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  4. Let f(x ) = ax (a > 0) be written as f( x) = f1(x ) + f2(x) , where f1( x), is an even function of f2(x) is an odd function. Then f1(x + y) + f1(x - y) equals

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Q. 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 of $f_2(x)$ is an odd function. Then $f_1(x + y) + f_1(x - y)$ equals

JEE MainJEE Main 2019Relations and Functions

Solution:

$f\left(x\right) =a^{x}, a >0 $
$ f\left(x\right) = \frac{a^{x} +a^{-x}+a^{x} -a^{-x}}{2} $
$ \Rightarrow f_{1} \left(x\right)=\frac{a^{x}+a^{-x}}{2} $
$f_{2}\left(x\right)= \frac{a^{x}-a^{-x}}{2} $
$\Rightarrow f_{1}\left(x+y\right) + f_{1}\left(x-y\right) $
$ = \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}\left(x\right) \times2f_{1}\left(y\right) $
$= 2f_{1}\left(x\right)f_{1}\left(y\right) $