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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 10

  • question_answer
    If the product of the roots of the equation \[{{x}^{2}}-2\sqrt{2}kx+2{{e}^{2\log k}}-1=0\] is 31, then the roots of the equation are real for k =

    A) -4                    

    B) 1    

    C) 4                     

    D) 0

    Correct Answer: C

    Solution :

    [c] : Let \[\alpha ,\beta \] be the roots of \[{{x}^{2}}-2\sqrt{2}kx+2{{e}^{2\log k}}-1=0\]. \[\therefore \]\[\alpha \beta =2{{e}^{2\log k}}-1\]                         ...(i) Product of roots \[(\alpha \beta )\] is given 31 Hence from (i), \[2{{e}^{2\log k}}-1=31\] or \[2{{e}^{\log {{k}^{2}}}}-1=31\] or \[2{{k}^{2}}=32\]or\[{{k}^{2}}=16\] or \[k=\pm 4\]. As log(-4) is not defined so k = 4.


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