If $f(x)=\frac{e^{x}}{1+e^{x}}, \mathrm{I}_{1}=\int_{\mathrm{f}(-\mathrm{a})}^{\mathrm{f}(\mathrm{a})} \mathrm{xg}\{\mathrm{x}(1-\mathrm{x})\} \mathrm{dx}$ and $\mathrm{I}_{2}=\int_{\mathrm{f}(-\mathrm{a})}^{\mathrm{f}(\mathrm{a})} \mathrm{g}\{\mathrm{x}(1-\mathrm{x})\}\mathrm{dx}$ then the value of $\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$ is

JEE Main · 2004 · 24 Apr 2004 · Offline · Definite Integrals

f (x) = \frac{e ^{x}}{1 + e ^{x}}, I_{1} = \int_{f (- a)}^{f (a)} xg {x (1 - x)} dx

and

I_{2} = \int_{f (- a)}^{f (a)} g {x (1 - x)} dx

then the value of

\frac{I _{2}}{I _{1}}

Answer: (A)

First note that

f (x) = \frac{e ^{x}}{1 + e ^{x}}

satisfies

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

, so the limits

f (- a)

and

f (a)

add to

1

and are symmetric about

\frac{1}{2}

. Applying the substitution

x \mapsto f (- a) + f (a) - x = 1 - x

I_{1}

replaces

x

1 - x

while leaving

x (1 - x)

unchanged, giving

I_{1} = \int_{f (- a)}^{f (a)} (1 - x) g {x (1 - x)} d x

. Adding this to the original

I_{1}

gives

2 I_{1} = \int_{f (- a)}^{f (a)} g {x (1 - x)} d x = I_{2}

, so

\frac{I _{2}}{I _{1}} = 2

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option