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

Correct option is A. 2f_1(x) f_1(y)
f(x) = a^x, a > 0
f(x) = a^x + a^-x + a^x - a^-x / 2
⇒ f_1(x) = (a^x + a^-x)/2
f_2(x) = (a^x - a^-x)/2
=a^(x+y) + a^(-x+y)/2 + a^(x-y) + a^(-x-y)/2
= f_1(x) × 2f_1(y)
=2f_1(x) f_1(y)