JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    If \[\lambda \in R\] is such that the sum of the cubes of the roots of the equation, \[{{x}^{2}}+(2-\lambda )x+(10-\lambda )=0\] is minimum, then the magnitude of the difference of the roots of this equation is [JEE Online 15-04-2018]

    A) \[20\]                           

    B) \[2\sqrt{5}\]    

    C) \[2\sqrt{7}\]                

    D) \[4\sqrt{2}\]

    Correct Answer: B

    Solution :

    By quadratic formula, the roots of this equation are: \[\alpha ,\beta =\frac{\lambda -2\pm \sqrt{4-4\lambda +{{\lambda }^{2}}-40+4\lambda }}{2}=\frac{\lambda -2\pm \sqrt{{{\lambda }^{2}}-36}}{2}\]The magnitude of the difference of the roots is clearly\[|\sqrt{{{\lambda }^{2}}-36}|\] We have, \[{{\alpha }^{3}}+{{\beta }^{3}}=\frac{{{(\lambda -2)}^{3}}}{4}+\frac{3(\lambda -2)({{\lambda }^{2}}-36)}{4}=\frac{(\lambda -2)(4{{\lambda }^{2}}-4\lambda -104)}{4}=(\lambda -2)({{\lambda }^{2}}-\lambda -26)\] This function attains its minimum value at \[\lambda =4\]. Thus, the magnitude of the difference of the roots is clearly \[|i\sqrt{20}|=2\sqrt{5}\]. So the correct answer is option B.


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