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

Correct Answer: 99

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,
\(f(\frac{1}{100}) + f(\frac{2}{100})+...+f(\frac{99}{100})\)
= \(\sum_{x=1}^{49} f\left(\frac{x}{100}\right) + f\left(\frac{1 - \frac{x}{100}}{100}\right) + f\left(\frac{1}{2}\right)\)
\(= 49 \times 2+1 = 99\)