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

**A.** 1 – M^{2}

**B.** 1 – 2M^{2}

**C.** 2M^{2} – 1

**D.** 2M^{2} + 1

## Show Answer

###
Answer-C

Solution-

Solution(By Apex Team)

$\begin{array}{l}\text{ Average of }\left(\Large\frac{x+\frac{1}{x}}{2}\right)=M\\
\text{ Put }x=1\\
\therefore\left(\Large\frac{1+\frac{1}{1}}{2}\right)=M\\
\Rightarrow M=1\\
\therefore\Large\frac{x^2+\frac{1}{x^2}}{2}\\
=\Large\frac{1^2+\frac{1}{1^2}}{2}\\
=1\end{array}$
Now check from the option
Option: (C) ${2M}^2 – 1$ (put M = 1)
= 2 × 1 – 1
= 1 (satisfied)

$\begin{array}{l}\text{According to the question}\\ \Rightarrow\left(\Large\frac{x+\frac{1}{x}}{2}\right)=M\\ \text{ Squaring the both sides }\\ \Rightarrow\left(\Large x^2+\frac{1}{x^2}\right)=(2M)^2-2\\ \Rightarrow\left(\Large x^2+\frac{1}{x^2}\right)=4M^2-2\\ \text{Required average}\\ \begin{array}{l}=\Large\frac{x^2+\frac{1}{x^2}}{2}\\ =\Large\frac{4M^2-2}{2}\\ =2M^2-1\end{array}\end{array}$

__Alternate Solution:__$\begin{array}{l}\text{According to the question}\\ \Rightarrow\left(\Large\frac{x+\frac{1}{x}}{2}\right)=M\\ \text{ Squaring the both sides }\\ \Rightarrow\left(\Large x^2+\frac{1}{x^2}\right)=(2M)^2-2\\ \Rightarrow\left(\Large x^2+\frac{1}{x^2}\right)=4M^2-2\\ \text{Required average}\\ \begin{array}{l}=\Large\frac{x^2+\frac{1}{x^2}}{2}\\ =\Large\frac{4M^2-2}{2}\\ =2M^2-1\end{array}\end{array}$