Let the function f: [0, 1] → ℝ be defined by f(x) = 4ˣ / (4ˣ + 2). Then the value of f(1/40) + f(2/4

Q: Why can't we just calculate each term individually? It's just 39 terms.

While technically possible, it's a trap! JEE problems are designed to test your ability to spot patterns, not your stamina for arithmetic. Calculating 39 terms would take forever and lead to calculation errors. Always look for symmetry first.

Q: How did you know to check $f(x) + f(1-x)$? Is there a rule?

Great question! Whenever you see a sum of a function where the arguments are in an arithmetic progression (like $1/40, 2/40, \dots$), and the function involves exponentials or trigonometric ratios, check for symmetry. Specifically, check $f(x) + f(a+b-x)$. Here, $a=0, b=1$, so $f(x) + f(1-x)$ is the natural first step.

Q: What happens to the middle term? Why is it left out?

The series has 39 terms. When you pair them up (1st with 39th, 2nd with 38th...), you have $39/2 = 19.5$ pairs. This means 19 pairs are formed, and one term remains unpaired. That term is the one where $k/40 = 1/2$, which is $20/40$.

Question

Let the function $f: [0, 1] 	o \mathbb{R}$ be defined by $f(x) = \frac{4^x}{4^x + 2}$. Then the value of $f\left(\frac{1}{40}ight) + f\left(\frac{2}{40}ight) + f\left(\frac{3}{40}ight) + \dots + f\left(\frac{39}{40}ight) - f\left(\frac{1}{2}ight)$ is ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing before a massive, daunting series of thirty-nine terms. At first glance, it looks like a nightmare of arithmetic:
$$f\left(\frac{1}{40}ight) + f\left(\frac{2}{40}ight) + \dots + f\left(\frac{39}{40}ight)$$

If you try to calculate each term individually, you will be trapped in a cycle of tedious calculations that lead nowhere. In the world of JEE Advanced, complexity is often just a mask for elegance.

The key here is to look for symmetry. When you see a function like $f(x) = \frac{4^x}{4^x + 2}$, your intuition should immediately scream "symmetry!" Let us investigate the behavior of this function when we pair $x$ with $1-x$.

### The Algebraic Dance

Let us test the sum $f(x) + f(1-x)$. We have $f(x) = \frac{4^x}{4^x + 2}$. Now, substitute $1-x$ into the function:
$$f(1-x) = \frac{4^{1-x}}{4^{1-x} + 2}$$

Using the exponent rule $a^{m-n} = \frac{a^m}{a^n}$, we can rewrite $4^{1-x}$ as $\frac{4}{4^x}$. So, the expression becomes:
$$f(1-x) = \frac{\frac{4}{4^x}}{\frac{4}{4^x} + 2}$$

To clear the fractions, multiply the numerator and denominator by $4^x$. This yields:
$$\frac{4}{4 + 2 \cdot 4^x}$$

Now, divide both the numerator and denominator by $2$ to obtain:
$$f(1-x) = \frac{2}{2 + 4^x}$$

Look at that! The denominator is now $4^x + 2$, which is identical to the denominator of $f(x)$. When we add $f(x) + f(1-x)$, we get:
$$\frac{4^x}{4^x + 2} + \frac{2}{4^x + 2} = \frac{4^x + 2}{4^x + 2} = 1$$

This is the "Aha!" moment. The entire series collapses into a beautiful, simple sum.

### Visualizing the Pairs

Now that we know $f(x) + f(1-x) = 1$, we can pair the terms in our series. The first term $f(1/40)$ pairs with the last term $f(39/40)$ because $\frac{1}{40} + \frac{39}{40} = 1$. Their sum is $1$.

The second term $f(2/40)$ pairs with $f(38/40)$, and so on. Since there are $39$ terms, we can form exactly $19$ such pairs, each summing to $1$. This leaves us with one lonely middle term: $f(20/40)$, which is $f(1/2)$.

So, our total sum $S$ is:
$$S = 19 	imes 1 + f\left(\frac{1}{2}ight) = 19 + f\left(\frac{1}{2}ight)$$

The question asks for $S - f(1/2)$. Substituting our value for $S$, we get:
$$\left(19 + f\left(\frac{1}{2}ight)ight) - f\left(\frac{1}{2}ight) = 19$$

The middle term cancels out perfectly, leaving us with the final answer of **19**.

### The Pro Tip

Remember this: whenever you encounter a function of the form $f(x) = \frac{a^x}{a^x + \sqrt{a}}$, the identity $f(x) + f(1-x) = 1$ is your best friend. It is a powerful tool that turns a mountain of work into a simple, elegant solution.

Keep this in your arsenal, and you will conquer any similar problem that comes your way!

Enter Numerical Value:

Visualized Solution

The Series

Symmetry Check

Setting up $f (1 - x)$

Exponent Rules

Clearing Fractions

Simplifying the Second Term

Adding the Fractions

Visualizing the Pairs

Counting the Pairs

Grouping the Series

Evaluating the Sum

The Final Expression

Conclusion

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Algebraic Dance

Visualizing the Pairs

The Pro Tip

Similar Questions

Let $f : R \to R$ be a function which satisfies $f (x + y) = f (x) + f (y) \forall x, y \in R$ . If $f (1) = 2$ and $g (n) = \sum_{k = 1}^{n - 1} f (k)$ , $n \in N$ then the value of $n$ , for which $g (n) = 20$ , is :

Consider a function $f : N \to R$ , satisfying $f (1) + 2 f (2) + 3 f (3) + \dots + x f (x) = x (x + 1) f (x); x \geq 2$ with $f (1) = 1$ . Then $\frac{1}{f ( 2022 )} + \frac{1}{f ( 2028 )}$ is equal to

The sum $\sum_{k = 1}^{20} (1 + 2 + 3 + .... + k)$ is _____________.

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Let $S_{n} = 1 \cdot (n - 1) + 2 \cdot (n - 2) + 3 \cdot (n - 3) + \dots + (n - 1) \cdot 1, n \geq 4$ . The sum $\sum_{n = 4}^{\infty} (\frac{2 S _{n}}{n !} - \frac{1}{( n - 2 )!})$ is equal to :

$\frac{1}{3 ^{2} - 1} + \frac{1}{5 ^{2} - 1} + \frac{1}{7 ^{2} - 1} + \dots + \frac{1}{( 201 ) ^{2} - 1}$ is equal to

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Enter Numerical Value:

Visualized Solution

The Series

Symmetry Check

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1−x)

Exponent Rules

Clearing Fractions

Simplifying the Second Term

Adding the Fractions

Visualizing the Pairs

Counting the Pairs

Grouping the Series

Evaluating the Sum

The Final Expression

Conclusion

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Algebraic Dance

Visualizing the Pairs

The Pro Tip

Similar Questions

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Setting up $f (1 - x)$

Let $f : R \to R$ be a function which satisfies $f (x + y) = f (x) + f (y) \forall x, y \in R$ . If $f (1) = 2$ and $g (n) = \sum_{k = 1}^{n - 1} f (k)$ , $n \in N$ then the value of $n$ , for which $g (n) = 20$ , is :

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