Official solution
f(x) = ax
As f1(x) is even function then
f1(x) = \[{{{f\left( x \right) + f\left( { - x} \right)} \over 2}}\]
= \[{{{a^x} + {a^{ - x}}} \over 2}\]
As f2(x) is odd function then
f2(x) = \[{{{f\left( x \right) - f\left( { - x} \right)} \over 2}}\]
= \[{{{a^x} - {a^{ - x}}} \over 2}\]
Now,
f1
(x + y) + f1
(x -- y)
= \[{{{a^{x + y}} + {a^{ - \left( {x + y} \right)}} + {a^{x + y}} - {a^{ - \left( {x - y} \right)}}} \over 2}\]
Also f1
(x)f1
(y) = \[\left( {{{{a^x} + {a^{ - x}}} \over 2}} \right)\left( {{{{a^y} + {a^{ - y}}} \over 2}} \right)\]
= \[{{{a^{x + y}} + {a^{x - y}} + {a^{ - x + y}} + {a^{ - x - y}}} \over 4}\]
= \[{{{{f_1}\left( {x + y} \right) + {f_2}\left( {x - y} \right)} \over 2}}\]
\[ \therefore \] f1
(x + y) + f1
(x -- y) = 2f1
(x)f1
(y)