Class 9 Mathematics - Introduction to Polynomials - Algebraic expressions Medium Quiz

Level 26 • 2/50 questions • 35 seconds per question.

Level readiness 2/50 Questions
Time Left 01:10 35 sec/question
RewardsCoins + XP
ModeClassic Quiz
Share
This level needs 48 more active questions. Admin panel me same class, subject, difficulty aur level_no 26 par question add karein.
Question 1 / 2 0 score
Answered 0/2 Correct 0 Time 01:10

कौन-सा व्यंजक (z) में बहुपद है?

Which expression is a polynomial in (z)?

Explanation opens after your attempt
Correct Answer

C. \(3z^4-z^2+8\)

Explanation

Simple Explanation

\(3z^4-z^2+8\) में \(z\) की घातें \(4\), \(2\) और \(0\) हैं, जो सभी अऋणात्मक पूर्णांक हैं। इसलिए यह \(z\) में बहुपद है। \(z^{-1}+2\) तथा \(\frac{6}{z}+1\) में \(z\) की ऋणात्मक घात आती है, जबकि \(\sqrt{z}+5\) में \(z\) की घात \(\frac12\) है; अतः ये बहुपद नहीं हैं। परीक्षा टिप: बहुपद में चर की घात केवल \(0,1,2,\ldots\) हो सकती है। / In \(3z^4-z^2+8\), the powers of \(z\) are \(4\), \(2\), and \(0\), all of which are non-negative integers. Therefore, it is a polynomial in \(z\). In \(z^{-1}+2\) and \(\frac{6}{z}+1\), \(z\) has a negative power, while \(\sqrt{z}+5\) contains \(z^{\frac12}\); hence, these are not polynomials. Exam tip: powers of the variable in a polynomial must be \(0,1,2,\ldots\).

Open Question Page
Ask Friends

\(x^2+\frac{1}{x}+4\) बहुपद क्यों नहीं है?

Why is \(x^2+\frac{1}{x}+4\) not a polynomial?

Explanation opens after your attempt
Correct Answer

B. क्योंकि \(\frac{1}{x}=x^{-1}\) में \(x\) की घात ऋणात्मक हैBecause \(\frac{1}{x}=x^{-1}\) has a negative exponent of \(x\)

Explanation

Simple Explanation

बहुपद में चर की घातें केवल अशून्य पूर्णांक हो सकती हैं, जैसे \(0,1,2,\ldots\)। यहाँ \(\frac{1}{x}=x^{-1}\) है, इसलिए \(x\) की घात \(-1\) है और यह व्यंजक बहुपद नहीं है। \(x^2\) तथा अचर \(4\) बहुपद के वैध पद हैं; तीन पद होना भी कोई समस्या नहीं है। परीक्षा टिप: चर यदि हर में हो, तो उसे ऋणात्मक घात मानकर जाँचें। / In a polynomial, the exponents of a variable must be non-negative integers such as \(0,1,2,\ldots\). Here, \(\frac{1}{x}=x^{-1}\), so the exponent of \(x\) is \(-1\); hence the expression is not a polynomial. The term \(x^2\) and the constant \(4\) are valid polynomial terms, and having three terms is not a restriction. Exam tip: If a variable occurs in the denominator, rewrite it using a negative exponent and check it.

Open Question Page
Ask Friends
FAQs

Class 9 Mathematics Quiz FAQs

How many questions are in this quiz?

This level is designed for 50 active questions. Currently 2 questions are available for the selected class and difficulty.

Is there a timer in this quiz?

Yes, the timer uses 35 seconds per question for Medium difficulty and shows the total remaining time on the page.

Can I open each question separately?

Yes, every question has its own SEO-friendly page with answer, explanation and related practice links.