JEE Main 17 March 2021 Shift 2 Solved Paper

The given circuit, with the output of the respective gates is as given below.The Boolean expression of the output is
Y=(A⋅B)⋅(A+B)
⇒‌‌Y=(A+B)⋅(A+B)
(using de-Morgan's theorem, x⋅y=x+y )
⇒‌‌Y=A⋅B+B⋅A
This represents the Boolean expression for XOR gate.
Alternate solution
This question can also be verified from the following truth table.

A	B	P=A⋅B	Q=A+B	Y=P⋅Q
0	0	1	0	0
0	1	1	1	1
1	0	1	1	1
1	1	0	1	0

Hence, the output of this circuit represents the output of a XOR gate.