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 $)$
$=1011$

Answer 2

Step 1: Given function 

The given function is:

\[ f(x) = \frac{2^{2x^2}}{2x+2} = \frac{4x}{4x+2}. \] 

Step 2: Adding \( f(x) \) and \( f(1 - x) \)

Observe that:

\[ f(x) + f(1 - x) = \frac{4x}{4x + 2} + \frac{4(1 - x)}{4(1 - x) + 2}. \] 

Step 3: Simplifying the second term We simplify the second term:

 

\[ f(x) + f(1 - x) = \frac{4x}{4x + 2} + \frac{4(1 - x)}{4(1 - x) + 2}. \] 

Step 4: Further simplification: Now, we simplify further:

 

\[ f(x) + f(1 - x) = \frac{4x}{4x + 2} + \frac{4}{4x + 2} + 2(4x). \] 

Step 5: Final simplification :Finally, simplifying the sum gives us:

 

\[ f(x) + f(1 - x) = \frac{4x}{4x + 2} + \frac{2}{2 + 4x}. \] 
We see that the final expression is:

 

\[ f(x) + f(1 - x) = 1. \] 

Step 6: Summation 
Consider the summation:

 

\[ f(1) + f(2) + \ldots + f(2023). \] Using the property \( f(x) + f(1 - x) = 1 \), we observe that pairs of terms add up to 1. The total number of terms is 2022, so there are \( \frac{2022}{2} = 1011 \) pairs. 

Step 7: Final sum
The sum of all the pairs is:

 

\[ \text{Sum} = 1 + 1 + \ldots + 1 \quad (1011 \text{ terms}) = 1011. \]