व्यंजक \(7x^6-4x^3+x-11\) की डिग्री क्या है?
What is the degree of \(7x^6-4x^3+x-11\)?
#polynomial
#degree
#highest_power
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A (3)
B (4)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
The highest power is (6). The degree of a polynomial is found from the highest power.
Step 2
Why this answer is correct
The correct answer is C. (6). The highest power is (6). The degree of a polynomial is found from the highest power.
Step 3
Exam Tip
सबसे बड़ी घात (6) है। बहुपद की डिग्री सबसे बड़ी घात से मिलती है।
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व्यंजक ((a-3)x-4 +2x-2 -5) की डिग्री (4) कब होगी?
When will the degree of ((a-3)x-4 +2x-2 -5) be (4)?
#polynomial
#degree
#parameter
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A जब (a=3) / When (a=3)
B जब \(a\neq3\) / When \(a\neq3\)
C जब (a=0) / When (a=0)
D हर मान पर / For every value
Explanation opens after your attempt
Correct Answer
B. जब \(a\neq3\) / When \(a\neq3\)
Step 1
Concept
The degree remains (4) only when the coefficient of \(x^4\) is not zero. Hence \(a-3\neq0\).
Step 2
Why this answer is correct
The correct answer is B. जब \(a\neq3\) / When \(a\neq3\). The degree remains (4) only when the coefficient of \(x^4\) is not zero. Hence \(a-3\neq0\).
Step 3
Exam Tip
डिग्री (4) तभी रहेगी जब \(x^4\) का गुणांक शून्य न हो। इसलिए \(a-3\neq0\) होना चाहिए।
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यदि (k=2), तो ((k-2)x-5 +3x-2 -1) की डिग्री क्या होगी?
If (k=2), what will be the degree of ((k-2)x-5 +3x-2 -1)?
#polynomial
#degree
#zero_coefficient
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A (5)
B (3)
C (2)
D (0)
Explanation opens after your attempt
Step 1
Concept
Putting (k=2) makes the coefficient of \(x^5\) equal to (0). The remaining highest power is (2).
Step 2
Why this answer is correct
The correct answer is C. (2). Putting (k=2) makes the coefficient of \(x^5\) equal to (0). The remaining highest power is (2).
Step 3
Exam Tip
(k=2) रखने पर \(x^5\) का गुणांक (0) हो जाता है। बची हुई सबसे बड़ी घात (2) है।
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कौन सा व्यंजक बहुपद है लेकिन रेखीय बहुपद नहीं है?
Which expression is a polynomial but not a linear polynomial?
#polynomial
#classification
#linear
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A (4x-9)
B \(x^2+1\)
C \(\frac{1}{x}+2\)
D \(\sqrt{x}+1\)
Explanation opens after your attempt
Correct Answer
B. \(x^2+1\)
Step 1
Concept
\(x^2+1\) is a polynomial and its degree is (2). A linear polynomial has degree (1).
Step 2
Why this answer is correct
The correct answer is B. \(x^2+1\). \(x^2+1\) is a polynomial and its degree is (2). A linear polynomial has degree (1).
Step 3
Exam Tip
\(x^2+1\) बहुपद है और इसकी डिग्री (2) है। रेखीय बहुपद की डिग्री (1) होती है।
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व्यंजक \(0x^7+4x^3-2x+6\) की डिग्री क्या है?
What is the degree of \(0x^7+4x^3-2x+6\)?
#polynomial
#degree
#zero_coefficient
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A (7)
B (4)
C (3)
D (0)
Explanation opens after your attempt
Step 1
Concept
The term \(0x^7\) becomes zero. In the remaining polynomial, the highest power is (3).
Step 2
Why this answer is correct
The correct answer is C. (3). The term \(0x^7\) becomes zero. In the remaining polynomial, the highest power is (3).
Step 3
Exam Tip
\(0x^7\) पद शून्य हो जाता है। शेष बहुपद में सबसे बड़ी घात (3) है।
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कौन सा व्यंजक (x) में बहुपद है और उसकी डिग्री (3) है?
Which expression is a polynomial in (x) and has degree (3)?
#polynomial
#cubic
#valid_powers
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A \(2x^3+\frac{1}{x}\)
B \(5x^2+7x-1\)
C \(-4x^3+\sqrt{5}x+8\)
D \(x^{\frac{3}{2}}+1\)
Explanation opens after your attempt
Correct Answer
C. \(-4x^3+\sqrt{5}x+8\)
Step 1
Concept
The third expression is a polynomial because the powers are (3), (1), and (0). The highest power is (3).
Step 2
Why this answer is correct
The correct answer is C. \(-4x^3+\sqrt{5}x+8\). The third expression is a polynomial because the powers are (3), (1), and (0). The highest power is (3).
Step 3
Exam Tip
तीसरा व्यंजक बहुपद है क्योंकि घातें (3), (1) और (0) हैं। सबसे बड़ी घात (3) है।
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बहुपद \(-8x^5+3x^2-4x+9\) का अग्रणी गुणांक क्या है?
What is the leading coefficient of \(-8x^5+3x^2-4x+9\)?
#polynomial
#leading_coefficient
#degree
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A (-8)
B (5)
C (3)
D (9)
Explanation opens after your attempt
Step 1
Concept
The leading term is \(-8x^5\). Therefore the leading coefficient is (-8).
Step 2
Why this answer is correct
The correct answer is A. (-8). The leading term is \(-8x^5\). Therefore the leading coefficient is (-8).
Step 3
Exam Tip
अग्रणी पद \(-8x^5\) है। इसलिए अग्रणी गुणांक (-8) है।
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बहुपद \(5x^4+0x^3-7x+2\) में \(x^3\) का गुणांक क्या है?
What is the coefficient of \(x^3\) in \(5x^4+0x^3-7x+2\)?
#polynomial
#coefficient
#missing_term
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A (5)
B (0)
C (-7)
D (2)
Explanation opens after your attempt
Step 1
Concept
The coefficient of the \(x^3\) term is clearly (0). A zero-coefficient term does not change the degree.
