A) \[3\]
B) \[\frac{1}{3}\]
C) \[1\]
D) \[\frac{4}{3}\]
Correct Answer: B
Solution :
[b] We have $a{{x}^{2}}+bx+c=3x-y$ ? (1) $a{{y}^{2}}+by+c=y-3x$ ? (2) Given ${{b}^{2}}-4ac\le 0$ Case I: $a{{x}^{2}}+bx+c\ge 0\forall x\in R$ and $a{{y}^{2}}+by+c\ge 0\forall y\in R$ $\therefore \,\,\,\,3x-y\ge 0$ and $y-3x\ge 0$ $\therefore \,\,\,3x-y=0$ $\therefore \,\,\,\,\frac{x}{y}=\frac{1}{3}$ Case II: $a{{x}^{2}}+bx+c\le 0,\,\forall x\in R$ and $a{{y}^{2}}+by+c\le 0\,\forall y\in R$ $\therefore \,\,\,3x-y\le 0$ and $y-3x\le 0$ \[\therefore \,\,\,\,\,\,3x-y=0\] \[\therefore \,\,\,\,\,\frac{x}{y}=\frac{1}{3}\]You need to login to perform this action.
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