JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Semiconductors

The truth table for the circuit given below is :

1. (A)A | B | Y
   0 | 0 | 0
   0 | 1 | 1
   1 | 0 | 1
   1 | 1 | 0✓ Correct
2. (B)A | B | Y
   0 | 0 | 0
   1 | 0 | 0
   1 | 1 | 0
   0 | 1 | 1
3. (C)A | B | Y
   0 | 0 | 0
   1 | 0 | 1
   0 | 1 | 0
   1 | 1 | 0
4. (D)A | B | Y
   0 | 0 | 0
   1 | 1 | 1
   1 | 0 | 1
   0 | 1 | 1

Answer: (A)

Solution

Tracing the two gates feeding the final OR gate, the upper branch produces $A\bar{B}$ and the lower branch produces $\bar{A}B$, and the OR gate combines them:

$$Y=A\bar{B}+\bar{A}B$$

This is exactly the exclusive-OR operation — Y is high only when A and B differ:

$A=0,B=0\Rightarrow Y=0$
$A=0,B=1\Rightarrow Y=1$
$A=1,B=0\Rightarrow Y=1$
$A=1,B=1\Rightarrow Y=0$

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option