For 0<c<b<a, let (a+b–2c)x^2+(b+c–2a)x+(c+a–2b)=0 and α ≠ 1 be one of its root. Then, among the two statements (I) If α ∈ (-1,0), then b cannot be…

Given: $f(x)=(a+b–2c){x}^{2}+(b+c–2a)x+(c+a–2b)$f(x)=(a+b–2c)x2+(b+c–2a)x+(c+a–2b)

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

$$\Rightarrow f(1)=0$$⇒f(1)=0

Using product of roots,

$$\Rightarrow \alpha \cdot 1=\frac{c+a-2b}{a+b-2c}$$⇒α⋅1=a+b−2cc+a−2b​

$$\Rightarrow \alpha =\frac{c+a-2b}{a+b-2c}$$⇒α=a+b−2cc+a−2b​

If, $-1<\alpha <0$−1<α<0

Statement-I:

$$\Rightarrow -1<\frac{c+a-2b}{a+b-2c}<0$$⇒−1<a+b−2cc+a−2b​<0

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

So, $b$b cannot be G.M. between $a$a and $c$c. ${\text{as}A.M\geq G.M}$asA.M≥G.M

So, statement I is correct.

Statement-II:

$$\Rightarrow 0<\alpha <1$$⇒0<α<1

$$\Rightarrow 0<\frac{c+a-2b}{a+b-2c}<1$$⇒0<a+b−2cc+a−2b​<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, $b$b may be the G.M. between $a$a and $c$c. ${\text{as}A.M\geq G.M}$asA.M≥G.M

So, statement II is also correct.