JEE Advance - Mathematics (2024 - Paper 1 Online - No. 6)
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?
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.
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.
Explanation
$$\begin{aligned} \text{(A)}\quad & \mathrm{ax}^2+2 \mathrm{bxy}+\mathrm{cy}^2>0 \forall(\mathrm{x}, \mathrm{y}) \in \mathbb{R}^2-\{(0,0)\} \\ & \Rightarrow \mathrm{ax}+2 \mathrm{bxy}+\mathrm{cy}^2 \text { must represent pair of imaginary lines and } \mathrm{a}, \mathrm{c}>0 . \\ & \Rightarrow \mathrm{b}^2<\mathrm{ac} \end{aligned}$$
$$\mathrm{(B)}\quad\mathrm{b}^2<3 \times \frac{1}{12} \Rightarrow|2 \mathrm{~b}|<1$$
$$\mathrm{(C)}\quad$$ since $$\mathrm{b}^2 \neq \mathrm{ac}$$
$$\Rightarrow \mathrm{ax}+\mathrm{by}=1 \text { and } \mathrm{bx}+\mathrm{cy}=-1$$
are not parallel lines.
$$\mathrm{(D)}\quad\mathrm{ac}+\mathrm{a}+\mathrm{c}>\mathrm{b}^2 \Rightarrow$$ lines are not parallel.
$$\Rightarrow$$ Options (B), (C), (D) are Correct.
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