Problem 78 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(g(x)=x f(x)\), where \(f(x)=\left\\{\begin{array}{ll}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.\). At \(x=0\), (A) \(g\) is differentiable but \(g^{\prime}\) is not continuous (B) \(\mathrm{g}\) is differentiable while \(f\) is not (C) both \(f\) and \(g\) are differentiable (D) \(g\) is differentiable and \(g^{\prime}\) is continuous

Step-by-Step Solution

Answer

(A) g is differentiable but g' is not continuous.

Step 1: Formulate g(x) at x=0

We first need to analyze the function \( g(x) = x f(x) \). The function \( f(x) \) is given as \( f(x) = x \sin\left( \frac{1}{x} \right) \) for \( x eq 0 \) and \( f(0) = 0 \). Therefore, at \( x=0 \), we have \( g(0) = 0 \cdot f(0) = 0 \).

Step 2: Compute the derivative of g(x)

To find whether \( g \) is differentiable at \( x=0 \), we take the limit of \( \frac{g(x) - g(0)}{x - 0} \) as \( x \to 0 \). This is \( \lim_{x \to 0} \frac{x^2 \sin\left( \frac{1}{x} \right)}{x} = \lim_{x \to 0} x \sin\left( \frac{1}{x} \right) \). As \( x \to 0 \), \( x \sin\left( \frac{1}{x} \right) \to 0 \) by the squeeze theorem. So, \( g'(0) = 0 \), showing that \( g \) is differentiable at \( x = 0 \).

Step 3: Check continuity of g'(x)

For \( x eq 0 \), calculating the derivative of \( g(x) = x^2 \sin\left( \frac{1}{x} \right) \) using the product rule gives \( g'(x) = 2x \sin\left( \frac{1}{x} \right) - \cos\left( \frac{1}{x} \right) \). As \( x \to 0 \), this expression does not approach any particular value, indicating that \( g'(x) \) is not continuous at \( x = 0 \).

Step 4: Analyze differentiability of f(x)

We know \( f(x) = x \sin\left( \frac{1}{x} \right) \). Establishing the differentiability of \( f(x) \) at \( x = 0 \) involves a similar limit process as in Step 2. The limit \( \lim_{x \to 0} \frac{x \sin\left( \frac{1}{x} \right)}{x - 0} = 0 \) shows \( f \) is differentiable at \( x = 0 \). However, \( f'(0) = 0 \), and \( f' \) is not continuous at \( x = 0 \) as \( f'(x) = \sin\left( \frac{1}{x} \right) - \frac{\cos\left( \frac{1}{x} \right)}{x} \) oscillates wildly.

Key Concepts

Understanding the Squeeze Theorem

The Squeeze Theorem is a powerful concept in calculus that helps us determine the limit of a function. It's particularly useful when a function is difficult to evaluate directly. The theorem states that if you have three functions, say \( f(x) \), \( g(x) \), and \( h(x) \), and \( f(x) \leq g(x) \leq h(x) \) for all x in some interval except possibly at a point \( a \), then if \( \lim_{x \to a}f(x) = \lim_{x \to a}h(x) = L \), it follows that \( \lim_{x \to a}g(x) = L \) as well.
This theorem was used in the original exercise to show that the derivative of function \( g(x) \) at \( x = 0 \) is zero. Specifically, \( x \sin\left( \frac{1}{x} \right) \) oscillates between \( -|x| \) and \( |x| \). This means that as \( x \) approaches 0, the values of \( x \sin\left( \frac{1}{x} \right) \) are squeezed closer and closer to zero, thus \( g'(0) = 0 \).
The conclusion drawn from the Squeeze Theorem helped confirm that the function \( g \) is differentiable at \( x = 0 \). In other scenarios where functions show erratic behavior, this theorem can also be a reliable tool to establish limits.

Exploring Continuity of Derivatives

Continuity of derivatives is an essential property in calculus, but it's not always guaranteed for every function. A derivative is continuous at a given point if the limit of the derivative as we approach that point equals the derivative's value at the point. In simpler terms, a derivative is smooth without jumps or gaps around that point.
In the original exercise, we concluded that although \( g \) is differentiable at \( x = 0 \), its derivative \( g' \) is not continuous there. This is because the derivative, calculated using the product rule, yields \( g'(x) = 2x \sin\left( \frac{1}{x} \right) - \cos\left( \frac{1}{x} \right) \). As \( x \to 0 \), this expression fails to settle at a specific value, oscillating instead.
This highlights the fact that just because a function is differentiable at a point, it doesn't imply that the derivative is continuous there. Such functions demonstrate unique oscillating behaviors where traditional derivative application lacks the expected continuity.

Applying the Product Rule

The product rule is a fundamental rule for differentiation applicable when we have a function formed by multiplying two or more simpler functions. It allows us to find the derivative of a product of two functions, \( u(x) \) and \( v(x) \). The rule states:

• The derivative of \( u(x)v(x) \) is \( u'(x)v(x) + u(x)v'(x) \).

In our exercise, \( g(x) = x^2 \sin\left( \frac{1}{x} \right) \) can be seen as a product of \( u(x) = x^2 \) and \( v(x) = \sin\left( \frac{1}{x} \right) \).
Applying the product rule, we found:

• \( g'(x) = 2x \sin\left( \frac{1}{x} \right) - \cos\left( \frac{1}{x} \right) \).

This method shows how differentiating complex expressions can be systematically broken down using foundational rules. Using a detailed application of the product rule allows us to explore more about the behavior of the function in the vicinity of specific troublesome points like \( x = 0 \).