KVPY Sample Paper KVPY Stream-SX Model Paper-5

If \[{{I}_{1}}=\int_{0}^{1}{\frac{{{e}^{x}}}{x+3}dx}\] and \[{{I}_{2}}=\int_{0}^{1}{\frac{{{x}^{3}}}{{{e}^{{{x}^{4}}}}(4-{{x}^{4}})}dx,}\] then \[\frac{{{I}_{1}}}{{{I}_{2}}}\] is

A) \[4\]

B) \[4e\]

C) \[3e\]

D) \[2e\]

Correct Answer: B

Solution :

[b]

We have, \[{{I}_{1}}=\int_{0}^{1}{\frac{{{e}^{x}}}{x+3}dx}\]

\[{{I}_{2}}=\int_{0}^{1}{\frac{{{x}^{3}}}{{{e}^{{{x}^{4}}}}(4-{{x}^{4}})}dx}\]

In \[{{I}_{2}}\]put \[{{x}^{4}}=t\] \[\Rightarrow \] \[4{{x}^{3}}\,\,\,dx=dt\]

\[\therefore \] \[{{I}_{2}}=\frac{1}{4}\int_{0}^{1}{\frac{dt}{{{e}^{t}}(4-t)}}\]

\[{{I}_{2}}=\frac{1}{4}\int_{0}^{1}{\frac{dt}{{{e}^{1-t}}(4-(1-t)}}\] \[\left[ \because \int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}} \right]\]

\[{{I}_{2}}=\frac{1}{4}\int_{0}^{1}{\frac{{{e}^{t}}}{e\,(3+t)}dt}\]

\[{{I}_{2}}=\frac{1}{4e}{{I}_{1}}\]\[\Rightarrow \]\[\frac{{{I}_{1}}}{{{I}_{2}}}=4e\]