# 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: Option 1) $2f_{1}(x)f_{1}(y)$ Option 2) $2f_{1}(x+y)f_{1}(x-y)$ Option 3) $2f_{1}(x)f_{2}(y)$ Option 4) $2f_{1}(x+y)f_{2}(x-y)$

V Vakul

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

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

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

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

Now,

$f_{1}(x+y)+f_{2}(x-y)$

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

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

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

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

Option 1)

$2f_{1}(x)f_{1}(y)$

Option 2)

$2f_{1}(x+y)f_{1}(x-y)$

Option 3)

$2f_{1}(x)f_{2}(y)$

Option 4)

$2f_{1}(x+y)f_{2}(x-y)$

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