Question

Let $f\left(x\right)=a^x(a>0)$ be written as $f\left(x\right)=f_1\left(x\right)+f_2\left(x\right)$, where $f_1\left(x\right)$ is an even function and $f_2\left(x\right)$ is an odd function. Then $f_1(x+y)+f_1(x-y)$ equals :

• A. $2f_1\left(x\right)f_1\left(y\right)$
• B. $2f_1(x+y)f_1(x-y)$
• C. $2f_1\left(x\right)f_2\left(y\right)$
• D. $2f_1(x+y)f_2(x-y)$

Answer 1

The correct option is A $2f_1\left(x\right)f_1\left(y\right)$

Every function $f\left(x\right)$ can be represented as
$f\left(x\right)=\frac{f\left(x\right)+f(-x)}{2}+\frac{f\left(x\right)-f(-x)}{2}$,
where $\frac{f\left(x\right)+f(-x)}{2}$ is even function and $\frac{f\left(x\right)-f(-x)}{2}$ is odd function.

$f\left(x\right)=a^x$
So, $f_1\left(x\right)=\frac{a^x+a^{-x}}{2}$

$\therefore f_1(x+y)+f_1(x-y)$
$=\frac{a^{x+y}+a^{-(x+y)}}{2}+\frac{a^{x-y}+a^{-(x-y)}}{2}$
$=\frac{a^x(a^y+a^{-y})+a^{-x}(a^y+a^{-y})}{2}$
$=\frac{(a^x+a^{-x})(a^y+a^{-y})}{2}$
$=2f_1\left(x\right)f_1\left(y\right)$