\(x^2+\sqrt{x^2+4}\) के बारे में सही कथन कौन-सा है?

Which statement is correct about \(x^2+\sqrt{x^2+4}\)?

Explanation opens after your attempt
Correct Answer

B. यह बहुपद नहीं है क्योंकि चर मूल के अंदर हैIt is not a polynomial because the variable is inside a root

Step 1

Concept

\(\sqrt{x^2+4}\) has the variable inside a root. In the class (9) definition, such a term is not treated as a polynomial term.

Step 2

Why this answer is correct

The correct answer is B. यह बहुपद नहीं है क्योंकि चर मूल के अंदर है / It is not a polynomial because the variable is inside a root. \(\sqrt{x^2+4}\) has the variable inside a root. In the class (9) definition, such a term is not treated as a polynomial term.

Step 3

Exam Tip

\(\sqrt{x^2+4}\) में चर मूल के अंदर है। कक्षा (9) की परिभाषा में ऐसा पद बहुपद पद नहीं माना जाता।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

\(x^2+\sqrt{x^2+4}\) के बारे में सही कथन कौन-सा है? / Which statement is correct about \(x^2+\sqrt{x^2+4}\)?

Correct Answer: B. यह बहुपद नहीं है क्योंकि चर मूल के अंदर है / It is not a polynomial because the variable is inside a root. Explanation: \(\sqrt{x^2+4}\) में चर मूल के अंदर है। कक्षा (9) की परिभाषा में ऐसा पद बहुपद पद नहीं माना जाता। / \(\sqrt{x^2+4}\) has the variable inside a root. In the class (9) definition, such a term is not treated as a polynomial term.

Which concept should I revise for this Mathematics MCQ?

\(\sqrt{x^2+4}\) has the variable inside a root. In the class (9) definition, such a term is not treated as a polynomial term.

What exam hint can help solve this Mathematics question?

\(\sqrt{x^2+4}\) में चर मूल के अंदर है। कक्षा (9) की परिभाषा में ऐसा पद बहुपद पद नहीं माना जाता।