Problem 76 - A Complete Resource Book in Mathematics for JEE Main

Question

A function \(f: R \rightarrow R\), where \(R\) is the set of real numbers satisfies the equation \(f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)+f(0)}{3}\) for all \(x, y\) in \(R\). If the function \(f\) is differentiable at \(x=\) 0 , then \(f\) is (A) linear (B) quadratic (C) cubic (D) biquadratic

Step-by-Step Solution

Answer

(A) linear

Step 1: Analyze the Functional Equation

The given functional equation is \( f\left(\frac{x+y}{3}\right) = \frac{f(x) + f(y) + f(0)}{3} \). This equation holds for all \(x, y \in \mathbb{R}\). Our goal is to determine the form of \( f \).

Step 2: Test Simple Values

First, substitute \(x = y = 0\) into the equation:\[ f\left(\frac{0 + 0}{3}\right) = \frac{f(0) + f(0) + f(0)}{3} \]This simplifies to \( f(0) = f(0) \), which is trivially true and gives no new information.

Step 3: Find a General Form for f(x)

Let's substitute \( y = 0 \) in the equation:\[ f\left(\frac{x+0}{3}\right) = \frac{f(x) + f(0) + f(0)}{3} \]Simplifying, we get:\[ f\left(\frac{x}{3}\right) = \frac{f(x) + 2f(0)}{3} \]

Step 4: Assume a Form for f(x) and Verify

Assume \(f(x) = ax + b\) where \(f\) is linear. For a linear function, \(f(x) = ax + b\), we have:\[f\left(\frac{x}{3}\right) = a\frac{x}{3} + b\] and from our earlier work:\[\frac{f(x) + 2f(0)}{3} = \frac{ax+b + 2b}{3} = \frac{ax + 3b}{3}\]Equating both results from \(f\left(\frac{x}{3}\right)\):\[a\frac{x}{3} + b = \frac{ax + 3b}{3}\]we see this equation holds. Therefore, \(f(x)\) is likely linear.

Step 5: Use Differentiability Condition

The problem states that \( f \) is differentiable at \( x = 0 \). For a function to be differentiable at a point, it must be linear around that point. Since our candidate \( f(x) = ax + b \) is linear, it aligns with differentiability at \( x = 0 \). Therefore, \( f(x) = ax + b \) is consistent with all given conditions.

Key Concepts

Functional Equations

Functional equations are equations in which the unknowns are functions rather than simple variables. As in our exercise, the goal is to find the specific function that satisfies the given mathematical relationship for all inputs. The functional equation provided is \[ f\left(\frac{x+y}{3}\right) = \frac{f(x) + f(y) + f(0)}{3} \]. When analyzing such equations, one common strategy is to test simple values like zero or perform substitutions that simplify the equation. For example, examining the structure when \(x = y = 0\) yielded no new result because it reduced to a basic truth \(f(0) = f(0)\).

• Functional equations can represent complex relationships like symmetries or periodic patterns.
• Strategies in solving these usually involve substitution, testing simple inputs, and symmetry consideration.

In this exercise, substitutions with known values helped simplify the functional form of \(f(x)\). This strategic substitution leads to insights about the possible forms the function can take, such as linearity.

Real Analysis

Real analysis is the branch of mathematics that deals with the study of real numbers, sequences, series, and functions. It is focused on rigorous proofs and exploring the properties of real-valued functions. In our particular problem, real analysis principles assist in the exploration of whether a function described by a functional equation can be differentiable, and if so, what form it might take. Differentiability, a core idea in real analysis, deals with the function's ability to be represented closely by a linear approximation around some point (e.g., \(x = 0\) in this exercise). Differentiability at a point implies a function can be approximated by a tangent line at that region:

• A function is differentiable at a point if it is smooth and has no sharp changes in direction at that point.
• If a function is differentiable everywhere, it is continuous everywhere. However, continuity does not imply differentiability.

Therefore, the differentiability condition at \(x = 0\) helps indicate that the function could be linear overall, validating the method of assuming \(f(x) = ax + b\) to be the function form.

Linear Functions

Linear functions are among the simplest types of functions, having the form \(f(x) = ax + b\). They produce a straight line when graphed. Recognizing or assuming linearity is a common step in solving functional equations, especially when differentiability is involved. The exercise's functional equation was ultimately shown to suggest linear behavior based on its solution. Key characteristics of linear functions include:

• Constant rate of change: The slope \(a\) is constant, meaning the function changes at a steady rate for each unit increase in \(x\).
• Simplicity and predictability: The graph of a linear function is straightforward, making it easy to understand its behavior.

Given the problem's conditions and differentiability at \(x=0\), it supports the conclusion that \(f(x)\) is indeed linear. This often makes linear functions a preferred assumption when beginning to solve such equations, highlighting the simplicity and elegance of linear solutions.