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

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

Explanation opens after your attempt
Correct Answer

A. \( \mathbb{R}-{-1,2} \)

Step 1

Concept

The first function needs \(x\neq-1\), and the second needs \(x\neq2\). The domain of the sum is the common part of these conditions.

Step 2

Why this answer is correct

The correct answer is A. \( \mathbb{R}-{-1,2} \). The first function needs \(x\neq-1\), and the second needs \(x\neq2\). The domain of the sum is the common part of these conditions.

Step 3

Exam Tip

पहले फलन में \(x\neq-1\) और दूसरे में \(x\neq2\) चाहिए। जोड़ का डोमेन इन दोनों शर्तों का साझा भाग है।

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Mathematics Answer, Explanation and Revision Hints

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

Correct Answer: A. \( \mathbb{R}-{-1,2} \). Explanation: पहले फलन में \(x\neq-1\) और दूसरे में \(x\neq2\) चाहिए। जोड़ का डोमेन इन दोनों शर्तों का साझा भाग है। / The first function needs \(x\neq-1\), and the second needs \(x\neq2\). The domain of the sum is the common part of these conditions.

Which concept should I revise for this Mathematics MCQ?

The first function needs \(x\neq-1\), and the second needs \(x\neq2\). The domain of the sum is the common part of these conditions.

What exam hint can help solve this Mathematics question?

पहले फलन में \(x\neq-1\) और दूसरे में \(x\neq2\) चाहिए। जोड़ का डोमेन इन दोनों शर्तों का साझा भाग है।