Problem 38 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(f(x)\) is a differentiable function in the interval \(((0, \infty)\) such that \(\mathrm{f}(\mathrm{a})=1\) and \(\lim _{\mathrm{t} \rightarrow \mathrm{x}} \frac{\mathrm{t}^{2} \mathrm{f}(\mathrm{x})-\mathrm{x}^{2} \mathrm{f}(\mathrm{t})}{\mathrm{t}-\mathrm{x}}=1\), for each \(\mathrm{x}>0\). then \(\mathrm{f}\left(\frac{3}{2}\right)\) is equal to: (a) \(\frac{23}{18}\) (b) \(\frac{13}{6}\) (c) \(\frac{25}{9}\) (d) \(\frac{31}{18}\)

Step-by-Step Solution

Answer

The correct choice for \( f\left(\frac{3}{2}\right) \) is not on the list. Re-evaluate calculations or the function choice.

Step 1: Interpret the Limit Condition

Given the condition \( \lim_{t \rightarrow x} \frac{t^2 f(x) - x^2 f(t)}{t-x} = 1 \), recognize that this expression resembles the definition of the derivative. Specifically, it implies that the derivative of some function involving \( f(x) \) is equal to 1 for each \( x > 0 \).

Step 2: Set up a Suitable Function

Consider the function \( g(t) = t^2 f(t) \). Its derivative according to the limit condition is \( g'(x) = 2x f(x) + x^2 f'(x) = 1 \). This results from setting up the derivative \( g'(x) = \lim_{t \rightarrow x} \frac{g(t) - g(x)}{t-x} \).

Step 3: Solve for f(x) and f'(x) Relationship

From \( 2x f(x) + x^2 f'(x) = 1 \), rearrange to solve for \( f'(x) \):\[ f'(x) = \frac{1 - 2x f(x)}{x^2} \].

Step 4: Use Initial Condition

The problem states \( f(a) = 1 \). Initially, we can use this condition as \( f(x) = 1 \) is assumed for simplicity, giving \( f'(x) = 0 \). This assumption simplifies and helps verify solutions or adjustments based on conditions.

Step 5: Integrate to Find General Form of f(x)

The differential equation obtained should be solved to find the form of \( f(x) \) by integration. First, consider a particular solution where \( f(x) = \frac{c}{x^2} + \frac{1}{2x} \) by simplifying boundary conditions or modifying possible candidates that meet original conditions.

Step 6: Use All Given Conditions to Finalize \( f(x) \)

Given \( f(a) = 1 \), for some specific value (e.g., \( a = 1 \)), solve for constants that hold it true or adjust candidates in the form \( f(x) = \frac{x+2}{2x} \) by satisfying conditions in any interval.

Step 7: Calculate f(\frac{3}{2})

Calculate \( f\left(\frac{3}{2}\right) \) using the determined function form \( f(x) = \frac{x+2}{2x} \). Substituting \( x = \frac{3}{2} \) gives:\[ f\left(\frac{3}{2}\right) = \frac{\frac{3}{2}+2}{2 \cdot \frac{3}{2}} = \frac{\frac{3}{2} + \frac{4}{2}}{3} = \frac{\frac{7}{2}}{3} = \frac{7}{6} \].

Key Concepts

Calculus Limits

Limits are foundational in understanding calculus. They describe the behavior of functions as inputs approach a certain value, essentially predicting the function's output value. In the context of the exercise, the limit is expressed as \( \lim_{t \rightarrow x} \frac{t^2 f(x) - x^2 f(t)}{t-x} = 1 \). This sort of expression is not just any limit, but one used to find derivatives, offering deep insights into the function's behavior near specific points.To better comprehend this, consider limits as a way to "zoom in" on the function's path near a point. This helps determine slopes of tangents, critical for establishing changes at specific points.

• If \( \lim_{x \to c} f(x) = L \), then as \( x \to c \), \( f(x) \) becomes arbitrarily close to \( L \).
• Limits are integral for defining derivatives, optimizing functions, and solving complex equations.

As you grasp limits, it becomes easier to predict and understand continuity, abrupt changes, and evaluate derivatives, setting the stage for advanced calculus topics.

Function Derivatives

Function derivatives tell us the rate at which a function changes. In simpler terms, they provide the slope of the function at any point—in essence, how steeply a curve is rising or falling. In the context of our exercise, derivatives arise from recognizing the limit expression, which is akin to the derivative definition.Seeing this, the differentiation of the function \( g(t) = t^2 f(t) \) gives us a glimpse into derivative rules:

• The derivative of \( g(t) \) is found using the rule: \( g'(x) = 2x f(x) + x^2 f'(x) = 1 \).
• It involves a combination of the product and chain rules, which help untangle rates of change across multiplied and nested functions.

By expressing \( f'(x) \) in terms of \( f(x) \), we can solve differentiable equations to understand the function's pattern and predict future positions. This exercise exemplifies the power of derivatives to analyze and model real-life scenarios.

Differential Equations

Differential equations are central to modeling natural phenomena. These equations relate functions with their derivatives, essentially linking an element's present state to its rate of change.In our problem, we encounter a first-order differential equation: \( 2x f(x) + x^2 f'(x) = 1 \).This type of equation allows us to decipher how a function evolves over time or under varying conditions.When examining differential equations:

• Simplification is often needed before integrating. In this problem, simplifying identifies a form like \( f(x) = \frac{x+2}{2x} \), which obeys given constraints.
• Boundary conditions like \( f(a) = 1 \) are crucial for solving the equation uniquely, guiding the function towards the exact solution.

Through manipulation and solving of differential equations, we arrive at expressions characterizing the behavior of complex variable-dependent systems. This highlights their utility in physics, engineering, and beyond.