Mathematics - Straight Lines Question with Solution | TestHub

\(\begin{aligned} & a x^2+2 b x y+c y^2>0 \\ & y, x \in \mathbb{R}-\{(0,0)\} \\ & \Rightarrow c\left(\frac{y}{x}\right)^2+2 b\left(\frac{y}{x}\right)+a>0 \\ & 4 b^2-4 a c<0 \\ & \Rightarrow b^22 \times 6\) $\therefore$ option A is incorrect $\begin{array}{rl}& \text{(B) if }\left(3,b,\frac{1}{12}\right)\in S\\ & \Rightarrow b^2<3\cdot \frac{1}{12}\\ & \Rightarrow b^2<\frac{1}{4}\\ & \Rightarrow 4b^2<1\\ & \Rightarrow |2b|<1\text{ option }B\text{ is correct }\end{array}$ $\begin{array}{rl}& \text{(C) }ax+by=1\\ & bx+cy=-1\\ & D=\left|\begin{array}{ll}a& b\\ b& c\end{array}\right|=ac-b^2\ne 0\end{array}$ $\therefore$ unique solution option C is correct. \(\begin{aligned} & \text {(D) }(a+1) \mathrm{x}+\mathrm{by}=0 \\ & \mathrm{bx}+(\mathrm{c}+1) \mathrm{y}=0 \\ & D=\left|\begin{array}{cc} (a+1) & b \\ b & (c+1) \end{array}\right| \\ & =(a+1)(c+1)-b^2 \\ & \Rightarrow a c-b^2+a+c+1 \\ & b^20 \end{aligned}\) $\therefore$ unique solution $\therefore$ option D is correct.