Let $\mathbb{R}^{2}$ denote $\mathbb{R} \times \mathbb{R}$. Let $S = \{(a, b, c) : a, b, c \in \mathbb{R}$ and $ax^{2} + 2bxy + cy^{2} > 0$ for all $(x, y) \in \mathbb{R}^{2} - \{(0, 0)\}\}$. Then which of the following statements is (are) TRUE?

Let R² denote R × R. Let S = {(a, b, c) : a, b, c ∈ R and ax² + 2bxy + cy² > 0 for all (x, y) ∈ R² - {(0, 0)}}. Then which of the following statements is (are) TRUE?

$\text{Given: } S = \{(a,b,c) : a,b,c \in \mathbb{R}$

ax² + 2bxy + cy² > 0 ∀ (x,y) ≠ (0,0)}

$\text{To find: which of } (1), (2), (3), (4) \text{ are TRUE}$

$Q(x,y) = ax^{2} + 2bxy + cy^{2}$ $Q = a\left(x + \dfrac{b}{a}y\right)^{2} + \left(c - \dfrac{b^{2}}{a}\right)y^{2}$ $\text{In } S \iff a > 0 \ \text{ and } \ ac - b^{2} > 0$ $(\Rightarrow c > 0)$ $1.\ \text{check } (2, 7/2, 6)$ $2.\ \text{check } (3, b, 1/12)$ $3, 4.\ \text{check the two linear systems}$ $ac - b^{2} = 2\cdot 6 - \left(\dfrac{7}{2}\right)^{2} = 12 - \dfrac{49}{4} = -\dfrac{1}{4} < 0$ $\therefore (1) \text{ is FALSE}$ $3\cdot\dfrac{1}{12} - b^{2} > 0 \ \Rightarrow \ b^{2} < \dfrac{1}{4}$ $\Rightarrow |b| < \dfrac{1}{2} \ \Rightarrow \ |2b| < 1 \quad \therefore (2) \text{ TRUE}$ $ax + by = 1, \qquad by + cy = -1 \ \Rightarrow \ (b+c)y = -1$ $\begin{vmatrix} a & b \\ 0 & b+c \end{vmatrix} = a(b+c)$ $(4,-1,1) \in S: \ ac - b^{2} = 4 - 1 = 3 > 0, \ \text{but } b + c = 0$ $0 = -1 \ \text{impossible} \quad \therefore (3) \text{ is FALSE}$ $\begin{vmatrix} a+1 & b \\ b & c+1 \end{vmatrix} = (a+1)(c+1) - b^{2}$ $= (ac - b^{2}) + a + c + 1 > 0$ $\therefore (4) \text{ is TRUE}$ $\text{True statements: } (2) \ \text{and} \ (4)$