For 0 < c < b < a$0 < c < b < a$, let (a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0$(a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0$ and alpha neq 1$\alpha \neq 1$ be one of its root. Then, among the two statements (I) If alpha in (-1,0)$\alpha \in (-1,0)$, then b$b$ cannot be the geometric mean of a$a$ and c$c$ (II) If alpha in (0,1)$\alpha \in (0,1)$, then b$b$ may be the geometric mean of a$a$ and c$c$

### Core Logic For the denominator x^2 - 8x + 32$x^2 - 8x + 32$, D = 64 - 128 < 0$D = 64 - 128 < 0$ and a = 1 > 0$a = 1 > 0$. Thus, x^2 - 8x + 32 > 0 forall x in mathbbR$x^2 - 8x + 32 > 0 \forall x \in \mathbb{R}$. ### Step 1: Constraint on Numerator Since the denominator is always positive, the numerator must be strictly negative for all x in mathbbR$x \in \mathbb{R}$. ax^2 + 2(a + 1)x + 9a + 4 < 0 quad forall x in mathbbR$$ax^2 + 2(a + 1)x + 9a + 4 < 0 \quad \forall x \in \mathbb{R}$$ This requires a < 0$a < 0$ and D < 0$D < 0$. ### Step 2: Conclusion Since a$a$ must be strictly less than 0$0$, there are no *positive* integral values of a$a$ that satisfy the condition. Hence, S$S$ is an empty set. Number of elements is 0. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations