Functions Question 3
Question 3 - 24 January - Shift 2
If $f(x)=\frac{2^{2 x}}{2^{2 x}+2}, x \in R$,
then $f(\frac{1}{2023})+f(\frac{2}{2023})+\ldots \ldots .+f(\frac{2022}{2023})$ is
equal to
(1) 2011
(2) 1010
(3) 2010
(4) 1011
Show Answer
Answer: (4)
Solution:
Formula: Operations on functions
$f(x)=\frac{4^{x}}{4^{x}+2}$
$f(x)+f(1-x)=\frac{4^{x}}{4^{x}+2}+\frac{4^{1-x}}{4^{1-x}+2}$
$=\frac{4^{x}}{4^{x}+2}+\frac{4}{4+2(4^{x})}$
$=\frac{4^{x}}{4^{x}+2}+\frac{2}{2+4^{x}}$
$=1$
$\Rightarrow f(x)+f(1-x)=1$
Now $f(\frac{1}{2023})+f(\frac{2}{2023})+f(\frac{3}{2023})+\ldots \ldots+$
$ +f(1-\frac{3}{2023})+f(1-\frac{2}{2023})+f(1-\frac{1}{2023}) $
Now sum of terms equidistant from beginning and end is 1
Sum $=1+1+1+$ +1 (1011 times)
$=1011$
