If the function f (x) defined on R satisfies: F (x) = F & nbsp; (4-x) and F & nbsp; (2-x) + F & nbsp; (X-2) = 0, then the value of F & nbsp; (2008) is______ ．

If the function f (x) defined on R satisfies: F (x) = F & nbsp; (4-x) and F & nbsp; (2-x) + F & nbsp; (X-2) = 0, then the value of F & nbsp; (2008) is______ ．

∵ F & nbsp; (2-x) + F & nbsp; (X-2) = 0 ∵ f (X-2) = - f (2-x) replace x with x + 2 to get f (x) = - f (- x), so f (x) is an odd function ∵ f (x) = F & nbsp; (4-x) f (x) = - f (x-4) replace x with x + 4 to get f (x + 4) = - f (x) so f (x + 4) = f (x-4) function

For the functions f (x) and G (x) with the same domain D, if there is a function H (x) = KX + B (k, B is constant), for any given positive number m, there is a corresponding x0 D, such that when x D and x > x0, there is always a line L: y = KX + B, which is called the "fractional asymptote" of the curves y = f (x) and y = g (x)
① F (x) = x & # 178;, G (x) = radical X; ② f (x) = 10 ^ (- x) + 2, G (x) = (2x-3) / X;
③f(x)=(x²+1)/x,g(x)=(xlnx+1)/lnx; ④f(x)=2x²/(2x+1),g(x)=2(x-1-e^(-x)).
Where, the curve y = f (x) and y = g (x) have "sub asymptote"
A．①④ B.②③ C.②④ D.③④
Why is there a fractional asymptote if and only if f (x) - G (x) → 0 when x →∞

If and only if f (x) and G (x) have a fractional asymptote when x →∞, f (x) - G (x) → 0. For (1) f (x) = X2, G (x) = x, when x ＞ 1, it does not conform, so (1) does not exist; for (2) f (x) = 10-x + 2, G (x) = 2x-3, x, there must be a fractional asymptote, because when, f (x) - G (x) → 0

If f (x) satisfies f (x + a) = - f (x), then f (x) is a periodic function with period 2A (a > 0),

Let f (x) = 2x + A, x > 2x + A2, X ≤ 2. If the range of F (x) is r, then the range of constant a is ()
A. （-∞，-1]∪[2，+∞）B. [-1，2]C. （-∞，-2]∪[1，+∞）D. [-2，1]

When x ＞ 2, y = 2x + a ＞ 4 + A; when x ≤ 2, y = x + A2 ≤ 2 + A2 ∵ f (x) has a range of R, A2 + 2 ≥ a + 4. By solving the inequality, a ≥ 2 or a ≤ - 1 is selected