Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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If (x-2+(m+1)x+16=0) has equal roots, what are the values of (m)?
A. (7) और (-9)(7) and (-9)
समान जड़ों के लिए ((m+1)2-64=0) होगा। इसलिए \(m+1=\pm8\), अतः (m=7) या (m=-9)। / For equal roots, ((m+1)2-64=0). Thus \(m+1=\pm8\), so (m=7) or (m=-9).
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If the roots of \(x^2+ax+b=0\) are \(3+\sqrt{2}\) and \(3-\sqrt{2}\), what is (a+b)?
B. (1)
जड़ों का योग (6) है, इसलिए (a=-6)। गुणनफल (7) है, इसलिए (b=7), अतः (a+b=1)। / The sum of roots is (6), so (a=-6). The product is (7), so (b=7), hence (a+b=1).
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If \(\alpha,\beta\) are the roots of \(x^2-4x-2=0\), what is \(\alpha^2+4\beta+\alpha\beta\)?
C. (16)
क्योंकि \(\alpha\) जड़ है, \(\alpha^2=4\alpha+2\) होगा और \(\alpha\beta=-2\) है। व्यंजक (4\alpha+2+4\beta-2=4\(\alpha+\beta\)=16) बनता है। / Since \(\alpha\) is a root, \(\alpha^2=4\alpha+2\), and \(\alpha\beta=-2\). The expression becomes (4\alpha+2+4\beta-2=4\(\alpha+\beta\)=16).
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What is the positive difference between the roots of \(x^2-2qx+q^2-16=0\)?
C. (8)
समीकरण ((x-q)2-16=0) है। इसलिए जड़ें (q+4) और (q-4) हैं, जिनका अंतर (8) है। / The equation is ((x-q)2-16=0). Thus the roots are (q+4) and (q-4), whose difference is (8).
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If both roots of \(x^2-5x+c=0\) are real and positive, what is the correct condition on (c)?
B. \(0<c\le\frac{25}{4}\)
योग (5) धनात्मक है और दोनों जड़ों के लिए गुणनफल (c>0) चाहिए। वास्तविक जड़ों के लिए \(25-4c\ge0\), इसलिए \(0<c\le\frac{25}{4}\)। / The sum (5) is positive and product (c>0) is needed for both roots. For real roots, \(25-4c\ge0\), so \(0<c\le\frac{25}{4}\).
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If \(\alpha,\beta\) are the roots of \(x^2-3x-10=0\), what is (\(\alpha-4\)\(\beta-4\))?
C. (-6)
(\(\alpha-4\)\(\beta-4\)=\alpha\beta-4\(\alpha+\beta\)+16) है। \(\alpha+\beta=3\) और \(\alpha\beta=-10\), इसलिए मान (-6) है। / We use (\(\alpha-4\)\(\beta-4\)=\alpha\beta-4\(\alpha+\beta\)+16). Since \(\alpha+\beta=3\) and \(\alpha\beta=-10\), the value is (-6).
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C. हर (t) परFor every (t)
इन दोनों जड़ों का योग \(\frac{4t+2}{3}\) और गुणनफल (\frac{t(t+2)}{3}) है। ये दिए गए समीकरण के \(-\frac{b}{a}\) और \(\frac{c}{a}\) से मेल खाते हैं। / The sum of these two roots is \(\frac{4t+2}{3}\), and the product is (\frac{t(t+2)}{3}). These match \(-\frac{b}{a}\) and \(\frac{c}{a}\) of the given equation.
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What are the roots of (x-2-2(a+1)x+a-2+2a=0)?
A. (a) और (a+2)(a) and (a+2)
जड़ों का योग (2a+2) और गुणनफल \(a^2+2a\) है। ये (a) और (a+2) से मिलते हैं। / The sum of roots is (2a+2) and the product is \(a^2+2a\). These match (a) and (a+2).
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C. \(\frac{73}{24}\)
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\) है। यहाँ \(\alpha^2+\beta^2=73\) और \(\alpha\beta=24\), इसलिए मान \(\frac{73}{24}\) है। / We use \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\). Here \(\alpha^2+\beta^2=73\) and \(\alpha\beta=24\), so the value is \(\frac{73}{24}\).
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C. \(s^2\ge4\)
\(\tan\theta\cdot\cot\theta=1\) और \(\tan\theta+\cot\theta=s\) है। वास्तविक मानों के लिए \(s^2-4\ge0\), इसलिए \(s^2\ge4\)। / We have \(\tan\theta\cdot\cot\theta=1\) and \(\tan\theta+\cot\theta=s\). For real values, \(s^2-4\ge0\), so \(s^2\ge4\).
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