व्यंजक \(\frac{x+1}{x}\) बहुपद क्यों नहीं है?

Why is \(\frac{x+1}{x}\) not a polynomial?

Explanation opens after your attempt
Correct Answer

B. क्योंकि सरल करने पर \(\frac{1}{x}\) वाला पद आता हैBecause after simplifying, a \(\frac{1}{x}\) term appears

Step 1

Concept

\(\frac{x+1}{x}=1+\frac{1}{x}\), which contains \(x^{-1}\). Negative powers are not valid in a polynomial.

Step 2

Why this answer is correct

The correct answer is B. क्योंकि सरल करने पर \(\frac{1}{x}\) वाला पद आता है / Because after simplifying, a \(\frac{1}{x}\) term appears. \(\frac{x+1}{x}=1+\frac{1}{x}\), which contains \(x^{-1}\). Negative powers are not valid in a polynomial.

Step 3

Exam Tip

\(\frac{x+1}{x}=1+\frac{1}{x}\), जिसमें \(x^{-1}\) आता है। ऋणात्मक घात बहुपद में मान्य नहीं होती।

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Mathematics Answer, Explanation and Revision Hints

व्यंजक \(\frac{x+1}{x}\) बहुपद क्यों नहीं है? / Why is \(\frac{x+1}{x}\) not a polynomial?

Correct Answer: B. क्योंकि सरल करने पर \(\frac{1}{x}\) वाला पद आता है / Because after simplifying, a \(\frac{1}{x}\) term appears. Explanation: \(\frac{x+1}{x}=1+\frac{1}{x}\), जिसमें \(x^{-1}\) आता है। ऋणात्मक घात बहुपद में मान्य नहीं होती। / \(\frac{x+1}{x}=1+\frac{1}{x}\), which contains \(x^{-1}\). Negative powers are not valid in a polynomial.

Which concept should I revise for this Mathematics MCQ?

\(\frac{x+1}{x}=1+\frac{1}{x}\), which contains \(x^{-1}\). Negative powers are not valid in a polynomial.

What exam hint can help solve this Mathematics question?

\(\frac{x+1}{x}=1+\frac{1}{x}\), जिसमें \(x^{-1}\) आता है। ऋणात्मक घात बहुपद में मान्य नहीं होती।