\(x^4+\frac{1}{x^2}+x\) बहुपद क्यों नहीं है?

Why is \(x^4+\frac{1}{x^2}+x\) not a polynomial?

Explanation opens after your attempt
Correct Answer

C. क्योंकि \(\frac{1}{x^2}=x^{-2}\) हैBecause \(\frac{1}{x^2}=x^{-2}\)

Step 1

Concept

In \(\frac{1}{x^2}\), the power of (x) is (-2). Due to the negative power, it is not a polynomial.

Step 2

Why this answer is correct

The correct answer is C. क्योंकि \(\frac{1}{x^2}=x^{-2}\) है / Because \(\frac{1}{x^2}=x^{-2}\). In \(\frac{1}{x^2}\), the power of (x) is (-2). Due to the negative power, it is not a polynomial.

Step 3

Exam Tip

\(\frac{1}{x^2}\) में (x) की घात (-2) है। ऋणात्मक घात के कारण यह बहुपद नहीं है।

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FAQs

Mathematics Answer, Explanation and Revision Hints

\(x^4+\frac{1}{x^2}+x\) बहुपद क्यों नहीं है? / Why is \(x^4+\frac{1}{x^2}+x\) not a polynomial?

Correct Answer: C. क्योंकि \(\frac{1}{x^2}=x^{-2}\) है / Because \(\frac{1}{x^2}=x^{-2}\). Explanation: \(\frac{1}{x^2}\) में (x) की घात (-2) है। ऋणात्मक घात के कारण यह बहुपद नहीं है। / In \(\frac{1}{x^2}\), the power of (x) is (-2). Due to the negative power, it is not a polynomial.

Which concept should I revise for this Mathematics MCQ?

In \(\frac{1}{x^2}\), the power of (x) is (-2). Due to the negative power, it is not a polynomial.

What exam hint can help solve this Mathematics question?

\(\frac{1}{x^2}\) में (x) की घात (-2) है। ऋणात्मक घात के कारण यह बहुपद नहीं है।