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

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

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R})

Step 1

Concept

Since \(x^2+1>0\) for all real (x), the denominator is never zero. Therefore the domain is \(\mathbb{R}\).

Step 2

Why this answer is correct

The correct answer is A. \(\mathbb{R}). Since \(x^2+1>0\) for all real (x), the denominator is never zero. Therefore the domain is \(\mathbb{R}\).

Step 3

Exam Tip

क्योंकि \(x^2+1>0\) सभी वास्तविक (x) के लिए है, हर कभी शून्य नहीं होता। इसलिए प्रांत \(\mathbb{R}\) है।

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Mathematics Answer, Explanation and Revision Hints

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

Correct Answer: A. \(\mathbb{R}). Explanation: क्योंकि \(x^2+1>0\) सभी वास्तविक (x) के लिए है, हर कभी शून्य नहीं होता। इसलिए प्रांत \(\mathbb{R}\) है। / Since \(x^2+1>0\) for all real (x), the denominator is never zero. Therefore the domain is \(\mathbb{R}\).

Which concept should I revise for this Mathematics MCQ?

Since \(x^2+1>0\) for all real (x), the denominator is never zero. Therefore the domain is \(\mathbb{R}\).

What exam hint can help solve this Mathematics question?

क्योंकि \(x^2+1>0\) सभी वास्तविक (x) के लिए है, हर कभी शून्य नहीं होता। इसलिए प्रांत \(\mathbb{R}\) है।