यदि \(A={x:x\in\mathbb{N},;x\le 12}\), \(B={x:x\in A,;x सम है}\), और (C={x:\(x\in A\),;x अभाज्य है\(}), तो (B\cup C) में कितने तत्व हैं\)?
If \(A={x:x\in\mathbb{N},;x\le 12}\), \(B={x:x\in A,;x is even}\), and (C={x:\(x\in A\),;x is prime\(}), how many elements are in (B\cup C)\)?
Explanation opens after your attempt
A. (9)
Concept
Here \(B=\{2,4,6,8,10,12\}\) and \(C=\{2,3,5,7,11\}\). Counting common (2) once gives (9) elements.
Why this answer is correct
The correct answer is A. (9). Here \(B=\{2,4,6,8,10,12\}\) and \(C=\{2,3,5,7,11\}\). Counting common (2) once gives (9) elements.
Exam Tip
\(B=\{2,4,6,8,10,12\}\) और \(C=\{2,3,5,7,11\}\) हैं। सामान्य (2) को एक बार गिनकर कुल (9) तत्व मिलते हैं।
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