Let f: R → R be a function such that f(x+y) = f(x) + f(y), ∀ x, y ∈ R. If f(x) is differentiable at

Q: Wait, why does differentiability at $x=0$ imply differentiability everywhere?

That is a fantastic question! The functional equation $f(x+y) = f(x) + f(y)$ is a very special structure known as Cauchy's Functional Equation. Because the derivative at any arbitrary point $x$ depends only on the limit of $\frac{f(h)}{h}$ as $h \to 0$, and this limit is fixed by the condition at $x=0$, the derivative $f'(x)$ becomes independent of $x$. Essentially, the 'local' behavior at the origin dictates the 'global' behavior of the entire function.

Q: Can we just assume $f(x) = cx$ from the start?

While your intuition is spot on—the function is indeed linear—in a JEE Advanced context, you must prove it. Simply assuming $f(x) = cx$ is a guess. The rigorous path is to use the First Principle of Derivatives to show that the derivative is constant, which forces the function to be of the form $f(x) = cx$.

Q: Why can't we just use energy conservation or some other physics trick?

This is a pure mathematical problem, not a physics one! However, think of it like a constant velocity problem. If the derivative (the 'velocity' of the function's growth) is constant everywhere, then the 'position' (the function value) must change linearly with time (or $x$). It is the same logic as $v = \frac{dx}{dt} = \text{constant}$.

Question

Let $f: R 	o R$ be a function such that $f(x+y) = f(x) + f(y), \forall x, y \in R$. If $f(x)$ is differentiable at $x=0$, then

Options:

(A) $f(x)$ is differentiable only in a finite interval containing zero

(B) $f(x)$ is continuous $\forall x \in R$

(C) $f'(x)$ is constant $\forall x \in R$

(D) $f(x)$ is differentiable except at finitely many points

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the edge of a vast mathematical landscape. You are presented with a mysterious rule: $f(x+y) = f(x) + f(y)$.

This is the famous **Cauchy Functional Equation**. It tells us that the function respects the structure of addition, meaning if you add two inputs, the outputs add up as well.

### The Origin

Every great journey begins at the start. In the world of functions, that is the origin, $x=0$.

Let us test our rule by setting $x=0$ and $y=0$:
$$f(0+0) = f(0) + f(0)$$

This simplifies to $f(0) = 2f(0)$. Subtracting $f(0)$ from both sides, we find that **$f(0) = 0$**.

This is a profound realization. Our function is anchored at the origin and passes directly through $(0,0)$.

### The First Principle

Now, we bring in the heavy artillery of calculus: the First Principle of Derivatives. We are told that $f(x)$ is differentiable at $x=0$, meaning the slope at the origin exists.

Let us write it down:
$$f'(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h}$$

Since we discovered that $f(0) = 0$, this simplifies to:
$$f'(0) = \lim_{h 	o 0} \frac{f(h)}{h}$$

Let us call this finite limit $L$. This constant $L$ is the 'DNA' of our function; it defines the steepness at the start.

### The Generalization

Now, let us look at any arbitrary point $x$ on the curve. By definition, the derivative is:
$$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$$

Here is where the magic happens. We use our original rule, $f(x+h) = f(x) + f(h)$, and substitute it into the numerator:
$$f'(x) = \lim_{h 	o 0} \frac{f(x) + f(h) - f(x)}{h}$$

Look closely—the $f(x)$ terms cancel out completely. We are left with:
$$f'(x) = \lim_{h 	o 0} \frac{f(h)}{h} = L$$

Therefore, $f'(x) = L$ for every single $x$ in the real number line.

### The Grand Conclusion

We have proven that the derivative of our function is a constant $L$ everywhere. Geometrically, a function with a constant derivative is a straight line.

Since it also passes through the origin, it must be of the form:
**$$f(x) = Lx$$**

Because the derivative exists everywhere, the function is differentiable for all $x \in \mathbb{R}$. As differentiability is a stronger condition than continuity, we have arrived at the truth: **$f(x)$ is continuous for all $x \in \mathbb{R}$**, and its derivative is constant.

You have successfully decoded the mystery of the additive function!

Animated Solution for Mathematics - Differentiation: Let $f : R \to R$ be a function such that $f (x + y) = f (x) + f (y), \forall x, y \in R$ . If $f (x)$ is differentiable at $x = 0$ , then

Select Answer:

Visualized Solution

Understanding the Functional Equation

Finding $f (0)$

Value at Origin

Derivative at $x = 0$

Defining the Constant $L$

Derivative at Arbitrary $x$

Applying the Functional Equation

Simplifying the Limit

Constant Derivative Everywhere

Final Conclusion

The Sigma Insight: Differentiability of a Function

Analyzing the Setup

The Origin

The First Principle

The Generalization

The Grand Conclusion

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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→R be a function such that f(x+y)=f(x)+f(y),∀x,y∈R. If f(x) is differentiable at x=0, then