If f(x)=frac{2^{2 x}}{2^{2 x}+2}, x in R then | vedclass

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Question

$f(x)=\frac{4^x}{4^x+2}$

$f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2}$

$=\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^x\right)}$

$=\frac

Vedclass · Accepted Answer

$1011$

Vedclass · Answer

$f(x)=\frac{4^x}{4^x+2}$

$f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2}$

$=\frac{4^x}{4^x+2}+\frac{4}{4+2\left(4^xight)}$

$=\frac{4^x}{4^x+2}+\frac{2}{2+4^x}$

$=1$

$\Rightarrow f(x)+f(1-x)=1$

$	ext { Now } f \left(\frac{1}{2023}ight)+ f \left(\frac{2}{2023}ight)+ f \left(\frac{3}{2023}ight)+\ldots \ldots .+$

$\ldots \ldots \ldots . .+ f \left(1-\frac{3}{2023}ight)+ f \left(1-\frac{2}{2023}ight)+ f \left(1-\frac{1}{2023}ight)$

Now sum of terms equidistant from beginning and end is 1

$	ext { Sum }=1+1+1+\ldots \ldots \ldots+1 	ext { (1011 times) }$

$=1011$

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

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