Problem 110 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Assertion: Let \(f(x+y)=f(x) f(y)\) for all \(x, y\), where \(f(0) \neq 0 .\) If \(f^{\prime}(0)=2\), then \(f(x)=A e^{2 x}\), where \(A\) is a constant. Reason: \(f^{\prime}(x)=f(x)\)

Step-by-Step Solution

Answer

The function is \( f(x) = e^{2x} \).

Step 1: Use the given functional equation

Start with the functional equation given: \( f(x+y) = f(x) f(y) \). By substituting \( y = 0 \), we have \( f(x+0) = f(x)f(0) \), which simplifies to \( f(x) = f(x)f(0) \). This implies that \( f(0) = 1 \) (since \( f(0) eq 0 \) by the problem statement).

Step 2: Differentiate the functional equation

Consider differentiating the given equation \( f(x+y) = f(x) f(y) \) with respect to \( y \) at \( y = 0 \). This gives: \( \frac{d}{dy}f(x+y)|_{y=0} = \frac{d}{dy}[f(x)f(y)]|_{y=0} \). The left-hand side becomes \( f'(x) \). The right-hand side differentiates as \( f(x)f'(0) \). Thus, we have the equation \( f'(x) = f(x)f'(0) \).

Step 3: Solve the differential equation

Since \( f'(x) = f(x) f'(0) \), the solution is a first-order linear differential equation of the form \( \frac{f'(x)}{f(x)} = f'(0) \). Integrate both sides with respect to \( x \): \( \int \frac{1}{f(x)} df(x) = \int f'(0) dx \), resulting in \( \ln |f(x)| = f'(0)x + C \), where \( C \) is the integration constant. Exponentiate to find \( |f(x)| = e^{f'(0)x + C} \). Since \( f(x) \) is never zero, \( f(x) = Ae^{f'(0)x} \) where \( A = e^C \).

Step 4: Use initial condition to find specific solution

We know that \( f'(0) = 2 \), so substituting this value into the expression \( f(x) = A e^{f'(0)x} \), we have \( f(x) = A e^{2x} \). Since \( f(0) = 1 \), substitute \( x = 0 \) into \( f(x) = A e^{2x} \) to get \( 1 = A e^{0} = A \). Therefore, \( A = 1 \).

Step 5: Verify the assertion

The assertion claims \( f(x) = A e^{2x} \), where \( A = 1 \). Therefore, \( f(x) = e^{2x} \), which matches with the solution from the functional equation and differential equation we solved. Thus, the assertion \( f(x) = e^{2x} \) is correct.

Key Concepts

Differential Equations

Differential equations are a fundamental concept in mathematics and are used to describe the relationship between a function and its derivatives. They often model real-world systems that involve rates of change. In this problem, we were given the equation \( f'(x) = f(x)f'(0) \), which represents a differential equation.

• This differential equation is of first order, meaning it involves the first derivative of the function.
• The equation directly relates the rate of change of \( f(x) \) to its value, highlighting a significant property of exponential functions.
• Solving this differential equation involves integrating both sides, a common technique in calculus that simplifies the differentiation back to an equation for \( f(x) \).

To solve, we integrate both sides with respect to \( x \):\[ \int \frac{1}{f(x)} df(x) = \int f'(0) \, dx\]This form allows us to eventually solve for \( f(x) \) in a closed equation, linking its behavior to its sliding rates of change by constructing a map of how \( f(x) \) will evolve over any domain of \( x \). By understanding this approach, we can model complex systems ranging from population growth to thermal dynamics.

Exponential Functions

Exponential functions are a critical part of our problem's solution since \( f(x) = Ae^{2x} \) is an exponential equation. An exponential function can be generally expressed in the form \( f(x) = a e^{bx} \), where \( a \) is a constant multiplier, \( e \) is the base of natural logarithms (approximately 2.718), and \( b \) is a constant that affects the rate of growth or decay.

• Exponential functions are characterized by their constant percentage rate of change, which makes them appear in various fields like finance for inflation modeling, biology for population growth, and physics for radioactive decay.
• By understanding \( b \) as the rate at which the function grows or shrinks, we gain insight into the speed of changes that \( f(x) \) undergoes.
• In our solution, knowing \( f'(0) = 2 \) guided us to determine the function's rate, leading to the exponential term \( e^{2x} \), showing a consistent doubling effect outside typical linear growth models.

This demonstrates why exponential functions are a powerful tool in modeling changes that escalate or decline at consistent proportional rates rather than fixed amounts.

Calculus Differentiation

Calculus differentiation is the process of finding the derivative of a function, revealing how a function changes at any point. This is a key tool in understanding the behavior of functions like in our problem here.

• The derivative \( f'(x) \) shows the instantaneous rate of change of \( f(x) \) and gives us insights into the behavior of \( f(x) \) like its increasing or decreasing nature, and the slope of a curve at any point.
• For this exercise, differentiating the given functional equation \( f(x+y) = f(x) f(y) \) allowed us to derive \( f'(x) = f(x)f'(0) \), which gives a specific form due to the nature of the problem.
• This differentiation step is crucial as it relates to the structure of the original functional form to a more usable form, especially when tackling differential equations.

Through differentiation, we translated a functional relationship into a differential one, simplifying it, which made solving the problem possible. Understanding and applying differentiation is essential for effectively analyzing and solving real-world problems involving change.