Step 2
Why this answer is correct
The correct answer is B. (0). The coefficient of the \(x^3\) term is clearly (0). A zero-coefficient term does not change the degree.
Step 3
Exam Tip
\(x^3\) पद का गुणांक स्पष्ट रूप से (0) है। शून्य गुणांक वाला पद डिग्री नहीं बदलता।
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किस मान पर ((m+1)x-3 +5x-6) रेखीय बहुपद बन जाएगा?
For which value will ((m+1)x-3 +5x-6) become a linear polynomial?
#polynomial
#linear
#parameter
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A (m=1)
B (m=-1)
C (m=0)
D किसी भी (m) पर / For any (m)
Explanation opens after your attempt
Step 1
Concept
To become linear, the coefficient of \(x^3\) must be (0). So (m+1=0) gives (m=-1).
Step 2
Why this answer is correct
The correct answer is B. (m=-1). To become linear, the coefficient of \(x^3\) must be (0). So (m+1=0) gives (m=-1).
Step 3
Exam Tip
रेखीय बनने के लिए \(x^3\) का गुणांक (0) होना चाहिए। इसलिए (m+1=0) से (m=-1)।
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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)?
#polynomial
#absolute_value
#non_polynomial
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A हां क्योंकि \(\sqrt{x^2}=x\) हमेशा होता है / Yes because \(\sqrt{x^2}=x\) always
B नहीं क्योंकि \(\sqrt{x^2}=|x|\) होता है / No because \(\sqrt{x^2}=|x|\)
C हां क्योंकि इसमें \(x^2\) है / Yes because it has \(x^2\)
D नहीं क्योंकि इसमें (1) है / No because it has (1)
Explanation opens after your attempt
Correct Answer
B. नहीं क्योंकि \(\sqrt{x^2}=|x|\) होता है / No because \(\sqrt{x^2}=|x|\)
Step 1
Concept
For real (x), \(\sqrt{x^2}=|x|\). The term (|x|) is not a polynomial term.
Step 2
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.
Step 3
Exam Tip
वास्तविक (x) के लिए \(\sqrt{x^2}=|x|\) होता है। (|x|) बहुपद पद नहीं है।
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कौन सा विकल्प शून्य बहुपद और अशून्य नियतांक बहुपद के बारे में सही है?
Which option is correct about the zero polynomial and a non-zero constant polynomial?
#polynomial
#zero_polynomial
#constant_degree
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A दोनों की डिग्री (0) होती है / Both have degree (0)
B शून्य बहुपद की डिग्री परिभाषित नहीं होती और अशून्य नियतांक की डिग्री (0) होती है / The zero polynomial degree is not defined and a non-zero constant has degree (0)
C दोनों बहुपद नहीं हैं / Both are not polynomials
D शून्य बहुपद की डिग्री (1) होती है / The zero polynomial has degree (1)
Explanation opens after your attempt
Correct Answer
B. शून्य बहुपद की डिग्री परिभाषित नहीं होती और अशून्य नियतांक की डिग्री (0) होती है / The zero polynomial degree is not defined and a non-zero constant has degree (0)
Step 1
Concept
The degree of the zero polynomial is not defined. A non-zero constant polynomial has degree (0).
Step 2
Why this answer is correct
The correct answer is B. शून्य बहुपद की डिग्री परिभाषित नहीं होती और अशून्य नियतांक की डिग्री (0) होती है / The zero polynomial degree is not defined and a non-zero constant has degree (0). The degree of the zero polynomial is not defined. A non-zero constant polynomial has degree (0).
Step 3
Exam Tip
शून्य बहुपद की डिग्री परिभाषित नहीं होती। अशून्य नियतांक बहुपद की डिग्री (0) होती है।
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यदि (a=0), तो \(ax^4+2x^3-x+1\) की डिग्री क्या होगी?
If (a=0), what will be the degree of \(ax^4+2x^3-x+1\)?
#polynomial
#degree
#parameter_zero
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A (4)
B (3)
C (1)
D (0)
Explanation opens after your attempt
Step 1
Concept
When (a=0), the \(x^4\) term disappears. The remaining highest power is (3).
Step 2
Why this answer is correct
The correct answer is B. (3). When (a=0), the \(x^4\) term disappears. The remaining highest power is (3).
Step 3
Exam Tip
(a=0) होने पर \(x^4\) पद हट जाता है। बची हुई सबसे बड़ी घात (3) है।
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कौन सा व्यंजक बहुपद नहीं है लेकिन देखने में एकपदी जैसा है?
Which expression is not a polynomial but looks like a monomial?
#polynomial
#monomial_like
#square_root
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A \(9x^2\)
B \(-5\sqrt{x}\)
C (7x)
D (11)
Explanation opens after your attempt
Correct Answer
B. \(-5\sqrt{x}\)
Step 1
Concept
\(-5\sqrt{x}=-5x^{\frac{1}{2}}\). Due to a fractional power of the variable, it is not a polynomial.
Step 2
Why this answer is correct
The correct answer is B. \(-5\sqrt{x}\). \(-5\sqrt{x}=-5x^{\frac{1}{2}}\). Due to a fractional power of the variable, it is not a polynomial.
Step 3
Exam Tip
\(-5\sqrt{x}=-5x^{\frac{1}{2}}\) है। चर की भिन्न घात होने से यह बहुपद नहीं है।
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किस विकल्प में (x) में बहुपद को मानक रूप में लिखा गया है?
Which option writes a polynomial in (x) in standard form?
#polynomial
#standard_form
#descending_powers
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A \(7-2x+x^5+3x^2\)
B \(x^5+3x^2-2x+7\)
C \(3x^2+x^5+7-2x\)
D \(7+x^5-2x+3x^2\)
Explanation opens after your attempt
Correct Answer
B. \(x^5+3x^2-2x+7\)
Step 1
Concept
In standard form, terms are written in descending powers. \(x^5+3x^2-2x+7\) follows this order.
