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If $f(x)=\frac{2^{2 x}}{2^{2 x}+2}, x \in R$ then $f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+\ldots \ldots . .+f\left(\frac{2022}{2023}\right)$ is equal to
$2011$
$1010$
$2010$
$1011$
Solution
$f(x)=\frac{4^x}{4^x+2}$
$f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2}$
$=\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^x\right)}$
$=\frac{4^x}{4^x+2}+\frac{2}{2+4^x}$
$=1$
$\Rightarrow f(x)+f(1-x)=1$
$\text { Now } f \left(\frac{1}{2023}\right)+ f \left(\frac{2}{2023}\right)+ f \left(\frac{3}{2023}\right)+\ldots \ldots .+$
$\ldots \ldots \ldots . .+ f \left(1-\frac{3}{2023}\right)+ f \left(1-\frac{2}{2023}\right)+ f \left(1-\frac{1}{2023}\right)$
Now sum of terms equidistant from beginning and end is 1
$\text { Sum }=1+1+1+\ldots \ldots \ldots+1 \text { (1011 times) }$
$=1011$
