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

If (f(x)=x-2-1) and (g(x)=x-1), what are the rule and domain of \(\frac{f}{g}\)?

Explanation opens after your attempt
Correct Answer

A. \(x+1,\ \mathbb{R}-{1}\)

Step 1

Concept

\(\frac{x^2-1}{x-1}=x+1\), but (x=1) makes the original denominator zero. After cancellation, remember the removed point.

Step 2

Why this answer is correct

The correct answer is A. \(x+1,\ \mathbb{R}-{1}\). \(\frac{x^2-1}{x-1}=x+1\), but (x=1) makes the original denominator zero. After cancellation, remember the removed point.

Step 3

Exam Tip

\(\frac{x^2-1}{x-1}=x+1\), पर (x=1) मूल हर को शून्य करता है। रद्द करने के बाद भी हटाए गए बिंदु याद रखें।

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Mathematics Answer, Explanation and Revision Hints

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

Correct Answer: A. \(x+1,\ \mathbb{R}-{1}\). Explanation: \(\frac{x^2-1}{x-1}=x+1\), पर (x=1) मूल हर को शून्य करता है। रद्द करने के बाद भी हटाए गए बिंदु याद रखें। / \(\frac{x^2-1}{x-1}=x+1\), but (x=1) makes the original denominator zero. After cancellation, remember the removed point.

Which concept should I revise for this Mathematics MCQ?

\(\frac{x^2-1}{x-1}=x+1\), but (x=1) makes the original denominator zero. After cancellation, remember the removed point.

What exam hint can help solve this Mathematics question?

\(\frac{x^2-1}{x-1}=x+1\), पर (x=1) मूल हर को शून्य करता है। रद्द करने के बाद भी हटाए गए बिंदु याद रखें।