- Home
- Standard 12
- Mathematics
यदि $f(x)=\frac{2^{2 x}}{2^{2 x}+2}, x \in R$, है, तो $\mathrm{f}\left(\frac{1}{2023}\right)+\mathrm{f}\left(\frac{2}{2023}\right)+\ldots \ldots .+\mathrm{f}\left(\frac{2022}{2023}\right)$ बराबर है
$2011$
$1010$
$2010$
$1011$
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
$\text { Sum }=1+1+1+\ldots \ldots \ldots+1 \text { (1011 times) }$
$=1011$