यदि (f(x)=\frac{1}{x-2-16}) और (g(x)=x-2+1) हैं, तो (\left\(\frac{g}{f}\right\)(x)) का प्रांत क्या है?

If (f(x)=\frac{1}{x-2-16}) and (g(x)=x-2+1), what is the domain of (\left\(\frac{g}{f}\right\)(x))?

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R}\setminus{-4,4}\)

Step 1

Concept

For (f), \(x^2-16\ne 0\), so \(x\ne -4,4\), and (f(x)\ne 0) always. Hence only (-4) and (4) are excluded.

Step 2

Why this answer is correct

The correct answer is A. \(\mathbb{R}\setminus{-4,4}\). For (f), \(x^2-16\ne 0\), so \(x\ne -4,4\), and (f(x)\ne 0) always. Hence only (-4) and (4) are excluded.

Step 3

Exam Tip

(f) के लिए \(x^2-16\ne 0\), इसलिए \(x\ne -4,4\), और (f(x)\ne 0) हमेशा है। इसलिए केवल (-4) और (4) हटेंगे।

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Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=\frac{1}{x-2-16}) और (g(x)=x-2+1) हैं, तो (\left\(\frac{g}{f}\right\)(x)) का प्रांत क्या है? / If (f(x)=\frac{1}{x-2-16}) and (g(x)=x-2+1), what is the domain of (\left\(\frac{g}{f}\right\)(x))?

Correct Answer: A. \(\mathbb{R}\setminus{-4,4}\). Explanation: (f) के लिए \(x^2-16\ne 0\), इसलिए \(x\ne -4,4\), और (f(x)\ne 0) हमेशा है। इसलिए केवल (-4) और (4) हटेंगे। / For (f), \(x^2-16\ne 0\), so \(x\ne -4,4\), and (f(x)\ne 0) always. Hence only (-4) and (4) are excluded.

Which concept should I revise for this Mathematics MCQ?

For (f), \(x^2-16\ne 0\), so \(x\ne -4,4\), and (f(x)\ne 0) always. Hence only (-4) and (4) are excluded.

What exam hint can help solve this Mathematics question?

(f) के लिए \(x^2-16\ne 0\), इसलिए \(x\ne -4,4\), और (f(x)\ne 0) हमेशा है। इसलिए केवल (-4) और (4) हटेंगे।