JEE MAIN - Mathematics (2007 - No. 11)
Let $$F\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$$ where $$f\left( x \right) = \int\limits_l^x {{{\log t} \over {1 + t}}dt,} $$ Then $$F(e)$$ equals
$$1$$
$$2$$
$$1/2$$
$$0$$
Explanation
Given $$f\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$$
where $$f\left( x \right) = \int_1^x {{{\log \,t} \over {1 + t}}} \,dt$$
$$\therefore$$ $$F\left( e \right) = f\left( e \right) + f\left( {{1 \over e}} \right)$$
$$ \Rightarrow F\left( e \right)$$
$$ = \int_1^e {{{\log \,t} \over {1 + t}}dt + \int_1^{1/e} {{{\log t} \over {1 + t}}} } dt\,\,\,....\left( A \right)$$
Now for solving, $$I = \int_1^{1/e} {{{\log t} \over {1 + t}}dt} $$
$$\therefore$$ Put $${1 \over t} = z \Rightarrow - {1 \over {{t^2}}}dt = dz$$
$$ \Rightarrow dt = - {{dz} \over {{z^2}}}$$
and limit for $$t = 1 \Rightarrow z = 1$$ and for
$$t = 1/e \Rightarrow z = e$$
$$\therefore$$ $$I = \int_1^e {{{\log \left( {{1 \over z}} \right)} \over {1 + {1 \over z}}}} \left( { - {{dz} \over {{z^2}}}} \right)$$
$$ = \int_1^e {{{\left( {\log 1 - \log z} \right).z} \over {z + 1}}\left( { - {{dz} \over {{z^2}}}} \right)} $$
$$ = \int_1^e { - {{\log z} \over {\left( {z + 1} \right)}}\left( { - {{dz} \over z}} \right)} $$
[ as $$\log 1 = 0$$ ]
$$ = \int_1^e {{{\log z} \over {z\left( {z + 1} \right)}}} dz$$
$$\therefore$$ $$I = \int_1^e {{{\log \,t} \over {t\left( {t + 1} \right)}}dt} $$
$$\left[ {} \right.$$ By property $$\int_a^b {f\left( t \right)dt} $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \int_a^b {f\left( x \right)dx} $$ $$\left. {} \right]$$
Equation $$(A)$$ becomes
$$F\left( e \right) = \int_1^e {{{\log t} \over {1 + t}}dt + \int_1^e {{{\log t} \over {t\left( {1 + t} \right)}}} } dt$$
$$ = \int_1^e {{{t.\log t + \log t} \over {t\left( {1 + t} \right)}}} dt$$
$$ = \int_1^e {{{\left( {\log t} \right)\left( {t + 1} \right)} \over {t\left( {1 + t} \right)}}} $$
$$ \Rightarrow F\left( e \right) = \int_1^e {{{\log t} \over t}dt} $$
Let $$\log t = x$$
$$\therefore$$ $${1 \over t}dt = dx$$
$$\left[ {} \right.$$ for limit $$t = 1,x = 0$$
and $$t = e,x = \log \,e = 1$$ $$\left. {} \right]$$
$$\therefore$$ $$F\left( e \right) = \int_0^1 {x\,dx} $$
$$ \Rightarrow F\left( e \right) = \left[ {{{{x^2}} \over 2}} \right]_0^1$$
$$ \Rightarrow F\left( e \right) = {1 \over 2}$$
where $$f\left( x \right) = \int_1^x {{{\log \,t} \over {1 + t}}} \,dt$$
$$\therefore$$ $$F\left( e \right) = f\left( e \right) + f\left( {{1 \over e}} \right)$$
$$ \Rightarrow F\left( e \right)$$
$$ = \int_1^e {{{\log \,t} \over {1 + t}}dt + \int_1^{1/e} {{{\log t} \over {1 + t}}} } dt\,\,\,....\left( A \right)$$
Now for solving, $$I = \int_1^{1/e} {{{\log t} \over {1 + t}}dt} $$
$$\therefore$$ Put $${1 \over t} = z \Rightarrow - {1 \over {{t^2}}}dt = dz$$
$$ \Rightarrow dt = - {{dz} \over {{z^2}}}$$
and limit for $$t = 1 \Rightarrow z = 1$$ and for
$$t = 1/e \Rightarrow z = e$$
$$\therefore$$ $$I = \int_1^e {{{\log \left( {{1 \over z}} \right)} \over {1 + {1 \over z}}}} \left( { - {{dz} \over {{z^2}}}} \right)$$
$$ = \int_1^e {{{\left( {\log 1 - \log z} \right).z} \over {z + 1}}\left( { - {{dz} \over {{z^2}}}} \right)} $$
$$ = \int_1^e { - {{\log z} \over {\left( {z + 1} \right)}}\left( { - {{dz} \over z}} \right)} $$
[ as $$\log 1 = 0$$ ]
$$ = \int_1^e {{{\log z} \over {z\left( {z + 1} \right)}}} dz$$
$$\therefore$$ $$I = \int_1^e {{{\log \,t} \over {t\left( {t + 1} \right)}}dt} $$
$$\left[ {} \right.$$ By property $$\int_a^b {f\left( t \right)dt} $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \int_a^b {f\left( x \right)dx} $$ $$\left. {} \right]$$
Equation $$(A)$$ becomes
$$F\left( e \right) = \int_1^e {{{\log t} \over {1 + t}}dt + \int_1^e {{{\log t} \over {t\left( {1 + t} \right)}}} } dt$$
$$ = \int_1^e {{{t.\log t + \log t} \over {t\left( {1 + t} \right)}}} dt$$
$$ = \int_1^e {{{\left( {\log t} \right)\left( {t + 1} \right)} \over {t\left( {1 + t} \right)}}} $$
$$ \Rightarrow F\left( e \right) = \int_1^e {{{\log t} \over t}dt} $$
Let $$\log t = x$$
$$\therefore$$ $${1 \over t}dt = dx$$
$$\left[ {} \right.$$ for limit $$t = 1,x = 0$$
and $$t = e,x = \log \,e = 1$$ $$\left. {} \right]$$
$$\therefore$$ $$F\left( e \right) = \int_0^1 {x\,dx} $$
$$ \Rightarrow F\left( e \right) = \left[ {{{{x^2}} \over 2}} \right]_0^1$$
$$ \Rightarrow F\left( e \right) = {1 \over 2}$$
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