Question

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

• A. $\frac{1}{2}\left\{f\right(2x)+f(2y\left)\right\}$
• B. $\frac{1}{2}\left\{f\right(2x)-f(2y\left)\right\}$
• C. $\frac{1}{4}\left\{f\right(2x)+f(2y\left)\right\}$
• D. $\frac{1}{4}\left\{f\right(2x)-f(2y\left)\right\}$

Answer 1

The correct option is A

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

Given:
$f\left(x\right)=\frac{2^x+2^{-x}}{2}$
Now,
$f(x+y)f(x-y)$
$=\left(\frac{2^{x+y}+2^{-x-y}}{2}\right)\left(\frac{2^{x-y}+2^{-x+y}}{2}\right)$
$\Rightarrow f(x+y)f(x-y)=\frac{1}{2}$
$(\frac{2^{2x}+2^{-2x}}{2}+\frac{2^{2y}+2^{-2y}}{2})$
$\Rightarrow f(x+y)f(x-y)=\frac{1}{2}\left[f\right(2x+f\left(2y\right)\left)\right]$