Let f (x) = a^x (a 0) be written as f( x | vedclass

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Question

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

Vedclass · Accepted Answer

$2{f_1}\left( x \right){f_1}\left( y \right)$

Vedclass · Answer

${f_1}\left( x \right) = \frac{{{a^x} + {a^{ - x}}}}{2}$ 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)$

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

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