Problem 89 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f: R \rightarrow R\) be a function satisfying the condition \(f\left(\frac{x+y}{k}\right)=\frac{f(x)+f(y)}{k}\), where \(k \neq 0,2\). The function \(f(x)\) is differentiable on \(R\) and \(f^{\prime}(0)=m\). \(f^{\prime}(x)\) is equal to (A) \(m\) (B) \(2 m\) (C) \(m+1\) (D) 0

Step-by-Step Solution

Answer

The answer is (A) m.

Step 1: Recognize the Functional Equation

The given functional equation is \( f\left(\frac{x+y}{k}\right) = \frac{f(x) + f(y)}{k} \). This is a form of Cauchy's functional equation which is associated with linear functions.

Step 2: Seek a Differentiable Solution Form

Since \( f(x) \) is differentiable, we propose that \( f(x) \) could take the form of a linear function \( f(x) = mx \) based on the differentiability condition and the available information \( f'(0) = m \).

Step 3: Substitute the Linear Function Form

Substitute \( f(x) = mx \) into the functional equation: \[ f\left(\frac{x+y}{k}\right) = m\left(\frac{x+y}{k}\right) \] and \[ \frac{f(x) + f(y)}{k} = \frac{mx + my}{k} = m\left(\frac{x+y}{k}\right) \]. This equality holds, supporting the hypothesis that \( f(x) = mx \).

Step 4: Differentiate the Hypothesized Function

Differentiate \( f(x) = mx \) with respect to \( x \) to find \( f'(x) \). Since \( f(x) = mx \), it follows that \( f'(x) = m \).

Step 5: Conclude the Solution Based on Derivative

Given that \( f'(0) = m \) and the derivative \( f'(x) = m \) holds for all \( x \), the correct choice for \( f'(x) \) across \( R \) is consistent with option (A).

Key Concepts

Differentiability

Differentiability is a fundamental concept in calculus that deals with the smoothness of a function. A function is said to be differentiable at a point if it has a derivative at that point, meaning it can be locally linearized. For a function to be differentiable, it must also be continuous. If a function is differentiable, it implies that small changes in input near that point lead to predictable changes in the output.
In the context of Cauchy's functional equation, differentiability helps in assuming a simple form for the solution, like a linear function. When a function is differentiable across its entire domain, it signifies that we can consistently compute its derivative across that domain. In the given exercise, the differentiability of the function combined with the information about the derivative at zero helps determine the overall form of the function.

Linear Functions

Linear functions are algebraic equations involving terms with at most a first degree. They follow the form of a straight line, expressed as \( f(x) = mx + b \). In particular, when dealing with differentiable functions satisfying functional equations, the assumption of linearity often simplifies the problem.
In the case of Cauchy's functional equation in the exercise, we assume possible solutions are of the linear form due to the differentiability condition provided, \( f'(0) = m \). By postulating \( f(x) = mx \), we confirm this through substitution into the functional equation. The linearity aspect implies that the function's rate of change is constant across its domain, which matches the given condition \( f'(0) = m \).

Functional Equations

Functional equations are equations in which the unknowns are functions, and they involve the values of these functions at some points. Cauchy's functional equation is a classic example where it looks for functions that satisfy specific operations, typically involving linearity, additivity, or homogeneity.
The exercise presents a version of Cauchy's functional equation, \( f\left(\frac{x+y}{k}\right) = \frac{f(x) + f(y)}{k} \). Such equations often lead to linear solutions, especially under conditions like differentiability. Solving them generally involves testing potential forms that satisfy these conditions, leading us to deduce the function's form from the given functional properties. This problem-solving process includes verifying hypothesized forms by substitution.

Mathematical Problem Solving

Mathematical problem solving is the process of finding solutions to mathematical questions and challenges in a structured manner. It involves understanding the problem, exploring possible solution methods, implementing those methods, and verifying the results.
The exercise reflects mathematical problem solving by starting with identifying that we have a functional equation. By understanding concepts of differentiability, linearity, and functional equations, we explore appropriate function classes, like linear functions, that might satisfy both the equation and the differentiability condition \( f'(0) = m \). The problem-solving process ultimately concludes by differentiating the chosen function form \( f(x) = mx \) to confirm that \( f'(x) = m \). This methodical approach helps ensure that the entire domain complies with the derived solution, reaffirming the solution's correctness.