If the average of x and frac{1}{x} (x neq 0) | vedclass

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(C) Given that the average of $x$ and $\frac{1}{x}$ is $M$.$\frac{x + \frac{1}{x}}{2} = M$Multiplying both sides by $2$,we get:$x + \frac{1}{x} = 2M$N

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$2M^{2}-1$

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(C) Given that the average of $x$ and $\frac{1}{x}$ is $M$.$\frac{x + \frac{1}{x}}{2} = M$Multiplying both sides by $2$,we get:$x + \frac{1}{x} = 2M$Now,we need to find the average of $x^{2}$ and $\frac{1}{x^{2}}$,which is given by:$	ext{Average} = \frac{x^{2} + \frac{1}{x^{2}}}{2}$Using the algebraic identity $(a + b)^{2} = a^{2} + b^{2} + 2ab$,we can write:$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2(x)(\frac{1}{x}) = (x + \frac{1}{x})^{2} - 2$Substituting the value of $(x + \frac{1}{x}) = 2M$:$x^{2} + \frac{1}{x^{2}} = (2M)^{2} - 2 = 4M^{2} - 2$Therefore,the required average is:$\frac{4M^{2} - 2}{2} = 2M^{2} - 1$

Age (in years)	$8, 20, 26, 29$
Number of People	$3, 2, m, 1$

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

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