यदि \(U=\mathbb{R}\) और \(A={x:x^2-2x-8\le 0}\), तो (A') क्या है?
If \(U=\mathbb{R}\) and \(A={x:x^2-2x-8\le 0}\), what is (A')?
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,-2\)\cup\(4,\infty\))
Concept
\(x^2-2x-8\le 0\Rightarrow -2\le x\le 4\). Its complement is (x<-2) or (x>4), that is (\(-\infty,-2\)\cup\(4,\infty\)).
Why this answer is correct
The correct answer is A. (\(-\infty,-2\)\cup\(4,\infty\)). \(x^2-2x-8\le 0\Rightarrow -2\le x\le 4\). Its complement is (x<-2) or (x>4), that is (\(-\infty,-2\)\cup\(4,\infty\)).
Exam Tip
\(x^2-2x-8\le 0\Rightarrow -2\le x\le 4\)। इसका पूरक (x<-2) या (x>4), यानी (\(-\infty,-2\)\cup\(4,\infty\)) है।
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