किस विकल्प में व्यंजक मूल रूप में बहुपद नहीं है क्योंकि चर हर में है?
Which option is not a polynomial in its original form because the variable is in the denominator?
Explanation opens after your attempt
Correct Answer
B. \(\frac{x^2+4}{x+1}\)
Concept
In \(\frac{x^2+4}{x+1}\), the variable is in the denominator. Such an original expression does not satisfy the definition of a polynomial.
Why this answer is correct
The correct answer is B. \(\frac{x^2+4}{x+1}\). In \(\frac{x^2+4}{x+1}\), the variable is in the denominator. Such an original expression does not satisfy the definition of a polynomial.
Exam Tip
\(\frac{x^2+4}{x+1}\) में चर हर में है। ऐसा मूल व्यंजक बहुपद की परिभाषा को पूरा नहीं करता।
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