Solve que 76

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{e}}^{2\mathrm{x}-1}}{1+{\mathrm{e}}^{2\mathrm{x}-1}} \\ \mathrm{f}\left(1-\mathrm{x}\right)=\frac{{\mathrm{e}}^{2\left(1-\mathrm{x}\right)-1}}{1+{\mathrm{e}}^{2\left(1-\mathrm{x}\right)-1}} \\ =\frac{{\mathrm{e}}^{2-2\mathrm{x}-1}}{1+{\mathrm{e}}^{2-2\mathrm{x}-1}} \\ =\frac{{\mathrm{e}}^{1-2\mathrm{x}}}{1+{\mathrm{e}}^{1-2\mathrm{x}}} \\ =\frac{{\mathrm{e}}^{-\left(2\mathrm{x}-1\right)}}{1+{\mathrm{e}}^{-\left(2\mathrm{x}-1\right)}} \\ =\frac{{\displaystyle \frac{1}{{\mathrm{e}}^{\left(2\mathrm{x}-1\right)}}}}{1+\frac{1}{{\mathrm{e}}^{\left(2\mathrm{x}-1\right)}}} \\ \mathrm{f}\left(1-\mathrm{x}\right)=\frac{1}{1+{\mathrm{e}}^{2\mathrm{x}-1}} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(1-\mathrm{x}\right)=\frac{{\mathrm{e}}^{2\mathrm{x}-1}}{1+{\mathrm{e}}^{2\mathrm{x}-1}}+\frac{1}{1+{\mathrm{e}}^{2\mathrm{x}-1}} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(1-\mathrm{x}\right)=\frac{1+{\mathrm{e}}^{2\mathrm{x}-1}}{1+{\mathrm{e}}^{2\mathrm{x}-1}} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(1-\mathrm{x}\right)=1 \\ \mathrm{x}=\frac{1}{2009} \\ \Rightarrow \mathrm{f}\left(\frac{1}{2009}\right)+\mathrm{f}\left(1-\frac{1}{2009}\right)=1 \\ \Rightarrow \boxed{\mathrm{f}\left(\frac{1}{2009}\right)+\mathrm{f}\left(\frac{2008}{2009}\right)=1} \\ \mathrm{Put} \mathrm{x}=\frac{2}{2009} \\ \mathrm{f}\left(\frac{2}{2009}\right)+\mathrm{f}\left(1-\frac{2}{2009}\right)=1 \\ \Rightarrow \boxed{\mathrm{f}\left(\frac{2}{2009}\right)+\mathrm{f}\left(\frac{2007}{2009}\right)=1} \\ \mathrm{Put} \mathrm{x}=\frac{3}{2009} \\ \mathrm{f}\left(\frac{3}{2009}\right)+\mathrm{f}\left(1-\frac{3}{2009}\right)=1 \\ \Rightarrow \boxed{\mathrm{f}\left(\frac{3}{2009}\right)+\mathrm{f}\left(\frac{2006}{2009}\right)=1} \\ . \\ . \\ \mathrm{Put} \mathrm{x}=\frac{1004}{2009} \\ \Rightarrow \mathrm{f}\left(\frac{1004}{2009}\right)+\mathrm{f}\left(1-\frac{1004}{2009}\right)=1 \\ \Rightarrow \boxed{\mathrm{f}\left(\frac{1004}{2009}\right)+\mathrm{f}\left(\frac{1005}{2009}\right)=1} \\ \mathrm{Adding} \mathrm{all} \mathrm{equations} \mathrm{we} \mathrm{get}, \\ \mathrm{f}\left(\frac{1}{2009}\right)+\mathrm{f}\left(\frac{2}{2009}\right)+\mathrm{f}\left(\frac{3}{2009}\right)+...\mathrm{f}\left(\frac{1004}{2009}\right)+\mathrm{f}\left(\frac{1005}{2009}\right)......+\mathrm{f}\left(\frac{2006}{2009}\right)+\mathrm{f}\left(\frac{2007}{2009}\right)+\mathrm{f}\left(\frac{2008}{2009}\right)=1+1+1.... 1004 \mathrm{times} \\ =1004 \left(\mathrm{Answer}\right) \end{gathered}$$

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