Question:
For Boolean variables A and B, \( A \oplus B = A\bar{B} + \bar{A}B \). Then, which of the following statements is/are correct?
For Boolean variables A and B, \( A \oplus B = A\bar{B} + \bar{A}B \). Then, which of the following statements is/are correct?
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XOR is a "difference detector". It outputs 1 only when the two inputs are different:
- \( A \oplus A \): Inputs are identical, so output is always 0 (Statement B is correct).
- \( A \oplus \bar{A} \): Inputs are always opposite, so output is always 1 (Statement D is correct).
- \( 1 \oplus A \): Output is the invert of \( A \) (Statement A is correct).
This conceptual shortcut avoids algebraic derivations.
- \( A \oplus A \): Inputs are identical, so output is always 0 (Statement B is correct).
- \( A \oplus \bar{A} \): Inputs are always opposite, so output is always 1 (Statement D is correct).
- \( 1 \oplus A \): Output is the invert of \( A \) (Statement A is correct).
This conceptual shortcut avoids algebraic derivations.
Updated On: May 28, 2026
- \( 1 \oplus A = \bar{A} \)
- \( A \oplus A = 0 \)
- \( 0 \oplus A = 0 \)
- \( A \oplus \bar{A} = 1 \)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem defines the XOR (exclusive OR) operation \( \oplus \) for Boolean variables and asks us to identify the correct statements from the given algebraic identities.
Step 2: Key Formula or Approach:
We will evaluate the XOR expression \( X \oplus Y = X\bar{Y} + \bar{X}Y \) for each statement using basic Boolean algebra laws:
- Identity law: \( X \cdot 1 = X \) and \( X \cdot 0 = 0 \)
- Complementarity law: \( X \cdot \bar{X} = 0 \) and \( X + \bar{X} = 1 \)
- Involution (Double Negation): \( \overline{(\bar{X})} = X \)
Step 3: Detailed Explanation:
Let's check each statement:
- Option (A): \( 1 \oplus A \)
\[ 1 \oplus A = 1 \cdot \bar{A} + \bar{1} \cdot A \]
Since \( \bar{1} = 0 \):
\[ 1 \oplus A = \bar{A} + 0 \cdot A = \bar{A} \]
This statement is correct.
- Option (B): \( A \oplus A \)
\[ A \oplus A = A \cdot \bar{A} + \bar{A} \cdot A \]
Since \( A \cdot \bar{A} = 0 \):
\[ A \oplus A = 0 + 0 = 0 \]
This statement is correct.
- Option (C): \( 0 \oplus A \)
\[ 0 \oplus A = 0 \cdot \bar{A} + \bar{0} \cdot A \]
Since \( \bar{0} = 1 \):
\[ 0 \oplus A = 0 + 1 \cdot A = A \]
The option states \( 0 \oplus A = 0 \), which is incorrect.
- Option (D): \( A \oplus \bar{A} \)
\[ A \oplus \bar{A} = A \cdot \overline{(\bar{A})} + \bar{A} \cdot A \]
Since \( \overline{(\bar{A})} = A \):
\[ A \oplus \bar{A} = A \cdot A + 0 = A + \bar{A} = 1 \]
This statement is correct.
Step 4: Final Answer:
The correct statements are (A), (B), and (D).
The problem defines the XOR (exclusive OR) operation \( \oplus \) for Boolean variables and asks us to identify the correct statements from the given algebraic identities.
Step 2: Key Formula or Approach:
We will evaluate the XOR expression \( X \oplus Y = X\bar{Y} + \bar{X}Y \) for each statement using basic Boolean algebra laws:
- Identity law: \( X \cdot 1 = X \) and \( X \cdot 0 = 0 \)
- Complementarity law: \( X \cdot \bar{X} = 0 \) and \( X + \bar{X} = 1 \)
- Involution (Double Negation): \( \overline{(\bar{X})} = X \)
Step 3: Detailed Explanation:
Let's check each statement:
- Option (A): \( 1 \oplus A \)
\[ 1 \oplus A = 1 \cdot \bar{A} + \bar{1} \cdot A \]
Since \( \bar{1} = 0 \):
\[ 1 \oplus A = \bar{A} + 0 \cdot A = \bar{A} \]
This statement is correct.
- Option (B): \( A \oplus A \)
\[ A \oplus A = A \cdot \bar{A} + \bar{A} \cdot A \]
Since \( A \cdot \bar{A} = 0 \):
\[ A \oplus A = 0 + 0 = 0 \]
This statement is correct.
- Option (C): \( 0 \oplus A \)
\[ 0 \oplus A = 0 \cdot \bar{A} + \bar{0} \cdot A \]
Since \( \bar{0} = 1 \):
\[ 0 \oplus A = 0 + 1 \cdot A = A \]
The option states \( 0 \oplus A = 0 \), which is incorrect.
- Option (D): \( A \oplus \bar{A} \)
\[ A \oplus \bar{A} = A \cdot \overline{(\bar{A})} + \bar{A} \cdot A \]
Since \( \overline{(\bar{A})} = A \):
\[ A \oplus \bar{A} = A \cdot A + 0 = A + \bar{A} = 1 \]
This statement is correct.
Step 4: Final Answer:
The correct statements are (A), (B), and (D).
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