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

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Question

Average of $x$ and $\frac{1}{x}=m$

$\Rightarrow \frac{x+\frac{1}{x}}{2}=m \Rightarrow x+\frac{1}{x}=2 m$

Now $x^{2}$ and $\frac{1}{x^{2}}$

$\

Vedclass · Accepted Answer

$2 M ^{2}-1$

Vedclass · Answer

Average of $x$ and $\frac{1}{x}=m$

$\Rightarrow \frac{x+\frac{1}{x}}{2}=m \Rightarrow x+\frac{1}{x}=2 m$

Now $x^{2}$ and $\frac{1}{x^{2}}$

$\Rightarrow \frac{x^{2}+\frac{1}{x^{2}}}{2}=\frac{\left(x+\frac{1}{x}\right)^{2}-2}{2}=\frac{4 m^{2}-2}{2}=2 m^{2}-1$

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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