A) \[{{l}_{3}}>{{l}_{4}}\]
B) \[{{l}_{3}}={{l}_{4}}\]
C) \[{{l}_{1}}>{{l}_{2}}\]
D) \[{{l}_{2}}>{{l}_{1}}\]
Correct Answer: C
Solution :
\[{{l}_{1}}=\int_{0}^{1}{{{2}^{{{x}^{2}}}}}dx,{{l}_{2}}=\int_{0}^{1}{{{2}^{{{x}^{3}}}}}dx,\] \[{{l}_{3}}=\int_{1}^{2}{{{2}^{{{x}^{2}}}}}dx\]and\[{{l}_{4}}=\int_{1}^{2}{{{2}^{{{x}^{3}}}}}dx\] \[\because \]\[{{2}^{{{x}^{3}}}}<{{2}^{{{x}^{2}}}}\] \[\therefore \]\[0<x<1\] and \[{{2}^{{{x}^{3}}}}>{{2}^{{{x}^{2}}}}\] \[\therefore \] \[x>1\] \[\therefore \] \[\int_{0}^{1}{{{2}^{{{x}^{3}}}}}dx<\int_{0}^{1}{{{2}^{{{x}^{2}}}}}dx\] and \[\int_{1}^{2}{{{2}^{{{x}^{3}}}}}dx>\int_{1}^{2}{{{2}^{{{x}^{2}}}}}dx\] \[\therefore \,\,\,{{l}_{2}}<{{l}_{1}}\] and \[{{l}_{4}}>{{l}_{3}}\]
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