JEE Main 2026 · 22 January · evening · Single correct · +4 / −1
The correct truth table for the given input data of the following logic gate is :
- (a) Inputs | Output; A | B | C | D | Y; 1 | 1 | 0 | 1 | 1; 0 | 0 | 1 | 1 | 0; 1 | 0 | 1 | 0 | 0; 1 | 1 | 1 | 1 | 1
Inputs Output A B C D Y 1 1 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 1 1 - (b) Inputs | Output; A | B | C | D | Y; 1 | 1 | 0 | 1 | 0; 0 | 0 | 1 | 1 | 1; 1 | 0 | 1 | 0 | 1; 1 | 1 | 1 | 1 | 1
Inputs Output A B C D Y 1 1 0 1 0 0 0 1 1 1 1 0 1 0 1 1 1 1 1 1 - (c) Inputs | Output; A | B | C | D | Y; 1 | 1 | 0 | 1 | 1; 0 | 0 | 1 | 1 | 0; 1 | 0 | 1 | 0 | 1; 1 | 1 | 1 | 1 | 0
Inputs Output A B C D Y 1 1 0 1 1 0 0 1 1 0 1 0 1 0 1 1 1 1 1 0 - (d) Inputs | Output; A | B | C | D | Y; 1 | 1 | 0 | 1 | 0; 0 | 0 | 1 | 1 | 0; 1 | 0 | 1 | 0 | 1; 1 | 1 | 1 | 1 | 1
Inputs Output A B C D Y 1 1 0 1 0 0 0 1 1 0 1 0 1 0 1 1 1 1 1 1
Answer and solution
Answer (a) Inputs | Output; A | B | C | D | Y; 1 | 1 | 0 | 1 | 1; 0 | 0 | 1 | 1 | 0; 1 |…
Solution
Physics sir · 3 steps
Two NOTs around the last AND turn this into AB + C'D', and that checks every row in seconds.
- Read the gates. The top branch is AND then NOT, so it gives A̅B̅. The bottom branch is an OR, so it gives C + D. These two go into an AND, and a final NOT gives Y.
Y = A̅B̅·(C+D)̅ = AB + C̅+̅D̅ = AB + C̅ D̅
- Use the rule. By De Morgan's theorem, Y = 1 whenever A = B = 1, or whenever C = D = 0. In every other case Y = 0.
A B C D AB C + D Y = AB + C̅+̅D̅ 1 1 0 1 1 1 1 0 0 1 1 0 1 0 1 0 1 0 0 1 0 1 1 1 1 1 1 1 - So. The output column reads 1, 0, 0, 1 from top to bottom. Only option (a) matches this, so (a) is correct.
The idea to keep. When a NOT follows an AND whose inputs are already inverted, apply De Morgan first. A short Boolean expression lets you check each row at a glance.
Answer from the NTA final answer key. Solution by PrepWiser.