फलन (f(x)=\sqrt{x-2-16}) का प्रांत ज्ञात कीजिए।

Find the domain of (f(x)=\sqrt{x-2-16}).

Explanation opens after your attempt
Correct Answer

A. (\(-\infty,-4]\cup[4,\infty\))

Step 1

Concept

For the square root \(x^2-16\ge 0\), so \(x^2\ge 16\). Hence \(x\le -4\) or \(x\ge 4\).

Step 2

Why this answer is correct

The correct answer is A. (\(-\infty,-4]\cup[4,\infty\)). For the square root \(x^2-16\ge 0\), so \(x^2\ge 16\). Hence \(x\le -4\) or \(x\ge 4\).

Step 3

Exam Tip

वर्गमूल के लिए \(x^2-16\ge 0\), इसलिए \(x^2\ge 16\)। अतः \(x\le -4\) या \(x\ge 4\)।

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Mathematics Answer, Explanation and Revision Hints

फलन (f(x)=\sqrt{x-2-16}) का प्रांत ज्ञात कीजिए। / Find the domain of (f(x)=\sqrt{x-2-16}).

Correct Answer: A. (\(-\infty,-4]\cup[4,\infty\)). Explanation: वर्गमूल के लिए \(x^2-16\ge 0\), इसलिए \(x^2\ge 16\)। अतः \(x\le -4\) या \(x\ge 4\)। / For the square root \(x^2-16\ge 0\), so \(x^2\ge 16\). Hence \(x\le -4\) or \(x\ge 4\).

Which concept should I revise for this Mathematics MCQ?

For the square root \(x^2-16\ge 0\), so \(x^2\ge 16\). Hence \(x\le -4\) or \(x\ge 4\).

What exam hint can help solve this Mathematics question?

वर्गमूल के लिए \(x^2-16\ge 0\), इसलिए \(x^2\ge 16\)। अतः \(x\le -4\) या \(x\ge 4\)।