यदि \(U={x:x\in \mathbb{Z},-8\le x\le 8}\) और \(A={x:x^2-4x\le 0}\), तो (A') क्या है?
If \(U={x:x\in \mathbb{Z},-8\le x\le 8}\) and \(A={x:x^2-4x\le 0}\), what is (A')?
Explanation opens after your attempt
A. ({-8,-7,-6,-5,-4,-3,-2,-1,5,6,7,8})
Concept
\(x^2-4x\le 0\Rightarrow 0\le x\le 4\), so \(A=\{0,1,2,3,4\}\). The remaining integers of (U) form (A').
Why this answer is correct
The correct answer is A. ({-8,-7,-6,-5,-4,-3,-2,-1,5,6,7,8}). \(x^2-4x\le 0\Rightarrow 0\le x\le 4\), so \(A=\{0,1,2,3,4\}\). The remaining integers of (U) form (A').
Exam Tip
\(x^2-4x\le 0\Rightarrow 0\le x\le 4\), इसलिए \(A=\{0,1,2,3,4\}\)। (U) के बाकी पूर्णांक (A') बनाते हैं।
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