Difficulty: Easy
Correct Answer: Correct
Explanation:
Introduction / Context: The XOR gate is widely used for parity generation, adders (sum bit), and change detection. This item verifies the canonical XOR condition in two-input form.
Given Data / Assumptions:
Concept / Approach: XOR evaluates to 1 if and only if its inputs differ. It is often read as “either A or B, but not both.” Therefore, stating that the output is HIGH when inputs are unequal is accurate.
Step-by-Step Solution:
1) Recall definition: XOR(A,B) = (A * NOT(B)) + (NOT(A) * B).2) For unequal inputs (01 or 10), one product term is 1 → output = 1.3) For equal inputs (00 or 11), both product terms are 0 → output = 0.4) Conclusion: the statement is correct.Verification / Alternative check: Truth table: 00→0, 01→1, 10→1, 11→0. In full adders, XOR generates the sum bit, further reinforcing the “unequal → 1” behavior.
Why Other Options Are Wrong: “Incorrect” would describe XNOR, not XOR. The options “Only when both inputs are LOW/HIGH” refer to equality cases, which produce 0 for XOR. “Only for TTL families” confuses logic family with function; XOR behavior is universal across families.
Common Pitfalls: Mixing XOR with inclusive OR. Inclusive OR is 1 for 01, 10, and 11; XOR excludes the 11 case, hence “exclusive.”
Final Answer: Correct
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