Key Concepts and Formulas

• Roots of a Quadratic Equation: For a quadratic equation $Ax^2 + Bx + C = 0$, if $x=1$ is a root, then $A+B+C=0$. If $\alpha$ and $\beta$ are the roots, then $\alpha + \beta = -B/A$ and $\alpha \beta = C/A$.
• Arithmetic Mean (AM) and Geometric Mean (GM): For two positive numbers $a$ and $c$, $AM = \frac{a+c}{2}$ and $GM = \sqrt{ac}$.
• Inequality Properties: When multiplying or dividing inequalities, pay close attention to the sign of the multiplier/divisor. For $0 < c < b < a$, expressions like $a+b-2c$ are positive.
• AM-GM Inequality: For non-negative numbers, $AM \ge GM$. Equality holds if and only if the numbers are equal.

Step-by-Step Solution

Step 1: Analyze the Quadratic Equation and Identify a Root

Let the given quadratic equation be $f(x) = (a+b-2c)x^2 + (b+c-2a)x + (c+a-2b) = 0$. We check if $x=1$ is a root by substituting $x=1$ into the equation: $f(1) = (a+b-2c)(1)^2 + (b+c-2a)(1) + (c+a-2b)$ $f(1) = a+b-2c + b+c-2a + c+a-2b$ $f(1) = (a-2a+a) + (b+b-2b) + (-2c+c+c)$ $f(1) = 0 + 0 + 0 = 0$ Since $f(1)=0$, $x=1$ is a root of the quadratic equation.

Step 2: Find the Other Root ($\alpha$) in Terms of Coefficients

Let the roots of the quadratic equation be $\alpha$ and $1$. From Vieta's formulas, the product of the roots is given by the constant term divided by the leading coefficient: $\alpha \cdot 1 = \frac{c+a-2b}{a+b-2c}$ $\alpha = \frac{c+a-2b}{a+b-2c}$ We are given that $\alpha \neq 1$. This is consistent because if $\alpha = 1$, then $\frac{c+a-2b}{a+b-2c} = 1$, which implies $c+a-2b = a+b-2c$, leading to $3c = 3b$, or $c=b$. However, we are given $c < b$, so $\alpha \neq 1$.

Step 3: Analyze Statement (I): If $\alpha \in (-1, 0)$, then $b$ cannot be the geometric mean of $a$ and $c$.

We are given $-1 < \alpha < 0$. Substituting the expression for $\alpha$: $-1 < \frac{c+a-2b}{a+b-2c} < 0$ We know that $0 < c < b < a$. This implies $a > c$. Consider the denominator: $a+b-2c$. Since $a>c$ and $b>c$, $a+b > 2c$, so $a+b-2c > 0$. We can split the inequality into two parts:

Part 1: $\frac{c+a-2b}{a+b-2c} < 0$ Since the denominator is positive, the numerator must be negative: $c+a-2b < 0$ $a+c < 2b$ $b > \frac{a+c}{2}$ This implies that $b$ is greater than the arithmetic mean of $a$ and $c$.

Part 2: $\frac{c+a-2b}{a+b-2c} > -1$ Multiply both sides by the positive denominator $(a+b-2c)$: $c+a-2b > -(a+b-2c)$ $c+a-2b > -a-b+2c$ Rearrange the terms to isolate $b$: $a+a+b-b > 2c-c+2b$ $2a > c+2b$ $2a-c > 2b$ $b < \frac{2a-c}{2}$

So, if $\alpha \in (-1, 0)$, we have $\frac{a+c}{2} < b < \frac{2a-c}{2}$.

Now, let's consider the condition that $b$ is the geometric mean of $a$ and $c$, i.e., $b = \sqrt{ac}$. If $b = \sqrt{ac}$, then from Part 1, we must have $\sqrt{ac} > \frac{a+c}{2}$. For positive numbers $a$ and $c$, the AM-GM inequality states that $\frac{a+c}{2} \ge \sqrt{ac}$, with equality if and only if $a=c$. Since we are given $c < a$, it must be that $\frac{a+c}{2} > \sqrt{ac}$. Therefore, the condition $b > \frac{a+c}{2}$ derived from $\alpha < 0$ implies that $b$ cannot be equal to $\sqrt{ac}$, because $\sqrt{ac}$ is always less than $\frac{a+c}{2}$ when $a \neq c$. Thus, if $\alpha \in (-1, 0)$, then $b$ cannot be the geometric mean of $a$ and $c$. Statement (I) is true.

