व्यंजक \(4x^2+\sqrt{x^2}+1\) को (x) में बहुपद मानना सही है या नहीं?
Is it correct to treat \(4x^2+\sqrt{x^2}+1\) as a polynomial in (x)?
Explanation opens after your attempt
Correct Answer
B. नहीं क्योंकि \(\sqrt{x^2}=|x|\) होता हैNo because \(\sqrt{x^2}=|x|\)
Concept
For real (x), \(\sqrt{x^2}=|x|\). The term (|x|) is not a polynomial term.
Why this answer is correct
The correct answer is B. नहीं क्योंकि \(\sqrt{x^2}=|x|\) होता है / No because \(\sqrt{x^2}=|x|\). For real (x), \(\sqrt{x^2}=|x|\). The term (|x|) is not a polynomial term.
Exam Tip
वास्तविक (x) के लिए \(\sqrt{x^2}=|x|\) होता है। (|x|) बहुपद पद नहीं है।
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