Let f : R → R be a function defined f(x) = 2e²ˣ / (e²ˣ + e). Then f(1/100) + f(2/100) + f(3/100) + .

Q: Why did we choose to evaluate $f(x) + f(1-x)$? Is there a general rule for this?

Great question! In JEE problems involving sums of functions, whenever you see inputs that sum to a constant (like $1$ in this case), it is a massive signal to check the symmetry of the function. Specifically, evaluating $f(x) + f(a-x)$ often leads to a constant, which simplifies the entire summation into a simple multiplication.

Q: Wait, why is the answer $99$ and not $98$? Did I miscount the pairs?

You are very observant! We have $99$ terms in total. When we pair them up, we get $49$ pairs (each summing to $2$), which accounts for $98$ terms. The $99$th term is the middle term, $f\left(\frac{50}{100}\right) = f\left(\frac{1}{2}\right)$, which remains unpaired. We must calculate this middle term separately and add it to the sum of the pairs.

Q: Can I solve this using integration or some other calculus method?

While the function looks like it could be part of an integral, this is a discrete series problem. Using calculus would be like using a sledgehammer to crack a nut! Stick to the algebraic symmetry; it is much faster, more elegant, and less prone to calculation errors.

Question

Let $f : \mathbb{R} 	o \mathbb{R}$ be a function defined $f(x) = \frac{2e^{2x}}{e^{2x} + e}$. Then $f\left(\frac{1}{100}ight) + f\left(\frac{2}{100}ight) + f\left(\frac{3}{100}ight) + \dots + f\left(\frac{99}{100}ight)$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Wall of Numbers: A Journey into Symmetry

Imagine you are staring at a long, intimidating line of numbers. You have to sum up $f\left(\frac{1}{100}ight) + f\left(\frac{2}{100}ight) + \dots + f\left(\frac{99}{100}ight)$, where $f(x) = \frac{2e^{2x}}{e^{2x} + e}$.

At first glance, it looks like a nightmare. Calculating each term individually would take hours and likely lead to errors.

In the world of JEE Advanced, whenever you see a problem that looks like a brute-force calculation, there is almost always a hidden, elegant shortcut. Today, we are going to find that shortcut.

### Phase 1: The Mirror Insight

Look closely at the inputs. We have $\frac{1}{100}$ at the start and $\frac{99}{100}$ at the end.

When we add them, we get $\frac{1}{100} + \frac{99}{100} = 1$. This is not a coincidence; it is a mirror.

If we look at the second term, $\frac{2}{100}$, and the second-to-last term, $\frac{98}{100}$, they also sum to $1$. This symmetry is our golden ticket, implying we only need to understand how the function behaves when we swap $x$ for $1-x$.

### Phase 2: The Algebraic Dance

Let's perform the substitution to see what $f(1-x)$ looks like. We take our original function $f(x) = \frac{2e^{2x}}{e^{2x} + e}$ and replace every $x$ with $(1-x)$:

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

Now, let's simplify the exponent. Since $2(1-x) = 2-2x$, we use the laws of exponents to write $e^{2-2x} = \frac{e^2}{e^{2x}}$.

Substituting this into our expression, we get:

$$f(1-x) = \frac{2\left(\frac{e^2}{e^{2x}}ight)}{\frac{e^2}{e^{2x}} + e}$$

To clear the fractions, we multiply the numerator and the denominator by $e^{2x}$:

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

If we factor out an $e$ from the denominator, we get $e(e + e^{2x})$. Canceling the $e$ from the numerator and denominator leaves us with:

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

### Phase 3: The Magic Sum

Now, let's add $f(x)$ and $f(1-x)$ together. Since we have the same denominator, the addition is straightforward:

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

Factor out the $2$ in the numerator to get $2(e^{2x} + e)$. The term $(e^{2x} + e)$ cancels out perfectly with the denominator, leaving us with a constant: $2$.

This is the beauty of the problem. Every pair of terms that sums to $1$ in the input will sum to $2$ in the output.

### Phase 4: The Lonely Survivor

We have $99$ terms in total. We can pair the first with the last, the second with the second-to-last, and so on.

This gives us $49$ pairs, which accounts for $98$ terms. Since each pair sums to $2$, the sum of these pairs is $49 	imes 2 = 98$.

However, we have $99$ terms, meaning there is one term left in the middle: $f\left(\frac{50}{100}ight)$, which is $f\left(\frac{1}{2}ight)$. Let's calculate this middle term:

$$f\left(\frac{1}{2}ight) = \frac{2e^{2(1/2)}}{e^{2(1/2)} + e} = \frac{2e}{e + e} = \frac{2e}{2e} = 1$$

### Conclusion

The total sum is the sum of our $49$ pairs plus the middle term: $98 + 1 = 99$.

We didn't need to calculate $99$ complex exponential values. We just needed to see the symmetry, trust the algebra, and account for the middle term.

That is the heart of JEE mathematics—finding the simple, elegant truth hidden beneath a complex exterior. The final answer is **99**.

Animated Solution for Mathematics - Functions: Let $f : R \to R$ be a function defined $f (x) = \frac{2 e ^{2 x}}{e ^{2 x} + e}$ . Then $f (\frac{1}{100}) + f (\frac{2}{100}) + f (\frac{3}{100}) + \dots + f (\frac{99}{100})$ is equal to ______.

Enter Numerical Value:

Visualized Solution

The Function and the Series

Spotting the Symmetry

Setting up $f (1 - x)$

Simplifying the Exponent

Clearing the Denominators

Final Form of $f (1 - x)$

The Magic Sum: $f (x) + f (1 - x)$

Proving the Constant Sum

Pairing the Series Terms

Counting the Pairs

The Middle Term

Evaluating $f (1/2)$

Simplifying $f (1/2)$

The Final Sum

The Sigma Insight: Classification of Functions

The Wall of Numbers

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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