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JCECE Engineering JCECE Engineering Solved Paper-2013

  • question_answer
    Total number of values of\['a'\], so that \[{{x}^{2}}-x-a=0\] has integral roots, where \[a\in N\] and\[6\le a\le 100\], is equal to

    A) \[2\]                                     

    B) \[4\]

    C) \[8\]                                     

    D) \[6\]

    Correct Answer: B

    Solution :

    \[{{x}^{2}}-x-a=0,\,\,D=1+4a=odd\] \[D\]must be perfect square of some odd integer. Let\[D={{(2\lambda +1)}^{2}}\Rightarrow 1+4a=1+4{{\lambda }^{2}}+4\lambda \] \[\Rightarrow \]               \[a=\lambda (\lambda +1),\]as\[a\in [6,\,\,100]\] \[\Rightarrow \]               \[a=6,\,\,12,\,\,20,\,\,30,\,\,42,\,\,56,\,\,72,\,\,90\] Thus, \['a'\] can attain 8 different values.


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