यदि (f(x)=\sqrt{9-x-2}) और (g(x)=\frac{1}{x}) हैं, तो (fg) का डोमेन क्या होगा?

If (f(x)=\sqrt{9-x-2}) and (g(x)=\frac{1}{x}), what is the domain of (fg)?

Explanation opens after your attempt
Correct Answer

A. ( [-3,3]-{0} )

Step 1

Concept

For \(\sqrt{9-x^2}\), \(-3\le x\le3\), and for \(\frac{1}{x}\), \(x\neq0\). Hence the domain is ( [-3,3]-{0} ).

Step 2

Why this answer is correct

The correct answer is A. ( [-3,3]-{0} ). For \(\sqrt{9-x^2}\), \(-3\le x\le3\), and for \(\frac{1}{x}\), \(x\neq0\). Hence the domain is ( [-3,3]-{0} ).

Step 3

Exam Tip

\(\sqrt{9-x^2}\) के लिए \(-3\le x\le3\) और \(\frac{1}{x}\) के लिए \(x\neq0\) है। अतः डोमेन ( [-3,3]-{0} ) है।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=\sqrt{9-x-2}) और (g(x)=\frac{1}{x}) हैं, तो (fg) का डोमेन क्या होगा? / If (f(x)=\sqrt{9-x-2}) and (g(x)=\frac{1}{x}), what is the domain of (fg)?

Correct Answer: A. ( [-3,3]-{0} ). Explanation: \(\sqrt{9-x^2}\) के लिए \(-3\le x\le3\) और \(\frac{1}{x}\) के लिए \(x\neq0\) है। अतः डोमेन ( [-3,3]-{0} ) है। / For \(\sqrt{9-x^2}\), \(-3\le x\le3\), and for \(\frac{1}{x}\), \(x\neq0\). Hence the domain is ( [-3,3]-{0} ).

Which concept should I revise for this Mathematics MCQ?

For \(\sqrt{9-x^2}\), \(-3\le x\le3\), and for \(\frac{1}{x}\), \(x\neq0\). Hence the domain is ( [-3,3]-{0} ).

What exam hint can help solve this Mathematics question?

\(\sqrt{9-x^2}\) के लिए \(-3\le x\le3\) और \(\frac{1}{x}\) के लिए \(x\neq0\) है। अतः डोमेन ( [-3,3]-{0} ) है।