Question

The roots $\alpha$, $\beta$ of the quadratic equation $ax^2 + bx + c = 0, a \neq 0$ lie in the interval [0, 1], then?

• A. the maximum value of $\frac{(a-b)(2a-b)}{a(a-b+c)}$ is 3
• B. the minimum value of $\frac{(a-b)(2a-b)}{a(a-b+c)}$ is 2
• C. if $\frac{(a-b)(2a-b)}{a(a-b+c)}$ is at its maximum value, then $\alpha^2 + \beta^2$ can be 2
• D. if $\frac{(a-b)(2a-b)}{a(a-b+c)}$ is at its maximum value, then $\alpha^2 + \beta^2$ can be 1

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

The expression $E = \frac{(a-b)(2a-b)}{a(a-b+c)}$ can be rewritten in terms of the sum of roots $S = \alpha + \beta$ and the product of roots $P = \alpha \beta$ as $E = \frac{(1+S)(2+S)}{1+S+P}$. The condition that the roots $\alpha, \beta$ lie in the interval [0, 1] implies the following constraints on $S$ and $P$:

1. $0 \le S \le 2$
2. $0 \le P \le 1$
3. $P \le S^2/4$ (from the discriminant condition for real roots)
4. $P \ge S-1$ (from the condition that roots lie in [0,1])

Combining these, the feasible region for $(S, P)$ is $\max(0, S-1) \le P \le \min(1, S^2/4)$.

To find the maximum value of $E$, we need to minimize the denominator $1+S+P$. This occurs when $P$ is at its minimum, $P_{min} = \max(0, S-1)$.

• For $0 \le S < 1$, $P_{min} = 0$, so $E = \frac{(1+S)(2+S)}{1+S+0} = 2+S$. The maximum value in this range approaches $2+1=3$ as $S \to 1$.
• For $1 \le S \le 2$, $P_{min} = S-1$, so $E = \frac{(1+S)(2+S)}{1+S+(S-1)} = \frac{(1+S)(2+S)}{2S}$. At $S=1$, $E = \frac{(2)(3)}{2} = 3$. At $S=2$, $E = \frac{(3)(4)}{4} = 3$. The function $\frac{(1+S)(2+S)}{2S}$ has a minimum at $S=\sqrt{2}$ and maximums at $S=1$ and $S=2$.

The overall maximum value of $E$ is 3. This occurs when $(S, P) = (1, 0)$ (roots are 0 and 1) or $(S, P) = (2, 1)$ (roots are 1 and 1). So, option A is correct.

To find the minimum value of $E$, we need to maximize the denominator $1+S+P$. This occurs when $P$ is at its maximum, $P_{max} = S^2/4$ (since $S^2/4 \le 1$ for $S \in [0,2]$). $E = \frac{(1+S)(2+S)}{1+S+S^2/4} = \frac{(1+S)(2+S)}{(1+S/2)^2} = \frac{4(1+S)(2+S)}{(2+S)^2} = \frac{4(1+S)}{2+S}$. Let $g(S) = \frac{4(1+S)}{2+S}$. This function is increasing for $S \in [0, 2]$ ($g'(S) = \frac{4}{(2+S)^2} > 0$). The minimum value occurs at $S=0$, where $g(0) = \frac{4(1)}{2} = 2$. This occurs when $(S, P) = (0, 0)$ (roots are 0 and 0). The overall minimum value of $E$ is 2. So, option B is correct.

Now consider options C and D, which are conditional statements about $\alpha^2 + \beta^2$ when $E$ is at its maximum value of 3. The maximum value of $E=3$ occurs in two cases:

1. $S=1, P=0$: The roots are $\alpha=0, \beta=1$. In this case, $\alpha^2 + \beta^2 = 0^2 + 1^2 = 1$.
2. $S=2, P=1$: The roots are $\alpha=1, \beta=1$. In this case, $\alpha^2 + \beta^2 = 1^2 + 1^2 = 2$.

Therefore, if $E$ is at its maximum value, $\alpha^2 + \beta^2$ can be 1 (Option D is correct) and $\alpha^2 + \beta^2$ can be 2 (Option C is correct).

All four options are correct.