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AIEEE 2007 Solved Paper

Section: Mathematics

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Question : 103 of 118

Marks: +1, -0

Let

F(x)=f(x)+f\left(\frac{1}{x}\right)

f(x)=\int\limits_{l}^{x} \frac{\log t}{1+t}\, dt

F(e)

$1$
$2$
$\frac{1}{2}$
$0$

Solution:

Given

f(x)=f(x)+f\left(\frac{1}{x}\right)

,

where

f(x)=\int\limits_{1}^{x} \frac{\log t}{1+t}\, dt

\therefore F(e)=f(e)+f\left(\frac{1}{e}\right)

\Rightarrow F(e)

=\int\limits_{1}^{e} \frac{\log t}{1+t}\, dt+\int\limits_{1}^{\frac{1}{e}} \frac{\log t}{1+t}\, dt \dots (A)

Now for solving,

I=\int\limits_{1}^{\frac{1}{e}} \frac{\log t}{1+t}\, dt

∴ Put

\frac{1}{t}=z \Rightarrow -\frac{1}{t^{2}}\, dt = dz

\Rightarrow dt = -\frac{dz}{z^{2}}

and limit for

t=1 \Rightarrow z=1

and for

t=\frac{1}{e} \Rightarrow z=e

I=\int\limits_{1}^{e} \frac{\log \left(\frac{1}{z}\right)}{1+\frac{1}{z}} \left(-\frac{dz}{z^{2}}\right)

=\int\limits_{1}^{e} \frac{(\log 1 - \log z) z}{z+1} \left(-\frac{dz}{z^{2}}\right)

=\int\limits_{1}^{e} \frac{\log z}{z+1} \left(-\frac{dz}{z}\right)

\left[\text{ as } \log 1=0\right]

=\int\limits_{1}^{e} \frac{\log z}{z(z+1)}\, dz

\therefore I=\int\limits_{1}^{e} \frac{\log t}{t(t+1)}\, dt

[ By property

\int\limits_{a}^{b} f(t)\, dt

=\int\limits_{a}^{b} f(x)\, dx

]

Equation

(A)

becomes

F(e)=\int\limits_{1}^{e} \frac{\log t}{1+t}\, dt+\int\limits_{1}^{e} \frac{\log t}{t(1+t)}\, dt

=\int\limits_{1}^{e} \frac{t\cdot \log t + \log t}{t(1+t)}\, dt

=\int\limits_{1}^{e} \frac{(\log t)(t+1)}{t(1+t)}

\Rightarrow F(e)=\int\limits_{1}^{e} \frac{\log t}{t}\, dt

Let

\log t = x

\therefore \frac{1}{t}\, dt = dx

[

for limit

t=1, x=0

and

t=e, x=\log e=1]

\therefore F(e)=\int\limits_{0}^{1} x\, dx

\Rightarrow F(e)= \left[\frac{x^{2}}{2}\right]_{0}^{1}

\Rightarrow F(e)=\frac{1}{2}

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