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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
$2{f_1}\left( x \right){f_2}\left( y \right)$
$2{f_1}\left( x \right){f_1}\left( y \right)$
$2{f_1}\left( {x + y} \right){f_2}\left( {x - y} \right)$
$2{f_1}\left( {x + y} \right){f_1}\left( {x - y} \right)$
Solution
${f_1}\left( x \right) = \frac{{{a^x} + {a^{ – x}}}}{2}$ and ${f_2}\left( x \right) = \frac{{{a^x} – {a^{ – x}}}}{2}$
${f_1}\left( {x + y} \right) + {f_1}\left( {x – y} \right)$
$ = \frac{1}{2}\left( {{a^{x + y}} + {a^{ – x – y}} + {a^{x – y}} + {a^{ – x + y}}} \right)$
$ = \frac{1}{2}\left( {{a^x}\left( {{a^y} + {a^{ – y}}} \right) + {a^{ – x}}\left( {{a^y} + {a^{ – y}}} \right)} \right)$
$ = 2.\left( {\frac{{{a^x} + {a^{ – x}}}}{2}} \right)\left( {\frac{{{a^y} + {a^{ – y}}}}{2}} \right)$
$ = 2{f_1}\left( x \right){f_1}\left( y \right)$
