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

Answer: (A)

Both (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: