Set values of $K$ for which roots of the quadratic $x^2 - (2K - 1) x + K(K - 1) = 0$ are
- both less than $2$ is $K \in (2, \infty)$
- of opposite sign is $K \in (-\infty, 0) \cup (1, \infty)$
- of same sign is $K \in (-\infty, 0) \cup (1, \infty)$
- both greater than $2$ is $K \in (2, \infty)$
Roots of x^2 - (2K-1)x + K(K-1) = 0 are (2K-1 +/- sqrt((2K-1)^2 - 4K(K-1))) / 2. Discriminant = 4K^2 - 4K + 1 - 4K^2 + 4K = 1. Roots are (2K-1+1)/2 = K and (2K-1-1)/2 = K-1. For same sign, their product K(K-1) > 0. This holds when K > 1 or K < 0.
For the roots of the quadratic equation to have the same sign, their product must be strictly positive. The product of the roots is given by the constant term divided by the leading coefficient, which equals K(K - 1) / 1. Solving the inequality K(K - 1) > 0 yields the intervals K < 0 or K > 1, meaning the set is (-infinity, 0) union (1, infinity). Therefore, the correct result is of same sign is K in (-infinity, 0) union (1, infinity).