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

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

Explanation opens after your attempt
Correct Answer

A. (2x), \(x\ne 0\)

Step 1

Concept

The \(\frac{1}{x}\) terms cancel, but (x=0) remains outside the domain. Keep original restrictions even after simplification.

Step 2

Why this answer is correct

The correct answer is A. (2x), \(x\ne 0\). The \(\frac{1}{x}\) terms cancel, but (x=0) remains outside the domain. Keep original restrictions even after simplification.

Step 3

Exam Tip

\(\frac{1}{x}\) वाले पद कटते हैं, पर (x=0) प्रांत से बाहर रहता है। अभिव्यक्ति सरल होने पर भी मूल प्रतिबंध रखें।

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Mathematics Answer, Explanation and Revision Hints

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

Correct Answer: A. (2x), \(x\ne 0\). Explanation: \(\frac{1}{x}\) वाले पद कटते हैं, पर (x=0) प्रांत से बाहर रहता है। अभिव्यक्ति सरल होने पर भी मूल प्रतिबंध रखें। / The \(\frac{1}{x}\) terms cancel, but (x=0) remains outside the domain. Keep original restrictions even after simplification.

Which concept should I revise for this Mathematics MCQ?

The \(\frac{1}{x}\) terms cancel, but (x=0) remains outside the domain. Keep original restrictions even after simplification.

What exam hint can help solve this Mathematics question?

\(\frac{1}{x}\) वाले पद कटते हैं, पर (x=0) प्रांत से बाहर रहता है। अभिव्यक्ति सरल होने पर भी मूल प्रतिबंध रखें।