For $0<c<b<a$, let $(a+b–2c)x^2+(b+c–2a)x+(c+a–2b)=0$ and $\alpha \ne 1$ be one of its root. Then, among the two statements

(I) If $\alpha \in \left(-1, 0\right)$, then $b$ cannot be the geometric mean of $a$ and $c$.

(II) If $\alpha \in \left(0, 1\right)$, then $b$ may be the geometric mean of $a$ and $c$.

Given: $f\left(x\right)=(a+b–2c) x^2+(b+c–2a) x+(c+a–2b)$

$\Rightarrow f\left(1\right)=a+b–2c+b+c–2a+c+a–2b=0$

$\Rightarrow f\left(1\right)=0$

Using product of roots,

$\Rightarrow \alpha ·1=\frac{c+a-2b}{a+b-2c}$

$\Rightarrow \alpha =\frac{c+a-2b}{a+b-2c}$

If, $-1<\alpha <0$

Statement-I:

$\Rightarrow -1<\frac{c+a-2b}{a+b-2c}<0$

$\Rightarrow b+c<2a$ and $b>\frac{a+c}{2}$

So, $b$ cannot be G.M. between $a$ and $c$. $\left\{\text{as} A.M\ge G.M\right\}$

So, statement I is correct.

Statement-II:

$\Rightarrow 0<\alpha <1$

$\Rightarrow 0<\frac{c+a-2b}{a+b-2c}<1$

$\Rightarrow \frac{c+a-2b}{a+b-2c}>0 \& \frac{c+a-2b}{a+b-2c}<1$

$\Rightarrow \frac{c+a-2b}{a+b-2c}>0 \& \frac{c+a-2b-a-b+2c}{a+b-2c}<0$

$\Rightarrow \frac{c+a-2b}{a+b-2c}>0 \& \frac{3c-3b}{a+b-2c}<0$

$\Rightarrow b<\frac{a+c}{2} \& b>c$  

Therefore, $b$ may be the G.M. between $a$ and $c$. $\left\{\text{as} A.M\ge G.M\right\}$

So, statement II is also correct.