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
B. क्योंकि \(\sqrt{x}\) की घात \(\frac{1}{2}\) हैBecause \(\sqrt{x}\) has power \(\frac{1}{2}\)
Concept
\(\sqrt{x}=x^{\frac{1}{2}}\). A fractional power is not valid in the definition of a polynomial.
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.
Exam Tip
\(\sqrt{x}=x^{\frac{1}{2}}\) है। भिन्न घात बहुपद की परिभाषा में मान्य नहीं होती।
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