यदि (f(x)=\frac{1}{x-2-9}) और (g(x)=x-2) हैं, तो (\left\(\frac{g}{f}\right\)(x)) का प्रांत क्या है?

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

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R}\setminus{-3,3}\)

Step 1

Concept

For (f), \(x^2-9\ne 0\), so \(x\ne \pm3\), and (f(x)) is never (0). Hence only (-3) and (3) are excluded.

Step 2

Why this answer is correct

The correct answer is A. \(\mathbb{R}\setminus{-3,3}\). For (f), \(x^2-9\ne 0\), so \(x\ne \pm3\), and (f(x)) is never (0). Hence only (-3) and (3) are excluded.

Step 3

Exam Tip

(f) के लिए \(x^2-9\ne 0\), इसलिए \(x\ne \pm3\), और (f(x)) कभी (0) नहीं होता। अतः केवल (-3) और (3) हटेंगे।

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Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=\frac{1}{x-2-9}) और (g(x)=x-2) हैं, तो (\left\(\frac{g}{f}\right\)(x)) का प्रांत क्या है? / If (f(x)=\frac{1}{x-2-9}) and (g(x)=x-2), what is the domain of (\left\(\frac{g}{f}\right\)(x))?

Correct Answer: A. \(\mathbb{R}\setminus{-3,3}\). Explanation: (f) के लिए \(x^2-9\ne 0\), इसलिए \(x\ne \pm3\), और (f(x)) कभी (0) नहीं होता। अतः केवल (-3) और (3) हटेंगे। / For (f), \(x^2-9\ne 0\), so \(x\ne \pm3\), and (f(x)) is never (0). Hence only (-3) and (3) are excluded.

Which concept should I revise for this Mathematics MCQ?

For (f), \(x^2-9\ne 0\), so \(x\ne \pm3\), and (f(x)) is never (0). Hence only (-3) and (3) are excluded.

What exam hint can help solve this Mathematics question?

(f) के लिए \(x^2-9\ne 0\), इसलिए \(x\ne \pm3\), और (f(x)) कभी (0) नहीं होता। अतः केवल (-3) और (3) हटेंगे।