व्यंजक \(x^2+\frac{1}{\sqrt{x}}\) बहुपद क्यों नहीं है?
Why is \(x^2+\frac{1}{\sqrt{x}}\) not a polynomial?
Explanation opens after your attempt
Correct Answer
B. क्योंकि \(\frac{1}{\sqrt{x}}\) में चर की घात \(-\frac{1}{2}\) हैBecause \(\frac{1}{\sqrt{x}}\) has variable power \(-\frac{1}{2}\)
Concept
\(\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}\). Negative and fractional powers are not valid in a polynomial.
Why this answer is correct
The correct answer is B. क्योंकि \(\frac{1}{\sqrt{x}}\) में चर की घात \(-\frac{1}{2}\) है / Because \(\frac{1}{\sqrt{x}}\) has variable power \(-\frac{1}{2}\). \(\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}\). Negative and fractional powers are not valid in a polynomial.
Exam Tip
\(\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}\) होता है। ऋणात्मक और भिन्न घात बहुपद में मान्य नहीं होती।
Login to save your score, XP, coins and progress.