Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
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What is the degree of \(0x^9+6x^5-4x^2+3\)?
B. (5)
\(0x^9\) शून्य पद है। बची हुई सबसे बड़ी घात (5) है। / \(0x^9\) is a zero term. The remaining highest power is (5).
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When will the degree of ((a+4)x-6+3x-2-8) be (6)?
C. जब \(a\neq-4\)When \(a\neq-4\)
डिग्री (6) के लिए \(x^6\) का गुणांक शून्य नहीं होना चाहिए। इसलिए \(a+4\neq0\) होना चाहिए। / For degree (6), the coefficient of \(x^6\) must not be zero. So \(a+4\neq0\).
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If (k=5), what will be the degree of ((k-5)x-7+4x-3-x+2)?
C. (3)
(k=5) रखने पर \(x^7\) का गुणांक (0) हो जाता है। शेष बहुपद में सबसे बड़ी घात (3) है। / Putting (k=5) makes the coefficient of \(x^7\) equal to (0). In the remaining polynomial, the highest power is (3).
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For which value will ((m+6)x-4-2x+9) become a linear polynomial?
B. (m=-6)
रेखीय बनने के लिए \(x^4\) का गुणांक (0) होना चाहिए। (m+6=0) से (m=-6) मिलता है। / To become linear, the coefficient of \(x^4\) must be (0). From (m+6=0), we get (m=-6).
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Why is \(3x^2+\sqrt{2x-1}\) not a polynomial?
C. क्योंकि (2x-1) मूल चिह्न के अंदर हैBecause (2x-1) is inside a radical sign
\(\sqrt{2x-1}\) बहुपद की मान्य घातों वाला पद नहीं है। इसलिए पूरा व्यंजक बहुपद नहीं है। / \(\sqrt{2x-1}\) is not a term with valid polynomial powers. Therefore the whole expression is not a polynomial.
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If (p(x)=5x-5+0x-4-3x+6), what is the coefficient of \(x^4\)?
C. (0)
\(x^4\) वाला पद \(0x^4\) है। इसलिए \(x^4\) का गुणांक (0) है। / The term containing \(x^4\) is \(0x^4\). Therefore the coefficient of \(x^4\) is (0).
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Which option is a quadratic polynomial in (x)?
A. \(3x^2-\pi x+\sqrt{7}\)
पहले व्यंजक में सबसे बड़ी घात (2) है और सभी घातें मान्य हैं। इसलिए यह द्विघात बहुपद है। / In the first expression, the highest power is (2), and all powers are valid. So it is a quadratic polynomial.
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Why is \(\frac{x^5+3x^2}{x^2}\) not considered a polynomial in its original form?
A. क्योंकि चर हर में हैBecause the variable is in the denominator
मूल रूप में हर में \(x^2\) है। बहुपद रूप के लिए सरल करते समय डोमेन शर्त भी ध्यान रखनी होती है। / In the original form, \(x^2\) is in the denominator. While simplifying to polynomial form, the domain condition must also be considered.
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What is obtained by simplifying \(\frac{x^5+3x^2}{x^2}\) for \(x\neq0\)?
A. \(x^3+3\)
\(x\neq0\) पर \(\frac{x^5}{x^2}=x^3\) और \(\frac{3x^2}{x^2}=3\) होता है। इसलिए सरल रूप \(x^3+3\) है। / For \(x\neq0\), \(\frac{x^5}{x^2}=x^3\) and \(\frac{3x^2}{x^2}=3\). So the simplified form is \(x^3+3\).
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What is the degree of the simplified form \(x^3+3\)?
C. (3)
\(x^3+3\) में सबसे बड़ी घात (3) है। इसलिए इसकी डिग्री (3) है। / In \(x^3+3\), the highest power is (3). Therefore its degree is (3).
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