question_answer

If \[{{I}_{1}}=\int_{0}^{1}{{{e}^{-x}}{{\cos }^{2}}xdx;}\,\,\,\,{{I}_{2}}=\int_{0}^{1}{{{e}^{-{{x}^{2}}}}{{\cos }^{2}}xdx}\,\,\,and\,\,\,{{I}_{3}}=\int_{0}^{1}{{{e}^{-{{x}^{2}}}}dx;}\]then                           [JEE Online 15-04-2018 (II)]

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

B) \[{{I}_{3}}>{{I}_{1}}>{{I}_{2}}\] 

C) \[{{I}_{2}}>{{I}_{1}}>{{I}_{3}}\] 

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

Correct Answer: D

Solution :

Given: \[{{I}_{1}}=\int_{0}^{1}{{{e}^{-x}}{{\cos }^{2}}xdx;}\] \[{{I}_{2}}=\int_{0}^{1}{{{e}^{-{{x}^{2}}}}{{\cos }^{2}}xdx}\] and \[{{I}_{3}}=\int_{0}^{1}{{{e}^{-{{x}^{3}}}}dx}\] For \[x\in (0,1),x>{{x}^{2}}\Rightarrow -x<-{{x}^{2}}\] \[\Rightarrow {{x}^{2}}>{{x}^{3}}\] \[\Rightarrow -{{x}^{2}}<-{{x}^{3}}\] \[\Rightarrow {{e}^{-{{x}^{2}}}}<{{e}^{-{{x}^{3}}}}\] And \[{{e}^{-x}}<{{e}^{-{{x}^{2}}}}\] \[\Rightarrow {{e}^{-x}}<{{e}^{-{{x}^{2}}}}<{{e}^{-{{x}^{3}}}}\] \[\Rightarrow {{e}^{-{{x}^{3}}}}<{{e}^{-{{x}^{2}}}}>{{e}^{-x}}\] \[\Rightarrow {{I}_{3}}>{{I}_{2}}>{{I}_{1}}\] Green line denotes \[\int_{{}}^{{}}{(x)={{e}^{-x}}{{\cos }^{2}}x}\] Blue line denotes \[g(x)={{e}^{-{{x}^{3}}}}{{\cos }^{2}}x\] Red line denotes \[h(x)={{e}^{-{{x}^{3}}}}\] Also, from the graph we get the same result. Hence, option D is correct.