JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Analyzing +4 | -1

f(x)=\frac{e^x}{1+e^x}, I_1=\int\limits_{{f(-a)}}^{f(a)} x g\{x(1-x)\} d x

and

I_2=\int\limits_{{f(-a)}}^{f(a)} g\{x(1-x)\} d x

then the value of

\frac{I_2}{I_1}

View answer and explanation

Correct option: D

Option D (2) is correct. By using the property

\int_a^b f(x) dx = \int_a^b f(a+b-x) dx

and the given function's property

f(x) + f(-x) = 1

, we find that

2I_1 = I_2

, hence

I_2/I_1 = 2

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