Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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Showing 501-510 of 649 questions.
For which value of (k) will \(x^2-6x+k\) have real and irrational zeroes?
C. (k=10)
यहाँ (D=36-4k) है। (k=10) पर (D=-4) नहीं बल्कि (D=36-40=-4), इसलिए यह सही नहीं है। / Here (D=36-4k). For (k=10), (D=36-40=-4), so this is not correct.
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A. (5)
स्थिर पद गुणनफल है और (\(4+\sqrt{11}\)\(4-\sqrt{11}\)=16-11=5)। संयुग्मी गुणनफल में बीच का अपरिमेय भाग हट जाता है। / The constant term is the product, and (\(4+\sqrt{11}\)\(4-\sqrt{11}\)=16-11=5). In conjugate products, the irrational middle part cancels.
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A. (-4)
दूसरा शून्यक \(2-\sqrt{13}\) होगा, इसलिए योग (4) है। एकक बहुपद में (x) का गुणांक (-4) होगा। / The other zero will be \(2-\sqrt{13}\), so the sum is (4). In a monic polynomial, the coefficient of (x) will be (-4).
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If \(\alpha=5+\sqrt{6}\) and \(\beta=5-\sqrt{6}\), what is the value of \(\alpha^2+\beta^2\)?
A. (62)
\(\alpha+\beta=10\) और \(\alpha\beta=25-6=19\), इसलिए \(\alpha^2+\beta^2=100-38=62\)। पहचान (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) उपयोग करें। / \(\alpha+\beta=10\) and \(\alpha\beta=25-6=19\), so \(\alpha^2+\beta^2=100-38=62\). Use (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta).
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If (p(x)=x-2-10x+17), what is the difference between its zeroes?
A. \(4\sqrt{2}\)
शून्यक \(5\pm2\sqrt{2}\) हैं, इसलिए अंतर \(4\sqrt{2}\) है। संयुग्मी शून्यकों में अंतर मूल भाग का दोगुना होता है। / The zeroes are \(5\pm2\sqrt{2}\), so the difference is \(4\sqrt{2}\). For conjugate zeroes, the difference is twice the radical part.
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If (p(x)=3x-2-18x+21), what type of zeroes does it have?
B. वास्तविक और अपरिमेयReal and irrational
सामान्य गुणनखंड हटाने पर \(x^2-6x+7\) मिलता है और (D=36-28=8)। (D) धनात्मक अपूर्ण वर्ग है, इसलिए शून्यक वास्तविक अपरिमेय हैं। / After removing the common factor, we get \(x^2-6x+7\), and (D=36-28=8). Since (D) is positive and not a perfect square, the zeroes are real irrational.
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A. \(-3\sqrt{3}\)
\(\sqrt{12}=2\sqrt{3}\), इसलिए योग \(3\sqrt{3}\) है। एकक बहुपद में (x) का गुणांक शून्यकों के योग का ऋणात्मक होता है। / \(\sqrt{12}=2\sqrt{3}\), so the sum is \(3\sqrt{3}\). In a monic polynomial, the coefficient of (x) is the negative of the sum of zeroes.
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Which polynomial has zeroes \(3+\sqrt{2}\) and \(3-\sqrt{2}\)?
A. \(x^2-6x+7\)
योग (6) और गुणनफल (9-2=7) है। बहुपद \(x^2-6x+7\) होगा। / The sum is (6) and the product is (9-2=7). The polynomial is \(x^2-6x+7\).
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If (p(x)=x-2-\(\sqrt{5}+\sqrt{7}\)x+\sqrt{35}), which is the correct pair of zeroes?
A. \(\sqrt{5}\) और \(\sqrt{7}\)\(\sqrt{5}\) and \(\sqrt{7}\)
योग \(\sqrt{5}+\sqrt{7}\) और गुणनफल \(\sqrt{35}\) है। ये दोनों \(\sqrt{5}\) और \(\sqrt{7}\) से मिलते हैं। / The sum is \(\sqrt{5}+\sqrt{7}\) and the product is \(\sqrt{35}\). Both match \(\sqrt{5}\) and \(\sqrt{7}\).
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If (p(x)=x-2-2\sqrt{10}x+10), which statement about its zeroes is correct?
A. दोनों शून्यक \(\sqrt{10}\) हैंBoth zeroes are \(\sqrt{10}\)
(p(x)=\(x-\sqrt{10}\)2), इसलिए शून्यक दो बार \(\sqrt{10}\) है। पूर्ण वर्ग रूप से समान शून्यक तुरंत मिलते हैं। / (p(x)=\(x-\sqrt{10}\)2), so the zero \(\sqrt{10}\) occurs twice. A perfect-square form quickly gives equal zeroes.
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