A) 1
B) 2
C) 1/2
D) 0
Correct Answer: C
Solution :
| [c] Given \[F(x)=f(x)+f\left( \frac{1}{x} \right)\], where |
| \[f(x)=\int_{1}^{x}{\frac{\log \,\,t}{1+t}dt}\] |
| \[\therefore F(e)=f(e)+f\left( \frac{1}{e} \right)\] |
| \[\Rightarrow F(e)=\int_{1}^{e}{\frac{\log t}{1+t}dt+\int_{1}^{1/e}{\frac{\log \,\,t}{1+t}dt...(A)}}\] |
| 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 \[t=1/e\Rightarrow z=e\] |
| \[\therefore \,\,\,\,\,I=\int_{1}^{e}{\frac{\log \left( \frac{1}{z} \right)}{1+\frac{1}{z}}\left( -\frac{dz}{{{z}^{2}}} \right)}\] |
| \[=\int_{1}^{e}{\frac{(\log 1-\log z).z}{z+1}\left( -\frac{dz}{{{z}^{2}}} \right)}\] |
| \[=\int_{1}^{e}{-\frac{\log z}{(z+1)}\left( -\frac{dz}{z} \right)=\int_{1}^{e}{\frac{\log z}{z(z+1)}dz}}\] |
| \[\therefore I=\int_{1}^{e}{\frac{\log t}{t(t+1)}dt}\] |
| 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 \[\log \,\,t=x\therefore \frac{1}{t}dt=dx\] |
| \[[for\,\,\lim it\,\,t=1,x=0\,\,and\,\,t=e,x=\log \,\,e=1]\] |
| \[\therefore \,\,\,\,\,\,F(e)=\int_{0}^{1}{x}\,\,dx\,\,\,\,\,\,\,\,\,\,F(e)=\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{1}\Rightarrow F(e)=\frac{1}{2}\] |
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