कौन-सा व्यंजक (x) में बहुपद नहीं है क्योंकि \(\sqrt{x^2+1}\) मौजूद है?
Which expression is not a polynomial in (x) because \(\sqrt{x^2+1}\) is present?
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{x^2+1}+x\)
Concept
\(\sqrt{x^2+1}\) has the variable inside a root. In the class (9) definition, it is not treated as a polynomial term.
Why this answer is correct
The correct answer is C. \(\sqrt{x^2+1}+x\). \(\sqrt{x^2+1}\) has the variable inside a root. In the class (9) definition, it is not treated as a polynomial term.
Exam Tip
\(\sqrt{x^2+1}\) में चर मूल के अंदर है। कक्षा (9) की परिभाषा में यह बहुपद पद नहीं माना जाता।
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