यदि (f(x)=\sqrt{x-2-4}) और (g(x)=\frac{1}{x-5}) हैं, तो (f+g) का डोमेन क्या होगा?

If (f(x)=\sqrt{x-2-4}) and (g(x)=\frac{1}{x-5}), what is the domain of (f+g)?

Explanation opens after your attempt
Correct Answer

A. ( \(-\infty,-2]\cup[2,\infty\)-{5} )

Step 1

Concept

For \(\sqrt{x^2-4}\), \(x\le-2\) or \(x\ge2\), and for (g), \(x\neq5\). Both conditions must be applied together.

Step 2

Why this answer is correct

The correct answer is A. ( \(-\infty,-2]\cup[2,\infty\)-{5} ). For \(\sqrt{x^2-4}\), \(x\le-2\) or \(x\ge2\), and for (g), \(x\neq5\). Both conditions must be applied together.

Step 3

Exam Tip

\(\sqrt{x^2-4}\) के लिए \(x\le-2\) या \(x\ge2\), और (g) के लिए \(x\neq5\) है। दोनों शर्तें साथ लगेंगी।

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Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=\sqrt{x-2-4}) और (g(x)=\frac{1}{x-5}) हैं, तो (f+g) का डोमेन क्या होगा? / If (f(x)=\sqrt{x-2-4}) and (g(x)=\frac{1}{x-5}), what is the domain of (f+g)?

Correct Answer: A. ( \(-\infty,-2]\cup[2,\infty\)-{5} ). Explanation: \(\sqrt{x^2-4}\) के लिए \(x\le-2\) या \(x\ge2\), और (g) के लिए \(x\neq5\) है। दोनों शर्तें साथ लगेंगी। / For \(\sqrt{x^2-4}\), \(x\le-2\) or \(x\ge2\), and for (g), \(x\neq5\). Both conditions must be applied together.

Which concept should I revise for this Mathematics MCQ?

For \(\sqrt{x^2-4}\), \(x\le-2\) or \(x\ge2\), and for (g), \(x\neq5\). Both conditions must be applied together.

What exam hint can help solve this Mathematics question?

\(\sqrt{x^2-4}\) के लिए \(x\le-2\) या \(x\ge2\), और (g) के लिए \(x\neq5\) है। दोनों शर्तें साथ लगेंगी।