Question

If $f(x)=\frac{2^{2 x}}{2^{2 x}+2}, x \in R$, then $f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+\ldots+f\left(\frac{2022}{2023}\right)$ is equal to

• A. 1011
• B. 2010
• C. 1010
• D. 2011

Answer 1

The Correct Option is A

The correct answer is (A) : 1011
$f(x)=\frac{4^x}{4^x+2}$
$f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2}$
$=\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^x\right)}$
$=\frac{4^x}{4^x+2}+\frac{2}{2+4^x}$
$=1$
$\Rightarrow f(x)+f(1-x)=1$
Now $f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+f\left(\frac{3}{2023}\right)+\dots \dots +$
$\dots \dots \dots +f\left(1-\frac{3}{2023}\right)+f\left(1-\frac{2}{2023}\right)+f\left(1-\frac{1}{2023}\right)$
Now sum of terms equidistant from beginning and end is $1$
Sum $=1+1+1+\dots \dots \dots .+1(1011$ times $)$
$=1011$