Question

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

Answer 1

The correct answer is 99
$f(x) = \frac{2e^{2x}}{e^{2x} + e^x}$ and $f(1-x) = \frac{2e^{2-2x}}{e^{2-2x} + e^{1-x}}$
∴$$ \frac{f(x)+f(1-x)}{2} = 1
i.e. f(x)+f(1-x) = 2
Therefore,
$f(\frac{1}{100}) + f(\frac{2}{100})+...+f(\frac{99}{100})$
= $\sum_{x=1}^{49} f\left(\frac{x}{100}\right) + f\left(\frac{1 - \frac{x}{100}}{100}\right) + f\left(\frac{1}{2}\right)$
$= 49 \times 2+1 = 99$