Problem 52 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f: R \rightarrow R\) be a function such that \(f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}, f(0)=0\) and \(f^{\prime}(0)=3\) Then, (A) \(f(x)\) is a quadratic function (B) \(f(x)\) is continuous but not differentiable (C) \(f(x)\) is differentiable in \(R\) (D) \(f(x)\) is bounded in \(R\)

Step-by-Step Solution

Answer

(C) \( f(x) \) is differentiable in \( R \).

Step 1: Understanding the Functional Equation

The given functional equation is \( f\left(\frac{x+y}{3}\right) = \frac{f(x) + f(y)}{3} \). We are looking for a function with this property along with \( f(0) = 0 \) and \( f'(0) = 3 \). Apply the substitution \( x = y = 0 \) to show that \( f(0) = 0 \) is consistent.

Step 2: Trying Common Functions

Consider simple functions like linear (\( f(x) = cx \)) and quadratic functions (\( f(x) = ax^2 + bx + c \)). For a linear function \( f(x) = cx \), the given equation would hold if \( 3cx = c(x+y) \), which is true, thus linear functions satisfy the equation. Also, it needs further checking with other types.

Step 3: Checking Differentiability

For arbitrarily many substitutions in the functional equation, checking differentiability would mean checking if derivative conditions are consistent with \( f'(0)=3 \). For linear case \( f(x)= 3x \), since it is globally linear, it is infinitely differentiable, satisfying differentiability requirements in \( R \).

Step 4: Verifying with Calculus

Assuming \( f(x) = 3x \), check \( f^{\prime}(0) = 3 \) and \( f^{\prime}(x) = 3 \) for \( x eq 0\) ensuring differentiability across \( R \). Since this derivative is uniform, \( f(x) = 3x \) is linear, so not quadratic or bounded across \( R \).

Step 5: Conclusion

From these checks, the function \( 3x \) satisfies all conditions: continuity, differentiability, but not quadratic or bounded. Thus, the correct choice is \( C \): \( f(x) \) is differentiable in \( R \).

Key Concepts

Functional Equations

Functional equations involve finding functions that satisfy certain conditions or equations. In this exercise, we have a specific functional equation: \( f\left(\frac{x+y}{3}\right) = \frac{f(x) + f(y)}{3} \). This means that the function behaves in a symmetrically balanced way when inputs are averaged. This type of problem often requires testing simple functions like linear or quadratic functions to see if they satisfy the equation.

For this problem, by substituting different values of \(x\) and \(y\), we can deduce potential forms of \( f(x) \). One common approach to solving functional equations is to try basic function forms such as constant, linear, and quadratic. Here, substituting \( x = y = 0 \) verifies that \( f(0) = 0 \), which is consistent with the functional equation and given conditions.

By exploring these basic function forms, we may find that a linear function like \( f(x) = 3x \) satisfies all the given conditions of the functional equation along with being consistent with the differentiability condition \( f'(0) = 3 \). Thus, understanding and manipulating functional equations is key to solving such problems.

Continuity in Calculus

Continuity is a fundamental concept in calculus that describes how a function behaves without any interruptions. A function \( f(x) \) is continuous at a point \( a \) if the limit of \( f(x) \) as \( x \) approaches \( a \) is equal to \( f(a) \). This means you can draw \( f(x) \) at that point without lifting your pencil from the paper.

In the context of this exercise, we explore continuity as part of verifying the properties of the function suggested by the functional equation. The test function \( f(x) = 3x \) is indeed continuous everywhere because it is a linear function. Linear functions are inherently continuous, as they do not have any sudden jumps or breaks at any point on the real number line.

Being continuous is crucial for examining differentiability, as a function must be continuous to be differentiable. Thus, understanding continuity helps us confirm the behavior of \( f(x) \) over its entire domain.

Derivatives

A derivative represents the rate of change of a function concerning its input. For a function \( f(x) \), the derivative, denoted \( f'(x) \), describes how \( f(x) \) changes as \( x \) changes.

In the given problem, we know \( f'(0) = 3 \), indicating the instantaneous rate of change of \( f(x) \) at \( x = 0 \) is 3. This provides a crucial piece of information about the linear behavior of the function around this point.

Derivatives help confirm not only the linear nature of our function but also affirm differentiability across the domain. Since \( f(x) = 3x \) is linear everywhere, its derivative \( f'(x) = 3 \) for all \( x \), demonstrating that \( f(x) \) is not only continuous but also differentiable across all real numbers. Differentiability implies that the slope of the tangent line is defined everywhere, which is a key characteristic of the function.

Quadratic Functions

Quadratic functions take the general form \( f(x) = ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants. These functions typically have a characteristic parabolic shape and are important in understanding different levels of polynomial equations.

In this exercise, we needed to test whether \( f(x) \) could potentially be a quadratic function. Based on the functional equation and conditions given, particularly \( f'(0) = 3 \), exploring a quadratic function like \( ax^2 + bx + c \) can help us test compatibility.

Ultimately, we find that \( f(x) = 3x \) fits the functional equation without requiring any quadratic component \( (ax^2) \). Despite the possibility, the equation's restrictions and the calculated derivative imply a linear solution rather than a quadratic one. This means that for this particular function, introducing a quadratic term is unnecessary.