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

A. 1 – M2

B. 1 – 2M2

C. 2M2 – 1

D. 2M2 + 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)
Alternate Solution:
$\begin{array}{l}\text{According to the question}\\ \Rightarrow\left(\Large\frac{x+\frac{1}{x}}{2}\right)=M\\ \text{ Squaring the both sides }\\ \Rightarrow\left(\Large x^2+\frac{1}{x^2}\right)=(2M)^2-2\\ \Rightarrow\left(\Large x^2+\frac{1}{x^2}\right)=4M^2-2\\ \text{Required average}\\ \begin{array}{l}=\Large\frac{x^2+\frac{1}{x^2}}{2}\\ =\Large\frac{4M^2-2}{2}\\ =2M^2-1\end{array}\end{array}$

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