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f(x-1)=g(x), g(1)=f(0)=0, g(x) is an odd function on r, g(-1)=-g(1)=0, f(-2)=g(-1)=0

f(x) is an even function on r, f(2)=0, g(3)=0, g(-3)=0, f(-4)=0, f(4)=0,..

f(2008) = 0 for b

Solution: f(x-1)=g(x)......Get.

f(-x-1)= g(-x)……

Because g(x) is an odd function, i.e. g(x) -g(-x), which is obtained by .

f(x-1)＝-f(-x-1)……

Because f(x) is an even function on r, so.

f(-x-1)＝f(x+1)……

By the way. f(x-1)+ f(x+1)＝0………Substituting x 1 gives f(0)+f(2)=0.........Because f(0) 0.........

From f(2)=0, f(0), f(2), f(4) ...... can be obtained by analogy of recyclingf(2n) is 0

where n is an integer).

So f(2008)=0, choose b

Up in the air. According to the title, it can only be concluded that f(odd)=0 f(even) is uncertain.

(1) 2-(x+3)/(x+1)>=0

x+3)/(x+1)<=2

x<-1 x+3>=2x+2

So x<=1 is x<-1

x>-1 x+3<=2x+2 x>=1 i.e. x>=1, so a=(2)(x-a-1)(x-a)<0b={x|a=1 or a+1<=-1

a>=1 or a<=-2

by f(1-x) = f(1+x).

If f(x) = f(y).

then x+y=2 (2).

f(x1)=f(x2)=f(x3)=0

Obtained from (2).

x1+x2=2

x2+x3=2

x3+x1=2

Add and divide by 2

Get x1 + x2 + x3 = 3

Because the curve is getting slower and slower, the slope of the tangent is getting smaller and smaller, so if the slope of 2 is greater than 3, according to the image you draw, f3 and f2 are larger than the slope of 2, and choose c

2 f (x) 3x 2 6a According to f (x), the monotonicity of f(x) can be judged to be an increase or decrease, and then the extreme value is discussed, and the range of a is calculated3