Prove the following
Question:

Let $f(\mathrm{x})=\mathrm{a}^{\mathrm{x}}(\mathrm{a}>0)$ be written as $f(\mathrm{x})=f_{1}(\mathrm{x})+f_{2}(\mathrm{x})$, where $f_{1}(\mathrm{x})$ is an even function of $f_{2}(x)$ is an odd function. Then $f_{1}(\mathrm{x}+\mathrm{y})+f_{1}(\mathrm{x}-\mathrm{y})$ equals

1. $2 f_{1}(\mathrm{x}) f_{1}(\mathrm{y})$

2. $2 f_{1}(\mathrm{x}) f_{2}(\mathrm{y})$

3. $2 f_{1}(\mathrm{x}+\mathrm{y}) f_{2}(\mathrm{x}-\mathrm{y})$

4. $2 f_{1}(\mathrm{x}+\mathrm{y}) f_{1}(\mathrm{x}-\mathrm{y})$

Correct Option: 1

Solution:

$f(x)=a^{x}, a>0$

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

$\Rightarrow f_{1}(\mathrm{x})=\frac{\mathrm{a}^{\mathrm{x}}+\mathrm{a}^{-\mathrm{x}}}{2}$

$f_{2}(\mathrm{x})=\frac{\mathrm{a}^{\mathrm{x}}-\mathrm{a}^{-\mathrm{x}}}{2}$

$\Rightarrow f_{1}(x+y)+f_{1}(x-y)$

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

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

$=f_{1}(\mathrm{x}) \times 2 f_{1}(\mathrm{y})$

$=2 f_{1}(\mathrm{x}) f_{1}(\mathrm{y})$