Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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If the zeroes of (p(x)=x-2+px+q) are (4) and (-5), what is (p+q)?
A. (-19)
योग (-1) है, इसलिए (p=1), और गुणनफल (-20) है, इसलिए (q=-20)। अतः (p+q=-19) है। / The sum is (-1), so (p=1), and the product is (-20), so (q=-20). Hence (p+q=-19).
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If the sum of zeroes of (p(x)=x-2-2(k+1)x+k-2) is (10), what is (k)?
A. (4)
शून्यकों का योग (2(k+1)) है, इसलिए (2(k+1)=10) और (k=4)। सूत्र लगाते समय ऋण चिह्न संभालें। / The sum of zeroes is (2(k+1)), so (2(k+1)=10) and (k=4). Handle the negative sign carefully in the formula.
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If the sum of zeroes of (p(x)=3x-2+(m-1)x-2) is (0), what is (m)?
A. (1)
योग \(-\frac{m-1}{3}\) है, इसे (0) रखने पर (m-1=0) मिलता है। शून्य योग में (x) का गुणांक (0) हो जाता है। / The sum is \(-\frac{m-1}{3}\); setting it to (0) gives (m-1=0). When the sum is zero, the coefficient of (x) becomes zero.
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If the product of zeroes of (p(x)=kx-2-6x+4) is (2), what is (k)?
A. (2)
गुणनफल \(\frac{4}{k}\) है, इसलिए \(\frac{4}{k}=2\) और (k=2)। गुणनफल में स्थिर पद को प्रमुख गुणांक से भाग देते हैं। / The product is \(\frac{4}{k}\), so \(\frac{4}{k}=2\) and (k=2). Product equals constant term divided by leading coefficient.
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If (p(x)=x-2-8x+15), what is the sum of squares of its zeroes?
A. (34)
यदि शून्यक \(\alpha,\beta\) हैं, तो \(\alpha+\beta=8\) और \(\alpha\beta=15\)। (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta=64-30=34)। / If the zeroes are \(\alpha,\beta\), then \(\alpha+\beta=8\) and \(\alpha\beta=15\). (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta=64-30=34).
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A. \(\frac{9}{4}\)
\(\alpha+\beta=\frac{9}{2}\) और \(\alpha\beta=2\) हैं। इसलिए \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{9}{4}\)। / \(\alpha+\beta=\frac{9}{2}\) and \(\alpha\beta=2\). Hence \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{9}{4}\).
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If the zeroes of (p(x)=x-2-5x+6) are \(\alpha,\beta\), what is (\(\alpha-\beta\)2)?
A. (1)
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=25-24=1)। इस पहचान से शून्यक निकाले बिना उत्तर मिल जाता है। / (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=25-24=1). This identity avoids finding the zeroes separately.
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A. \(x^2-8x+15\)
नए शून्यकों का योग \(\alpha+\beta+2=8\) और गुणनफल \(\alpha\beta+\alpha+\beta+1=15\) है। अतः बहुपद \(x^2-8x+15\) है। / The new sum is \(\alpha+\beta+2=8\) and product is \(\alpha\beta+\alpha+\beta+1=15\). Thus the polynomial is \(x^2-8x+15\).
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A. \(x^3-x^2-6x\)
शून्यक (-2,0,3) हैं, इसलिए बहुपद (x(x+2)(x-3)=x-3-x-2-6x) है। (x)-अक्ष काटने के बिंदु शून्यक बताते हैं। / The zeroes are (-2,0,3), so the polynomial is (x(x+2)(x-3)=x-3-x-2-6x). Intersections with the (x)-axis give zeroes.
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If (p(x)=x-3+x-2-10x+8) and (x=1) is a zero, what is the remaining quadratic factor?
A. \(x^2+2x-8\)
(x=1) शून्यक होने से (x-1) गुणनखंड है। भाग देने पर \(x^2+2x-8\) मिलता है। / Since (x=1) is a zero, (x-1) is a factor. Division gives \(x^2+2x-8\).
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