Question

A function $f\left(x\right)$ is given by $f\left(x\right)=\frac{5^x}{5^x+5},$ then the sum of the series $f\left(\frac{1}{20}\right)+f\left(\frac{2}{20}\right)+f\left(\frac{3}{20}\right)+.....+f\left(\frac{39}{20}\right)$ is equal to:

• A. $\frac{19}{2}$
• B. $\frac{49}{2}$
• C. $\frac{39}{2}$
• D. $\frac{29}{2}$

Answer 1

The correct option is C $\frac{39}{2}$

$f\left(x\right)=\frac{5^x}{5^x+5}\cdots \text{(i)}$
$\Rightarrow f(2-x)=\frac{5^{2-x}}{5^{2-x}+5}$
$\Rightarrow f(2-x)=\frac{5}{5^x+5}\cdots \text{(ii)}$
Adding equation $\text{(i) and (ii)}$
$f\left(x\right)+f(2-x)=1$
$f\left(\frac{1}{20}\right)+f\left(\frac{39}{20}\right)=1$
$f\left(\frac{2}{20}\right)+f\left(\frac{38}{20}\right)=1$
:
:
$f\left(\frac{19}{20}\right)+f\left(\frac{21}{20}\right)=1$
and $f\left(\frac{20}{20}\right)=f\left(1\right)=\frac{1}{2}$
$\therefore \left(\frac{1}{2}\right)+f\left(\frac{2}{20}\right)+f\left(\frac{3}{20}\right)+....+f\left(\frac{39}{20}\right)=19+\frac{1}{2}=\frac{39}{2}$