We are given a quadratic function $f(x) = ax^2 + bx + c$, where $a, b, c$ are real numbers. The maximum value of $f(x)$ on the interval $-1 \le x \le 1$ is $M$, and the minimum value is $m$. We are given that $M = 1$ and $m = -1$. We need to find the coordinates of the vertex of the parabola $y = f(x)$ when $a = 1$, and determine the conditions on $b$ for $|-b/A| > 1$, which leads to $|b| > A$. Then we need to find the relationship between $|f(1) - f(-1)|$ and $|b|$ and $M - m$. Finally, we need to find the pairs $(b, c)$ that satisfy the given conditions.

1. Problem Description

We are given a quadratic function $f(x) = ax^2 + bx + c$, where $a, b, c$ are real numbers. The maximum value of $f(x)$ on the interval $-1 \le x \le 1$ is $M$, and the minimum value is $m$. We are given that $M = 1$ and $m = -1$. We need to find the coordinates of the vertex of the parabola $y = f(x)$ when $a = 1$, and determine the conditions on $b$ for $|-b/A| > 1$, which leads to $|b| > A$. Then we need to find the relationship between $|f(1) - f(-1)|$ and $|b|$ and $M - m$. Finally, we need to find the pairs $(b, c)$ that satisfy the given conditions.

2. Solution Steps

(1) When $a = 1$, $f(x) = x^2 + bx + c$. The vertex of the parabola $y = f(x)$ is at $x = -\frac{b}{2}$. The $y$-coordinate of the vertex is $f(-\frac{b}{2}) = (-\frac{b}{2})^2 + b(-\frac{b}{2}) + c = \frac{b^2}{4} - \frac{b^2}{2} + c = c - \frac{b^2}{4}$.

Thus, the vertex is $(-\frac{b}{2}, c - \frac{b^2}{4})$. Therefore, A = 2 and B =

4. If $|-\frac{b}{A}| > 1$, then $|-\frac{b}{2}| > 1$, which means $|\frac{b}{2}| > 1$, so $|b| > 2$. Thus A =

2. Now we calculate $|f(1) - f(-1)| = |(1^2 + b(1) + c) - ((-1)^2 + b(-1) + c)| = |1 + b + c - (1 - b + c)| = |2b| = 2|b|$.

Since $M = 1$ and $m = -1$, $M - m = 1 - (-1) = 2$.

Thus, $|f(1) - f(-1)| = 2|b|$, and we are given $|f(1) - f(-1)| = C |b| D M - m (= 2)$. Therefore $C=2$. Also, we have $2|b|=2|b| D 2$. Then, $D$ represents the equality. Thus $D$ corresponds to option 0 which is $=$.

Since $|b| > 2$, and $M = 1, m = -1$ for $-1 \le x \le 1$, let us test the choices for (b, c).

Consider the case $|b| > 2$.

If $(b, c) = (1 + 2\sqrt{3}, 1 + 2\sqrt{3})$ (Option 0), $|b| = |1+2\sqrt{3}| > 2$. Then $f(x) = x^2 + (1+2\sqrt{3})x + (1+2\sqrt{3})$.

If $(b, c) = (1 - 2\sqrt{3}, 1 - 2\sqrt{3})$ (Option 1), $|b| = |1-2\sqrt{3}| \approx |1-2(1.732)| \approx |1-3.464| \approx 2.464 > 2$. Then $f(x) = x^2 + (1-2\sqrt{3})x + (1-2\sqrt{3})$.

If $(b, c) = (-1 + 2\sqrt{3}, 1 - 2\sqrt{3})$ (Option 2), $|b| = |-1+2\sqrt{3}| \approx |-1+3.464| \approx 2.464 > 2$. Then $f(x) = x^2 + (-1+2\sqrt{3})x + (1-2\sqrt{3})$.

If $(b, c) = (-1 - 2\sqrt{3}, 1 + 2\sqrt{3})$ (Option 3), $|b| = |-1-2\sqrt{3}| \approx |-1-3.464| \approx 4.464 > 2$. Then $f(x) = x^2 + (-1-2\sqrt{3})x + (1+2\sqrt{3})$.

The vertex is at $x = -\frac{b}{2}$. Since $|b| > 2$, $|-\frac{b}{2}| > 1$. This means the vertex is outside the interval $[-1, 1]$. The maximum and minimum must occur at the endpoints $x=-1$ and $x=1$.

For option 3, $f(1) = 1 + (-1-2\sqrt{3}) + (1+2\sqrt{3}) = 1$, and $f(-1) = 1 - (-1-2\sqrt{3}) + (1+2\sqrt{3}) = 1 + 1 + 2\sqrt{3} + 1 + 2\sqrt{3} = 3 + 4\sqrt{3} > 1$. Also, $f(-1) > 1$ and we have $f(x) = x^2 - (1+2\sqrt{3})x + (1+2\sqrt{3})$

In this case the minimum must be at x=

1. $f(1) = 1$. Thus $f(-1) > 1$. So $x= -1$ is neither maximum or minimum.

For option 1, $f(1) = 1 + (1-2\sqrt{3}) + (1-2\sqrt{3}) = 3-4\sqrt{3} \approx 3-4(1.732) = 3 - 6.928 = -3.928 < -1$.

$f(-1) = 1 - (1-2\sqrt{3}) + (1-2\sqrt{3}) = 1 - 1 + 2\sqrt{3} + 1 - 2\sqrt{3} = 1$

The vertex is at $x = \frac{2\sqrt{3} - 1}{2} = \sqrt{3} - 0.5 \approx 1.732 - 0.5 = 1.232 > 1$.

Try $(b, c) = (2 + 2\sqrt{2}, 2 + 2\sqrt{2})$. Then $|b| = 2 + 2\sqrt{2} > 2$. Also $f(x) = x^2 + (2 + 2\sqrt{2})x + (2+2\sqrt{2})$

$f(1) = 1 + 2 + 2\sqrt{2} + 2+2\sqrt{2} = 5 + 4\sqrt{2} > 1$.

Consider $M=1$ and $m=-1$. Let E = 1, F =

2. $E < F$

3. Final Answer

A: 2

B: 4

C: 2

D: 0

E: 1

F: 2