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 511-520 of 600 questions.
If \(q=\sqrt{7}+3\), what is the value of \(q^2-6q\)?
A. (-2)
\(q^2=16+6\sqrt{7}\) और \(6q=18+6\sqrt{7}\) है। घटाने पर (-2) मिलता है। / \(q^2=16+6\sqrt{7}\) and \(6q=18+6\sqrt{7}\). Subtracting gives (-2).
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What is the value of (\(\sqrt{7}+\sqrt{5}\)2-\(\sqrt{7}-\sqrt{5}\)2)?
A. \(4\sqrt{35}\)
पहचान ((a+b)2-(a-b)2=4ab) लगाएं। यहाँ मान \(4\sqrt{35}\) है। / Use ((a+b)2-(a-b)2=4ab). Here the value is \(4\sqrt{35}\).
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If \(t=\sqrt{23}+4\), what is the value of \(t+\frac{7}{t}\)?
A. \(2\sqrt{23}\)
\(\frac{7}{\sqrt{23}+4}=\sqrt{23}-4\) है क्योंकि हर (23-16=7) बनता है। इसलिए योग \(2\sqrt{23}\) है। / \(\frac{7}{\sqrt{23}+4}=\sqrt{23}-4\) because the denominator becomes (23-16=7). So the sum is \(2\sqrt{23}\).
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What is the value of \(\frac{\sqrt{98}-\sqrt{50}}{\sqrt{2}}\)?
B. (2)
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए अंश \(2\sqrt{2}\) है। भाग देने पर (2) मिलता है। / \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so the numerator is \(2\sqrt{2}\). Dividing gives (2).
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A. (16), \(2\sqrt{5}\)
(cd=21-5=16) और \(c-d=2\sqrt{5}\) है। संयुग्मी युग्म में दोनों मान अलग-अलग निकालें। / (cd=21-5=16) and \(c-d=2\sqrt{5}\). Find both values separately in a conjugate pair.
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What is the simplified form of \(\sqrt{539}-\sqrt{275}+\sqrt{99}\)?
B. \(7\sqrt{11}\)
\(\sqrt{539}=7\sqrt{11}\), \(\sqrt{275}=5\sqrt{11}\) और \(\sqrt{99}=3\sqrt{11}\) है। इसलिए परिणाम \(5\sqrt{11}\) नहीं, \(7\sqrt{11}-5\sqrt{11}+3\sqrt{11}=5\sqrt{11}\) है। / \(\sqrt{539}=7\sqrt{11}\), \(\sqrt{275}=5\sqrt{11}\), and \(\sqrt{99}=3\sqrt{11}\). So the result is \(7\sqrt{11}-5\sqrt{11}+3\sqrt{11}=5\sqrt{11}\).
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What is the rationalised form of \(\frac{6}{\sqrt{19}-\sqrt{10}}\)?
C. (\frac{6\(\sqrt{19}+\sqrt{10}\)}{9})
संयुग्मी से गुणा करने पर हर (19-10=9) बनता है। इसलिए रूप (\frac{6\(\sqrt{19}+\sqrt{10}\)}{9}) है। / Multiplying by the conjugate makes the denominator (19-10=9). So the form is (\frac{6\(\sqrt{19}+\sqrt{10}\)}{9}).
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If \(x=\sqrt{17}+\sqrt{6}\), what is the value of \(x^2-23\)?
C. \(2\sqrt{102}\)
\(x^2=17+6+2\sqrt{102}=23+2\sqrt{102}\) है। इसलिए \(x^2-23=2\sqrt{102}\) है। / \(x^2=17+6+2\sqrt{102}=23+2\sqrt{102}\). So \(x^2-23=2\sqrt{102}\).
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What is the square of \(\sqrt{13+\sqrt{42}}\) equal to?
D. \(13+\sqrt{42}\)
वर्गमूल का वर्ग अंदर की संख्या देता है। इसलिए (\left\(\sqrt{13+\sqrt{42}}\right\)2=13+\sqrt{42}) है। / The square of a square root gives the number inside. So (\left\(\sqrt{13+\sqrt{42}}\right\)2=13+\sqrt{42}).
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B. (15)
क्षेत्रफल (\(\sqrt{22}+\sqrt{7}\)\(\sqrt{22}-\sqrt{7}\)=22-7=15) है। संयुग्मी आयामों से परिमेय क्षेत्रफल मिल सकता है। / Area is (\(\sqrt{22}+\sqrt{7}\)\(\sqrt{22}-\sqrt{7}\)=22-7=15). Conjugate dimensions can give a rational area.
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