Let f : R - {0} → R be a function satisfying f(x/y) = f(x)/f(y) for all x, y, f(y) ≠ 0. If f'(1) = 2

Q: Wait, why are we allowed to treat $y$ as a constant during differentiation? Isn't $y$ a variable?

This is the beauty of partial differentiation! When we differentiate with respect to $x$, we are essentially asking, 'How does the function change if I only nudge $x$, while keeping everything else frozen?' By treating $y$ as a constant, we are isolating the rate of change relative to $x$. It is a standard, rigorous technique in functional equations to transform an algebraic relation into a differential one. Think of it as taking a 'slice' of the function's behavior.

Q: Why do we set $y = x$ in the final step? Is that just a lucky guess?

It is not a guess; it is a strategic necessity! We have an equation involving $f'(\frac{x}{y})$ and we know the value of $f'(1)$. To bridge the gap between what we have and what we know, we must make the argument of the derivative match. Since $\frac{x}{y} = 1$ when $x = y$, setting $y = x$ is the only logical path to substitute the known value of $2024$ into our equation.

Q: Can't I just assume $f(x) = x^n$ and solve for $n$? It seems much faster.

While that 'guess-and-check' method might work for simple power functions, it is dangerous in JEE Advanced. It assumes the function must be a power function, which isn't always true. The method of partial differentiation we used is universal—it works regardless of the function's form. Always prioritize the rigorous, analytical path over the shortcut; it builds the mathematical intuition you need for the tougher problems.

Question

Let $f : \mathbb{R} - \{0\} 	o \mathbb{R}$ be a function satisfying $f(\frac{x}{y}) = \frac{f(x)}{f(y)}$ for all $x, y, f(y) 
e 0$. If $f'(1) = 2024$, then

Options:

(A) $xf'(x) - 2024f(x) = 0$

(B) $xff'(x) + 2024f(x) = 0$

(C) $xf'(x) + f(x) = 2024$

(D) $xf'(x) - 2023f(x) = 0$

Sigma Percentile · Accepted Answer

### The Symphony of Functional Equations

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are decoding a hidden law of nature.

Functional equations are the 'DNA' of mathematics. They don't tell us what a function *is* explicitly; they tell us how a function *behaves*.

When we see an equation like $f(\frac{x}{y}) = \frac{f(x)}{f(y)}$, we are looking at a fundamental symmetry. Our mission is to translate this symmetry into the language of calculus—the language of change.

### Phase 1: Finding the Anchor

Before we rush into the complex machinery of derivatives, we must find our footing. Every function has a 'base state,' a point of reference. In this problem, that point is $x = 1$.

Imagine we substitute $x = 1$ and $y = 1$ into our given equation:

$$f\left(\frac{1}{1}ight) = \frac{f(1)}{f(1)}$$

Since the problem guarantees $f(y) 
eq 0$, we can safely cancel the terms. This leaves us with the elegant result: $f(1) = 1$.

This is our anchor. It is the solid ground upon which we will build the rest of our derivation. Never underestimate the power of testing simple values; they often reveal the secrets that complex algebra hides.

### Phase 2: The Calculus Hammer

Now, we need to bridge the gap between the function $f(x)$ and its derivative $f'(x)$. We have an algebraic relation, but we need a differential one. The tool for this job is partial differentiation.

We treat $y$ as a constant—a fixed number, like $5$ or $\pi$—and we differentiate both sides of the equation with respect to $x$.

On the left-hand side, we have $f(\frac{x}{y})$. Applying the chain rule, the derivative becomes:

$$\frac{\partial}{\partial x} \left[f\left(\frac{x}{y}ight)ight] = f'\left(\frac{x}{y}ight) \cdot \frac{1}{y}$$

On the right-hand side, since $f(y)$ is treated as a constant, we simply differentiate $f(x)$:

$$\frac{\partial}{\partial x} \left[\frac{f(x)}{f(y)}ight] = \frac{f'(x)}{f(y)}$$

Equating these two, we arrive at our golden key:

$$f'\left(\frac{x}{y}ight) \cdot \frac{1}{y} = \frac{f'(x)}{f(y)}$$

Take a moment to appreciate this. We have successfully transformed a static relationship into a dynamic one. We are now looking at how the rate of change of the function at one point relates to the rate of change at another.

### Phase 3: The Strategic Substitution

We are almost at the finish line. We know that $f'(1) = 2024$. Looking at our equation, we have a term $f'(\frac{x}{y})$.

To utilize our known value, we need the argument $\frac{x}{y}$ to become $1$. This is the 'Aha!' moment. By setting $y = x$, we force the argument to be $1$.

Let us substitute $y = x$ into our equation:

$$f'\left(\frac{x}{x}ight) \cdot \frac{1}{x} = \frac{f'(x)}{f(x)}$$

This simplifies beautifully to:

$$f'(1) \cdot \frac{1}{x} = \frac{f'(x)}{f(x)}$$

Substituting our known value $f'(1) = 2024$, we get:

$$\frac{2024}{x} = \frac{f'(x)}{f(x)}$$

### The Final Elegance

We have arrived at a first-order differential equation. To match the options provided, we simply rearrange the terms. Cross-multiplying gives us:

$$2024 f(x) = x f'(x)$$

Or, moving everything to one side:

**$$x f'(x) - 2024 f(x) = 0$$**

And there it is. We started with a mysterious functional equation and, through the systematic application of calculus, we uncovered the differential law governing the function.

This is the essence of JEE Advanced physics and mathematics: taking a complex, abstract problem and breaking it down into logical, manageable steps. You have mastered the technique; now, go forth and apply this logic to the next challenge!

Animated Solution for Mathematics - Differentiation: Let $f : R - {0} \to R$ be a function satisfying $f (\frac{x}{y}) = \frac{f ( x )}{f ( y )}$ for all $x, y, f (y) \neq = 0$ . If $f^{'} (1) = 2024$ , then

Select Answer:

Visualized Solution

Introduction to the Functional Equation

Finding the Value of $f (1)$

Partial Differentiation Strategy

Applying the Chain Rule

Equating the Derivatives

Strategic Substitution of $y = x$

Simplifying After Substitution

Substituting the Given Derivative Value

Final Rearrangement

The Sigma Insight: Techniques of Differentiation

The Symphony of Functional Equations

Phase 1

Phase 2

Phase 3

The Final Elegance

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R−{0}→R be a function satisfying f(yx​)=f(y)f(x)​ for all x,y,f(y)=0. If f′(1)=2024, then