Problem 59 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f(x)=\left\\{\begin{array}{c}(x-1) \sin \frac{1}{x-1} \text { if } x \neq 1 \\ 0 \quad \text { if } x=1\end{array} \quad[\mathbf{2 0 0 8 ]}\right.\) Then which one of the following is true? (a) \(f\) is neither differentiable at \(x=0\) nor at \(x=1\) (b) \(f\) is differentiable at \(x=0\) and at \(x=1\) (c) \(f\) is differentiable at \(x=0\) but not at \(x=1\) (d) \(f\) is differentiable at \(x=1\) but not at \(x=0\)

Step-by-Step Solution

Answer

(d) \(f\) is differentiable at \(x=1\) but not at \(x=0\).

Step 1: Understanding the function

The function given is a piecewise function. It takes the value \((x-1) \sin \frac{1}{x-1}\) for \(x eq 1\) and the value \(0\) when \(x = 1\). Our task is to analyze the differentiability of \(f\) at \(x=0\) and \(x=1\).

Step 2: Checking behavior at x=1

To determine if \(f\) is differentiable at \(x=1\), we first check continuity. We calculate the limit \(\lim_{x \to 1} (x-1) \sin \frac{1}{x-1}\). This limit simplifies similarly to the famous limit \(\lim_{u \to \infty} \frac{\sin u}{u} = 0\). The expression tends to 0 as \(x \to 1\), proving continuity as \(f(1) = 0\). To check differentiability, we examine the derivative limit \(\lim_{h \to 0} \frac{f(1+h) - f(1)}{h}\). Substituting in \(f(x)\), we have \(\lim_{h \to 0} h \sin \frac{1}{h}\). This is \(0\) by the squeeze theorem. Therefore, \(f\) is differentiable at \(x=1\) and \(f'(1) = 0\).

Step 3: Checking behavior at x=0

Since \(f(x)= (x-1) \sin \frac{1}{x-1}\) in a neighborhood around, but not equal to \(x = 0\), we analyze it. For differentiability, continuity at \(x=0\) must hold first. As \(xeq1\), around zero, the function output wanders undefinedly because the sine function oscillates rapidly. Since we are given that \(f(0)\) is not explicitly defined, continuity nor differentiability at \(x=0\) isn't clear without the output around zero. Thus, \(f\) is neither continuous nor differentiable at \(x=0\).

Step 4: Conclusion based on analysis

From the above analysis, \(f\) is differentiable at \(x=1\) (and continuous there by the limit) but not continuous or differentiable at \(x=0\) due to the absence of a directly usable form at zero.

Key Concepts

Piecewise Functions

A piecewise function is a mathematical expression defined by multiple sub-functions, each applying to a different part of the domain. These functions are crucial for representing scenarios where a formula changes across intervals.

• Mathematically, a piecewise function is represented as different expressions over different intervals of the independent variable.
• For example, the function given in our exercise: \[ f(x) = \begin{cases}(x-1) \sin \frac{1}{x-1}, & \text{if } x eq 1 \ 0, & \text{if } x = 1 \end{cases}\] clearly shows two different expressions depending on the value of \(x\).

To evaluate such functions, it's essential to understand how each piece behaves over its specified interval. In the exercise, determining differentiability requires analyzing each part separately where the differentiability conditions may vary.

Continuity

Continuity of a function at a point means there is no interruption or jump at that point. For a function to be continuous at a specific point \(x = a\), three conditions must be satisfied:

• The function \(f(x)\) is defined at \(x = a\).
• The limit \(\lim_{x \to a} f(x)\) exists.
• \(\lim_{x \to a} f(x) = f(a)\).

In the original exercise, showing continuity at \(x = 1\) was crucial. The function approached \(0\) as \(x\) went to \(1\), which matches the defined value \(f(x) = 0\) at that point, confirming continuity. For \(x = 0\), the lack of a defined value for \(f(x)\) and the oscillating nature of the function imply discontinuity, making it impossible to satisfy all continuity conditions.

Squeeze Theorem

The Squeeze Theorem is a handy tool for evaluating limits, especially when dealing with oscillating functions. It states that if a function \(f(x)\) is sandwiched between two other functions \(g(x)\) and \(h(x)\), and the limits of \(g(x)\) and \(h(x)\) as \(x\) approaches \(a\) are equal, then the limit of \(f(x)\) as \(x\) approaches \(a\) is the same. Mathematically, it is expressed as:\[\text{If } g(x) \leq f(x) \leq h(x) \text{ for all } x \text{ in some interval around } a, \\text{and } \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L, \text{ then } \lim_{x \to a} f(x) = L.\]In the step-by-step solution, the Squeeze Theorem was effectively used to show that the function was differentiable at \(x = 1\). By controlling the oscillation of \(h \sin \frac{1}{h}\) as \(h\) approaches \(0\), it was deduced through this theorem that the limit is \(0\). This deduction played a pivotal role in proving that \(f(x)\) is differentiable at that point, giving us \(f'(1) = 0\).