फलन \(y=\sqrt{x^2-9}\) के ग्राफ का डोमेन क्या है?

What is the domain of the graph of \(y=\sqrt{x^2-9}\)?

Explanation opens after your attempt
Correct Answer

A. (\(-\infty,-3]\cup[3,\infty\))

Step 1

Concept

For the square root \(x^2-9\ge 0\) is required. This gives \(x\le -3\) or \(x\ge 3\).

Step 2

Why this answer is correct

The correct answer is A. (\(-\infty,-3]\cup[3,\infty\)). For the square root \(x^2-9\ge 0\) is required. This gives \(x\le -3\) or \(x\ge 3\).

Step 3

Exam Tip

वर्गमूल के लिए \(x^2-9\ge 0\) चाहिए। इससे \(x\le -3\) या \(x\ge 3\) मिलता है।

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Mathematics Answer, Explanation and Revision Hints

फलन \(y=\sqrt{x^2-9}\) के ग्राफ का डोमेन क्या है? / What is the domain of the graph of \(y=\sqrt{x^2-9}\)?

Correct Answer: A. (\(-\infty,-3]\cup[3,\infty\)). Explanation: वर्गमूल के लिए \(x^2-9\ge 0\) चाहिए। इससे \(x\le -3\) या \(x\ge 3\) मिलता है। / For the square root \(x^2-9\ge 0\) is required. This gives \(x\le -3\) or \(x\ge 3\).

Which concept should I revise for this Mathematics MCQ?

For the square root \(x^2-9\ge 0\) is required. This gives \(x\le -3\) or \(x\ge 3\).

What exam hint can help solve this Mathematics question?

वर्गमूल के लिए \(x^2-9\ge 0\) चाहिए। इससे \(x\le -3\) या \(x\ge 3\) मिलता है।