If the average of $x$ and $\frac{1}{x}$ ($x\ne 0$) is $m$, then the average of $x^2$ and $\frac{1}{x^2}$ is :

Step 1: Express the given average as an equation
The average of $x$ and $\frac{1}{x}$ is $m$:

$$\frac{x+\frac{1}{x}}{2}=m⟹x+\frac{1}{x}=2m$$

Step 2: Square both sides of the equation

$${\left(x+\frac{1}{x}\right)}^2=(2m{)}^2$$

Using the algebraic identity $(a+b{)}^2=a^2+2ab+b^2$:

$$x^2+2(x)\left(\frac{1}{x}\right)+\frac{1}{x^2}=4m^2$$

$$x^2+2+\frac{1}{x^2}=4m^2$$

$$x^2+\frac{1}{x^2}=4m^2-2$$

**Step 3: Find the average of $x^2$ and $\frac{1}{x^2}$ **

$$\text{Average}=\frac{x^2+\frac{1}{x^2}}{2}$$

Substitute the value of $x^2+\frac{1}{x^2}$ from Step 2:

$$\text{Average}=\frac{4m^2-2}{2}=\frac{2(2m^2-1)}{2}=2m^2-1$$