Let f(x) be a real differentiable function such that f(0)=1 and f(x+y)=f(x)f'(y)+f'(x)f(y) for all x

Q: Wait, why do we differentiate with respect to $y$ and not $x$?

That is a fantastic question! You could actually differentiate with respect to $x$ as well, and you would get the same result. The choice is arbitrary. We chose $y$ simply to follow the structure of the equation $f(x+y)$. The goal is to isolate $f'(x)$ by setting $y=0$ after differentiation, which works regardless of which variable you choose to differentiate.

Q: Is $f''(0)$ really a constant? How do we know it doesn't depend on $x$?

Yes, it is absolutely a constant! Remember, $f(x)$ is a specific, fixed function. Therefore, $f''(x)$ is also a fixed function. When we evaluate this function at a specific point, like $x=0$, the result $f''(0)$ is just a single, unchanging number. It does not vary with $x$.

Q: Why can't we just guess $f(x) = e^{ax}$ from the start?

While guessing can sometimes lead to the right answer in a competition setting, it is dangerous! In JEE Advanced, you must prove the form of the function. By deriving the differential equation $f'(x) = k f(x)$, you are not just guessing; you are proving that the function *must* be an exponential. This rigorous approach is what separates a lucky guesser from a true physicist.

Question

Let $f(x)$ be a real differentiable function such that $f(0)=1$ and $f(x+y)=f(x)f^{\prime}(y)+f^{\prime}(x)f(y)$ for all $x, y\in R$. Then $\sum_{n=1}^{100}\log_{e}f(n)$ is equal to:

Options:

(A) 2525

(B) 5220

(C) 2384

(D) 2406

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to the beautiful world of functional equations. Today, we are not just solving a problem; we are uncovering the hidden identity of a function.

We are given a real differentiable function $f(x)$ that satisfies the elegant relationship $f(x+y) = f(x)f'(y) + f'(x)f(y)$, with the initial condition $f(0)=1$. Our mission is to find $f(x)$ and then evaluate the sum $\sum_{n=1}^{100} \log_e f(n)$.

### The Detective Work

Every great mystery begins with a single clue. Here, our clue is the initial condition $f(0)=1$.

In functional equations, whenever you see a condition at zero, your instinct should be to test the equation at that point. Let us substitute $x=0$ and $y=0$ into our given equation:

$$f(0+0) = f(0)f'(0) + f'(0)f(0)$$

This simplifies to $f(0) = 2f(0)f'(0)$. Since we know $f(0)=1$, we can substitute this value in:

$$1 = 2(1)f'(0) \Rightarrow f'(0) = \frac{1}{2}$$

We have our first breakthrough! We now know the slope of the function at the origin is exactly $\frac{1}{2}$.

### The Transformation

Now, we need to find the general form of $f(x)$. To do this, we must transform our functional equation into a differential equation.

We differentiate the original equation $f(x+y) = f(x)f'(y) + f'(x)f(y)$ with respect to $y$, treating $x$ as a constant. Using the chain rule on the left and the product rule on the right, we get:

$$f'(x+y) = f(x)f''(y) + f'(x)f'(y)$$

This looks more complex, but watch what happens when we set $y=0$. The variable $y$ vanishes, leaving us with an equation purely in terms of $x$:

$$f'(x) = f(x)f''(0) + f'(x)f'(0)$$

Substituting our known value $f'(0) = \frac{1}{2}$, we obtain:

$$f'(x) = f(x)f''(0) + \frac{1}{2}f'(x)$$

Rearranging the terms to group $f'(x)$, we find:

$$\frac{1}{2}f'(x) = f(x)f''(0) \Rightarrow f'(x) = [2f''(0)]f(x)$$

### The Revelation

We have arrived at a classic first-order linear differential equation: $f'(x) = k f(x)$, where $k = 2f''(0)$. The solution to this equation is well-known to be $f(x) = C e^{kx}$.

Using our initial condition $f(0)=1$, we find $C=1$. To find $k$, we differentiate $f(x) = e^{kx}$ to get $f'(x) = k e^{kx}$.

At $x=0$, $f'(0) = k = \frac{1}{2}$. Thus, our mystery function is revealed: **$f(x) = e^{x/2}$**.

### The Final Summation

Now, we turn to the final objective: $\sum_{n=1}^{100} \log_e f(n)$. Substituting our function $f(n) = e^{n/2}$, the expression becomes:

$$\log_e(e^{n/2}) = \frac{n}{2}$$

Our massive sum simplifies to:

$$\sum_{n=1}^{100} \frac{n}{2} = \frac{1}{2} \sum_{n=1}^{100} n$$

Using the sum formula $\sum_{i=1}^N i = \frac{N(N+1)}{2}$ with $N=100$, we get:

$$\frac{1}{2} 	imes \left( \frac{100 	imes 101}{2} ight) = \frac{5050}{2} = 2525$$

We have successfully decoded the function and conquered the sum. The final answer is **2525**.

Animated Solution for Mathematics - Differential Equations: Let $f (x)$ be a real differentiable function such that $f (0) = 1$ and $f (x + y) = f (x) f^{'} (y) + f^{'} (x) f (y)$ for all $x, y \in R$ . Then $\sum_{n = 1}^{100} lo g_{e} f (n)$ is equal to:

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Visualized Solution

Understanding the Functional Equation

Finding $f^{'} (0)$ - Substitution

Calculating the Value of $f^{'} (0)$

Differentiating with Respect to $y$

Setting $y = 0$ to Form an Equation

Simplifying to a Differential Equation

Defining the Constant $k$

Solving the Differential Equation

Finding the Constant $k$

Evaluating $lo g_{e} f (n)$

Setting Up the Summation

Calculating the Final Sum

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Detective Work

The Transformation

The Revelation

The Final Summation

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Visualized Solution

Understanding the Functional Equation

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(0) - Substitution

Calculating the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(0)

Differentiating with Respect to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Setting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=0 to Form an Equation