JEE Mathematics 2025 Question with Solution

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

$$S = \sum_{k=1}^{81} f\left(\frac{k}{82}\right)$$

Find: the value of $S$.

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

First, compute

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

Therefore,

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

Now pair the terms of the sum as

$$f\left(\frac{k}{82}\right) + f\left(\frac{82-k}{82}\right) = f\left(\frac{k}{82}\right) + f\left(1 - \frac{k}{82}\right) = 1$$

for $k = 1,2,\dots,40$.

So,

$$S = \sum_{k=1}^{40} \left[f\left(\frac{k}{82}\right) + f\left(\frac{82-k}{82}\right)\right] + f\left(\frac{41}{82}\right)$$

Hence,

$$S = 40 + f\left(\frac{1}{2}\right)$$

Now evaluate the middle term:

$$f\left(\frac{1}{2}\right) = \frac{2^{1/2}}{2^{1/2} + \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2} + \sqrt{2}} = \frac{1}{2}$$

Thus,

$$S = 40 + \frac{1}{2} = \frac{81}{2}$$

Therefore, the correct option is B.

Given: $f(x) = \frac{2^x}{2^x + \sqrt{2}}$ with $x \in \mathbb{R}$.

Find:

$$\sum_{k=1}^{81} f\left(\frac{k}{82}\right)$$

Observe that the arguments $\frac{k}{82}$ and $\frac{82-k}{82}$ add up to $1$. So the key step is to prove

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

Now,

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

Multiplying numerator and denominator by $2^x$ gives

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

Hence,

$$\begin{aligned} f(x) + f(1-x) &= \frac{2^x}{2^x + \sqrt{2}} + \frac{\sqrt{2}}{\sqrt{2} + 2^x} \\ &= 1 \end{aligned}$$

Therefore each pair contributes $1$:

$$f\left(\frac{k}{82}\right) + f\left(\frac{82-k}{82}\right) = 1$$

There are $40$ such pairs among the $81$ terms, and the remaining term is at $k=41$:

$$\frac{41}{82} = \frac{1}{2}$$

So,

$$S = 40 + f\left(\frac{1}{2}\right)$$

Now,

$$f\left(\frac{1}{2}\right) = \frac{\sqrt{2}}{\sqrt{2}+\sqrt{2}} = \frac{1}{2}$$

Thus,

$$S = 40 + \frac{1}{2} = \frac{81}{2}$$

Therefore, the value of the sum is $\frac{81}{2}$ and the correct option is B.