JEE Main & Advanced Sample Paper JEE Main - Mock Test - 43

  • question_answer
    If the curves \[y=\frac{{{\log }_{e}}x}{x}\] and \[y=\lambda {{x}^{2}}\] (where \[\lambda \] is constant) touch each other, then the value of \[\lambda \]is

    A) \[\frac{1}{e}\]               

    B)        \[\frac{1}{2e}\]                    

    C) \[\frac{1}{3e}\]    

    D)        \[\frac{1}{4e}\]

    Correct Answer: C

    Solution :

    [c] Curves touch each other. Solving equations of curves, we get \[\frac{{{\log }_{e}}x}{x}=\lambda {{x}^{2}}\] \[\Rightarrow \,\,\,\,\,{{\log }_{e}}x=\lambda {{x}^{3}}\]                       ?..(1) Comparing their derivatives, we get \[\frac{1-{{\log }_{e}}x}{{{x}^{2}}}=2\lambda x\] \[\Rightarrow \,\,\,1-\lambda {{x}^{3}}=2\lambda {{x}^{3}}\] \[\Rightarrow \,\,\,3\lambda {{x}^{3}}=1\] \[\Rightarrow \,\,\,x={{\left( \frac{1}{3\lambda } \right)}^{\frac{1}{3}}}={{(3\lambda )}^{\frac{-1}{3}}}\] Putting the value of x into (1), we get             \[\frac{-1}{3}{{\log }_{e}}(3\lambda )=\lambda .\frac{1}{3\lambda }\]             \[\Rightarrow \,\,\,\,{{\log }_{e}}(3\lambda )=1\]             \[\Rightarrow \,\,\,\,\lambda =\frac{1}{3e}\]              


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