Problem 23 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Define \(f(x)\) as the product of two real functions \([2011 \mathrm{RS}]\) \(f_{1}(x)=x, x \in R\), and \(f_{2}(x)= \begin{cases}\sin \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{cases}\) as follows: \(f(x)=\left\\{\begin{array}{cl}f_{1}(x) \cdot f_{2}(x), & \text { if } x=0 \\\ 0 & \text { if } x=0\end{array}\right.\) Statement \(-1: f(x)\) is continuous on \(\mathrm{R}\). Statement \(-2: f_{1}(x)\) and \(f_{2}(x)\) are continuous on \(\mathrm{R}\). (a) Statement \(-1\) is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement- \(1 .\) (b) Statement- 1 is true, Statement- 2 is true; Statement- 2 is NOT a correct explanation for Statement-1 (c) Statement-1 is true, Statement- 2 is false (d) Statement- 1 is false, Statement- 2 is true

Step-by-Step Solution

Answer

(b) Statement-1 is true, Statement-2 is true; Statement-2 is NOT a correct explanation for Statement-1.

Step 1: Evaluate the Continuity of f_1(x)

Let's start by analyzing the function \( f_1(x) = x \). This function is continuous on all real numbers \( \mathbb{R} \) because it is a linear function. Linear functions are continuous everywhere by definition.

Step 2: Analyze the Continuity of f_2(x)

Now, consider the function \( f_2(x) = \begin{cases} \sin \frac{1}{x}, & \text{if } x eq 0 \ 0, & \text{if } x = 0 \end{cases} \). This function is not continuous at \( x = 0 \) because the limit \( \lim_{{x \to 0}} \sin\frac{1}{x} \) does not exist, due to the oscillating nature of the sine function as \( x \to 0 \). For \( x eq 0 \), \( f_2(x) \) is the composition of continuous functions and is hence continuous.

Step 3: Evaluate the Continuity of f(x)

Given the definition of \( f(x) \), for \( x eq 0 \), we have \( f(x) = f_1(x) \cdot f_2(x) = x \cdot \sin \frac{1}{x} \). We need to check continuity at \( x = 0 \). The function is defined as 0 at \( x=0 \), and\[ \lim_{{x \to 0}} x \cdot \sin\frac{1}{x} = 0 \] because \( |x \cdot \sin\frac{1}{x}| \leq |x| \) and \( \lim_{{x \to 0}} |x| = 0 \). Hence, \( f(x) \) is continuous on \( \mathbb{R} \).

Step 4: Analyze Statements and Choose Answer

Statement -1 asserts that \( f(x) \) is continuous on \( \mathbb{R} \), which is true as just shown. Statement -2 asserts that both \( f_1(x) \) and \( f_2(x) \) are continuous on \( \mathbb{R} \), which is false. \( f_2(x) \) is not continuous at \( x = 0 \). So, both statements -1 and -2 are true, but -2 is not a correct explanation for -1. The answer is (b).

Key Concepts

Real Functions

Real functions are mathematical expressions that are defined to return real numbers for any real number input. In our exercise, we deal with two such functions: \( f_1(x) = x \) and \( f_2(x) = \sin \frac{1}{x} \). Real functions can be continuous or discontinuous, smooth or complex. They serve as a foundation for calculus, enabling us to explore more advanced concepts like limits and continuity.
Real functions are evaluated based on their behavior across their domains. For \( f_1(x) = x \), which is a simple linear function, the output equates directly to the input. On the other hand, \( f_2(x) = \sin \frac{1}{x} \) oscillates as \( x \to 0 \), reflecting more complexity in real functions' behavior.

Limits

The concept of limits helps us determine the behavior of functions as the input approaches a particular value. Limits are essential for understanding function continuity and differentiability. In the provided exercise, we consider the limit \( \lim_{{x \to 0}} x \cdot \sin\frac{1}{x} \).
This expression demonstrates the powerful utility of limits. Because \( \sin\frac{1}{x} \) oscillates between -1 and 1, it seems unstable at first glance. However, when multiplied by \( x \), which approaches zero, the combined magnitude of \( x \cdot \sin\frac{1}{x} \) is bounded by \( |x| \), resulting in a limit of zero as \( x \to 0 \).
Analyzing limits involves:

• Identifying points of interest, such as where the input approaches zero.
• Using mathematical properties to determine if a function approaches a certain value.

This technique is crucial for calculus students learning function behavior.

Linear Functions

Linear functions form a primary class of real functions, characterized by their straight-line graphs. The defining equation for a linear function is \( f(x) = ax + b \), where \( a \) and \( b \) are constants.
In the context of the given problem, \( f_1(x) = x \) is a linear function that passes through the origin with a slope of 1. Linear functions are continuous and differentiable across all real numbers \( \mathbb{R} \). This means there are no breaks, jumps, or holes in their graphs.
Understanding linear functions involves:

• Recognizing their constant rate of change (slope).
• Being able to graph these functions as a constant line.
• Understanding that any linear combination of \( ax + b \) maintains linear properties.

This foundational knowledge lays the groundwork for learning more complex functions.

Oscillating Functions

Oscillating functions are those that exhibit repetitive variations above and below a central value. In our context, \( f_2(x) = \sin \frac{1}{x} \) is an example of such a function when \( x eq 0 \).
The function oscillates between -1 and 1 rapidly as \( x \to 0 \), exhibiting a property called non-convergence. This oscillation prevents the existence of a limit as \( x \to 0 \) for \( \sin \frac{1}{x} \). However, the product \( x \cdot \sin \frac{1}{x} \) converges due to the damping effect of multiplying by \( x \).
Key points about oscillating functions include:

• They often lack a simple limit due to their oscillation.
• Frequency of oscillation can greatly increase as input decreases.
• They often contrast sharply with linear functions, which do not experience such fluctuations.

Understanding these concepts helps demystify the non-linear behavior of many mathematical phenomena.