Problem 27 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)\) be a differentiable function such that \(f(x y)=f(x)\) \(+f(y)\) for all \(x\) and \(y\), then \(f(e)+f(1 / e)=\) (A) 1 (B) 0 (C) \(-1\) (D) None of these

Step-by-Step Solution

Answer

The answer is (B) 0.

Step 1: Understand the functional equation

The problem gives us the equation \( f(xy) = f(x) + f(y) \) which is a standard form of a functional equation. The function is differentiable, which is additional information that might help us later.

Step 2: Test the case of f(1)

Set \( y = 1 \) in the equation \( f(xy) = f(x) + f(y) \) to get \( f(x \cdot 1) = f(x) + f(1) \), simplifying to \( f(x) = f(x) + f(1) \). This implies that \( f(1) = 0 \).

Step 3: Applying assumption to specific cases

From Step 2, know that \( f(1) = 0 \) and consider \( f(e) + f(1/e) \). Set \( x = e \) and \( y = 1/e \) in the original equation: \( f(e \cdot 1/e) = f(e) + f(1/e) \). This simplifies to \( f(1) = f(e) + f(1/e) \).

Step 4: Substitute into the equation

Since \( f(1) = 0 \) (from Step 2), substitute into the last step: \( 0 = f(e) + f(1/e) \). So, we conclude that \( f(e) + f(1/e) = 0 \).

Key Concepts

Differentiable Functions

In calculus, a differentiable function is one that has a derivative at each point in its domain. This means that the function's graph is smooth, without any sharp corners or discontinuities, allowing us to compute a slope at any point. Differentiability is a stronger condition than continuity. A function can be continuous without being differentiable, but if a function is differentiable, it must be continuous as well.

For the given problem, knowing that the function \( f(x) \) is differentiable adds depth to the functional equation \( f(xy) = f(x) + f(y) \). Differentiability implies certain properties, like the existence of a tangent line at any point, which can provide us with additional tools, such as using derivatives, to analyze the behavior of \( f(x) \) and solve the equation more effectively. Understanding differentiability helps to confirm that the function behaves well under calculus operations.

Functional Equation Solutions

Functional equations involve finding functions that satisfy a given form of equation under certain conditions. This exercise deals with the functional equation \( f(xy) = f(x) + f(y) \), which is a classic example of Cauchy's additive functional equation. Solutions to such equations can often take simple forms based on underlying properties, such as continuity and differentiability, which are present in our problem.

In solving such equations, we test particular values and discover properties such as \( f(1) = 0 \). This insight comes from substituting specific values into the equation, like setting \( y = 1 \). From there, substituting more complex values, like \( x = e \) and \( y = 1/e \), shows how the function behaves over a wide range of inputs, culminating in the key result: \( f(e) + f(1/e) = 0 \). Solving functional equations often involves creative substitution and a deep understanding of the function's properties generated from the given conditions.

Mathematical Problem-Solving

Solving mathematical problems, particularly those involving equations and abstract functions, involves a systematic approach. The problem-solving process begins with understanding what is being asked and identifying the type of problem—here, a functional equation involving a differentiable function.

A successful strategy for tackling such problems includes:

• Breaking down the problem into smaller, more manageable parts, as demonstrated in the step-by-step solution.
• Using known properties, like setting specific values or using given conditions, to simplify the problem.
• Employing logical deductions to make sense of the information gathered through substitutions or transformations.

In this exercise, each step uncovers more about the function's properties and narrows down the potential behavior of \( f(x) \). Finally, piecing together insights derived from logical deductions and mathematical properties culminates in the solution that \( f(e) + f(1/e) = 0 \). This problem exemplifies the importance of a structured approach in mathematical problem-solving.