Step 4: Analyze Statement (II): If $\alpha \in (0, 1)$, then $b$ may be the geometric mean of $a$ and $c$.

We are given $0 < \alpha < 1$. Substituting the expression for $\alpha$: $0 < \frac{c+a-2b}{a+b-2c} < 1$ Again, we know $a+b-2c > 0$. We split the inequality into two parts:

Part 1: $\frac{c+a-2b}{a+b-2c} > 0$ Since the denominator is positive, the numerator must be positive: $c+a-2b > 0$ $a+c > 2b$ $b < \frac{a+c}{2}$ This implies that $b$ is less than the arithmetic mean of $a$ and $c$.

Part 2: $\frac{c+a-2b}{a+b-2c} < 1$ Multiply both sides by the positive denominator $(a+b-2c)$: $c+a-2b < a+b-2c$ Rearrange the terms: $c+2c < b+2b$ $3c < 3b$ $c < b$ This inequality $c < b$ is already given in the problem statement ($0 < c < b < a$), so it does not impose any new constraints.

So, if $\alpha \in (0, 1)$, we have $c < b < \frac{a+c}{2}$.

Now, we need to determine if $b$ may be the geometric mean of $a$ and $c$, i.e., if $b = \sqrt{ac}$ is possible under the condition $c < b < \frac{a+c}{2}$. We are looking for a scenario where $c < \sqrt{ac} < \frac{a+c}{2}$. From the AM-GM inequality, for $a \neq c$, we know that $\sqrt{ac} < \frac{a+c}{2}$. So, the condition we need to satisfy is $c < \sqrt{ac}$. This is true if $c^2 < ac$, which simplifies to $c < a$ (since $c>0$). This is already given in the problem statement ($c < a$).

Therefore, it is possible for $b$ to be the geometric mean of $a$ and $c$ when $\alpha \in (0, 1)$, provided that $b = \sqrt{ac}$ satisfies $c < \sqrt{ac} < \frac{a+c}{2}$. This is true for any pair of distinct positive numbers $a$ and $c$. For example, let $a=4$ and $c=1$. Then $GM = \sqrt{4 \cdot 1} = 2$ and $AM = \frac{4+1}{2} = 2.5$. We have $c < GM < AM$, i.e., $1 < 2 < 2.5$. If we choose $b=2$, then $c < b < \frac{a+c}{2}$ is satisfied. This choice of $a, b, c$ leads to $\alpha \in (0,1)$. Thus, statement (II) is true.

Step 5: Conclude Based on the Truthfulness of the Statements

Statement (I) is true. Statement (II) is true.

Since both statements are true, the correct option is (B).

Common Mistakes & Tips

• Sign Errors in Inequalities: Always be cautious when manipulating inequalities. Use the given condition $0 < c < b < a$ to determine the signs of expressions like $a+b-2c$.
• Misinterpreting "May be": For statement (II), showing that there exists at least one scenario where $b$ can be the geometric mean is sufficient to prove it "may be" true.
• Forgetting AM-GM: The AM-GM inequality is fundamental for comparing $\sqrt{ac}$ and $\frac{a+c}{2}$.

Summary

We analyzed the given quadratic equation and found that $x=1$ is a root, allowing us to express the other root $\alpha$ in terms of the coefficients. By carefully analyzing the inequalities derived from the given ranges of $\alpha$ and the condition $0 < c < b < a$, we determined the validity of statements (I) and (II). Statement (I) was proven to be true by showing that the condition for $\alpha \in (-1,0)$ implies $b > \frac{a+c}{2}$, which contradicts $b=\sqrt{ac}$ due to the AM-GM inequality. Statement (II) was shown to be true by demonstrating that the condition for $\alpha \in (0,1)$ leads to $c < b < \frac{a+c}{2}$, a range where $b$ can indeed be the geometric mean of $a$ and $c$. Both statements are true.

The final answer is $\boxed{A}$