For 0 < c < b < a$0 < c < b < a$, let (a + b - 2c)x² + (b + c - 2a)x + (c + a - 2b) = 0$(a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0$ and α ≠ 1$\alpha \neq 1$ be one of its root. Then, among the two statements (I) If α in (-1,0)$\alpha \in (-1,0)$, then b$b$ cannot be the geometric mean of a$a$ and c$c$ (II) If α 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 x² - 8x + 32$x^2 - 8x + 32$, D = 64 - 128 < 0$D = 64 - 128 < 0$ and a = 1 > 0$a = 1 > 0$. Thus, x² - 8x + 32 > 0 x in R$x^2 - 8x + 32 > 0 \forall x \in \mathbb{R}$.

Step 1: Constraint on Numerator

Since the denominator is always positive, the numerator must be strictly negative for all x in R$x \in \mathbb{R}$.

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.

Chapter Mix

Class 11 Maths: Quadratic Equations