Step 2
Why this answer is correct
The correct answer is B. \(x^5+3x^2-2x+7\). In standard form, terms are written in descending powers. \(x^5+3x^2-2x+7\) follows this order.
Step 3
Exam Tip
मानक रूप में पद घटती घातों में लिखे जाते हैं। \(x^5+3x^2-2x+7\) इसी क्रम में है।
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बहुपद \(2x^6-5x^2+4\) में \(x^5\) का गुणांक क्या है?
What is the coefficient of \(x^5\) in \(2x^6-5x^2+4\)?
#polynomial
#missing_term
#coefficient_zero
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A (2)
B (5)
C (0)
D (4)
Explanation opens after your attempt
Step 1
Concept
The \(x^5\) term is absent in this polynomial. So the coefficient of \(x^5\) is taken as (0).
Step 2
Why this answer is correct
The correct answer is C. (0). The \(x^5\) term is absent in this polynomial. So the coefficient of \(x^5\) is taken as (0).
Step 3
Exam Tip
इस बहुपद में \(x^5\) पद अनुपस्थित है। इसलिए \(x^5\) का गुणांक (0) माना जाता है।
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किस विकल्प में चर की घात के कारण व्यंजक बहुपद नहीं है?
In which option is the expression not a polynomial because of the variable power?
#polynomial
#invalid_power
#non_polynomial
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A \(3x^5-2x+1\)
B \(4x^2+\pi x-6\)
C \(x^{-\frac{1}{3}}+7\)
D \(11x^0-5x\)
Explanation opens after your attempt
Correct Answer
C. \(x^{-\frac{1}{3}}+7\)
Step 1
Concept
In \(x^{-\frac{1}{3}}\), the power is both negative and fractional. Such a power is not valid in a polynomial.
Step 2
Why this answer is correct
The correct answer is C. \(x^{-\frac{1}{3}}+7\). In \(x^{-\frac{1}{3}}\), the power is both negative and fractional. Such a power is not valid in a polynomial.
Step 3
Exam Tip
\(x^{-\frac{1}{3}}\) में घात ऋणात्मक और भिन्न दोनों है। बहुपद में ऐसी घात मान्य नहीं होती।
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यदि (p(x)=0x-4 +0x-3 +5x-2), तो (p(x)) किस प्रकार का बहुपद है?
If (p(x)=0x-4 +0x-3 +5x-2), what type of polynomial is (p(x))?
#polynomial
#classification
#zero_coefficient
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A त्रिघात बहुपद / Cubic polynomial
B द्विघात बहुपद / Quadratic polynomial
C रेखीय बहुपद / Linear polynomial
D शून्य बहुपद / Zero polynomial
Explanation opens after your attempt
Correct Answer
C. रेखीय बहुपद / Linear polynomial
Step 1
Concept
Higher-power terms with zero coefficients disappear. The remaining (5x-2) is a linear polynomial.
Step 2
Why this answer is correct
The correct answer is C. रेखीय बहुपद / Linear polynomial. Higher-power terms with zero coefficients disappear. The remaining (5x-2) is a linear polynomial.
Step 3
Exam Tip
शून्य गुणांक वाले उच्च घात पद हट जाते हैं। बचा हुआ (5x-2) रेखीय बहुपद है।
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व्यंजक \(\frac{x^3+1}{x}\) बहुपद क्यों नहीं है?
Why is \(\frac{x^3+1}{x}\) not a polynomial?
#polynomial
#rational_expression
#denominator
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A क्योंकि सरल करने पर \(x^2+\frac{1}{x}\) मिलता है / Because simplifying gives \(x^2+\frac{1}{x}\)
B क्योंकि इसमें \(x^3\) है / Because it has \(x^3\)
C क्योंकि इसमें (1) है / Because it has (1)
D क्योंकि अंश में दो पद हैं / Because the numerator has two terms
Explanation opens after your attempt
Correct Answer
A. क्योंकि सरल करने पर \(x^2+\frac{1}{x}\) मिलता है / Because simplifying gives \(x^2+\frac{1}{x}\)
Step 1
Concept
\(\frac{x^3+1}{x}=x^2+\frac{1}{x}\). In \(\frac{1}{x}\), the power of (x) is (-1).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि सरल करने पर \(x^2+\frac{1}{x}\) मिलता है / Because simplifying gives \(x^2+\frac{1}{x}\). \(\frac{x^3+1}{x}=x^2+\frac{1}{x}\). In \(\frac{1}{x}\), the power of (x) is (-1).
Step 3
Exam Tip
\(\frac{x^3+1}{x}=x^2+\frac{1}{x}\) है। \(\frac{1}{x}\) में (x) की घात (-1) है।
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किस विकल्प में बहुपद की डिग्री गलत बताई गई है?
In which option is the degree of the polynomial stated incorrectly?
#polynomial
#degree
#error_identification
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A \(4x^3+1\) की डिग्री (3) / Degree of \(4x^3+1\) is (3)
B (9x-2) की डिग्री (1) / Degree of (9x-2) is (1)
C (5) की डिग्री (5) / Degree of (5) is (5)
D \(x^2+x+1\) की डिग्री (2) / Degree of \(x^2+x+1\) is (2)
Explanation opens after your attempt
Correct Answer
C. (5) की डिग्री (5) / Degree of (5) is (5)
Step 1
Concept
A non-zero constant polynomial has degree (0). The degree of (5) is not (5).
Step 2
Why this answer is correct
The correct answer is C. (5) की डिग्री (5) / Degree of (5) is (5). A non-zero constant polynomial has degree (0). The degree of (5) is not (5).
Step 3
Exam Tip
अशून्य नियतांक बहुपद की डिग्री (0) होती है। (5) की डिग्री (5) नहीं होती।
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व्यंजक (\(b^2-1\)x-2 +x+4) रेखीय बहुपद कब बनेगा?
When will (\(b^2-1\)x-2 +x+4) become a linear polynomial?
