\(\frac{x^2+1}{x}=x+\frac{1}{x}\) में \(\frac{1}{x}\) की घात (-1) है। इसलिए यह बहुपद नहीं है। / \(\frac{x^2+1}{x}=x+\frac{1}{x}\) contains \(\frac{1}{x}\), whose power is (-1). Therefore, it is not a polynomial.
Mathematics Answer, Explanation and Revision Hints
कौन-सा व्यंजक (x) में बहुपद नहीं है क्योंकि उसमें (x) हर में है? / Which expression is not a polynomial in (x) because it has (x) in the denominator?
Correct Answer: C. \(\frac{x^2+1}{x}\). Explanation: \(\frac{x^2+1}{x}=x+\frac{1}{x}\) में \(\frac{1}{x}\) की घात (-1) है। इसलिए यह बहुपद नहीं है। / \(\frac{x^2+1}{x}=x+\frac{1}{x}\) contains \(\frac{1}{x}\), whose power is (-1). Therefore, it is not a polynomial.
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