यदि (f(x)=\sqrt{x-2-1}) और (g(x)=\sqrt{4-x-2}) हैं, तो ((f+g)(x)) का प्रांत क्या है?

If (f(x)=\sqrt{x-2-1}) and (g(x)=\sqrt{4-x-2}), what is the domain of ((f+g)(x))?

Explanation opens after your attempt
Correct Answer

A. \([-2,-1]\cup[1,2]\)

Step 1

Concept

First \(x^2-1\ge0\) gives \(|x|\ge1\), and \(4-x^2\ge0\) gives \(|x|\le2\). The intersection is \([-2,-1]\cup[1,2]\).

Step 2

Why this answer is correct

The correct answer is A. \([-2,-1]\cup[1,2]\). First \(x^2-1\ge0\) gives \(|x|\ge1\), and \(4-x^2\ge0\) gives \(|x|\le2\). The intersection is \([-2,-1]\cup[1,2]\).

Step 3

Exam Tip

पहले \(x^2-1\ge0\) से \(|x|\ge1\), और \(4-x^2\ge0\) से \(|x|\le2\)। प्रतिच्छेद \([-2,-1]\cup[1,2]\) है।

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Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=\sqrt{x-2-1}) और (g(x)=\sqrt{4-x-2}) हैं, तो ((f+g)(x)) का प्रांत क्या है? / If (f(x)=\sqrt{x-2-1}) and (g(x)=\sqrt{4-x-2}), what is the domain of ((f+g)(x))?

Correct Answer: A. \([-2,-1]\cup[1,2]\). Explanation: पहले \(x^2-1\ge0\) से \(|x|\ge1\), और \(4-x^2\ge0\) से \(|x|\le2\)। प्रतिच्छेद \([-2,-1]\cup[1,2]\) है। / First \(x^2-1\ge0\) gives \(|x|\ge1\), and \(4-x^2\ge0\) gives \(|x|\le2\). The intersection is \([-2,-1]\cup[1,2]\).

Which concept should I revise for this Mathematics MCQ?

First \(x^2-1\ge0\) gives \(|x|\ge1\), and \(4-x^2\ge0\) gives \(|x|\le2\). The intersection is \([-2,-1]\cup[1,2]\).

What exam hint can help solve this Mathematics question?

पहले \(x^2-1\ge0\) से \(|x|\ge1\), और \(4-x^2\ge0\) से \(|x|\le2\)। प्रतिच्छेद \([-2,-1]\cup[1,2]\) है।