If the average of x and 1/x (x 0) is M, then the average of x2 and 1x^2 is :

Answer: Option C

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

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