#polynomial
#linear
#parameter
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A जब (b=1) या (b=-1) / When (b=1) or (b=-1)
B जब (b=0) / When (b=0)
C जब (b=2) / When (b=2)
D कभी नहीं / Never
Explanation opens after your attempt
Correct Answer
A. जब (b=1) या (b=-1) / When (b=1) or (b=-1)
Step 1
Concept
To become linear, the coefficient of \(x^2\) must be (0). From \(b^2-1=0\), we get \(b=\pm1\).
Step 2
Why this answer is correct
The correct answer is A. जब (b=1) या (b=-1) / When (b=1) or (b=-1). To become linear, the coefficient of \(x^2\) must be (0). From \(b^2-1=0\), we get \(b=\pm1\).
Step 3
Exam Tip
रेखीय बनने के लिए \(x^2\) का गुणांक (0) होना चाहिए। \(b^2-1=0\) से \(b=\pm1\) मिलता है।
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यदि \(c\neq0\), तो \(cx^5+2x^2-7\) की डिग्री क्या होगी?
If \(c\neq0\), what will be the degree of \(cx^5+2x^2-7\)?
#polynomial
#degree
#nonzero_coefficient
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A (0)
B (2)
C (5)
D (7)
Explanation opens after your attempt
Step 1
Concept
When \(c\neq0\), the \(x^5\) term remains present. Therefore the highest power will be (5).
Step 2
Why this answer is correct
The correct answer is C. (5). When \(c\neq0\), the \(x^5\) term remains present. Therefore the highest power will be (5).
Step 3
Exam Tip
\(c\neq0\) होने पर \(x^5\) पद मौजूद रहेगा। इसलिए सबसे बड़ी घात (5) होगी।
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कौन सा व्यंजक बहुपद है और उसमें \(x^2\) का गुणांक (0) है?
Which expression is a polynomial and has coefficient of \(x^2\) equal to (0)?
#polynomial
#coefficient
#missing_term
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A \(x^3+4x-2\)
B \(x^2+4x-2\)
C \(\frac{1}{x^2}+4\)
D \(x^{\frac{1}{2}}+4x\)
Explanation opens after your attempt
Correct Answer
A. \(x^3+4x-2\)
Step 1
Concept
\(x^3+4x-2\) is a polynomial and it has no \(x^2\) term. So the coefficient of \(x^2\) is (0).
Step 2
Why this answer is correct
The correct answer is A. \(x^3+4x-2\). \(x^3+4x-2\) is a polynomial and it has no \(x^2\) term. So the coefficient of \(x^2\) is (0).
Step 3
Exam Tip
\(x^3+4x-2\) बहुपद है और उसमें \(x^2\) पद नहीं है। इसलिए \(x^2\) का गुणांक (0) है।
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व्यंजक \(2x^2+3|x|+1\) बहुपद है या नहीं?
Is \(2x^2+3|x|+1\) a polynomial or not?
#polynomial
#absolute_value
#non_polynomial
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A हां क्योंकि इसमें \(x^2\) है / Yes because it has \(x^2\)
B नहीं क्योंकि (|x|) बहुपद पद नहीं है / No because (|x|) is not a polynomial term
C हां क्योंकि इसमें तीन पद हैं / Yes because it has three terms
D हां क्योंकि इसमें नियतांक है / Yes because it has a constant
Explanation opens after your attempt
Correct Answer
B. नहीं क्योंकि (|x|) बहुपद पद नहीं है / No because (|x|) is not a polynomial term
Step 1
Concept
The term (|x|) is not written using non-negative integer powers of (x). Therefore it is not a polynomial.
Step 2
Why this answer is correct
The correct answer is B. नहीं क्योंकि (|x|) बहुपद पद नहीं है / No because (|x|) is not a polynomial term. The term (|x|) is not written using non-negative integer powers of (x). Therefore it is not a polynomial.
Step 3
Exam Tip
(|x|) को (x) की अऋणात्मक पूर्णांक घातों के रूप में नहीं लिखा जाता। इसलिए यह बहुपद नहीं है।
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किस विकल्प में बहुपद \(3x^4-2x^2+x-5\) का अग्रणी पद सही है?
Which option correctly gives the leading term of \(3x^4-2x^2+x-5\)?
#polynomial
#leading_term
#standard_form
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A \(3x^4\)
B \(-2x^2\)
C (x)
D (-5)
Explanation opens after your attempt
Correct Answer
A. \(3x^4\)
Step 1
Concept
The leading term is the term with the highest power. Here the highest power is (4).
Step 2
Why this answer is correct
The correct answer is A. \(3x^4\). The leading term is the term with the highest power. Here the highest power is (4).
Step 3
Exam Tip
अग्रणी पद सबसे बड़ी घात वाला पद होता है। यहां सबसे बड़ी घात (4) है।
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यदि (p(x)=(2d-6)x-4 +x-2 +1) की डिग्री (2) है, तो (d) का मान क्या होगा?
If the degree of (p(x)=(2d-6)x-4 +x-2 +1) is (2), what is the value of (d)?
#polynomial
#degree
#parameter_value
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A (2)
B (3)
C (4)
D (6)
Explanation opens after your attempt
Step 1
Concept
For degree (2), the coefficient of \(x^4\) must be (0). From (2d-6=0), (d=3).
Step 2
Why this answer is correct
The correct answer is B. (3). For degree (2), the coefficient of \(x^4\) must be (0). From (2d-6=0), (d=3).
Step 3
Exam Tip
डिग्री (2) के लिए \(x^4\) का गुणांक (0) होना चाहिए। (2d-6=0) से (d=3)।
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व्यंजक \(x^2+\frac{x}{x}\) को \(x\neq0\) पर सरल करने से क्या मिलता है?
What is obtained by simplifying \(x^2+\frac{x}{x}\) for \(x\neq0\)?
#polynomial
#simplification
#domain_note
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A \(x^2+x\)
B \(x^2+1\)
C (2x)
D \(x^3\)
Explanation opens after your attempt
Correct Answer
B. \(x^2+1\)
Step 1
Concept
For \(x\neq0\), \(\frac{x}{x}=1\). Therefore the simplified form is \(x^2+1\).
