फलन \(y=\sqrt{x^2-16}\) का डोमेन क्या है?
What is the domain of \(y=\sqrt{x^2-16}\)?
Explanation opens after your attempt
Correct Answer
A. (\(-\infty,-4]\cup[4,\infty\))
Concept
For the square root \(x^2-16\ge 0\) is required. Hence \(x\le -4\) or \(x\ge 4\).
Why this answer is correct
The correct answer is A. (\(-\infty,-4]\cup[4,\infty\)). For the square root \(x^2-16\ge 0\) is required. Hence \(x\le -4\) or \(x\ge 4\).
Exam Tip
वर्गमूल के लिए \(x^2-16\ge 0\) चाहिए। इसलिए \(x\le -4\) या \(x\ge 4\) है।
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