Actually, that function is continuous for any value of k. The only point you have to worry about, both for continuity and differentiability, is the point x=1. You can explicitly show that the function is continuous at 1 by computing the limit of f as x approaches 1 and show that it equals f(1). You'll see it doesn't matter what k is for continuity.
To worry about differentiability, all you need to check is that the slope of the function k(x-1) at x=1 matches up with the slope of x2-1 at x=1. Remember that for a function to be differentiable at a point, the limit of the difference quotient at that point must exist. By checking that the slopes match up, you are checking that the left and right handed limits of the difference quotient equal each other (i.e., checking that the limit exists). The function is differentiable everywhere else since it is defined there by polynomials.