Step 2
Why this answer is correct
The correct answer is B. \(x^2+1\). For \(x\neq0\), \(\frac{x}{x}=1\). Therefore the simplified form is \(x^2+1\).
Step 3
Exam Tip
\(x\neq0\) पर \(\frac{x}{x}=1\) होता है। इसलिए सरल रूप \(x^2+1\) है।
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कौन सा कथन \(x^2+\frac{x}{x}\) के बारे में सबसे सही है?
Which statement about \(x^2+\frac{x}{x}\) is the most accurate?
#polynomial
#domain
#definition
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A यह मूल रूप में बहुपद है क्योंकि डिग्री (2) है / It is originally a polynomial because degree is (2)
B यह \(x\neq0\) पर \(x^2+1\) के बराबर है पर मूल रूप में चर हर में है / It equals \(x^2+1\) for \(x\neq0\) but originally has variable in denominator
C यह शून्य बहुपद है / It is the zero polynomial
D यह रेखीय बहुपद है / It is a linear polynomial
Explanation opens after your attempt
Correct Answer
B. यह \(x\neq0\) पर \(x^2+1\) के बराबर है पर मूल रूप में चर हर में है / It equals \(x^2+1\) for \(x\neq0\) but originally has variable in denominator
Step 1
Concept
In the original expression, \(\frac{x}{x}\) has a variable in the denominator. The simplified form is taken with domain in mind.
Step 2
Why this answer is correct
The correct answer is B. यह \(x\neq0\) पर \(x^2+1\) के बराबर है पर मूल रूप में चर हर में है / It equals \(x^2+1\) for \(x\neq0\) but originally has variable in denominator. In the original expression, \(\frac{x}{x}\) has a variable in the denominator. The simplified form is taken with domain in mind.
Step 3
Exam Tip
मूल व्यंजक में \(\frac{x}{x}\) में चर हर में है। सरल रूप डोमेन को ध्यान में रखकर लिया जाता है।
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किस विकल्प में बहुपद की डिग्री (0) है?
In which option is the degree of the polynomial (0)?
#polynomial
#degree_zero
#constant
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A (0)
B (-13)
C (13x)
D \(x^2-13\)
Explanation opens after your attempt
Step 1
Concept
A non-zero constant polynomial has degree (0). The degree of the zero polynomial is not defined.
Step 2
Why this answer is correct
The correct answer is B. (-13). A non-zero constant polynomial has degree (0). The degree of the zero polynomial is not defined.
Step 3
Exam Tip
अशून्य नियतांक बहुपद की डिग्री (0) होती है। शून्य बहुपद की डिग्री परिभाषित नहीं होती।
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व्यंजक \(x^4+x^2+x^0\) की डिग्री क्या है?
What is the degree of \(x^4+x^2+x^0\)?
#polynomial
#zero_power
#degree
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A (0)
B (2)
C (4)
D (7)
Explanation opens after your attempt
Step 1
Concept
\(x^0=1\), and the highest power is (4). Therefore the degree is (4).
Step 2
Why this answer is correct
The correct answer is C. (4). \(x^0=1\), and the highest power is (4). Therefore the degree is (4).
Step 3
Exam Tip
\(x^0=1\) होता है और सबसे बड़ी घात (4) है। इसलिए डिग्री (4) है।
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कौन सा व्यंजक बहुपद नहीं है क्योंकि चर मूल चिह्न के अंदर है?
Which expression is not a polynomial because the variable is inside a radical sign?
#polynomial
#radical
#non_polynomial
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A \(3\sqrt{2}x+1\)
B \(2x^2-\sqrt{5}\)
C \(\sqrt{x+1}+2\)
D \(x^3+\sqrt{7}x\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{x+1}+2\)
Step 1
Concept
In \(\sqrt{x+1}\), the variable is inside a radical sign. It is not written using valid polynomial powers.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{x+1}+2\). In \(\sqrt{x+1}\), the variable is inside a radical sign. It is not written using valid polynomial powers.
Step 3
Exam Tip
\(\sqrt{x+1}\) में चर मूल चिह्न के अंदर है। इसे बहुपद की मान्य घातों में नहीं लिखा जाता।
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यदि \(a\neq0\) और (b=0), तो \(ax^4+bx^6+3\) की डिग्री क्या होगी?
If \(a\neq0\) and (b=0), what will be the degree of \(ax^4+bx^6+3\)?
#polynomial
#degree
#parameters
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A (0)
B (4)
C (6)
D (10)
Explanation opens after your attempt
Step 1
Concept
Since (b=0), the \(x^6\) term disappears, and since \(a\neq0\), the \(x^4\) term remains. So the degree is (4).
Step 2
Why this answer is correct
The correct answer is B. (4). Since (b=0), the \(x^6\) term disappears, and since \(a\neq0\), the \(x^4\) term remains. So the degree is (4).
Step 3
Exam Tip
(b=0) होने से \(x^6\) पद हट जाता है और \(a\neq0\) से \(x^4\) पद बचता है। इसलिए डिग्री (4) है।
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बहुपद \(2x^3-4x^2+6x-8\) में कुल कितने अशून्य पद हैं?
How many non-zero terms are there in \(2x^3-4x^2+6x-8\)?
#polynomial
#terms
#nonzero_terms
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A (2)
B (3)
C (4)
D (6)
Explanation opens after your attempt
Step 1
Concept
All four terms have non-zero coefficients. Therefore it has (4) non-zero terms.
Step 2
Why this answer is correct
The correct answer is C. (4). All four terms have non-zero coefficients. Therefore it has (4) non-zero terms.
Step 3
Exam Tip
चारों पदों के गुणांक अशून्य हैं। इसलिए इसमें (4) अशून्य पद हैं।
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किस विकल्प में दिया गया व्यंजक (x) में बहुपद है लेकिन (x) में नहीं बल्कि (y) में लिखा दिखता है?
Which option is a polynomial in (x) but appears to be written without (x) terms except a constant?
