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JEE Main & Advanced Sample Paper JEE Main Sample Paper-20

  • question_answer
    If\[\alpha \]and\[\beta \]are the real roots of the equation\[{{x}^{2}}-(k-2)x+({{k}^{2}}+3k+5)=0(k\in R)\]. Find the maximum and minimum values of\[({{\alpha }^{2}}+{{\beta }^{2}})\].

    A) \[18,\,\,50/9\]                  

    B) \[18,\,\,25/9\]

    C) \[27,\,\,50/9\]                  

    D)  None of these

    Correct Answer: A

    Solution :

     For real roots\[D\ge 0\]                 \[{{(k-2)}^{2}}-4({{k}^{2}}+3k+5)\ge 0\]                 \[({{k}^{2}}+4-4k)-4{{k}^{2}}-12k-20\ge 0\] \[-3{{k}^{2}}-16k-16\ge 0;\]  \[3{{k}^{2}}+16k+16\ge 0\]                 \[\left( k+\frac{4}{3} \right)(k+4)\le 0\] Now\[E={{\alpha }^{2}}+{{\beta }^{2}};\]    \[E={{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[E={{(k-2)}^{2}}-2({{k}^{2}}+3k+5)=-{{k}^{2}}-10k-6\] \[E=-({{k}^{2}}+10k+6)=-[{{(k+5)}^{2}}-19]=19-{{(k+5)}^{2}}\]\[\therefore {{E}_{\min }}\]occurs when\[k=-4/3\] \[\therefore \]\[{{E}_{\min }}=\frac{121}{9}=\frac{171-121}{9}=\frac{50}{9}\] \[{{E}_{\max }}\]occurs when\[k=-4\] \[{{E}_{\max }}=19-1=18\]


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