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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Self Evaluation Test - Complex Numbers and Quadratic Equations

  • question_answer
    If x be real and \[b<c,\]then \[\frac{{{x}^{2}}-bc}{2x-b-c}\] lies in

    A) \[(b,c)\] 

    B) \[[b,c]\]

    C) \[(-\infty ,b]\cup [c,\infty )\]

    D) \[(-\infty ,b)\cup (c,\infty )\]

    Correct Answer: C

    Solution :

    Let \[y=\frac{{{x}^{2}}-bc}{2x-b-c}\] \[\Rightarrow \,\,\,{{x}^{2}}-2yx+(b+c)y-bc=0\] \[\because  \,\,x\,\in \,R,\,\,\,so \,4{{y}^{2}}-4\left( b+c \right)y+4bc\ge 0\] \[\Rightarrow \,\, x\le b\,\,or\,\,x\ge c~~~\,\,\,\,\left( \because \,\,b<c \right)\]


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