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

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: