Problem 34 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f:[0,1] \rightarrow R\) be such that \(f(x y)=f(x) \cdot f(y)\), for all \(x, y \in[0,1]\), and \(f(0) \neq 0\). If \(y=y(x)\) satisfies the differential equation, \(\frac{d y}{d x}=f(x)\) with \(y(0)=1\), then \(y\left(\frac{1}{4}\right)+y\left(\frac{3}{4}\right)\) equal to: (a) 3 (b) 4 (c) 2 (d) 5

Step-by-Step Solution

Answer

The answer is (a) 3.

Step 1: Understand the given condition of f

The function \( f:[0,1] \rightarrow \mathbb{R} \) satisfies the functional equation: \( f(xy)=f(x) \cdot f(y) \). This implies that \( f \) is likely an exponential function. Also, it is given that \( f(0) eq 0 \). Using the property, substitute \( y=0 \), we get \( f(x \/cdot 0)=f(x) \cdot f(0) \). Therefore, \( f(0)=f(0) \cdot f(x) \), indicating that \( f(x) = 1 \). We confirm that \( f(x)=1 \) for all \( x \) because it satisfies \( f(xy) = f(x)f(y) \).

Step 2: Solve the differential equation

Now, solve the differential equation \( \frac{dy}{dx} = f(x) = 1 \) with the initial condition \( y(0) = 1 \). The equation simplifies to \( \frac{dy}{dx} = 1 \), the solution to which is \( y(x) = x + C \), where \( C \) is a constant. Using the initial condition \( y(0) = 1 \), we find \( C = 1 \). Thus, \( y(x) = x + 1 \).

Step 3: Evaluate the function y at specific x values

We need to calculate \( y\left(\frac{1}{4}\right) \) and \( y\left(\frac{3}{4}\right) \). Using \( y(x) = x + 1 \), we find: 1. \( y\left(\frac{1}{4}\right) = \frac{1}{4} + 1 = \frac{5}{4} \). 2. \( y\left(\frac{3}{4}\right) = \frac{3}{4} + 1 = \frac{7}{4} \).

Step 4: Find the sum of the function values

Add the values obtained from Step 3: \( y\left(\frac{1}{4}\right) + y\left(\frac{3}{4}\right) = \frac{5}{4} + \frac{7}{4} = \frac{12}{4} = 3 \).

Key Concepts

Differential Equations

Differential equations are a crucial concept in mathematics that helps us understand how quantities change over time or space. They typically involve an unknown function and its derivatives. In this exercise, our differential equation is given as \( \frac{d y}{d x} = f(x) \). The left side represents the rate of change of \( y \) with respect to \( x \), while the right side is the function \( f(x) \). This equation shows that the rate at which \( y \) changes is directly governed by the function \( f(x) \).
One of the main tasks in dealing with a differential equation is to solve it, meaning to find the function \( y(x) \). Solving a differential equation typically involves integration or other techniques which allow us to express \( y \) in terms of \( x \). Here, since \( f(x) = 1 \), this simplifies the equation to \( \frac{d y}{d x} = 1 \), making it easy to integrate. Integrating both sides leads us to \( y = x + C \), with \( C \) being an integration constant.
Differential equations are everywhere in science and engineering because they model a wide range of phenomena, from how heat diffuses through a material to how populations grow.

Exponential Function

Exponential functions are powerful mathematical tools that describe processes where a quantity grows or decays at a rate proportional to its current value. In this exercise, instead of directly dealing with an exponential function that we expect from \( f(xy) = f(x) \cdot f(y) \), we deduced that \( f(x) = 1 \), which is still a form of an exponential function since it can be seen as \( 1^x \).
Such functional equations, where a function multiplied by another equals the same function values multiplied, often hint at an exponential nature. Exponential functions are significant in many disciplines because they model processes such as population growth, radioactive decay, and interest in finance.
Understanding exponential functions and their properties is critical, as they offer insights into how processes unfold over time, as showcased by their widespread presence and utility in different scientific fields.

Initial Conditions

Initial conditions are initial values provided to solve differential equations uniquely. They specify the situation of the function at the beginning and are necessary to find a specific solution rather than a whole family of solutions. In this context, the initial condition given is \( y(0) = 1 \).
To solve our differential equation \( \frac{d y}{d x} = 1 \) using this initial condition, we integrate to get \( y = x + C \). Using \( y(0) = 1 \) helps us find the constant \( C \). By substituting, we have \( 1 = 0 + C \,\), giving \( C = 1 \). Thus, the particular solution is \( y(x) = x + 1 \).
Having the right initial conditions is crucial as it allows for the accurate prediction or modeling of future behavior based on known starting points. This is highly important in physics and engineering, where systems often depend heavily on their initial states.