Question:
Find the value of \(F = xy + x’z\) using distributive law.
वितरण नियम (distributive law) का उपयोग करके \(F = xy + x’z\) का मान ज्ञात कीजिए।
Options:
- (x + z)(y + z)
- (x + y’)(x + z)
- (x’ + y)(y + z)
- (x’ + y)(x + z)(y + z)
- (x + z)(y + z)
- (x + y’)(x + z)
- (x’ + y)(y + z)
- (x’ + y)(x + z)(y + z)
Correct Answer: d)
Explanation:
Using Boolean algebra distributive law \(A + BC = (A + B)(A + C)\):
1. Treat \(xy\) as a single term: \(xy + x’z = (xy + x’)(xy + z)\)
2. Distribute \(x’\) over \(xy\): \(xy + x’ = (x’ + x)(x’ + y) = 1 \cdot (x’ + y) = x’ + y\)
3. Distribute \(z\) over \(xy\): \(xy + z = (x + z)(y + z)\)
4. Multiplying these factors gives: \((x’ + y)(x + z)(y + z)\).
Therefore, Option (4) is the correct expansion.
बूलियन बीजगणित के वितरण नियम \(A + BC = (A + B)(A + C)\) का उपयोग करके:
1. \(xy\) को एक पद मानकर: \(xy + x’z = (xy + x’)(xy + z)\)
2. \(xy + x’ = (x’ + x)(x’ + y) = 1 \cdot (x’ + y) = x’ + y\)
3. \(xy + z = (x + z)(y + z)\)
4. दोनों का गुणनफल: \((x’ + y)(x + z)(y + z)\) प्राप्त होता है।
अतः सही उत्तर विकल्प (4) है।