F (x) is a function defined on R. for any x belonging to R, there is f (x + 4) = f (x) + 2F (2). If the image of F (X-2) is symmetric with respect to x = 2, and f (2011) = 2 Find f (1)

F (x) is a function defined on R. for any x belonging to R, there is f (x + 4) = f (x) + 2F (2). If the image of F (X-2) is symmetric with respect to x = 2, and f (2011) = 2 Find f (1)

From the symmetry of F (X-2) with respect to x = 2, we know that f (x) is symmetric with respect to x = 0, f (2) = f (- 2) + 2F (2) = f (2) + 2F (2) = 0
f(2011)=f(2007)=f(2003)=...=f(3)=f(-1)=f(1)=2

It is known that f (x) is a function defined on R. for any x R, f (x + 10) = f (x) + 2F (5). If the image of F (x-1) is symmetric with respect to the line x = 1 and f (- 3) = 2013, then f (2013) =

The image of the function f (x-1) is symmetric about the line x = 1. The image of the function f (x) is symmetric about the Y-axis of the line. That is to say, f (x) is an even function ∵ f (x + 10) = f (x) + 2F (5) let x = - 5 ∵ f (5) = f (- 5) + 2F (5)