Given that $m$ is a real number not less than $-1$, such that equation $x^2 + 2(m-2)x + m^2 - 3m + 3 = 0$ has two distinct real roots $x_1$ and $x_2$. Find the maximum value of $\frac{1}{2}\left(\frac{mx_1^2}{1-x_1} + \frac{mx_2^2}{1-x_2}\right)$
Step-by-Step Solution
Key Concept: Vieta's formulas combined with algebraic simplification converts a symmetric expression in roots into a function of the parameter $m$.
For two distinct real roots, $\Delta > 0 \Rightarrow (m-2)^2 - (m^2 - 3m + 3) > 0$, which simplifies to $m < 1$, so $m \in [-1, 1)$. The expression $m\left(\frac{x_1^2}{1-x_1} + \frac{x_2^2}{1-x_2}\right)$ is then simplified using Vieta's formulas and algebraic manipulation to get $2(m^2 - 3m + 1) \in (-2, 10]$.
Correct Answer: 5