Step-by-Step Solution
Key Concept: The discriminant and behavior of the quadratic function determine whether roots are real and their signs.
For option (A), the discriminant is $D = (2abcos C)^2 - 4a^2b^2 = -4a^2b^2sin^2C 0$ with the condition $\frac{a+c}{2} = b\sqrt{ac}$, which implies $b^2 > ac$, making both roots real and negative since $f(x) = ax^2 + abx + c > 0$ for all $x \geq 0$. For option (C), $f(x) = x^2 - (a+1)x - (a^2 + a)$ has $f(0) b$ means $b^2 - ac < 0$, so $D = 4(b^2 - ac) < 0$, giving no real roots.
Correct Answer: [A-T] [B-R, S] [C-P, S] [D-T]