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JEE Main & Advanced Sample Paper JEE Main Sample Paper-8

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


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