यदि \(U=\mathbb{R}\), \(A={x:x^2-4x-12<0}\), तो (A') क्या है?
If \(U=\mathbb{R}\), \(A={x:x^2-4x-12<0}\), what is (A')?
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,-2]\cup[6,\infty\))
Concept
\(x^2-4x-12<0\Rightarrow -2<x<6\). Its complement is \(x\le -2\) or \(x\ge 6\), that is (\(-\infty,-2]\cup[6,\infty\)).
Why this answer is correct
The correct answer is A. (\(-\infty,-2]\cup[6,\infty\)). \(x^2-4x-12<0\Rightarrow -2<x<6\). Its complement is \(x\le -2\) or \(x\ge 6\), that is (\(-\infty,-2]\cup[6,\infty\)).
Exam Tip
\(x^2-4x-12<0\Rightarrow -2<x<6\)। इसका पूरक \(x\le -2\) या \(x\ge 6\), यानी (\(-\infty,-2]\cup[6,\infty\)) है।
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