If the average of x and $$\frac{1}{x}$$ (x $$ \ne $$ 0) is M, then the average of x² and $$\frac{1}{{{x^2}}}$$ is :

According to the question,
Average of $$\frac{{x + \frac{1}{x}}}{2} = M$$
Put x = 1
$$\eqalign{ & \therefore \,\,\frac{{1 + \frac{1}{1}}}{2} = M \cr & \Rightarrow M = 1 \cr & \therefore \,\,\frac{{{x^2} + \frac{1}{{{x^2}}}}}{2} \cr & = \frac{{{1^2} + \frac{1}{{{1^2}}}}}{2} \cr & = 1 \cr} $$
Now check from the option
Option: (C) 2M² - 1 (put M = 1)
= 2 × 1 - 1
= 1 (satisfied)

Alternate :
According to the question,
$$\eqalign{ & \Rightarrow \frac{{x + \frac{1}{x}}}{2} = M \cr & \text{Squaring the both sides} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {(2M)^2} - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 4{M^2} - 2 \cr} $$
Required average
$$\eqalign{ & = \frac{{{x^2} + \frac{1}{{{x^2}}}}}{2} \cr & = \frac{{4{M^2} - 2}}{2} \cr & = 2{M^2} - 1 \cr} $$