2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-22

  • question_answer
    Let  \[{{I}_{1}}=\int\limits_{0}^{x}{{{e}^{tx}}.{{e}^{-{{t}^{2}}}}dt}\]and \[{{I}_{2}}=\int\limits_{0}^{x}{{{e}^{\frac{-{{t}^{2}}}{4}}}dt}\] where \[x\text{ }>\text{ }0\] then the value of \[\frac{{{I}_{1}}}{{{I}_{2}}}\] is

    A)  \[{{e}^{\frac{-{{x}^{2}}}{2}}}\]                              

    B)  \[{{e}^{\frac{{{x}^{2}}}{4}}}\]

    C)  \[{{e}^{\frac{-{{x}^{2}}}{4}}}\]                  

    D)  \[{{e}^{\frac{{{x}^{2}}}{4}}}\]

    Correct Answer: B

    Solution :

    \[{{I}_{1}}=\int\limits_{0}^{x}{{{e}^{t(x-t)}}dt}\] \[t(x-t)=-[{{t}^{2}}-tx]\,=-\,\left[ {{\left( t-\frac{x}{2} \right)}^{2}}-\frac{{{x}^{2}}}{4} \right]\] \[=\frac{{{x}^{2}}}{4}\,-{{\left( t-\frac{x}{2} \right)}^{2}}\] \[{{I}_{1}}=\int\limits_{0}^{x}{{{e}^{\frac{{{x}^{2}}}{4}{{\left( t-\frac{x}{2} \right)}^{2}}}}dt={{e}^{\frac{{{x}^{2}}}{4}}}\int\limits_{0}^{x}{{{e}^{\frac{-{{(2t-x)}^{2}}}{4}}}}}\] \[dt;\,2t-x=y\] \[\Rightarrow \,\,dt=\frac{dy}{2}\]\[=\frac{{{e}^{\frac{{{x}^{2}}}{4}}}}{2}\,\int\limits_{-x}^{x}{{{e}^{\frac{-{{y}^{2}}}{4}dy}}}\] \[{{I}_{1}}={{e}^{\frac{{{x}^{2}}}{4}}}\,\int\limits_{0}^{x}{\frac{-{{t}^{2}}}{4}dt\,={{e}^{\frac{{{x}^{2}}}{4}}}.\,{{I}_{2}}\,\Rightarrow \,\,\frac{{{I}_{1}}}{{{I}_{2}}}\,={{e}^{\frac{{{x}^{2}}}{4}}}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner