Let R 2 denote R × R . Let S= \ (a, b, c): a, b, c in R . and x 2 +2 b x y+y 2 0 for all .(x, y) in R 2 -\ (0,0)\ \ Then which of the following statements is (are) TRUE?

If (3, b, 1/12 ) in S , then 2 b <1

For any given (a, b, c) in S , then the system of linear equations begin aligned & a x+b y=1 \\& b x+c y=-1end aligned has a unique solution

For any given (a, b, c) in S , then the system of linear equations begin aligned & (a+1) x+b y=0 \\& b x+(c+1) y=0end aligned has a unique solution

Mathematics · Determinants

JEE Advanced 2024 — Paper 1 — Question 6

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

Option A:
$\left(2, \frac{7}{2}, 6\right) \in S$
Option B:
If $\left(3, b, \frac{1}{12}\right) \in S$ , then $|2 b|<1$
Correct
Option C:
For any given $(a, b, c) \in S$ , then the system of linear equations $\begin{aligned}& a x+b y=1 \\& b x+c y=-1\end{aligned}$ has a unique solution
Correct
Option D:
For any given $(a, b, c) \in S$ , then 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
Correct

Answer: B, C, D

Step-by-step solution

$a x^{2}+2 b x y+c y^{2}>0$ $a, c>0, b^{2}0$ hence unique solution.

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Determinants
Topic: System of Linear Equations using Determinants