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

Correct answer: B. If (3, b, (1)/(12)) ∈ S , then |2 b| 0 & y, x ∈ ℝ-\ (0,0)\ & c((y)/(x))^2+2 b((y)/(x))+a>0 & 4 b^2-4 a c 2 × 6) ( ) option A is incorrect ( aligned & (B) if (3, b, (1)/(12)) ∈ S & b^2 0 aligned ) ( ) unique solution ( ) option D is correct.

JEE Advanced 2024Paper 1

Exam: JEE Advanced
Year: 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Straight Lines

JEE Advanced 2024 · Paper 1 · Mathematics · Straight Lines

Let ℝ^2 denote ℝ × ℝ. Let S=\ (a, b, c): a, b, c ∈ ℝ and a x^2+2 b...

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 \text{ all } }

(x, y) \in

\mathbb{R}^2-

\{(0,0)\}\right\} .

Then which of the following statements is (are) TRUE?

Options

A $\left(2, \frac{7}{2}, 6\right) \in S$
BIf $\left(3, b, \frac{1}{12}\right) \in S$ , then $|2 b| < 1$ .
CFor 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.
DFor 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.

Correct answer

B. If $\left(3, b, \frac{1}{12}\right) \in S$ , then $|2 b| < 1$ .

Step-by-step 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^2<a c \end{aligned}

) (A) (\left(2, \frac{7}{2}, 6\right)) (\left(\frac{7}{2}\right)^2>2 \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^2<a c \Rightarrow a c \text{ is +ve } \\ & \Rightarrow a \text{ and care positive then }\left(a c-b^2\right)+a+c+1>0 \end{aligned}

) (\therefore) unique solution (\therefore) option D is correct.

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