#polynomial
#in_x
#constant_polynomial
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A \(7y^2+3\)
B (5)
C \(\frac{1}{x}+y\)
D \(\sqrt{x}+y\)
Explanation opens after your attempt
Step 1
Concept
With respect to (x), (5) is a constant polynomial. It can be considered to have power (0) of (x).
Step 2
Why this answer is correct
The correct answer is B. (5). With respect to (x), (5) is a constant polynomial. It can be considered to have power (0) of (x).
Step 3
Exam Tip
(x) के संदर्भ में (5) नियतांक बहुपद है। इसमें (x) की घात (0) मानी जा सकती है।
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कौन सा कथन \(\frac{x^2}{2}-\frac{3x}{4}+9\) के लिए सही है?
Which statement is correct for \(\frac{x^2}{2}-\frac{3x}{4}+9\)?
#polynomial
#fraction_coefficients
#degree
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A यह बहुपद नहीं है क्योंकि गुणांक भिन्न हैं / It is not a polynomial because coefficients are fractions
B यह बहुपद है और डिग्री (2) है / It is a polynomial and degree is (2)
C यह शून्य बहुपद है / It is the zero polynomial
D यह बहुपद नहीं है क्योंकि इसमें घटाव है / It is not a polynomial because it has subtraction
Explanation opens after your attempt
Correct Answer
B. यह बहुपद है और डिग्री (2) है / It is a polynomial and degree is (2)
Step 1
Concept
Fractional coefficients are real numbers. The variable powers are (2), (1), and (0), so the degree is (2).
Step 2
Why this answer is correct
The correct answer is B. यह बहुपद है और डिग्री (2) है / It is a polynomial and degree is (2). Fractional coefficients are real numbers. The variable powers are (2), (1), and (0), so the degree is (2).
Step 3
Exam Tip
भिन्न गुणांक वास्तविक संख्याएं हैं। चर की घातें (2), (1) और (0) हैं इसलिए डिग्री (2) है।
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किस विकल्प में सभी गुणांक वास्तविक हैं और चर की घातें मान्य हैं?
In which option are all coefficients real and variable powers valid?
#polynomial
#real_coefficients
#valid_powers
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A \(2x^2-\pi x+\sqrt{11}\)
B \(x^{-\pi}+1\)
C \(\sqrt{x}+2\)
D \(\frac{1}{x}+3\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2-\pi x+\sqrt{11}\)
Step 1
Concept
\(\pi\) and \(\sqrt{11}\) are real coefficients. The powers of (x) are (2), (1), and (0).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-\pi x+\sqrt{11}\). \(\pi\) and \(\sqrt{11}\) are real coefficients. The powers of (x) are (2), (1), and (0).
Step 3
Exam Tip
\(\pi\) और \(\sqrt{11}\) वास्तविक गुणांक हैं। (x) की घातें (2), (1) और (0) हैं।
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यदि (q(x)=x-2 +\sqrt{x}+1), तो (q(x)) बहुपद क्यों नहीं है?
If (q(x)=x-2 +\sqrt{x}+1), why is (q(x)) not a polynomial?
#polynomial
#square_root
#fractional_power
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A क्योंकि इसमें \(x^2\) है / Because it has \(x^2\)
B क्योंकि \(\sqrt{x}\) की घात \(\frac{1}{2}\) है / Because \(\sqrt{x}\) has power \(\frac{1}{2}\)
C क्योंकि इसमें (1) है / Because it has (1)
D क्योंकि इसमें तीन पद हैं / Because it has three terms
Explanation opens after your attempt
Correct Answer
B. क्योंकि \(\sqrt{x}\) की घात \(\frac{1}{2}\) है / Because \(\sqrt{x}\) has power \(\frac{1}{2}\)
Step 1
Concept
\(\sqrt{x}=x^{\frac{1}{2}}\). A fractional power is not valid in the definition of a polynomial.
Step 2
Why this answer is correct
The correct answer is B. क्योंकि \(\sqrt{x}\) की घात \(\frac{1}{2}\) है / Because \(\sqrt{x}\) has power \(\frac{1}{2}\). \(\sqrt{x}=x^{\frac{1}{2}}\). A fractional power is not valid in the definition of a polynomial.
Step 3
Exam Tip
\(\sqrt{x}=x^{\frac{1}{2}}\) है। भिन्न घात बहुपद की परिभाषा में मान्य नहीं होती।
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व्यंजक ((n-4)x-2 +(n+1)x+6) नियतांक बहुपद कब बनेगा?
When will ((n-4)x-2 +(n+1)x+6) become a constant polynomial?
#polynomial
#constant_polynomial
#parameter
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A जब (n=4) / When (n=4)
B जब (n=-1) / When (n=-1)
C जब (n=4) और (n=-1) साथ-साथ / When (n=4) and (n=-1) together
D कभी नहीं / Never
Explanation opens after your attempt
Correct Answer
D. कभी नहीं / Never
Step 1
Concept
To become constant, both (n-4=0) and (n+1=0) are needed. One value of (n) cannot satisfy both conditions.
Step 2
Why this answer is correct
The correct answer is D. कभी नहीं / Never. To become constant, both (n-4=0) and (n+1=0) are needed. One value of (n) cannot satisfy both conditions.
Step 3
Exam Tip
नियतांक बनने के लिए (n-4=0) और (n+1=0) दोनों चाहिए। एक ही (n) दोनों शर्तें पूरी नहीं कर सकता।
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कौन सा व्यंजक बहुपद है जिसकी डिग्री (1) नहीं है और (0) भी नहीं है?
Which expression is a polynomial whose degree is neither (1) nor (0)?
#polynomial
#degree
#classification
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A (7x-3)
B (12)
C \(x^4+2x\)
D \(\frac{1}{x}+1\)
Explanation opens after your attempt
Correct Answer
C. \(x^4+2x\)
Step 1
Concept
\(x^4+2x\) is a polynomial and its degree is (4). So its degree is neither (1) nor (0).
Step 2
Why this answer is correct
The correct answer is C. \(x^4+2x\). \(x^4+2x\) is a polynomial and its degree is (4). So its degree is neither (1) nor (0).
