Class 11 Mathematics Sets Complement of a Set and its properties Hard Level 21

यदि \(U=\mathbb{R}\) और \(A={x:|x+2|\le 6}\), तो (A') क्या है?

If \(U=\mathbb{R}\) and \(A={x:|x+2|\le 6}\), what is (A')?

Explanation opens after your attempt
Correct Answer

A. (\(-\infty,-8\)\cup\(4,\infty\))

Step 1

Concept

\(|x+2|\le 6\Rightarrow -8\le x\le 4\). Its complement is (x<-8) or (x>4), that is (\(-\infty,-8\)\cup\(4,\infty\)).

Step 2

Why this answer is correct

The correct answer is A. (\(-\infty,-8\)\cup\(4,\infty\)). \(|x+2|\le 6\Rightarrow -8\le x\le 4\). Its complement is (x<-8) or (x>4), that is (\(-\infty,-8\)\cup\(4,\infty\)).

Step 3

Exam Tip

\(|x+2|\le 6\Rightarrow -8\le x\le 4\)। इसका पूरक (x<-8) या (x>4), यानी (\(-\infty,-8\)\cup\(4,\infty\)) है।

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Mathematics Answer, Explanation and Revision Hints

यदि \(U=\mathbb{R}\) और \(A={x:|x+2|\le 6}\), तो (A') क्या है? / If \(U=\mathbb{R}\) and \(A={x:|x+2|\le 6}\), what is (A')?

Correct Answer: A. (\(-\infty,-8\)\cup\(4,\infty\)). Explanation: \(|x+2|\le 6\Rightarrow -8\le x\le 4\)। इसका पूरक (x<-8) या (x>4), यानी (\(-\infty,-8\)\cup\(4,\infty\)) है। / \(|x+2|\le 6\Rightarrow -8\le x\le 4\). Its complement is (x<-8) or (x>4), that is (\(-\infty,-8\)\cup\(4,\infty\)).

Which concept should I revise for this Mathematics MCQ?

\(|x+2|\le 6\Rightarrow -8\le x\le 4\). Its complement is (x<-8) or (x>4), that is (\(-\infty,-8\)\cup\(4,\infty\)).

What exam hint can help solve this Mathematics question?

\(|x+2|\le 6\Rightarrow -8\le x\le 4\)। इसका पूरक (x<-8) या (x>4), यानी (\(-\infty,-8\)\cup\(4,\infty\)) है।