Y=lg(x-3) translates 3 units to the right,

Symmetrize the image along the straight line x=3 to the left of the straight line x=3 (keep the right part) and get y=lg|x-3|.

Then, the part below the X axis is symmetrical with the top of the X axis along the X axis, and an image of y = | LG | X-3 ||| is obtained.

Add a little more (3, 1) and you will get an image of y=f(x). As shown in the figure.

As can be seen from the figure, the number of roots with f(x)=k is

( 1)k & lt； 0, 0;

(2) When k = 0,2;

(3)0 & lt； K<4 is in1;

(4) When k = 1, 5;

(5)k & gt； 4 in 1.

If [f (x)] 2+AF (x)+B = 0 has 9 different real roots, then f(x)=k 1 has 5, and f(x)=k2? There are four,

Therefore, k 1= 1, k2 >；; 0 and k2 ≠ 1,

That is, the root of the equation x 2+ax+b = 0 is x= 1, and the positive root is x ≠ 1.

So 1+a+b=0, then b= -a- 1,

X2+ax-a-1= (x-1) (x+a+1) gives -A- 1 >: 0 and-a-1≠/kloc-0.

Get an a

Choose D.