Problem 10 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the function \(f\) defined as \(f(x)=\frac{1}{x}-\frac{k-1}{e^{2 x}-1}\) \(x \neq 0\), is continuous at \(x=0\), then the ordered pair \((k, f(0))\) is equal to? \(\begin{array}{llll}\text { (a) }(3,1) & \text { (b) }(3,2) & \text { (c) }\left(\frac{1}{3}, 2\right) & \text { (d) }(2,1)\end{array}\)

Step-by-Step Solution

Answer

The ordered pair \((k, f(0))\) is \((3, 1)\).

Step 1: Explore Continuity Condition

For continuity at \(x = 0\), the limit of \(f(x)\) as \(x\) approaches 0 must equal \(f(0)\). This implies that \(\lim_{{x \to 0}} f(x) = f(0)\).

Step 2: Analyze Individual Terms

The function is given as \(f(x) = \frac{1}{x} - \frac{k-1}{e^{2x} - 1}\). Note that \(\frac{1}{x}\) can be problematic near \(x=0\), so we will focus on its behavior separately. The term \(\frac{k-1}{e^{2x} - 1}\) requires finding a limit that must exist as \(x\) approaches zero.

Step 3: Simplify the Exponential Term

Expand \(e^{2x}\) using its series expansion around 0: \(e^{2x} \approx 1 + 2x + \frac{(2x)^2}{2} + \ldots \). For small \(x\), \(e^{2x} - 1 \approx 2x\). Therefore, \(\frac{k-1}{e^{2x} - 1} \approx \frac{k-1}{2x}\).

Step 4: Combine Terms and Find Limit

Substitute the approximation into the expression, we get \(f(x) \approx \frac{1}{x} - \frac{k-1}{2x} = \frac{2-(k-1)}{2x} = \frac{3-k}{2x}\). For \(f(x)\) to have a finite limit as \(x \to 0\), the numerator must be zero.

Step 5: Solve for \(k\)

Set the numerator equal to zero: \(3-k = 0\). Solving this gives \(k = 3\).

Step 6: Determine \(f(0)\) Value

If \(k=3\), then \(f(x)\) simplifies to \(f(x) = 0\) as \(x\) approaches 0, so \(f(0) = 1\) for continuity. Thus the ordered pair \((k, f(0)) = (3, 1)\).

Key Concepts

Limits in Calculus

In the realm of calculus, limits serve as a foundational concept, especially when dealing with the continuity of functions. In simple terms, limits help us understand the behavior of a function as the input approaches a certain value. To determine whether a function is continuous at a specific point, like in our exercise as we approach zero, the limit of the function as it gets infinitesimally close to that point must equal the function's value at that point.

Consider the function \(f(x) = \frac{1}{x} - \frac{k-1}{e^{2x} - 1}\). To examine its continuity at \(x = 0\), we evaluate the limit of \(f(x)\) as \(x\) approaches 0. The critical aspect is ensuring that

• The limit \(\lim_{{x \to 0}} f(x)\) exists.
• The limit equals \(f(0)\).

This exercise displays a common challenge with rational functions, where evaluating limits near undefined points requires careful manipulation.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. These functions appear frequently in calculus, notably in problems involving growth and decay.

In the exercise, the exponential function \(e^{2x}\) plays a pivotal role. As we expand \(e^{2x}\) near zero using series expansion, we approximate it by its Taylor series:

• Near zero, \(e^{2x} \approx 1 + 2x + \frac{(2x)^2}{2} + \ldots\).
• For very small values of \(x\), higher-order terms become negligible.
• This ultimately simplifies our problem by allowing us to substitute and simplify \(e^{2x} - 1\) as approximately \(2x\).

Understanding these approximations is essential to examining the function's behavior and ensuring correct limit evaluations as \(x\) approaches zero.

Series Expansion

Series expansion is a technique often used to approximate functions around a certain point. In calculus, it is indispensable for simplifying complicated expressions to make limit evaluation feasible.

The series expansion of an exponential function, like \(e^{2x}\), helps by providing:

• A way to approximate the behavior of functions close to a point.
• A means to transform complex expressions into simpler, polynomial-like forms.

Using the expansion \(e^{2x} \approx 1 + 2x\) for small \(x\) makes it easier to manage terms like \(\frac{1}{x} - \frac{k-1}{e^{2x} - 1}\). This simplification process is key to finding limits and checking continuity precisely by removing ambiguity about behavior near zero.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to make calculations easier. This step is often essential in solving calculus problems, particularly when dealing with limits.

In this exercise, by substituting the approximation \(e^{2x} - 1 \approx 2x\), the problem simplifies to evaluating:

• \(f(x) \approx \frac{1}{x} - \frac{k-1}{2x} = \frac{3-k}{2x}\)

To have a finite limit as \(x \to 0\), the numerator must vanish. Thus, setting \(3-k=0\) solves for \(k = 3\).

This example demonstrates how algebra can streamline evaluating limits and finding the continuity of a function. It stresses the importance of simplifying expressions to reach a solution efficiently and accurately.