JEE Main & Advanced Mathematics Quadratic Equations Question Bank Critical Thinking

  • question_answer The values of a for which \[2{{x}^{2}}-2\,(2a+1)\,\,x+a(a+1)=0\] may have one root less than a and other root greater than a are given by [UPSEAT 2001]

    A) \[1>a>0\]

    B) \[-1<a<0\]

    C) \[a\ge 0\]

    D) \[a>0\,\,\text{or  }a<-1\]

    Correct Answer: D

    Solution :

    The given condition suggest that a lies between the roots. Let \[f(x)=2{{x}^{2}}-2(2a+1)x+a(a+1)\] For ?a? to lie between the roots we must have Discriminant \[\ge 0\] and\[f(a)<0\]. Now, Discriminant \[\ge 0\] \[\Rightarrow \,\,4{{(2a+1)}^{2}}-8a\,(a+1)\ge 0\] \[\Rightarrow \,\,8({{a}^{2}}+a+1/2)\ge 0\] which is always true. Also \[f(a)<0\Rightarrow 2{{a}^{2}}-2a(2a+1)+a(a+1)<0\] \[\Rightarrow -{{a}^{2}}-a<0\]\[\Rightarrow {{a}^{2}}+a>0\Rightarrow a\,(1+a)>0\] \[\Rightarrow \,\,a>0\] or \[a<-1\].

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