Problem 54 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{f}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) be two functions defined by \(f(x)=\left\\{\begin{array}{l}x \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, \quad, x=0\end{array}\right.\), and \(g(x)=x f(x)\) Statement I: \(f\) is a continuous function at \(\mathrm{x}=0\). Statement II: \(g\) is a differentiable function at \(x=0\). (a) Both statement I and II are false. (b) Both statement \(I\) and II are true. (c) Statement I is true, statement II is false. (d) Statement I is false, statement II is true.

Step-by-Step Solution

Answer

(b) Both statement I and II are true.

Step 1: Understanding the Function f(x) at x ≠ 0

The function \( f(x) = x \sin\left(\frac{1}{x}\right) \) is defined for \( x eq 0 \). This behavior around \( x = 0 \) is crucial for checking continuity. Here, \( \sin\left(\frac{1}{x}\right) \) oscillates between \(-1\) and \(1\), but since it is multiplied by \( x \), the product is squeezed towards zero as \( x \) approaches zero.

Step 2: Checking Continuity of f(x) at x = 0

Continuity at a point requires \( \lim_{{x \to 0}} f(x) = f(0) \). Compute the limit: \[ \lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} x \sin\left(\frac{1}{x}\right). \] As \( |\sin(1/x)| \leq 1 \), \( x \sin(1/x) \to 0 \) as \( x \to 0 \). Thus, \( \lim_{{x \to 0}} f(x) = 0 \), and since \( f(0) = 0 \), \( f \) is continuous at \( x = 0 \).

Step 3: Analyzing g(x) and Its Continuity

The function \( g(x) = x f(x) = x^2 \sin\left(\frac{1}{x}\right) \) for \( x eq 0 \) and \( g(0) = 0 \). The function \( g \) is continuous at \( x = 0 \) because \( \lim_{{x \to 0}} g(x) = \lim_{{x \to 0}} x^2 \sin\left(\frac{1}{x}\right) = 0 \), which equals \( g(0) \).

Step 4: Checking Differentiability of g(x) at x = 0

Differentiability requires finding \( \lim_{{h \to 0}} \frac{g(h) - g(0)}{h} = \lim_{{h \to 0}} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} = \lim_{{h \to 0}} h \sin\left(\frac{1}{h}\right) \). This limit evaluates to 0 as the absolute value is less than or equal to \( |h| \), and \( h \to 0 \). Hence, \( g \) is differentiable at \( x = 0 \).

Step 5: Concluding the Truth of the Statements

Statement I asserts that \( f \) is continuous at \( x = 0 \), which we have shown to be true. Statement II claims \( g \) is differentiable at \( x = 0 \), which is also true. Therefore, both statements are correct.

Key Concepts

Continuity

Continuity is a fundamental concept in calculus, describing a function's behavior at a specific point. For a function to be continuous at a point, say at \(x = 0\), three conditions must be met:

• The function \(f(x)\) must be defined at \(x = 0\), meaning \(f(0)\) must exist.
• A limit approaching that point from either side must be the same \(\lim_{{x \to 0}} f(x) = L\).
• The function's value at the point must equal the limit, so \(f(0) = \lim_{{x \to 0}} f(x)\).

In the provided exercise, the function \(f(x) = x \sin\left(\frac{1}{x}\right)\) is tested for continuity at \(x = 0\). Since \(f(0) = 0\) is defined, we check \(\lim_{{x \to 0}} f(x)\). Because \( |\sin\left(\frac{1}{x}\right)| \leq 1\), the term \(x \sin\left(\frac{1}{x}\right)\) approaches zero as \(x \to 0\). Thus, \(\lim_{{x \to 0}} f(x) = 0\), which matches \(f(0)\), confirming that \(f(x)\) is continuous at \(x = 0\). Continuity ensures no abrupt jumps or breaks in the function at that point.

Differentiability

Differentiability extends continuity, demanding a function not only be continuous but also smooth enough around a point to have a defined tangent. A function \(g(x)\) is differentiable at a point \(x = c\) if the following limit exists:
\[ \lim_{{h \to 0}} \frac{g(c + h) - g(c)}{h} \]
This represents the slope of the tangent line at \(x = c\).
In our example, the function \(g(x) = x^2 \sin\left(\frac{1}{x}\right)\) needs to be checked for differentiability at \(x = 0\). Given \(g(0) = 0\), we evaluate:
\[ \lim_{{h \to 0}} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} = \lim_{{h \to 0}} h \sin\left(\frac{1}{h}\right) \]
As \(|\sin\left(\frac{1}{h}\right)| \leq 1\), \(h \sin\left(\frac{1}{h}\right)\) trends towards zero with \(h\). Thus, \(g(x)\) is differentiable at \(x = 0\), as the limit exists and equals zero. Smoothness and a well-defined slope at this point mean the function’s graph does not have sharp corners.

Limit Evaluation

Limits evaluate the behavior of a function as the input nears a particular value. Grasping limit evaluation is essential for understanding continuity and differentiability. It’s about predicting the function's value based on nearby values, without necessarily observing the exact point. The notation \(\lim_{{x \to a}} f(x) = L\) implies that as \(x\) gets very close to \(a\), \(f(x)\) approaches \(L\). This process helps resolve issues where functions appear undefined.
For instance, consider \(f(x) = x \sin\left(\frac{1}{x}\right)\) as \(x\) approaches 0. Direct substitution isn’t possible because \(\sin\left(\frac{1}{x}\right)\) becomes indeterminate. However, using limits, we estimate this by observing \(|x|\times |\sin\left(\frac{1}{x}\right)| \leq |x|\). Both upper and lower bounds squeeze \(x\sin\left(\frac{1}{x}\right)\) to zero as \(x\) converges to zero.
Applying this strategy, we see how limits bridge gaps left by direct substitution. They enable evaluation of functions at points where they initially seem unsettling or undefined.