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.
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If \(\frac{x^{a}}{x^{b}}=x^{7}\) and (a+b=13), what is the value of (a)?
C. (10)
\(\frac{x^{a}}{x^{b}}=x^{a-b}=x^{7}\), इसलिए (a-b=7) और (a+b=13)। हल करने पर (a=10) मिलता है। / Since \(\frac{x^{a}}{x^{b}}=x^{a-b}=x^{7}\), we have (a-b=7) and (a+b=13). Solving gives (a=10).
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What is the value of (\left\(\frac{27}{8}\right\)^{-\frac{2}{3}})?
A. \(\frac{4}{9}\)
(\left\(\frac{27}{8}\right\)^{\frac{1}{3}}=\frac{3}{2}), इसलिए (\left\(\frac{27}{8}\right\)^{-\frac{2}{3}}=\left\(\frac{3}{2}\right\)^{-2}=\frac{4}{9})। परीक्षा में पहले घनमूल निकालें। / Since (\left\(\frac{27}{8}\right\)^{\frac{1}{3}}=\frac{3}{2}), (\left\(\frac{27}{8}\right\)^{-\frac{2}{3}}=\left\(\frac{3}{2}\right\)^{-2}=\frac{4}{9}). In exams, take the cube root first.
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If \(x=\sqrt{5}+2\), then \(\frac{1}{x}\) is equal to which expression?
A. \(\sqrt{5}-2\)
\(\frac{1}{\sqrt{5}+2}\cdot\frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2\)। परीक्षा में हर के संयुग्म का प्रयोग करें। / Rationalizing gives \(\frac{1}{\sqrt{5}+2}\cdot\frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2\). In exams, use the conjugate of the denominator.
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What is the value of (\left\(9^{2}\right\)^{3}\div 3^{10})?
B. \(3^{2}\)
(\left\(9^{2}\right\)^{3}=\(3^{2}\)^{6}=3^{12}), और \(3^{12}\div3^{10}=3^{2}\)। परीक्षा में (9) को \(3^{2}\) लिखें। / Here (\left\(9^{2}\right\)^{3}=\(3^{2}\)^{6}=3^{12}), and \(3^{12}\div3^{10}=3^{2}\). In exams, write (9) as \(3^{2}\).
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Which option gives the correct simplified form of (\left\(ab^{-2}\right\)^{3}\cdot a^{-1}b^{5})?
A. \(a^{2}b^{-1}\)
(\left\(ab^{-2}\right\)^{3}=a^{3}b^{-6}), फिर \(a^{-1}b^{5}\) से गुणा करने पर \(a^{2}b^{-1}\) मिलता है। परीक्षा में हर चर की घात अलग-अलग जोड़ें। / We have (\left\(ab^{-2}\right\)^{3}=a^{3}b^{-6}), and multiplying by \(a^{-1}b^{5}\) gives \(a^{2}b^{-1}\). In exams, add exponents separately for each variable.
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If \(x=3+\sqrt{2}\) and \(y=3-\sqrt{2}\), what is the value of \(x^{2}-y^{2}\)?
A. \(12\sqrt{2}\)
(x^{2}-y^{2}=(x-y)(x+y)), जहाँ \(x-y=2\sqrt{2}\) और (x+y=6), इसलिए मान \(12\sqrt{2}\) है। परीक्षा में पहचान से लंबी गणना बचती है। / Using (x^{2}-y^{2}=(x-y)(x+y)), we get \(x-y=2\sqrt{2}\) and (x+y=6), so the value is \(12\sqrt{2}\). In exams, identities reduce calculation.
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What is the simplified value of (\frac{\(2^{5}\)^{3}\cdot\(4^{-2}\)}{8^{2}})?
A. \(2^{5}\)
(\(2^{5}\)^{3}=2^{15}), \(4^{-2}=2^{-4}\), और \(8^{2}=2^{6}\), इसलिए कुल घात (15-4-6=5) है। परीक्षा में सभी आधार (2) में बदलें। / Here (\(2^{5}\)^{3}=2^{15}), \(4^{-2}=2^{-4}\), and \(8^{2}=2^{6}\), so the net exponent is (15-4-6=5). In exams, convert all bases to (2).
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A. (x)
\(\sqrt{x^{3}}=x^{\frac{3}{2}}\), इसलिए \(x^{\frac{3}{2}}\cdot x^{-\frac{1}{2}}=x^{1}=x\)। परीक्षा में मूल को भिन्न घात में बदलें। / Since \(\sqrt{x^{3}}=x^{\frac{3}{2}}\), \(x^{\frac{3}{2}}\cdot x^{-\frac{1}{2}}=x^{1}=x\). In exams, convert radicals to fractional exponents.
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What is the simplified form of (\left\(\frac{x^{-2}y^{3}}{x^{4}y^{-1}}\right\)^{-1})?
A. \(x^{6}y^{-4}\)
अंदर \(\frac{x^{-2}y^{3}}{x^{4}y^{-1}}=x^{-6}y^{4}\), और (-1) घात लेने पर \(x^{6}y^{-4}\) मिलता है। परीक्षा में भाग में घात घटती है। / Inside, \(\frac{x^{-2}y^{3}}{x^{4}y^{-1}}=x^{-6}y^{4}\), and raising to (-1) gives \(x^{6}y^{-4}\). In exams, subtract exponents during division.
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If \(a=2+\sqrt{3}\), what is the value of \(a^{2}+\frac{1}{a^{2}}\)?
A. (14)
\(\frac{1}{a}=2-\sqrt{3}\), इसलिए \(a+\frac{1}{a}=4\) और \(a^{2}+\frac{1}{a^{2}}=4^{2}-2=14\)। परीक्षा में (\left\(a+\frac{1}{a}\right\)^{2}) पहचान लगाएं। / Here \(\frac{1}{a}=2-\sqrt{3}\), so \(a+\frac{1}{a}=4\) and \(a^{2}+\frac{1}{a^{2}}=4^{2}-2=14\). In exams, use the identity (\left\(a+\frac{1}{a}\right\)^{2}).
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