JEE Advanced · Mathematics · 9. Straight Lines
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?
- A \(\left(2, \frac{7}{2}, 6\right) \in S\)
- B If \(\left(3, b, \frac{1}{12}\right) \in S\), then \(|2 b| < 1\).
- C 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. - D 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.
Answer & Solution
Correct Answer
(B) If \(\left(3, b, \frac{1}{12}\right) \in S\), then \(|2 b| < 1\).
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& a x^2+2 b x y+c y^2>0 \\
& y, x \in \mathbb{R}-\{(0,0)\} \\
& \Rightarrow c\left(\frac{y}{x}\right)^2+2 b\left(\frac{y}{x}\right)+a>0 \\
& 4 b^2-4 a c<0 \\
& \Rightarrow b^2\end{aligned}\)
(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.
\(\begin{aligned}
& \text {(D) }(a+1) \mathrm{x}+\mathrm{by}=0 \\
& \mathrm{bx}+(\mathrm{c}+1) \mathrm{y}=0 \\
& D=\left|\begin{array}{cc}
(a+1) & b \\
b & (c+1)
\end{array}\right| \\
& =(a+1)(c+1)-b^2 \\
& \Rightarrow a c-b^2+a+c+1 \\
& b^2& \Rightarrow a \text { and care positive then }\left(a c-b^2\right)+a+c+1>0
\end{aligned}\)
\(\therefore\) unique solution
\(\therefore\) option D is correct.
& 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\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.
\(\begin{aligned}
& \text {(D) }(a+1) \mathrm{x}+\mathrm{by}=0 \\
& \mathrm{bx}+(\mathrm{c}+1) \mathrm{y}=0 \\
& D=\left|\begin{array}{cc}
(a+1) & b \\
b & (c+1)
\end{array}\right| \\
& =(a+1)(c+1)-b^2 \\
& \Rightarrow a c-b^2+a+c+1 \\
& b^2& \Rightarrow a \text { and care positive then }\left(a c-b^2\right)+a+c+1>0
\end{aligned}\)
\(\therefore\) unique solution
\(\therefore\) option D is correct.
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