JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    The value of \[\alpha \] for which \[4\alpha \int\limits_{-1}^{2}{{{e}^{-\,\alpha \left| x \right|}}}\] dx=5, is [JEE MAIN Held on 07-01-2020 Evening]

    A) \[{{\log }_{e}}\left( \frac{4}{3} \right)\]       

    B) \[{{\log }_{e}}\left( \frac{3}{2} \right)\]

    C) \[{{\log }_{e}}2\]      

    D) \[{{\log }_{e}}\sqrt{2}\]

    Correct Answer: C

    Solution :

    [c] \[\because \,\,\,\,\,\,4\alpha \int_{-1}^{2}{{{e}^{-\alpha \left| x \right|}}\,dx=5}\] \[\Rightarrow \,\,\,4\alpha \left\{ \int_{-1}^{0}{{{e}^{\alpha x}}}dx+\int_{0}^{2}{{{e}^{-\alpha x}}dx} \right\}=5\] \[\,4\alpha \left\{ \left( \frac{{{e}^{\alpha x}}}{\alpha } \right)_{-1}^{0}+\left( \frac{{{e}^{-\alpha x}}}{-\alpha } \right)_{0}^{2} \right\}=5\] \[4\,(1{{e}^{\,\alpha }}{{e}^{2\alpha }}+1)=5\] \[4\,(2{{e}^{\,\alpha }}{{e}^{2\alpha }})=5\] \[4\,{{e}^{\,2\alpha }}+4{{e}^{\alpha }}-3=0\] \[(2{{e}^{\alpha }}+3)(2{{e}^{\alpha }}1)=0\] \[\therefore \,\,\,{{e}^{-\,\alpha }}=\frac{1}{2}\Rightarrow \,\alpha =\ln 2\]

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