Question

Let $f:R\to R$ be a function defined $f(x)=\frac{2e^{2x}}{e^{2x}+e}$. Then $f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+f\left(\frac{3}{100}\right)+\dots ..+f\left(\frac{99}{100}\right)$ is equal to______.

Answer 1

Answer: 99

$f(x)+f(1-x)=\frac{2e^{2x}}{e^{2x}+e}+\frac{2e^{2-2x}}{e^{2-ex}+e}$
$=\left[\frac{e^{2x}}{e^{2x}+e}+\frac{e^2}{e^2+e^{2x+1}}\right]$
$=2\left[\frac{e^{2x-1}}{e^{2x-1}+1}+\frac{1}{1+e^{2x-1}}\right]=2$
$f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+f\left(\frac{3}{100}\right)+\dots .+f\left(\frac{99}{100}\right)$
$=\left\{f\left(\frac{1}{100}\right)+f\left(\frac{99}{100}\right)\right\}+\left\{f\left(\frac{2}{100}\right)+f\left(\frac{98}{100}\right)\right\}+\dots .+f\left\{\left(\frac{49}{100}\right)+f\left(\frac{51}{100}\right)\right\}+f\left(\frac{1}{2}\right)$
$=(2+2+2+----49$ times $)+\frac{2e}{e+e}$
$=98+1=99$