Lf A-x2-1-x2-B-x-1-x then minimum value of A-B equals

The correct option is D 2√(2) A=(x 1x)^2+2 B=(x 1x) AB=(x 1x)+2x 1x Use AM amp; GM (x 1x)+2x 1x2≥√(2) (x 1x)+2x 1x≥ 2√(2) ...

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lf A=x2+1/x...

Question

lf $A = x^{2} + \frac{1}{x^{2}}, B = x - \frac{1}{x}$ then minimum value of $\frac{A}{B}$ equals

\sqrt{2}

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2 \sqrt{2}

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\frac{1}{\sqrt{2}}

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2

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lim x \to \infty (\frac{x^{2} - 1}{x + 1} - a x - b) = 2

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{sin}^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - . . . . .) + {cos}^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . . . .) = \frac{π}{2}

0 < | x | < \sqrt{2}

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