Question: Let $ F(x)=f(x)+f( \frac{1}{x} ), $ where $ f(x)=\int\limits_l^{x}{\frac{\log t}{1+t}dt} $ , Then $ F(e) $ equals

Options:

A) 1

B) 2

C) 1/2

D) 0

Answer:

Correct Answer: C

Solution:

[c] Given $ F(x)=f(x)+f( \frac{1}{x} ) $ , where $ f(x)=\int_1^{x}{\frac{\log t}{1+t}dt} $

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

$ \Rightarrow F(e)=\int_1^{e}{\frac{\log t}{1+t}dt+\int_1^{1/e}{\frac{\log t}{1+t}dt…(A)}} $ Now for solving, $ I=\int_1^{1/e}{\frac{\log t}{1+t}dt} $

$ \therefore $ 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=1/e\Rightarrow z=e $

$ \therefore ,I=\int_1^{e}{\frac{\log ( \frac{1}{z} )}{1+\frac{1}{z}}( -\frac{dz}{z^{2}} )} $ $ =\int_1^{e}{\frac{(\log 1-\log z).z}{z+1}( -\frac{dz}{z^{2}} )} $ $ =\int_1^{e}{-\frac{\log z}{(z+1)}( -\frac{dz}{z} )=\int_1^{e}{\frac{\log z}{z(z+1)}dz}} $

$ \therefore I=\int_1^{e}{\frac{\log t}{t(t+1)}dt} $ Equation [a] becomes: $ F(e)=\int_1^{e}{\frac{\log t}{1+t}dt+\int_1^{e}{\frac{\log t}{t(1+t)}dt}} $ $ =\int_1^{e}{\frac{t.\log t+\log t}{t(1+t)}dt=\int_1^{e}{\frac{(\log t)(t+1)}{t(1+t)}dt}} $

$ \Rightarrow F(e)=\int_1^{e}{\frac{\log t}{t}dt} $ Let $ \log t=x\therefore \frac{1}{t}dt=dx $ $ [for\lim itt=1,x=0andt=e,x=\log e=1] $

$ \therefore F(e)=\int_0^{1}{x}dxF(e)=[ \frac{x^{2}}{2} ]_0^{1}\Rightarrow F(e)=\frac{1}{2} $