JEE MAIN - Mathematics (2004 - No. 21)
If $$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}},{I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}dx} $$
and $${I_2} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {g\left\{ {x\left( {1 - x} \right)} \right\}dx} ,$$ then the value of $${{{I_2}} \over {{I_1}}}$$ is
and $${I_2} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {g\left\{ {x\left( {1 - x} \right)} \right\}dx} ,$$ then the value of $${{{I_2}} \over {{I_1}}}$$ is
$$1$$
$$-3$$
$$-1$$
$$2$$
Explanation
$$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}}$$
$$ \Rightarrow f\left( { - x} \right) = {{{e^{ - x}}} \over {1 + {e^{ - x}}}} = {1 \over {{e^x} + 1}}$$
$$\therefore$$ $$f\left( x \right) + f\left( { - x} \right) = 1\forall x$$
Now $${I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}} dx$$
$$ = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)} g\left\{ {x\left( {1 - x} \right)} \right\}dx$$
$$\left[ {} \right.$$ using $$\int\limits_a^b {f\left( x \right)} dx\,a$$
$$ = \int\limits_a^b {f\left( {a + b - x} \right)dx} $$ $$\left. \, \right]$$
$$ = {I_2} - {I_1} \Rightarrow 2{I_1} = {I_2}$$
$$\therefore$$ $${{{I_2}} \over {{I_1}}} = 2$$
$$ \Rightarrow f\left( { - x} \right) = {{{e^{ - x}}} \over {1 + {e^{ - x}}}} = {1 \over {{e^x} + 1}}$$
$$\therefore$$ $$f\left( x \right) + f\left( { - x} \right) = 1\forall x$$
Now $${I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}} dx$$
$$ = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)} g\left\{ {x\left( {1 - x} \right)} \right\}dx$$
$$\left[ {} \right.$$ using $$\int\limits_a^b {f\left( x \right)} dx\,a$$
$$ = \int\limits_a^b {f\left( {a + b - x} \right)dx} $$ $$\left. \, \right]$$
$$ = {I_2} - {I_1} \Rightarrow 2{I_1} = {I_2}$$
$$\therefore$$ $${{{I_2}} \over {{I_1}}} = 2$$
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