If $f\left( x \right) = \frac{{{e^x}}}{{1\,\, + \,\,{e^x}}},\,\,{I_1} = \mathop \smallint \nolimits_{f\left( { - a} \right)}^{f\left( a \right)} \,\,x{\rm{g}}\,\,\left[ {x\,\,\left( {1 - x} \right)} \right]dx\,\,\;{\rm{and}}$ and ${I_2} = \mathop \smallint \nolimits_{f\left( { - a} \right)}^{f\left( a \right)} \,\,{\rm{g}}\,\,\left[ {x\,\left( {1 - x} \right)} \right]\,\,dx$, then the value of $\frac{{{I_2}}}{{{I_1}}}$ is

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Hard

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Analyzing +4 | -1

f\left( x \right) = \frac{{{e^x}}}{{1\,\, + \,\,{e^x}}},\,\,{I_1} = \int \nolimits_{f\left( { - a} \right)}^{f\left( a \right)} \,\,x\mathrm{g}\,\,\left[ {x\,\,\left( {1 - x} \right)} \right]dx\,\,\;\mathrm{and}

and

{I_2} = \int \nolimits_{f\left( { - a} \right)}^{f\left( a \right)} \,\,\mathrm{g}\,\,\left[ {x\,\left( {1 - x} \right)} \right]\,\,dx

, then the value of

\frac{{{I_2}}}{{{I_1}}}

View answer and explanation

Correct option: A

\mathrm{2}

Option A (2) is correct because by using the property of definite integrals and the given function, the ratio simplifies to 2.