यदि \(U=\mathbb{R}\), \(A={x:x^2<16}\), तो (A') क्या होगा?
If \(U=\mathbb{R}\), \(A={x:x^2<16}\), what is (A')?
Explanation opens after your attempt
Correct Answer
A. \(\(-\infty,-4]\cup[4,\infty\))
Concept
\(x^2<16\Rightarrow -4<x<4\), so the complement has \(x\le -4\) or \(x\ge 4\). A strict inequality adds equality in the complement.
Why this answer is correct
The correct answer is A. \(\(-\infty,-4]\cup[4,\infty\)). \(x^2<16\Rightarrow -4<x<4\), so the complement has \(x\le -4\) or \(x\ge 4\). A strict inequality adds equality in the complement.
Exam Tip
\(x^2<16\Rightarrow -4<x<4\), इसलिए पूरक में \(x\le -4\) या \(x\ge 4\) होगा। सख्त असमानता पूरक में बराबरी जोड़ देती है।
Login to save your score, XP, coins and progress.