Question Statement
A 2-input XOR gate with inputs $A$ and $B$ produces output $Y = A \oplus B$. If input $B$ is held permanently at logic HIGH ($B = 1$), the circuit functions as
(A)
A buffer ($Y = A$)
(B)
An inverter / NOT gate ($Y = \bar{A}$)
(C)
A constant HIGH ($Y = 1$)
(D)
A constant LOW ($Y = 0$)
Step-by-Step KaTeX Derivation & Detailed Explanation
### Step-by-Step Derivation & Verification: 1. **XOR Gate Definition**: $Y = A \oplus B = A\bar{B} + \bar{A}B$ 2. **Evaluation for Fixed Input $B = 1$**: $Y = A\overline{(1)} + \bar{A}(1) = A(0) + \bar{A} = \bar{A}$ 3. **Conclusion**: The output is the logical inversion of input $A$. Option (B) is correct.