If f is a real valued differentiable function satisfying |f(x) - f(y)|

Q: Wait, why can we just divide by $|x - y|$? What if $x = y$?

That is a brilliant observation! You are absolutely right to be cautious. When $x = y$, the inequality becomes $0 \leq 0$, which is always true and gives us no information. However, for the derivative definition, we are interested in the limit as $y$ approaches $x$, meaning we consider $x \neq y$ in the neighborhood of $x$. We are analyzing the behavior of the function as we get infinitely close to a point, not at the point itself.

Q: Why does $|f'(x)| \leq 0$ imply that $f'(x) = 0$?

Remember the fundamental property of the absolute value function: $|a| \geq 0$ for any real number $a$. If we have a condition that says $|f'(x)| \leq 0$ AND we know $|f'(x)| \geq 0$, the only possible value that satisfies both conditions simultaneously is zero. It is a mathematical sandwich!

Q: Can I just assume $f(x) = 0$ at the start and check if it works?

In an exam setting, checking the answer is a great strategy! If you test $f(x) = 0$, you see that $|0 - 0| \leq (x - y)^2$, which is $0 \leq (x - y)^2$, which is true. However, for JEE Advanced, you must prove that this is the *only* possible function. Our derivation using the derivative definition provides that rigorous proof.

Question

If $f$ is a real valued differentiable function satisfying $|f(x) - f(y)| \leq (x - y)^2, x, y \in R$ and $f(0) = 0$, then $f(1)$ equals

Options:

(A) -1

(B) 0

(C) 2

(D) 1

Sigma Percentile · Accepted Answer

### The Beauty of Constraints: Unlocking the Functional Equation

Welcome, future engineers! Today, we are going to peel back the layers of a problem that, at first glance, might look like a daunting functional equation. By the end of this journey, you will see it for what it truly is: a beautiful, elegant application of the fundamental definition of a derivative.

### Phase 1: The Geometric Intuition

We are given the inequality $|f(x) - f(y)| \leq (x - y)^2$. This inequality tells us that the change in the function value is bounded by the square of the change in the input.

If you consider the slope of a secant line, defined as:

$$ \left| \frac{f(x) - f(y)}{x - y} ight| $$

This inequality suggests that as the points $x$ and $y$ get closer together, the slope must approach zero. This observation is the key to the entire problem.

### Phase 2: The Bridge to Calculus

To turn this algebraic constraint into a solvable form, we invoke the definition of the derivative. Recall the first principle:

$$ f'(x) = \lim_{h 	o 0} \frac{f(x + h) - f(x)}{h} $$

Let us perform a substitution by setting $y = x + h$, which implies $x - y = -h$. Substituting this into our original inequality, we obtain:

$$ |f(x) - f(x + h)| \leq (-h)^2 $$

Since the square of a negative is positive, this simplifies to $|f(x + h) - f(x)| \leq h^2$.

### Phase 3: The Squeeze

We are now close to the definition of the derivative. To introduce the denominator $h$, we divide both sides of the inequality by $|h|$ (assuming $h 
eq 0$):

$$ \left| \frac{f(x + h) - f(x)}{h} ight| \leq |h| $$

As $h$ approaches zero, the left side becomes the absolute value of the derivative, $|f'(x)|$. On the right side, $|h|$ approaches zero. By the **Squeeze Theorem**, we are forced to conclude that $|f'(x)| \leq 0$.

### Phase 4: The Final Revelation

We know that for any real number, the absolute value is always non-negative, meaning $|f'(x)| \geq 0$. However, our derived inequality states $|f'(x)| \leq 0$.

The only way for a number to be both greater than or equal to zero and less than or equal to zero is if it is exactly zero. Thus, we conclude:

$$ f'(x) = 0 $$

A function whose derivative is zero everywhere must be a constant function, $f(x) = C$. Using the initial condition $f(0) = 0$, we find that $C = 0$.

Therefore, **$f(x) = 0$** for all $x$. Evaluating this at $x = 1$ gives us the final result:

$$ **f(1) = 0** $$

You see? We did not need complex machinery. We simply listened to what the inequality was telling us about the slope. Keep practicing this conceptual thinking, and you will master the JEE!

Animated Solution for Mathematics - Differentiation: If $f$ is a real valued differentiable function satisfying $∣ f (x) - f (y) ∣ \leq (x - y)^{2}, x, y \in R$ and $f (0) = 0$ , then $f (1)$ equals

Select Answer:

Visualized Solution

Analyze the Given Constraint

The Derivative Definition

Setting up the Increment

Simplifying the Inequality

Dividing by $∣ h ∣$

Applying the Limit

Squeeze Theorem Result

Concluding the Derivative

Finding the Constant

The Exact Function

Evaluating $f (1)$

The Sigma Insight: Differentiability of a Function

The Beauty of Constraints

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $f$ be any function defined on $R$ and let it satisfy the condition : $∣ f (x) - f (y) ∣ \leq ∣ (x - y)^{2} ∣, \forall (x, y) \in R$ . If $f (0) = 1$ , then:

Let $f$ be a differentiable function from $R$ to $R$ such that $∣ f (x) - f (y) ∣ \leq 2∣ x - y ∣^{3/2}$ , for all $x, y \in R$ . If $f (0) = 1$ then $\int_{0}^{1} f^{2} (x) d x$ is equal to

Let $R$ be the set of real numbers and $f : R \to R$ be such that for all $x$ and $y$ in $R$ $∣ f (x) - f (y) ∣ \leq ∣ x - y ∣^{2}$ . Prove that $f (x)$ is a constant.

The function given by $y = ∣∣ x ∣ - 1∣$ is differentiable for all real numbers except the points

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

Let $f : R \to R$ satisfy the equation $f (x + y) = f (x) \cdot f (y)$ for all $x, y \in R$ and $f (x) \neq = 0$ for any $x \in R$ . If the function $f$ is differentiable at $x = 0$ and $f^{'} (0) = 3$ , then $lim_{h \to 0} \frac{1}{h} (f (h) - 1)$ is equal to ___

Let $f : R \to R$ be a function defined by $f (x) = max {x, x^{2}}$ . Let $S$ denote the set of all points in $R$ , where $f$ is not differentiable. Then :

Let $f (x)$ be a function satisfying the condition $f (- x) = f (x)$ for all real $x$ . If $f^{'} (0)$ exists, find its value.

The function $f (x) = (x^{2} - 1) ∣ x^{2} - 3 x + 2∣ + cos (∣ x ∣)$ is NOT differentiable at

Let $h (x) = min {x, x^{2}}$ , for every real number of $x$ , Then