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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 35

  • question_answer
    If \[{{b}^{2}}-4ac\le 0\](where \[a\ne 0\] and a, b, c, x, \[y\in R\]) satisfies the system \[a{{x}^{2}}+x(b-3)+c+y=0\] and \[a{{y}^{2}}+y(b-1)+c+3x=0,\] then value of  is

    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 $y-3x\le 0$ \[\therefore \,\,\,\,\,\,3x-y=0\] \[\therefore \,\,\,\,\,\frac{x}{y}=\frac{1}{3}\]           


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