यदि \(U={x:x\in\mathbb{N},x\le 96}\), \(A={x:x\in U,8\mid x}\) और \(B={x:x\in U,12\mid x}\), तो (n\(A'\cap B'\)) क्या है?
If \(U={x:x\in\mathbb{N},x\le 96}\), \(A={x:x\in U,8\mid x}\), and \(B={x:x\in U,12\mid x}\), what is (n\(A'\cap B'\))?
Explanation opens after your attempt
A. (80)
Concept
By De Morgan's law, (A'\cap B'=\(A\cup B\)'). Since (n\(A\cup B\)=12+8-4=16), the complement has (96-16=80) elements.
Why this answer is correct
The correct answer is A. (80). By De Morgan's law, (A'\cap B'=\(A\cup B\)'). Since (n\(A\cup B\)=12+8-4=16), the complement has (96-16=80) elements.
Exam Tip
डी मॉर्गन से (A'\cap B'=\(A\cup B\)') है। (n\(A\cup B\)=12+8-4=16), इसलिए पूरक में (96-16=80) सदस्य हैं।
Login to save your score, XP, coins and progress.
