The first step is to draw the functional diagram of lgx;

Step 1, move the above picture to the right by one unit and draw the functional diagram of lg(x- 1);

Step 2, fold the part of the original image whose function value is less than 0 to the top of the X axis along the X axis, and then draw the function diagram of |lg(x- 1)|;

Step 3, take the axis x= 1 as the axis of symmetry, and then draw the function diagram of | LG | X- 1 ||| |;

Step 4, add a point where x= 1 and f(x)=0;

Therefore, let f(x)=a,

Then:

1)a & gt； 0, there are four different real number solutions;

2)a=0, there are three different real number solutions;

3)a & lt； 0, no solution;

Therefore, the necessary and sufficient condition for F 2 (x)+BF (x)+C = 0 to have seven different real number solutions is that the equation X 2+BX+C = 0 has two roots, one is equal to 0 and the other is greater than 0. This time it should be B.

So choose C.