mbⁿ | m > 0 or n > 0}.
Comparing with the given choices, this matches Option D.
Answer:
(D)