यदि (q(x)=x-2+\sqrt{x}+1), तो (q(x)) बहुपद क्यों नहीं है?

If (q(x)=x-2+\sqrt{x}+1), why is (q(x)) not a polynomial?

Explanation opens after your attempt
Correct Answer

B. क्योंकि \(\sqrt{x}\) की घात \(\frac{1}{2}\) हैBecause \(\sqrt{x}\) has power \(\frac{1}{2}\)

Step 1

Concept

\(\sqrt{x}=x^{\frac{1}{2}}\). A fractional power is not valid in the definition of a polynomial.

Step 2

Why this answer is correct

The correct answer is B. क्योंकि \(\sqrt{x}\) की घात \(\frac{1}{2}\) है / Because \(\sqrt{x}\) has power \(\frac{1}{2}\). \(\sqrt{x}=x^{\frac{1}{2}}\). A fractional power is not valid in the definition of a polynomial.

Step 3

Exam Tip

\(\sqrt{x}=x^{\frac{1}{2}}\) है। भिन्न घात बहुपद की परिभाषा में मान्य नहीं होती।

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Mathematics Answer, Explanation and Revision Hints

यदि (q(x)=x-2+\sqrt{x}+1), तो (q(x)) बहुपद क्यों नहीं है? / If (q(x)=x-2+\sqrt{x}+1), why is (q(x)) not a polynomial?

Correct Answer: B. क्योंकि \(\sqrt{x}\) की घात \(\frac{1}{2}\) है / Because \(\sqrt{x}\) has power \(\frac{1}{2}\). Explanation: \(\sqrt{x}=x^{\frac{1}{2}}\) है। भिन्न घात बहुपद की परिभाषा में मान्य नहीं होती। / \(\sqrt{x}=x^{\frac{1}{2}}\). A fractional power is not valid in the definition of a polynomial.

Which concept should I revise for this Mathematics MCQ?

\(\sqrt{x}=x^{\frac{1}{2}}\). A fractional power is not valid in the definition of a polynomial.

What exam hint can help solve this Mathematics question?

\(\sqrt{x}=x^{\frac{1}{2}}\) है। भिन्न घात बहुपद की परिभाषा में मान्य नहीं होती।