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 391-400 of 600 questions.
If \(x=2+\sqrt{6}\), what is the value of \(x^2-4x\)?
B. (-2)
\(x^2=10+4\sqrt{6}\) और \(4x=8+4\sqrt{6}\) है। घटाने पर (2) नहीं, बल्कि (2) मिलता है। / \(x^2=10+4\sqrt{6}\) and \(4x=8+4\sqrt{6}\). Subtracting gives (2).
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What is the simplified form of \(\sqrt{245}-\sqrt{180}+\sqrt{80}\)?
C. \(7\sqrt{5}\)
\(\sqrt{245}=7\sqrt{5}\), \(\sqrt{180}=6\sqrt{5}\) और \(\sqrt{80}=4\sqrt{5}\) है। इसलिए परिणाम \(5\sqrt{5}\) नहीं, \(7\sqrt{5}-6\sqrt{5}+4\sqrt{5}=5\sqrt{5}\) है। / \(\sqrt{245}=7\sqrt{5}\), \(\sqrt{180}=6\sqrt{5}\), and \(\sqrt{80}=4\sqrt{5}\). So the result is \(7\sqrt{5}-6\sqrt{5}+4\sqrt{5}=5\sqrt{5}\).
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If \(r=\frac{1}{\sqrt{11}+3}\), which is the simplified form of (r)?
A. \(\frac{\sqrt{11}-3}{2}\)
संयुग्मी से गुणा करने पर हर (11-9=2) बनता है। इसलिए सरल रूप \(\frac{\sqrt{11}-3}{2}\) है। / Multiplying by the conjugate makes the denominator (11-9=2). So the simplified form is \(\frac{\sqrt{11}-3}{2}\).
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What is the value of (\(\sqrt{18}+\sqrt{50}\)2)?
A. (128)
\(\sqrt{18}=3\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए योग \(8\sqrt{2}\) है। इसका वर्ग (128) है। / \(\sqrt{18}=3\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so the sum is \(8\sqrt{2}\). Its square is (128).
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What is the rationalised form of \(\frac{2}{\sqrt{11}+\sqrt{7}}\)?
B. \(\frac{\sqrt{11}-\sqrt{7}}{2}\)
संयुग्मी से गुणा करने पर हर (11-7=4) बनता है। इसलिए (\frac{2\(\sqrt{11}-\sqrt{7}\)}{4}=\frac{\sqrt{11}-\sqrt{7}}{2}) है। / Multiplying by the conjugate makes the denominator (11-7=4). So (\frac{2\(\sqrt{11}-\sqrt{7}\)}{4}=\frac{\sqrt{11}-\sqrt{7}}{2}).
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If \(s=\sqrt{98}+\sqrt{200}\), what is the value of \(\frac{s}{\sqrt{2}}\)?
C. (17)
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{200}=10\sqrt{2}\), इसलिए \(s=17\sqrt{2}\) है। \(\sqrt{2}\) से भाग देने पर (17) मिलता है। / \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{200}=10\sqrt{2}\), so \(s=17\sqrt{2}\). Dividing by \(\sqrt{2}\) gives (17).
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If \(a=\sqrt{3}+\sqrt{5}\) and \(b=\sqrt{5}-\sqrt{3}\), what is the value of (ab)?
A. (2)
क्रम बदलने पर (ab=\(\sqrt{5}+\sqrt{3}\)\(\sqrt{5}-\sqrt{3}\)=5-3=2) है। संयुग्मी रूप पहचानें। / After rearranging (ab=\(\sqrt{5}+\sqrt{3}\)\(\sqrt{5}-\sqrt{3}\)=5-3=2). Identify the conjugate form.
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Which option will remain irrational?
C. \(\frac{\sqrt{45}}{\sqrt{9}}\)
पहले, दूसरे और चौथे विकल्प क्रमशः (5), (5) और (6) देते हैं। तीसरा \(\sqrt{5}\) देता है जो अपरिमेय है। / The first, second, and fourth options give (5), (5), and (6) respectively. The third gives \(\sqrt{5}\), which is irrational.
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If \(x=\sqrt{8+\sqrt{15}}\), what is the value of \(x^2-8\)?
A. \(\sqrt{15}\)
\(x^2=8+\sqrt{15}\) है। इसलिए \(x^2-8=\sqrt{15}\) होगा। / \(x^2=8+\sqrt{15}\). Therefore \(x^2-8=\sqrt{15}\).
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What is the value of (\(\sqrt{45}+\sqrt{5}\)\(\sqrt{45}-\sqrt{5}\))?
A. (40)
यह \(a^2-b^2\) रूप है, इसलिए मान (45-5=40) है। संयुग्मी गुणन में सीधे वर्गों का अंतर लें। / This is the \(a^2-b^2\) form, so the value is (45-5=40). In conjugate multiplication take the difference of squares directly.
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