Question

If $f\left(x\right)=\frac{2^x+2^{-x}}{2}$, then $f\left(x+y\right)·f\left(x-y\right)$ is equal to

• A. $\frac{1}{2}\left[f\left(x+y\right)+f\left(x-y\right)\right]$
• B. $\frac{1}{2}\left[f\left(2x\right)+f\left(2y\right)\right]$
• C. $\frac{1}{2}\left[f\left(x+y\right)·f\left(x-y\right)\right]$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{2}\left[f\left(2x\right)+f\left(2y\right)\right]$

Explanation for the correct option.

Find the value of the given expression.

For the function $f\left(x\right)=\frac{2^x+2^{-x}}{2}$, the expression $f\left(x+y\right)·f\left(x-y\right)$ can be written as:

$\begin{array}{rcl}f\left(x+y\right)·f\left(x-y\right)& =& \frac{2^{x+y}+2^{-\left(x+y\right)}}{2}\times \frac{2^{x-y}+2^{-\left(x-y\right)}}{2}\\ & =& \frac{1}{4}\left[\left(2^x{2}^y+2^{-x}{2}^{-y}\right)\left(2^x{2}^{-y}+2^{-x}{2}^y\right)\right]\mathbf{}\mathbf{}\left[a^{m+n}=a^m{a}^n\right]\\ & =& \frac{1}{4}\left[{\left(2^x\right)}^2+{\left(2^y\right)}^2+{\left(2^{-y}\right)}^2+{\left(2^{-x}\right)}^2\right]\mathbf{[}{\mathbf{a}}^{\mathbf{m}}{\mathbf{a}}^{\mathbf{-}\mathbf{m}}\mathbf{=}\mathbf{1}\mathbf{]}\\ & =& \frac{1}{4}\left[2^{2x}+2^{2y}+2^{-2y}+2^{-2x}\right]\\ & =& \frac{1}{2}\left[\frac{1}{2}\left(2^{2x}+2^{-2x}\right)+\frac{1}{2}\left(2^{2y}+2^{-2y}\right)\right]\\ & =& \frac{1}{2}\left[f\left(2x\right)+f\left(2y\right)\right]\mathbf{\left[}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(2^x+2^{-x}\right)\mathbf{\right]}\end{array}$

Hence, the correct option is B.