Y = f (x) is an even function defined on R, The image is symmetric with respect to x = 1. For any x 1, x 2 belongs to [0,1 / 2], and f (x 1 + x 2) = f (x 1) * f (x 2), and f (1) = a > 0 1. Find f (1 / 2) and f (1 / 4) 2. Prove that f (x) is a periodic function

Y = f (x) is an even function defined on R, The image is symmetric with respect to x = 1. For any x 1, x 2 belongs to [0,1 / 2], and f (x 1 + x 2) = f (x 1) * f (x 2), and f (1) = a > 0 1. Find f (1 / 2) and f (1 / 4) 2. Prove that f (x) is a periodic function

Let X1 = x2 = 1 / 2
Then f (1) = [f (1 / 2)] ^ 2 = a
∴f(1/2)=sqrt(a)
Sqrt means square root
Let X1 = x2 = 1 / 4
Then f (1 / 2) = [f (1 / 4)] ^ 2 = sqrt (a)
Ψ f (1 / 4) = the root of the fourth power a
∵ f (x) is an even function
∴f(x)=f(-x)
∵ f (x) image is symmetric about x = 1
∴f(x)=f(2-x)
∴f(2-x)=f(-x)
That is, f (x) = f (x + 2)
F (x) is a periodic function

It is known that f (x) is an even function on R, G (x) is an odd function on R, and G (x) = f (x-1). If G (1) = 2, then f (2012) = ()
A. 2B. 0C. -2D. ±2

∵ g (x) is an odd function on R, and G (x) = f (x-1), ∵ g (- x) = f (- x-1) = - f (x-1), ∵ f (x) is an even function on R, ∵ f (- x-1) = - f (x-1) = f (x + 1), then f (x + 2) = - f (x), that is, f (x + 4) = - f (x + 2) = f (x), then f (x) is a periodic function with period 4, then f (2012) = f (0) = f (1-1) = g (1) = 2, so select: a