Problem 11 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f(x)=\left\\{\begin{array}{cc}(x-1)^{\frac{1}{2-x}}, & x>1, x \neq 2 \\\ k, & x=2\end{array}\right.\) The value of \(k\) for which \(f\) is continuous at \(x=2\) is |Online April \(\mathbf{1 5}, \mathbf{2 0 1 8}]\) \(\begin{array}{ll}\text { (a) } e^{-2} & \text { (b) } e\end{array}\) (c) \(e^{-1}\) (d) 1

Step-by-Step Solution

Answer

The value of \( k \) for which \( f \) is continuous at \( x = 2 \) is \( e^{-1} \) (option c).

Step 1: Understand the Definition of Continuity

A function is continuous at a point if the limit of the function as it approaches the point is equal to the function's value at that point. For a function \( f(x) \) to be continuous at \( x = 2 \), we need \( \lim_{{x \to 2}} f(x) = f(2) \). In this problem, since \( f(2) = k \), we must have \( \lim_{{x \to 2}} f(x) = k \).

Step 2: Analyze the Piecewise Function

We know that the function \( f(x) = (x-1)^{\frac{1}{2-x}} \) when \( x > 1 \) and \( x eq 2 \). For continuity at \( x = 2 \), we need to find \( \lim_{{x \to 2}} (x-1)^{\frac{1}{2-x}} \).

Step 3: Simplify the Expression for Limit

Focus on the expression \( (x-1)^{\frac{1}{2-x}} \). Rewrite it as \( (x-1)^{-rac{1}{x-2}} \) to investigate the limit as \( x \to 2 \). This expression can be rewritten as \( e^{\ln((x-1)^{-rac{1}{x-2}})} = e^{\frac{-\ln(x-1)}{x-2}} \).

Step 4: Evaluate the Limit Using L'Hôpital's Rule

Evaluate \( \lim_{{x \to 2}} \frac{-\ln(x-1)}{x-2} \). This limit is of the form \( \frac{0}{0} \), allowing the use of L'Hôpital's Rule. Differentiating the numerator and denominator, we have:\( \lim_{{x \to 2}} \frac{-\frac{1}{x-1}}{1} = \lim_{{x \to 2}} \frac{-1}{x-1} \). As \( x \) approaches 2, this simplifies to \( -1 \).

Step 5: Solve for Limit and Value of k

Since \( \lim_{{x \to 2}} \frac{-\ln(x-1)}{x-2} = -1 \), substitute back into the exponential form to get \( \lim_{{x \to 2}} e^{\frac{-\ln(x-1)}{x-2}} = e^{-1} \). Therefore, for the function to be continuous at \( x = 2 \), the limit must equal \( k \): \( k = e^{-1} \).

Key Concepts

Understanding Piecewise Function

A piecewise function is a type of function defined by multiple sub-functions, each applying to a certain interval of the domain. This structure allows a single function to handle various behaviors depending on the input. In the exercise we discuss, the function is defined as:

• For values of \(x \) greater than 1 and not equal to 2, \(f(x) = (x-1)^{\frac{1}{2-x}}\).
• For \(x = 2\), the function's value is \(k\).

This demonstrates how the piecewise definition allows the function to adapt based on the input value.
The goal is to determine when the function is continuous, especially at the point where the behavior changes, specifically \(x=2\). To do this, the limits and values of the function must be critically examined to ensure both parts of the function blend seamlessly together.

Evaluating Limits to Ensure Continuity

When evaluating whether a function is continuous at a specific point, such as \(x=2\) in our example, it's crucial to find whether the limit of the function approaches the same value as the function's output at that point. This involves computing \(\lim_{{x \to 2}} (x-1)^{\frac{1}{2-x}}\).
We have the following requirements for continuity:

• The function value \(f(2) = k\) must equal the limit value \(\lim_{{x \to 2}} f(x)\).
• Thus, the computation of the limit needs meticulous handling to see if it matches \(k\).

Approaching this involves rewriting the initial expression in a form used for limit evaluation, a strategy that involves taking natural logarithms and applying further mathematical techniques. Correct evaluation facilitates comprehending how the segments of a piecewise function converge or diverge at critical points.

Using L'Hôpital's Rule

L'Hôpital's Rule is a handy tool for evaluating difficult limits that result in indeterminate forms, most commonly \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). In our case, the expression \(\frac{-\ln(x-1)}{x-2}\) seen earlier when trying to evaluate the limit \(\lim_{{x \to 2}} (x-1)^{-\frac{1}{x-2}}\) simplifies into such a form as \(x\) approaches 2.
The steps to apply L'Hôpital's Rule are:

• Differentiate the numerator, which is \(-\ln(x-1)\), into \(-\frac{1}{x-1}\).
• Differentiate the denominator, which is simply \(x-2\), into 1.

With these derivatives, re-evaluate the limit: \(\lim_{{x \to 2}} \frac{-1}{x-1} = -1\). Once the limit is found, substitute back into the exponential form and reassess the value, leading to the resolve that continuity at \(x=2\) will occur if \(k = e^{-1}\). This process underscores the resourcefulness of mathematical rules in solving intricate problems.

Exploring Exponential Functions

Exponential functions are one of the key components in this exercise, primarily when rearranging and evaluating complex expressions. The expression \((x-1)^{-\frac{1}{x-2}}\) can be reconfigured into exponential form as \(e^{\frac{-\ln(x-1)}{x-2}}\). This transformation leverages the property that any expression \(a^b\) can be rewritten as \(e^{b \ln a}\), simplifying complicated algebraic forms.
In this problem, solving requires one to:

• Convert expressions into exponential form to evaluate limits smoothly.
• Use properties of exponentials to interpret results and maintain consistency in calculations.

Understanding how to manipulate exponential forms paves the way to navigate through complex problems, ensuring calculations are accurate and results are easily interpreted. It is a testament to the way exponential functions provide powerful tools for both straightforward and advanced mathematical problems.