Question

If $f\left(x\right)=\frac{e^{2x-1}}{1+e^{2x-1}}$, then the values of $f\left(\frac{1}{2009}\right)+f\left(\frac{2}{2009}\right)+f\left(\frac{3}{2009}\right)+....+f\left(\frac{2008}{2009}\right)$

• A. 1002.5
• B. 1001.5
• C. 1003
• D. 1004

Answer 1

The correct option is D 1004

$f\left(x\right)=\frac{e^{2x-1}}{1+e^{2x-1}}.$

$f(1-x)=\frac{e^{-(2x-1)}}{e^{-(2x-1)}+1}=\frac{1}{e^{(2x-1)}+1}$ (Dividing numerator and denominator by $e^{-(2x-1)}$ )

$f\left(x\right)+f(1-x)=1$
$f\left(\frac{1}{2009}\right)+f\left(\frac{2008}{2009}\right)=1.f\left(\frac{2}{2009}\right)+f\left(\frac{2007}{2009}\right)=1f\left(\frac{3}{2009}\right)+f\left(\frac{2006}{2009}\right)=1.f\left(\frac{1004}{2009}\right)+f\left(\frac{1005}{2009}\right)=1.$

we have 1004 pairs and their summation is 1004