Practicing Success
Practicing Success
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If the average of x and \(\frac{1 }{x}\) \((x\neq 0)\) is M, then the average of \(x^{2} and \frac{1 }{x^{2}}\) is: |
\(1-m^{2}\) 1 + 2m \(2m^{2}-1\) \(2m^{2}+1\) |
\(2m^{2}-1\) |
\(\frac{x + \frac{1}{×}}{2}\) = m ⇒ x + \(\frac{1}{×}\) = 2m x2 + \(\frac{1}{×^2}\) = 4m2 - 2 Avg. = \(\frac{4m^2 - 2}{2}\) = 2m2 - 1 |
