Let R^2 denote R x R. Let S = {(a,b,c) : a,b,c ∈ R} and ax^2 + 2bxy + cy^2 > 0

Question

Let \(\mathbb{R}^{2} \) denote \(\mathbb{R} \times \mathbb{R}\). Let 

\(S=\left\{(a, b, c): a, b, c \in \mathbb{R}\right.\) and \(a x^{2}+2 b x y+c y^{2}>0\) for all \(\left.(x, y) \in \mathbb{R}^{2}-\{(0,0)\}\right\}\)

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 1

Correct option is B,C,D

\(a x^{2}+2 b x y+c y^{2}>0\)

\(y, x \in \mathbb{R}-\{(0,0)\}\)

\(\Rightarrow c\left(\frac{y}{x}\right)^{2}+2 b\left(\frac{y}{x}\right)+a>0\)

\(4 b^{2}-4 a c<0\)

\(\Rightarrow b^{2}<a c\)

(A) \(\left(2, \frac{7}{2}, 6\right)\)

\(\left(\frac{7}{2}\right)^{2}>2 \times 6\)

\(\therefore\) option A is incorrect

(B) if \(\left(3, b, \frac{1}{12}\right) \in S\)

\(\Rightarrow b^{2}<3 \cdot \frac{1}{12}\)

\(\Rightarrow b^{2}<\frac{1}{4}\)

\(\Rightarrow 4 b^{2}<1\)

\(\Rightarrow|2 b|<1\) option B is correct

(C) ax + by = 1 

bx + cy = – 1

\(D=\left|\begin{array}{ll}a & b \\ b & c\end{array}\right|=a c-b^{2} \neq 0\)

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

(D) (a + 1)x + by = 0 

bx + (c + 1)y = 0

\(D=\left|\begin{array}{cc}(a+1) & b \\ b & (c+1)\end{array}\right|\)

\(=(a+1)(c+1)-b^{2}\)

\(\Rightarrow a c-b^{2}+a+c+1\)

\(b^{2}<a c \Rightarrow a c \) is +ve

\(\Rightarrow a\) and c are positive then \(\left(a c-b^{2}\right)+a+c+1>0\) 

\(\therefore\) unique solution

\(\therefore\) option D is correct.