Chapter 1.4, Problem 35E

To determine

To select: the arithmetic mean of x and y from the given options.

Answer to Problem 35E

A. $\frac{y^2+1}{2y}$

Explanation of Solution

Given information:

  $\begin{array}{l}xy=1\\ x=\frac{1}{y}\end{array}$

The given options are as follows:

A. $\frac{y^2+1}{2y}$

B. $\frac{y+1}{2y}$

C. $\frac{y^2+2}{2y}$

D. $\frac{y^2+1}{y}$

E. $\frac{x^2+1}{y}$

Calculation

The arithmetic mean of x and y will be as follows:

  $\begin{array}{c}\text{Airthmetic mean}=\frac{x+y}{2}\\ =\frac{\left(\frac{1}{y}\right)+y}{2}\left[\text{Substituting value of }x\right]\\ =\frac{\left(\frac{1}{y}\right)+y\left(\frac{y}{y}\right)}{2}\left[\text{Finding common denominator}\right]\\ =\frac{\left(\frac{1}{y}\right)+\left(\frac{y^2}{y}\right)}{2}\left[\text{Multiply}\right]\\ =\frac{\frac{1+y^2}{y}}{2}\left[\text{Add fraction in numerator}\right]\\ =\frac{\frac{1+y^2}{y}}{2}\left[\text{Multiply }\frac{y}{y}\right]\\ =\frac{\frac{1+y^2}{y}}{2}\left(\frac{y}{y}\right)\left[\text{Organize}\right]\\ =\frac{1+y^2}{2y}\left[\text{Cancel}\right]\end{array}$

Thus, the arithmetic mean is $\frac{1+y^2}{2y}$ .