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JEE Main & Advanced Sample Paper JEE Main Sample Paper-16

  • question_answer
    Consider \[{{I}_{1}}=\int\limits_{0}^{1}{{{e}^{-x}}{{\cos }^{2}}x\,dx}\] \[{{I}_{2}}=\int\limits_{0}^{1}{{{e}^{-{{x}^{2}}}}{{\cos }^{2}}x\,dx},\int\limits_{0}^{1}{{{e}^{\frac{-{{x}^{2}}}{2}}}}{{\cos }^{2}}x\,dx\] \[{{I}_{4}}\int\limits_{0}^{1}{{{e}^{\frac{-{{x}^{2}}}{2}}}}\,dx\]then correct sequence is

    A) \[{{I}_{2}}>{{I}_{4}}>{{I}_{1}}>{{I}_{3}}\]              

    B) \[{{I}_{2}}<{{I}_{4}}<{{I}_{1}}<{{I}_{3}}\]

    C) \[{{I}_{1}}<{{I}_{2}}<{{I}_{3}}<{{I}_{4}}\]              

    D) \[{{I}_{1}}>{{I}_{2}}>{{I}_{3}}>{{I}_{4}}\]

    Correct Answer: C

    Solution :

    For\[0<x<1\Rightarrow x>{{x}^{2}}>\frac{{{x}^{2}}}{2}\] \[\Rightarrow \]\[{{e}^{-x}}<{{e}^{-{{x}^{2}}}}<{{e}^{-\frac{{{x}^{2}}}{2}}}\] \[\therefore \]\[{{e}^{-x}}{{\cos }^{2}}x<{{e}^{-{{x}^{2}}}}{{\cos }^{2}}x<{{e}^{\frac{-{{x}^{2}}}{2}}}{{\cos }^{2}}x<{{e}^{\frac{-{{x}^{2}}}{2}}}\] Hence,\[\int\limits_{0}^{1}{{{e}^{-x}}}{{\cos }^{2}}xdx<\int\limits_{0}^{1}{{{e}^{-{{x}^{2}}}}}{{\cos }^{2}}dx\] \[<\int\limits_{0}^{1}{{{e}^{\frac{-{{x}^{2}}}{2}}}}{{\cos }^{2}}x\,dx<\int\limits_{0}^{1}{{{e}^{\frac{-{{x}^{2}}}{2}}}}dx\]


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