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 \( c \)
(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 \( c \)
- Both (I) and (II) are true
- Neither (I) nor (II) is true
- Only (II) is true
- Only (I) is true
The Correct Option is A
Approach Solution - 1
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.
- 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 \(c\).
- Geometric Mean Condition:
- For \(b\) to be the geometric mean of \(a\) and \(c\), we have: \(b = \sqrt{a \cdot c}\)
- 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 \(\alpha\).
- 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.
- 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.
Approach Solution -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:
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