When 1

When x>2, f(x)= 1/2*f(x-2)

①y=f(x)-ln(x+ 1)

When 0

And y(0)=0, so x=0 is the zero point of the function y.

When 1

y( 1)= 1-LN2 >； 0，y(2)=-ln3 & lt； 0, so there is a zero on (1, 2).

When x>2, y =1/2 * f (x-2)-ln (x+1), because 0

So y < 0, that is, there is no zero on x>2.

To sum up, the function y has only two zeros, and the answer is wrong.

② let g (x) = f (x)-k/x (x >; 0)

When 0

When x>2, let x=2n+m, where n is a positive integer, 0.

Then g (x) = f (x)-k/x = f (2n+m)-k/(2n+m) = f (m)/2n-k/(2n+m) = (1-kloc-0/-m |). =0 holds.

Obviously k is a positive number, [1-|1-m |] (2n+m) < = k * 2 n, k > =(2n+m)/2^n*[ 1-| 1-m|]

Because when m= 1 and n= 1, the maximum value of (2n+m)/2n * [1-|1-m |] is 3/2, so k >: =3/2.

In a word, k & gt=3/2, and the answer is correct.

③ When 0

When x>2, f(x)= 1/2*f(x-2), let x=2n+m, where n is a positive integer, 0.

F (2n+m) = f (m)/2n = [1-|1-m |]/2n, and when m= 1, the maximum value is 1/2 n, obviously there is no minimum value.

So there is no minimum and the answer is wrong.

This is obviously correct.

So the last two conclusions are correct, so the answer is B.