Problem 62 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

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 (a) \(-1\) (b) 0 (c) 2 (d) 1

Step-by-Step Solution

Answer

(b) 0

Step 1: Understanding the Problem

We have a differentiable function \( f(x) \) such that \(|f(x) - f(y)| \leq (x-y)^2\) for all real numbers \(x\) and \(y\). The function also satisfies \( f(0) = 0 \) and we need to find \( f(1) \). This indicates that the function does not change much as \( x \) and \( y \) are close, suggesting that \( f(x) \) could be a constant or very restricted in behavior.

Step 2: Applying the Condition

Using the given condition, \(|f(x) - f(y)| \leq (x-y)^2\), set \( y = 0 \). This simplifies our condition to \(|f(x) - f(0)| \leq x^2\). Since \( f(0) = 0 \), we have \(|f(x)| \leq x^2\).

Step 3: Considering x = 1

Now, substituting \(x = 1\) into \(|f(x)| \leq x^2\), we get \(|f(1)| \leq 1^2 = 1\). This means \(f(1)\) can be any value between -1 and 1.

Step 4: Using Derivatives and Differentiability

Since the function is differentiable, consider the mean value theorem: \(\frac{f(x) - f(0)}{x - 0} = f'(c)\) for some \(c\) in \((0, x)\). Then based on our earlier conclusion, \(|f'(c)| < 2|c|\), implying \(f'(c) = 0\) since it must hold for each \(c \in (0, x)\). Thus, \(f(x)\) must be a constant function.

Step 5: Evaluating f(x) Based on Known Values

Given that \(f(x)\) is constant and \(f(0) = 0\), we conclude that \(f(x) = 0\) for all \(x\). Therefore, we have \( f(1) = 0 \).

Key Concepts

Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental concept in calculus. It connects the average rate of change of a function over an interval to the instantaneous rate of change (derivative) at some point within that interval. Specifically, if a function \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there exists at least one point \( c \) in \((a, b)\) for which:\[ f'(c) = \frac{f(b) - f(a)}{b - a}. \] This theorem essentially states that there is some point in the interval where the tangent to the curve has the same slope as the chord connecting the endpoints \( (a, f(a)) \) and \( (b, f(b)) \).

• The MVT helps in establishing relationships that allow for approximations.
• It provides a formal framework for understanding how a function progresses from one point to another.

In the context of the given problem, MVT plays a pivotal role by underlining the behavior of the function's derivative, which led to the conclusion that the function must be constant.

Absolute Value Inequality

Absolute value inequalities are inequalities that involve the absolute value function, which can be thought of as the distance of a number from zero on the number line. The absolute value expression \(|a - b|\leq c\) implies that the value of \((a - b)\) lies within a distance \(c\) from zero. It can be rewritten as two separate inequalities:

• \(a - b \leq c\)
• and \(a - b \geq -c\)

These expressions show that the difference \((a - b)\) is squeezed between \(-c\) and \(c\). In the original exercise, \(|f(x) - f(y)| \leq (x-y)^2\) implies that the fluctuation of the values \(f(x)\) and \(f(y)\) is tightly restrained within a squared distance of \(x\) and \(y\). This tight restriction suggests that \(f\) does not deviate significantly over small intervals, assisting in initially suspecting that \(f\) could exhibit constant behavior or limited variability across any interval.

Constant Functions

A constant function is one of the simplest types of functions, where the output value does not change regardless of the input. Mathematically, for a function \( f(x) = c \) where \( c \) is a constant, the derivative \( f'(x) \) is always zero. In terms of graphs, these functions present as horizontal lines. Given that \( f'(c) = 0 \) for all \( c \) in the set domain, constant functions do not exhibit any change with respect to their variables.

• Constant functions are stable and predictable across their entire domain.
• They have straightforward properties, making them easy to analyze mathematically.

In the exercise at hand, the use of derivatives and the Mean Value Theorem led us to conclude that the function must be constant. Since we knew \( f(0) = 0 \), it followed naturally that \( f(x) = 0 \) for any \( x \), thus \( f(1) = 0 \). This demonstrates how the problem's conditions narrow down the possibilities to a constant solution.