कौन-सा कथन \(x^2+\sqrt{x}+1\) के बारे में सही है?

Which statement is correct about \(x^2+\sqrt{x}+1\)?

Explanation opens after your attempt
Correct Answer

B. यह बहुपद नहीं है क्योंकि (\sqrt{x}=x^{12}) है / It is not a polynomial because \(\sqrt{x}=x^{1 / 2}\)

Step 1

Concept

In a polynomial, powers of the variable must be integers. \(\frac{1}{2}\) is not an integer.

Step 2

Why this answer is correct

The correct answer is B. यह बहुपद नहीं है क्योंकि \(\sqrt{x}=x^{1 / 2}\) है / It is not a polynomial because \(\sqrt{x}=x^{1 / 2}\). In a polynomial, powers of the variable must be integers. \(\frac{1}{2}\) is not an integer.

Step 3

Exam Tip

बहुपद में चर की घातें पूर्णांक होनी चाहिए। \(\frac{1}{2}\) पूर्णांक नहीं है।

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Mathematics Answer, Explanation and Revision Hints

कौन-सा कथन \(x^2+\sqrt{x}+1\) के बारे में सही है? / Which statement is correct about \(x^2+\sqrt{x}+1\)?

Correct Answer: B. यह बहुपद नहीं है क्योंकि \(\sqrt{x}=x^{1 / 2}\) है / It is not a polynomial because \(\sqrt{x}=x^{1 / 2}\). Explanation: बहुपद में चर की घातें पूर्णांक होनी चाहिए। \(\frac{1}{2}\) पूर्णांक नहीं है। / In a polynomial, powers of the variable must be integers. \(\frac{1}{2}\) is not an integer.

Which concept should I revise for this Mathematics MCQ?

In a polynomial, powers of the variable must be integers. \(\frac{1}{2}\) is not an integer.

What exam hint can help solve this Mathematics question?

बहुपद में चर की घातें पूर्णांक होनी चाहिए। \(\frac{1}{2}\) पूर्णांक नहीं है।