Step 3
Exam Tip
\(x^4+2x\) बहुपद है और इसकी डिग्री (4) है। इसलिए इसकी डिग्री (1) या (0) नहीं है।
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बहुपद \(3x^4-2x^2+5\) में \(x^3\) और (x) के गुणांक क्रमशः क्या हैं?
In \(3x^4-2x^2+5\), what are the coefficients of \(x^3\) and (x) respectively?
#polynomial
#missing_terms
#coefficients
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A (3) और (-2) / (3) and (-2)
B (0) और (0) / (0) and (0)
C (4) और (1) / (4) and (1)
D (5) और (0) / (5) and (0)
Explanation opens after your attempt
Correct Answer
B. (0) और (0) / (0) and (0)
Step 1
Concept
The \(x^3\) and (x) terms are absent in this polynomial. Therefore both coefficients are (0).
Step 2
Why this answer is correct
The correct answer is B. (0) और (0) / (0) and (0). The \(x^3\) and (x) terms are absent in this polynomial. Therefore both coefficients are (0).
Step 3
Exam Tip
इस बहुपद में \(x^3\) और (x) पद नहीं हैं। इसलिए दोनों के गुणांक (0) हैं।
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कौन सा व्यंजक बहुपद नहीं है क्योंकि चर घात में नहीं बल्कि हर के अंदर है?
Which expression is not a polynomial because the variable is inside the denominator?
#polynomial
#denominator
#non_polynomial
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A \(x^2+3x+1\)
B \(\frac{5}{x+2}\)
C \(4x^3-1\)
D \(\sqrt{2}x^2+7\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{5}{x+2}\)
Step 1
Concept
In \(\frac{5}{x+2}\), the variable is in the denominator. Such an expression does not satisfy the definition of a polynomial.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{5}{x+2}\). In \(\frac{5}{x+2}\), the variable is in the denominator. Such an expression does not satisfy the definition of a polynomial.
Step 3
Exam Tip
\(\frac{5}{x+2}\) में चर हर में है। ऐसा व्यंजक बहुपद की परिभाषा को पूरा नहीं करता।
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यदि (r=0), तो (\(r^2-r\)x-3 +(r+2)x-2 +1) की डिग्री क्या है?
If (r=0), what is the degree of (\(r^2-r\)x-3 +(r+2)x-2 +1)?
#polynomial
#degree
#parameter_substitution
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A (0)
B (2)
C (3)
D (5)
Explanation opens after your attempt
Step 1
Concept
Putting (r=0), the coefficient of \(x^3\) is (0) and the coefficient of \(x^2\) is (2). Therefore the degree is (2).
Step 2
Why this answer is correct
The correct answer is B. (2). Putting (r=0), the coefficient of \(x^3\) is (0) and the coefficient of \(x^2\) is (2). Therefore the degree is (2).
Step 3
Exam Tip
(r=0) रखने पर \(x^3\) का गुणांक (0) और \(x^2\) का गुणांक (2) होता है। इसलिए डिग्री (2) है।
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कौन सा व्यंजक (x) में शून्य बहुपद है?
Which expression is the zero polynomial in (x)?
#polynomial
#zero_polynomial
#identification
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A \(0x^5+0x^2+0\)
B \(0x^5+x\)
C \(x^0-1\)
D (5-5x)
Explanation opens after your attempt
Correct Answer
A. \(0x^5+0x^2+0\)
Step 1
Concept
All terms in the first expression have coefficient (0). Therefore it is the zero polynomial.
Step 2
Why this answer is correct
The correct answer is A. \(0x^5+0x^2+0\). All terms in the first expression have coefficient (0). Therefore it is the zero polynomial.
Step 3
Exam Tip
पहले व्यंजक के सभी पदों के गुणांक (0) हैं। इसलिए यह शून्य बहुपद है।
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किस विकल्प में बहुपद की डिग्री (2) है और अग्रणी गुणांक ऋणात्मक है?
Which option has degree (2) and a negative leading coefficient?
#polynomial
#degree
#leading_coefficient
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A \(3x^2-5x+1\)
B \(-4x^2+7x-2\)
C \(-2x^3+x+1\)
D (5x-9)
Explanation opens after your attempt
Correct Answer
B. \(-4x^2+7x-2\)
Step 1
Concept
\(-4x^2+7x-2\) has degree (2) and leading coefficient (-4). Both conditions are satisfied here.
Step 2
Why this answer is correct
The correct answer is B. \(-4x^2+7x-2\). \(-4x^2+7x-2\) has degree (2) and leading coefficient (-4). Both conditions are satisfied here.
Step 3
Exam Tip
\(-4x^2+7x-2\) की डिग्री (2) है और अग्रणी गुणांक (-4) है। दोनों शर्तें इसी में पूरी होती हैं।
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व्यंजक (x-2 (x+1)-x-3 +4) को सरल करने पर कौन सा बहुपद मिलता है?
Which polynomial is obtained by simplifying (x-2 (x+1)-x-3 +4)?
#polynomial
#simplification
#degree
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A \(x^3+x^2+4\)
B \(x^2+4\)
C \(2x^3+x^2+4\)
D (4)
Explanation opens after your attempt
Correct Answer
B. \(x^2+4\)
Step 1
Concept
(x-2 (x+1)=x-3 +x-2 ), then \(-x^3\) cancels \(x^3\). The simplified form is \(x^2+4\).
Step 2
Why this answer is correct
The correct answer is B. \(x^2+4\). (x-2 (x+1)=x-3 +x-2 ), then \(-x^3\) cancels \(x^3\). The simplified form is \(x^2+4\).
Step 3
Exam Tip
(x-2 (x+1)=x-3 +x-2 ), फिर \(-x^3\) से \(x^3\) कट जाता है। सरल रूप \(x^2+4\) है।
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सरल रूप \(x^2+4\) की डिग्री क्या है?
What is the degree of the simplified form \(x^2+4\)?
#polynomial
#degree
#simplified_form
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A (0)
B (1)
C (2)
D (4)
Explanation opens after your attempt
Step 1
Concept
In \(x^2+4\), the highest power is (2). Therefore its degree is (2).
