Given the function f(x) = 1/(x-1) + 1/(x-2) + ...+1/(x-2009), determine the number of real solutions of the equation f(x)=a.

We'll use Rolle's theorem to determine the number of real
solutions of the equation.

For this reason, we'll
differentiate with respect to x:

f'(x) = -1/(x-1)^2 -
1/(x-2)^2 - ... - 1/(x-2009)^2

We notice that the 1st
derivative is negative for any real value of x, therefore f(x) is decreasing over each
of these intervals (-infinite,1) ; (1,2) ; (2,3); .....;
(2009,+infinite).

If we'll draw the graph of the function
we can find the number of roots just counting the intercepting points between the graph
of the function and the line x = a.

If "a" belongs to the
interval (0 ; +infinite), then the equation has 2009 solutions. If  "a" belongs to the
interval (-infinite,0), then the equation has 2009 solutions,
also.

Therefore, the equation f(x)=a has 2009
solutions.