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.
Showing 321-330 of 600 questions.
What is the remainder when \(x^3-5x^2+8x-4\) is divided by (x-1)?
A. (0)
शेषफल प्रमेय से शेषफल (p(1)) है। (1-5+8-4=0) मिलता है। / By the remainder theorem, the remainder is (p(1)). We get (1-5+8-4=0).
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What is the remainder when \(3x^3-2x^2+x+7\) is divided by (x+2)?
A. (-27)
शेषफल (p(-2)) होगा। (3(-8)-2(4)-2+7=-27) है। / The remainder is (p(-2)). (3(-8)-2(4)-2+7=-27).
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If the remainder when (p(x)) is divided by (x-4) is (7), what is (p(4))?
B. (7)
शेषफल प्रमेय के अनुसार (x-4) से भाग देने पर शेषफल (p(4)) होता है। इसलिए (p(4)=7)। / By the remainder theorem, the remainder on division by (x-4) is (p(4)). Therefore, (p(4)=7).
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Which one is not a linear factor of \(x^4-1\)?
C. \(x^2+1\)
\(x^2+1\) रैखिक नहीं है क्योंकि उसकी घात (2) है। वास्तविक संख्याओं में इसे रैखिक गुणनखंड नहीं माना जाता। / \(x^2+1\) is not linear because its degree is (2). Over real numbers, it is not a linear factor.
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If (p(x)=x-2-2x+5), what is (p(1)), and is (1) a zero?
A. (4), नहीं(4), no
(p(1)=1-2+5=4) है इसलिए (1) शून्यक नहीं है। शून्यक तभी होगा जब मान (0) हो। / (p(1)=1-2+5=4), so (1) is not a zero. A zero must make the value (0).
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A. \(x^2-5x+6\)
मोनिक बहुपद (x-2-\(\alpha+\beta\)x+\alpha\beta) होता है। इसलिए \(x^2-5x+6\) सही है। / The monic polynomial is (x-2-\(\alpha+\beta\)x+\alpha\beta). Hence \(x^2-5x+6\) is correct.
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A. \(-\frac{3}{5}\)
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\) है। यहां \(\frac{3}{2}\div-\frac{5}{2}=-\frac{3}{5}\)। / \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here \(\frac{3}{2}\div-\frac{5}{2}=-\frac{3}{5}\).
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If \(\alpha\) and \(\beta\) are zeroes of \(x^2-7x+10\), what is \(\alpha^2+\beta^2\)?
A. (29)
(\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) होता है। (72-2(10)=29) है। / (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta). Thus (72-2(10)=29).
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If \(\alpha\) and \(\beta\) are zeroes of \(3x^2+2x-1\), what is (\(\alpha+\beta\)2)?
A. \(\frac{4}{9}\)
\(\alpha+\beta=-\frac{2}{3}\) है। इसलिए (\(\alpha+\beta\)2=\frac{4}{9}) होगा। / \(\alpha+\beta=-\frac{2}{3}\). Therefore, (\(\alpha+\beta\)2=\frac{4}{9}).
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If the zeroes of \(x^2+px+16\) are equal and negative, what is (p)?
A. (8)
बराबर ऋणात्मक शून्यक (-4) और (-4) होंगे क्योंकि गुणनफल (16) है। योग (-8) है इसलिए (p=8)। / The equal negative zeroes are (-4) and (-4) because the product is (16). The sum is (-8), so (p=8).
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