Let's understand the problem step by step:

We are given: $f\left(x\right)=a^x$ (where $a>0$), and it is expressed as the sum of an even function $f_1\left(x\right)$ and an odd function $f_2\left(x\right)$, so:

$f\left(x\right)=f_1\left(x\right)+f_2\left(x\right)$

Recall the properties:

• Even function: $f_1(-x)=f_1\left(x\right)$
• Odd function: $f_2(-x)=-f_2\left(x\right)$

Step 1: Express $f_1\left(x\right)$ and $f_2\left(x\right)$ in terms of $f\left(x\right)$.

For any function, we can write:

$f_1\left(x\right)=\frac{f\left(x\right)+f(-x)}{2}$ (even part)

$f_2\left(x\right)=\frac{f\left(x\right)-f(-x)}{2}$ (odd part)

Given $f\left(x\right)=a^x$, so $f(-x)=a^{-x}$.

Therefore:

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

$f_2\left(x\right)=\frac{a^x-a^{-x}}{2}$

Step 2: We need to find $f_1(x+y)+f_1(x-y)$.

Substitute the expression for $f_1$:

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

Combine the fractions:

$=\frac{1}{2}[a^{x+y}+a^{-x-y}+a^{x-y}+a^{-x+y}]$

Group terms:

$=\frac{1}{2}\left[\right(a^{x+y}+a^{x-y})+(a^{-x-y}+a^{-x+y}\left)\right]$

Factor:

$=\frac{1}{2}\left[a^x\right(a^y+a^{-y})+a^{-x}(a^{-y}+a^y\left)\right]$

Notice $a^y+a^{-y}$ is common:

$=\frac{1}{2}\left[\right(a^x+a^{-x}\left)\right(a^y+a^{-y}\left)\right]$

But from earlier, $f_1\left(x\right)=\frac{a^x+a^{-x}}{2}$ and $f_1\left(y\right)=\frac{a^y+a^{-y}}{2}$.

So, $(a^x+a^{-x})(a^y+a^{-y})=4f_1\left(x\right)f_1\left(y\right)$.

Therefore:

$f_1(x+y)+f_1(x-y)=\frac{1}{2}\times 4f_1\left(x\right)f_1\left(y\right)=2f_1\left(x\right)f_1\left(y\right)$

So, the correct answer is: $2f_1\left(x\right)f_1\left(y\right)$

Related Topics and Formulae

Even and Odd Functions: Any function can be decomposed into even and odd parts. For a function f(x):

Even part: $f_{even}\left(x\right)=\frac{f\left(x\right)+f(-x)}{2}$

Odd part: $f_{odd}\left(x\right)=\frac{f\left(x\right)-f(-x)}{2}$

Exponential Identities: Useful identities include $a^{x+y}=a^x{a}^y$ and $a^{-x}=\frac{1}{a^x}$.