Suppose f: R → (0, ∞) be a differentiable function such that 5f(x+y) = f(x) · f(y), ∀ x, y ∈ R. If f

Q: Wait, why do we define $g(x) = \frac{f(x)}{5}$? Can't we just solve for $f(x)$ directly?

You absolutely can, but defining $g(x)$ is a strategic move to simplify the algebra. When you have an equation like $5f(x+y) = f(x)f(y)$, the $5$ is an 'obstruction' to the standard exponential property. By scaling the function, we map it to a form we already know how to solve. It's like changing your perspective to make a difficult problem look familiar!

Q: I calculated the sum $\sum_{n=0}^5 f(n)$ and got 5 terms. Why is the answer based on 6 terms?

This is a classic 'off-by-one' error! Look closely at the summation index: $n$ goes from $0$ to $5$. If you count them: $0, 1, 2, 3, 4, 5$, that is exactly $6$ terms. Always double-check your bounds—starting at zero is a common trap in JEE problems.

Q: Is $f(x) = 5 \cdot a^x$ the only possible solution for this functional equation?

Given that $f$ is differentiable, yes. The equation $g(x+y) = g(x)g(y)$ is a version of Cauchy's functional equation. For continuous or differentiable functions, the only non-zero solutions are of the form $a^x$ (or $e^{cx}$). Since our function maps to $(0, \infty)$, the exponential form is the unique solution.

Question

Suppose $f: \mathbb{R} 	o (0, \infty)$ be a differentiable function such that $5f(x+y) = f(x) \cdot f(y), \forall x, y \in \mathbb{R}$. If $f(3) = 320$, then $\sum_{n=0}^5 f(n)$ is equal to :

Options:

(A) 6875

(B) 6575

(C) 6825

(D) 6528

Sigma Percentile · Accepted Answer

### The Riddle of the Functional Equation

Functional equations are the hidden gems of the JEE Advanced syllabus. They are not just about plugging in numbers; they are about uncovering the 'DNA' of a function.

When you see a problem like $5f(x+y) = f(x) \cdot f(y)$, your mathematical intuition should immediately fire. You are looking at a relationship where the sum of inputs transforms into the product of outputs, which is the hallmark of an exponential function.

### Phase 1: The Art of Normalization

At first glance, that coefficient of $5$ on the left side feels like a nuisance. It breaks the symmetry of the standard exponential equation $g(x+y) = g(x)g(y)$.

In mathematics, when something looks 'wrong,' it is usually an invitation to transform it. Let us define a new function $g(x) = \frac{f(x)}{5}$.

By rewriting $f(x)$ as $5g(x)$, we can substitute this into our original equation:

$$5(5g(x+y)) = (5g(x))(5g(y))$$

Suddenly, the equation breathes. The left side becomes $25g(x+y)$, and the right side becomes $25g(x)g(y)$. Dividing both sides by $25$, we arrive at the elegant, standard form:

$$g(x+y) = g(x)g(y)$$

### Phase 2: Unlocking the Exponential Form

We know that the general solution to $g(x+y) = g(x)g(y)$ is $g(x) = a^x$. This is a fundamental result in functional analysis.

Since we defined $f(x) = 5g(x)$, our original function must take the form $f(x) = 5 \cdot a^x$. We have successfully stripped away the complexity to reveal the core structure of the function.

Now, we use the boundary condition provided: $f(3) = 320$. Substituting $x=3$ into our derived form, we get:

$$5 \cdot a^3 = 320$$

$$a^3 = \frac{320}{5} = 64$$

$$a = 4$$

With $a=4$, our function is fully defined: $f(x) = 5 \cdot 4^x$. We have conquered the functional equation.

### Phase 3: The Summation Trap

The final task is to evaluate the sum $S = \sum_{n=0}^5 f(n)$. Substituting our function, we get:

$$S = \sum_{n=0}^5 5 \cdot 4^n = 5 \sum_{n=0}^5 4^n$$

Here is where many students stumble. We are summing from $n=0$ to $n=5$. This is a geometric progression with $6$ terms, not $5$.

The series is $4^0 + 4^1 + 4^2 + 4^3 + 4^4 + 4^5$. Using the sum formula for a G.P., $S_n = a \frac{r^n - 1}{r - 1}$, where $a=1$, $r=4$, and the number of terms is $6$:

$$S = 5 \cdot \frac{1(4^6 - 1)}{4 - 1}$$

Calculating $4^6$ gives us $4096$. Thus, the sum becomes:

$$S = 5 \cdot \frac{4096 - 1}{3} = 5 \cdot \frac{4095}{3}$$

Dividing $4095$ by $3$ yields $1365$. Finally, $5 	imes 1365 = 6825$.

The elegance of the result is satisfying—a perfect conclusion to a rigorous journey. The final answer is **6825**.

Animated Solution for Mathematics - Functions: Suppose $f : R \to (0, \infty)$ be a differentiable function such that $5 f (x + y) = f (x) \cdot f (y), \forall x, y \in R$ . If $f (3) = 320$ , then $\sum_{n = 0}^{5} f (n)$ is equal to :

Select Answer:

Visualized Solution

Analyze the Functional Equation

Normalize the Function

Substitute $g (x)$ into the Equation

Simplify to Standard Form

Identify the General Solution

Apply the Boundary Condition

Solve for the Base $a$

Set up the Summation

Expand the Series

Apply the GP Sum Formula

Evaluate and Final Calculation

The Sigma Insight: Classification of Functions

The Riddle of the Functional Equation

Phase 1

Phase 2

Phase 3

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