यदि (f(x)=\frac{1}{x-2-1}) और (g(x)=x+5) हैं, तो ((f-g)(x)) का प्रांत क्या है?

If (f(x)=\frac{1}{x-2-1}) and (g(x)=x+5), what is the domain of ((f-g)(x))?

Explanation opens after your attempt
Correct Answer

A. \(x\ne-1,1\)

Step 1

Concept

In \(\frac{1}{x^2-1}\), \(x^2-1\ne0\), so \(x\ne\pm1\). The polynomial (g(x)) does not restrict the domain.

Step 2

Why this answer is correct

The correct answer is A. \(x\ne-1,1\). In \(\frac{1}{x^2-1}\), \(x^2-1\ne0\), so \(x\ne\pm1\). The polynomial (g(x)) does not restrict the domain.

Step 3

Exam Tip

\(\frac{1}{x^2-1}\) में \(x^2-1\ne0\), इसलिए \(x\ne\pm1\)। बहुपद (g(x)) प्रांत को सीमित नहीं करता।

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Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=\frac{1}{x-2-1}) और (g(x)=x+5) हैं, तो ((f-g)(x)) का प्रांत क्या है? / If (f(x)=\frac{1}{x-2-1}) and (g(x)=x+5), what is the domain of ((f-g)(x))?

Correct Answer: A. \(x\ne-1,1\). Explanation: \(\frac{1}{x^2-1}\) में \(x^2-1\ne0\), इसलिए \(x\ne\pm1\)। बहुपद (g(x)) प्रांत को सीमित नहीं करता। / In \(\frac{1}{x^2-1}\), \(x^2-1\ne0\), so \(x\ne\pm1\). The polynomial (g(x)) does not restrict the domain.

Which concept should I revise for this Mathematics MCQ?

In \(\frac{1}{x^2-1}\), \(x^2-1\ne0\), so \(x\ne\pm1\). The polynomial (g(x)) does not restrict the domain.

What exam hint can help solve this Mathematics question?

\(\frac{1}{x^2-1}\) में \(x^2-1\ne0\), इसलिए \(x\ne\pm1\)। बहुपद (g(x)) प्रांत को सीमित नहीं करता।