Step 2
Why this answer is correct
The correct answer is C. (2). In \(x^2+4\), the highest power is (2). Therefore its degree is (2).
Step 3
Exam Tip
\(x^2+4\) में सबसे बड़ी घात (2) है। इसलिए इसकी डिग्री (2) है।
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कौन सा निष्कर्ष \(x^0+\sqrt{3}x^2-\frac{x}{5}\) के लिए सही है?
Which conclusion is correct for \(x^0+\sqrt{3}x^2-\frac{x}{5}\)?
#polynomial
#real_coefficients
#degree
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A यह बहुपद नहीं है क्योंकि इसमें \(\sqrt{3}\) है / It is not a polynomial because it has \(\sqrt{3}\)
B यह बहुपद है और डिग्री (2) है / It is a polynomial and degree is (2)
C यह शून्य बहुपद है / It is the zero polynomial
D यह बहुपद नहीं है क्योंकि इसमें भिन्न गुणांक है / It is not a polynomial because it has a fractional coefficient
Explanation opens after your attempt
Correct Answer
B. यह बहुपद है और डिग्री (2) है / It is a polynomial and degree is (2)
Step 1
Concept
\(\sqrt{3}\) and \(-\frac{1}{5}\) are real coefficients. The powers of (x) are (0), (2), and (1), so the degree is (2).
Step 2
Why this answer is correct
The correct answer is B. यह बहुपद है और डिग्री (2) है / It is a polynomial and degree is (2). \(\sqrt{3}\) and \(-\frac{1}{5}\) are real coefficients. The powers of (x) are (0), (2), and (1), so the degree is (2).
Step 3
Exam Tip
\(\sqrt{3}\) और \(-\frac{1}{5}\) वास्तविक गुणांक हैं। (x) की घातें (0), (2) और (1) हैं इसलिए डिग्री (2) है।
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यदि (s=2), तो ((s-2 )x-4 +(s+1)x-3 -5) की डिग्री क्या होगी?
If (s=2), what will be the degree of ((s-2 )x-4 +(s+1)x-3 -5)?
#polynomial
#degree
#parameter_substitution
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A (4)
B (3)
C (1)
D (0)
Explanation opens after your attempt
Step 1
Concept
Putting (s=2) makes the coefficient of \(x^4\) equal to (0) and the coefficient of \(x^3\) equal to (3). So the highest power is (3).
Step 2
Why this answer is correct
The correct answer is B. (3). Putting (s=2) makes the coefficient of \(x^4\) equal to (0) and the coefficient of \(x^3\) equal to (3). So the highest power is (3).
Step 3
Exam Tip
(s=2) रखने पर \(x^4\) का गुणांक (0) और \(x^3\) का गुणांक (3) हो जाता है। इसलिए सबसे बड़ी घात (3) है।
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व्यंजक \(x^3+\frac{2x^2}{x}\) को \(x\neq0\) पर सरल करने से कौन सा रूप मिलता है?
Which form is obtained by simplifying \(x^3+\frac{2x^2}{x}\) for \(x\neq0\)?
#polynomial
#simplification
#domain_note
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A \(x^3+2x\)
B \(3x^3\)
C \(x^2+2x\)
D \(x^3+\frac{2}{x}\)
Explanation opens after your attempt
Correct Answer
A. \(x^3+2x\)
Step 1
Concept
For \(x\neq0\), \(\frac{2x^2}{x}=2x\). Therefore the simplified form is \(x^3+2x\).
Step 2
Why this answer is correct
The correct answer is A. \(x^3+2x\). For \(x\neq0\), \(\frac{2x^2}{x}=2x\). Therefore the simplified form is \(x^3+2x\).
Step 3
Exam Tip
\(x\neq0\) पर \(\frac{2x^2}{x}=2x\) होता है। इसलिए सरल रूप \(x^3+2x\) है।
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किस विकल्प में बहुपद की डिग्री (3) है लेकिन \(x^2\) का गुणांक (0) है?
Which option has degree (3) but the coefficient of \(x^2\) is (0)?
#polynomial
#degree
#missing_term
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A \(2x^3+5x-1\)
B \(2x^3+x^2-1\)
C \(x^2+5x-1\)
D \(2x^3+x^2+5x\)
Explanation opens after your attempt
Correct Answer
A. \(2x^3+5x-1\)
Step 1
Concept
In \(2x^3+5x-1\), the highest power is (3) and the \(x^2\) term is absent. So the coefficient of \(x^2\) is (0).
Step 2
Why this answer is correct
The correct answer is A. \(2x^3+5x-1\). In \(2x^3+5x-1\), the highest power is (3) and the \(x^2\) term is absent. So the coefficient of \(x^2\) is (0).
Step 3
Exam Tip
\(2x^3+5x-1\) में सबसे बड़ी घात (3) है और \(x^2\) पद नहीं है। इसलिए \(x^2\) का गुणांक (0) है।
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व्यंजक \(\frac{x^4-x^2}{x^2}\) को \(x\neq0\) पर सरल करने से क्या मिलता है?
What is obtained by simplifying \(\frac{x^4-x^2}{x^2}\) for \(x\neq0\)?
#polynomial
#rational_expression
#simplification
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A \(x^2-1\)
B \(x^2+1\)
C \(x^4-1\)
D \(x^2-x\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-1\)
Step 1
Concept
Dividing by \(x^2\) gives \(\frac{x^4}{x^2}=x^2\) and \(\frac{x^2}{x^2}=1\). Therefore the simplified form is \(x^2-1\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-1\). Dividing by \(x^2\) gives \(\frac{x^4}{x^2}=x^2\) and \(\frac{x^2}{x^2}=1\). Therefore the simplified form is \(x^2-1\).
Step 3
Exam Tip
\(x^2\) से भाग देने पर \(\frac{x^4}{x^2}=x^2\) और \(\frac{x^2}{x^2}=1\) मिलता है। इसलिए सरल रूप \(x^2-1\) है।
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