For 0 (I) If α ∈ (- 1, 0) , then b cannot be the geometric mean of...

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

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

$⇒ f ( 1 ) = 0$

Using product of roots,

$⇒ \alpha \cdot 1 = \dfrac{c + a - 2b}{a + b - 2c}$

$⇒ \alpha = \dfrac{c + a - 2b}{a + b - 2c}$

If, $- 1 < \alpha < 0$

Statement-I:

$⇒ - 1 < \dfrac{c + a - 2b}{a + b - 2c} < 0$

⇒ b + c < 2a and $b > \dfrac{a + c}{2}$

So, $b$ cannot be G.M. between $a and c$. \left\{\text{as}A . M \ge G . M\right\}$</p><p>So, statement I is correct.</p><p>Statement-II:</p><p>$⇒ 0 < \alpha < 1$</p><p>$⇒ 0 < \dfrac{c + a - 2b}{a + b - 2c} < 1$</p><p>$⇒ \dfrac{c + a - 2b}{a + b - 2c} > 0 \text{$and$} \dfrac{c + a - 2b}{a + b - 2c} < 1$</p><p>$⇒ \dfrac{c + a - 2b}{a + b - 2c} > 0 \text{$and$} \dfrac{c + a - 2b - a - b + 2c}{a + b - 2c} < 0$</p><p>$⇒ \dfrac{c + a - 2b}{a + b - 2c} > 0 \text{$and$} \dfrac{3c - 3b}{a + b - 2c} < 0$</p><p>$⇒ b < \dfrac{a + c}{2} \text{and } b > c </p><p>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.