Question

For \( 0<c<b<a \), let \( (a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0 \) and \( \alpha \neq 1 \) be one of its roots. Then, among the two statements
(I) If \( \alpha \in (-1, 0) \), then \( b \) cannot be the geometric mean of \( a \) and \( c \)
(II) If \( \alpha \in (0, 1) \), then \( b \) may be the geometric mean of \( a \) and \( c \)

• A. Both (I) and (II) are true
• B. Neither (I) nor (II) is true
• C. Only (II) is true
• D. Only (I) is true

Answer 1

The Correct Option is A

To solve the given quadratic expression \((a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0\) with a root \(\alpha \neq 1\), where the parameters satisfy \(0 < c < b < a\), we need to analyze the statements given in the question. 

1. Statement Analysis:
   • Statement I: If \(\alpha \in (-1, 0)\), then \(b\) cannot be the geometric mean of \(a\) and \(c\).
   • Statement II: If \(\alpha \in (0, 1)\), then \(b\) may be the geometric mean of \(a\) and \(c\).
2. Geometric Mean Condition:
   • For \(b\) to be the geometric mean of \(a\) and \(c\), we have: \(b = \sqrt{a \cdot c}\)
3. Quadratic Roots:
   • The quadratic equation is given as \((a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0\).
   • If the quadratic is to have \(\alpha\) as a root, we look at the signs dependent on parameter values and conditions.
   • If \(b = \sqrt{a \cdot c}\) holds true, verify how it affects the root \(\alpha\).
4. Checking Statement I:
   • For \(\alpha \in (-1, 0)\), assume \(b = \sqrt{a \cdot c}\).
   • If \(\alpha\) is negative, the polarity of combinations in quadratic terms changes which does not satisfy the geometric mean condition across \(0 < c < b < a\).
   • Thus, Statement I is true.
5. Checking Statement II:
   • For \(\alpha \in (0, 1)\), again check the root values under \(b = \sqrt{a \cdot c}\).
   • If \(\alpha\) is positive and not equal to 1, statement for possibility under the geometric mean holds under subsets of parameter configurations.
   • Thus, Statement II is true.

Therefore, the conclusion of this analysis is that both Statement I and II are true.

Answer 2

Let:

\[ f(x) = (a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) \]

Given that \( \alpha = -1 \) is a root of \( f(x) \), we substitute \( \alpha \) into the equation:

\[ f(\alpha) = (a + b - 2c)(-1)^2 + (b + c - 2a)(-1) + (c + a - 2b) = 0 \]

This simplifies to:

\[ a + b - 2c - b - c + 2a + c + a - 2b = 0 \]

Rearranging terms:

\[ 0 = a + b - 2c \]

Now, consider the conditions:

• If \( \alpha < (-1, 0) \), then \( b < a < c \). In this case, we deduce that:
• If \( \alpha \in (0, 1) \), then conditions allow for: