Let $\mathbb{R}^2$ denote $\mathbb{R} \times \mathbb{R}$. Let $S=\left\{(a, b, c): a, b, c \in \mathbb{R}…

Let $\mathbb{R}^2$ denote $\mathbb{R} \times \mathbb{R}$. Let $S=\left\{(a, b, c): a, b, c \in \mathbb{R} \text { and } a x^2+2 b x y+c y^2>0 \text { for all }(x, y) \in \mathbb{R}^2-\{(0,0)\}\right\} .$ Then which of the following statements is (are) TRUE?
  1. $\left(2, \frac{7}{2}, 6\right) \in S$
  2. If $\left(3, b, \frac{1}{12}\right) \in S$, then $|2 b| < 1$.
  3. For any given $(a, b, c) \in S$, the system of linear equations $\begin{aligned}& a x+b y=1 \\& b x+c y=-1\end{aligned}$ has a unique solution.
  4. For any given $(a, b, c) \in S$, the system of linear equations $\begin{aligned}& (a+1) x+b y=0 \\& b x+(c+1) y=0\end{aligned}$ has a unique solution.

Solution

\(\begin{aligned} & a x^2+2 b x y+c y^2>0 \\ & y, x \in \mathbb{R}-\{(0,0)\} \\ & \Rightarrow c\left(\frac{y}{x}\right)^2+2 b\left(\frac{y}{x}\right)+a>0 \\ & 4 b^2-4 a c<0 \\ & \Rightarrow b^22 \times 6\) \(\therefore\) option A is incorrect \(\begin{aligned} & \text {(B) if }\left(3, b, \frac{1}{12}\right) \in S \\ & \Rightarrow b^2<3 \cdot \frac{1}{12} \\ & \Rightarrow b^2<\frac{1}{4} \\ & \Rightarrow 4 b^2<1 \\ & \Rightarrow|2 b|<1 \text { option } B \text { is correct } \end{aligned}\) \(\begin{aligned} & \text {(C) } a x+b y=1 \\ & b x+c y=-1 \\ & D=\left|\begin{array}{ll} a & b \\ b & c \end{array}\right|=a c-b^2 \neq 0 \end{aligned}\) \(\therefore\) unique solution option C is correct. \(\begin{aligned} & \text {(D) }(a+1) \mathrm{x}+\mathrm{by}=0 \\ & \mathrm{bx}+(\mathrm{c}+1) \mathrm{y}=0 \\ & D=\left|\begin{array}{cc} (a+1) & b \\ b & (c+1) \end{array}\right| \\ & =(a+1)(c+1)-b^2 \\ & \Rightarrow a c-b^2+a+c+1 \\ & b^20 \end{aligned}\) \(\therefore\) unique solution \(\therefore\) option D is correct.

Asked in: JEE Advanced 2024 (Paper 1)

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