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

  • question_answer
    If the roots of the equation \[{{x}^{2}}-ax+b=0\] are real and differ by a quantity which is less than \[c(c>0),\] then b lies between

    A) \[\frac{{{a}^{2}}-{{c}^{2}}}{4}\] and \[\frac{{{a}^{2}}}{4}\]

    B) \[\frac{{{a}^{2}}+{{c}^{2}}}{4}\] and \[\frac{{{a}^{2}}}{4}\]

    C) \[\frac{{{a}^{2}}-{{c}^{2}}}{2}\] and \[\frac{{{a}^{2}}}{4}\]

    D) None of these

    Correct Answer: A

    Solution :

    Given roots are real and distinct, then \[{{\operatorname{a}}^{2}}-4b>0\] \[\Rightarrow \,\,\,\operatorname{b}<{{a}^{2}}/4\] Again \[\alpha \] and \[\beta \] differ by a quantity less than \[c\left( c >0 \right)\] \[\Rightarrow \,\,\,\left| \alpha -\beta  \right|<c\,\,or\,\,{{(\alpha -\beta )}^{2}}<{{c}^{2}}\] \[\Rightarrow \,\,\,{{\left( \alpha +\beta  \right)}^{2}}-4\alpha \beta <{{c}^{2}} or {{a}^{2}}-4b<{{c}^{2}}\] or \[\frac{{{a}^{2}}-{{c}^{2}}}{4}<b\] \[\Rightarrow \,\,\,\frac{{{a}^{2}}-{{c}^{2}}}{4}<b<\frac{{{a}^{2}}}{4}\] by (1) and (2)


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