Output using DeMorgan's theorem can be written as,
Y=((A⋅A⋅B)⋅(A⋅B⋅B))=(A⋅A⋅B)+(A⋅B⋅B)=(A⋅A⋅B)+(A⋅B⋅B)=(A⋅(A+B))+((A+B)⋅B)=A⋅A+A⋅B+A⋅B+B⋅B=(A+B)⋅(A+B)
Which represents XOR gate.
Therefore, required truth table will be
[A0011B0101Y0110]
JEE Main 2023 — Physics Modern Physics
2023mcqeasy
The output Y for the inputs A and B of circuit is given by
Truth table of the shown circuit is :
Official previous-year question
Held on 30 Jan 2023 · Verified 6 Jul 2026.
Options
- A
[A0011B0101Y1110]
- B
[A0011B0101Y1001]
- C
[A0011B0101Y0111]
- D
[A0011B0101Y0110]
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