Given (XY)’+X’Y = Z,final (XZ)’ + X’Z
Given (XY)’+X’Y = Z,final (XZ)’ + X’Z
Answer: A. X’+Y — Solution: simplify step by step. Start with Z = (XY)' + X'Y. Apply De Morgan: (XY)' = X' + Y', so Z = X' + Y' + X'Y. Use absorption: X' + X'Y = X', so Z = X'…
- A.
X’+Y
- B.
X+Y’
- C.
X’+Y’
- D.
X+Y
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Correct answer: A
Solution: simplify step by step.
Start with Z = (XY)' + X'Y.
Apply De Morgan: (XY)' = X' + Y', so Z = X' + Y' + X'Y.
Use absorption: X' + X'Y = X', so Z = X' + Y'.
Now compute the target expression (XZ)' + X'Z using Z = X' + Y'.
Compute XZ: XZ = X(X' + Y') = XX' + XY' = XY' (since XX' = 0).
Then (XZ)' = (XY')' = X' + Y (by De Morgan).
Compute X'Z: X'Z = X'(X' + Y') = X' (since X' + X'Y' = X').
Combine: (XZ)' + X'Z = (X' + Y) + X' = X' + Y.
Answer: X' + Y
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