Let $f(x)=a^{x}(a>0)$ be written as $f(x)=f_{1}(x)+f_{2}(x)$, where $f_{1}(x)$ is an even function and $f_{2}(x)$ is an odd function. Then $f_{1}(x+y)+f_{1}(x-y)$ equals :
Correct Option: 1
Given function can be written as
$f(x)=a^{x}=\left(\frac{a^{x}+a^{-x}}{2}\right)+\left(\frac{a^{x}-a^{-x}}{2}\right)$
where $f_{1}(x)=\frac{a^{x}+a^{-x}}{2}$ is even function
$f_{2}(x)=\frac{a^{x}-a^{-x}}{2}$ is odd function
$\Rightarrow f_{1}(x+y)+f_{1}(x-y)$
$=\left(\frac{a^{x+y}+a^{-x-y}}{2}\right)+\left(\frac{a^{x-y}+a^{-x+y}}{2}\right)$
$=\frac{1}{2}\left[a^{x}\left(a^{y}+a^{-y}\right)+a^{-x}\left(a^{y}+a^{-y}\right)\right]$
$=\frac{\left(a^{x}+a^{-x}\right)\left(a^{y}+a^{-y}\right)}{2}=2 f_{1}(x) \cdot f_{1}(y)$
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