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

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

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

For \(\sqrt{x-2}\), \(x\ge2\), and the denominator needs \(x-5\neq0\). Hence \( [2,\infty\)-{5} ) is correct.

Step 2

Why this answer is correct

The correct answer is A. \( [2,\infty\)-{5} ). For \(\sqrt{x-2}\), \(x\ge2\), and the denominator needs \(x-5\neq0\). Hence \( [2,\infty\)-{5} ) is correct.

Step 3

Exam Tip

\(\sqrt{x-2}\) के लिए \(x\ge2\) और हर \(x-5\neq0\) चाहिए। इसलिए \( [2,\infty\)-{5} ) सही है।

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Mathematics Answer, Explanation and Revision Hints

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

Correct Answer: A. \( [2,\infty\)-{5} ). Explanation: \(\sqrt{x-2}\) के लिए \(x\ge2\) और हर \(x-5\neq0\) चाहिए। इसलिए \( [2,\infty\)-{5} ) सही है। / For \(\sqrt{x-2}\), \(x\ge2\), and the denominator needs \(x-5\neq0\). Hence \( [2,\infty\)-{5} ) is correct.

Which concept should I revise for this Mathematics MCQ?

For \(\sqrt{x-2}\), \(x\ge2\), and the denominator needs \(x-5\neq0\). Hence \( [2,\infty\)-{5} ) is correct.

What exam hint can help solve this Mathematics question?

\(\sqrt{x-2}\) के लिए \(x\ge2\) और हर \(x-5\neq0\) चाहिए। इसलिए \( [2,\infty\)-{5} ) सही है।