JEE Mathematics 2023 Question with Solution

Given: $f(x) = \frac{2^{2x}}{2^{2x}+2} = \frac{4^x}{4^x+2}$ and we need

$$S=f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+\cdots+f\left(\frac{2022}{2023}\right)$$

Find: The value of $S$.

Use the symmetry relation between $f(x)$ and $f(1-x)$:

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

Now simplify the second term by dividing numerator and denominator by $4^{1-x}$ or rewriting appropriately. This gives

$$f(x)+f(1-x)=1$$

So any two terms whose arguments add to $1$ have sum $1$.

In the required sum, the arguments are

$$\frac{1}{2023},\frac{2}{2023},\ldots,\frac{2022}{2023}$$

These pair as

$$\frac{1}{2023} \text{ with } \frac{2022}{2023},\quad \frac{2}{2023} \text{ with } \frac{2021}{2023},\quad \ldots$$

Each pair adds to $1$, so each corresponding function pair adds to $1$.

The total number of terms is $2022$, hence the number of pairs is

$$\frac{2022}{2}=1011$$

Therefore

$$S=1011$$

So, the correct option is D.

Given: The sum contains values of $f\left(\frac{k}{2023}\right)$ for $k=1,2,\ldots,2022$.

Find: The total sum quickly.

Notice that for every term

$$f\left(\frac{k}{2023}\right)$$

there is a matching term

$$f\left(1-\frac{k}{2023}\right)=f\left(\frac{2023-k}{2023}\right)$$

From the solution property,

$$f(x)+f(1-x)=1$$

Hence each symmetric pair contributes exactly $1$.

Since there are $2022$ terms, there are $1011$ such pairs. Therefore the sum is

$$1011$$

So, the correct option is D.