A) \[\pi /4\]
B) \[\pi /2\]
C) \[{{e}^{{{\pi }^{2}}/16}}\]
D) \[{{e}^{{{\pi }^{2}}/4}}\]
Correct Answer: A
Solution :
\[I=\int_{0}^{\pi /2}{\frac{{{e}^{{{x}^{2}}}}\,\,\,\,\,\,dx}{{{e}^{{{x}^{2}}}}+{{e}^{\left( \frac{\pi }{2}\,\,-x \right)}}^{2}}}\]and \[I=\int_{0}^{\pi /2}{\frac{{{e}^{{{\left( \frac{\pi }{2}-x \right)}^{2}}}}\,\,\,\,\,\,dx}{{{e}^{{{\left( \frac{\pi }{2}\,-x \right)}^{2}}}}+{{e}^{{{x}^{2}}}}}}\] \[\left[ \because \int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}} \right]\] \[\Rightarrow 2I=\int_{0}^{\pi /2}{1dx=(x)_{0}^{\pi /2}}\] \[\Rightarrow I=\frac{\pi }{4}\].
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