Problem 77 - A Complete Resource Book in Mathematics for JEE Main

Question

A function \(f:(0, \infty) \rightarrow R\) satisfies the equation \(f(x y)=2 f(x)-f\left(\frac{x}{y}\right)\) If \(f\) is differentiable on \(R\) and \(f(1)=0, f^{\prime}(1)=1\), then (A) \(f(y)=-f\left(\frac{1}{y}\right)\) (B) \(f^{\prime}(x)=\frac{1}{x}\) (C) \(f(x)=\ln x\) (D) \(f(x)=e^{x}\)

Step-by-Step Solution

Answer

(A), (B), and (C) are true.

Step 1: Verify the Function's Initial Condition

The problem statement provides that \(f(1) = 0\) and \(f'(1) = 1\). We'll need to verify that these hold for the solutions we consider. With the given information, if we list potential functions like \(f(x) = \ln x\), applying \(x = 1\) gives \(\ln 1 = 0\), which satisfies \(f(1) = 0\). For differentiability, \(f'(x) = 1/x\) at \(x=1\) gives \(f'(1) = 1\) which holds true.

Step 2: Substitute and Simplify Given Equation

Substitute \(f(x) = \ln x\) into the functional equation: \(f(xy) = \ln(xy) = \ln x + \ln y\). The equation provided is \(f(xy) = 2f(x) - f\left(\frac{x}{y}\right)\). Replace all \(f\) with \(\ln\), \(\ln(xy) = 2\ln x - \ln\left(\frac{x}{y}\right)\). Right hand side simplifies to \(\ln(x^2) - \ln(x/y) = \ln(x^2) - (\ln x - \ln y) = \ln(xy)\). Both are equal verifying \(f(x) = \ln x\) satisfies the equation.

Step 3: Check Option (A)

Option (A) is \(f(y) = -f\left(\frac{1}{y}\right)\). For \(f(x) = \ln x\), check \(\ln y\) and \(-\ln(1/y) = \ln y\). Both sides yield \(\ln y\), so (A) is true.

Step 4: Check Option (B)

Option (B) is \(f'(x) = \frac{1}{x}\). If \(f(x) = \ln x\), derivative \(f'(x) = 1/x\). This exactly matches the statement in (B), so (B) is true.

Step 5: Check Option (C)

Option (C) is \(f(x) = \ln x\). Substitution from Step 1 verifies this function satisfies the initial equality and differentiability conditions. Thus, (C) is true.

Step 6: Check Option (D)

Option (D) is \(f(x) = e^x\). Substituting this in the functional equation yields inconsistent results as \(e^{xy} eq 2e^x - e^{x/y}\) for all positive \(x, y\). Thus, (D) is false.

Key Concepts

Differentiable Functions

A differentiable function is a function that has a derivative at each point in its domain. In simpler terms, this means the function is smooth and has a defined slope everywhere within its domain. This concept is crucial in calculus because it allows us to find instantaneous rates of change and draw tangents to the function graph.

For a function to be differentiable, it must not have any sharp turns or discontinuities. The exercise involves a differentiable function, indicating that its behavior is predictable and well-understood using differentiation techniques.

In the given exercise, the function \(f(x) = \ln x\) serves as an example of a differentiable function. The derivative of \(\ln x\) is \(f'(x) = \frac{1}{x}\), and assessing differentiability verified the initial conditions provided.

Initial Conditions

Initial conditions are specific values provided for a function at particular points, which help in finding unique solutions to problems. In our exercise, two initial conditions are given: \(f(1) = 0\) and \(f'(1) = 1\). These conditions play a crucial role in determining the proper function that satisfies the given functional equation.

Initial conditions are used to:

• Confirm the validity of a function solution.
• Distinguish among possible general solutions of differential equations.

If a function does not match the assigned initial conditions, it may not be the correct solution to a problem, which is evident when analyzing options in our exercise.

Natural Logarithm

The natural logarithm, denoted as \(\ln x\), is a mathematical function commonly used in calculus and higher mathematics. It is the inverse function of the exponential function \(e^x\). The natural logarithm has the unique property where it transforms multiplication into addition: \(\ln(xy) = \ln x + \ln y\).

In our exercise, this property is essential in simplifying the functional equation \(f(xy) = 2f(x) - f(\frac{x}{y})\). It was verified that \(f(x) = \ln x\) satisfies the equation due to the inherent properties of logarithms. This highlights how logarithmic functions can drastically simplify complex expressions.

Understanding \(\ln x\) is vital, as it frequently appears in calculus problems due to its interesting mathematical properties and relationship to growth processes and decay.

Calculus

Calculus is a branch of mathematics that studies how things change. It consists of two major parts: differentiation and integration. The focus of our exercise lies primarily in the differentiation part, examining how functions change over intervals and specific points.

To solve the exercise provided, you must utilize calculus tools:

• Find derivatives to verify differentiability and rate of change, using \(f'(x) = \frac{1}{x}\) as a check.
• Simplify and verify functional equations with understanding of continuous functions like \(\ln x\).

The foundational skills acquired in calculus allow us to verify that only certain functions satisfy the problem's criteria. By leveraging calculus effectively, complex problems can be broken into manageable parts.