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

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

(B) \(g\) is differentiable while \(f\) is not.

Step 1: Understand the definitions of functions

The function \(f(x)\) is defined piecewise: **\(f(x) = x \sin\left(\frac{1}{x}\right)\) when \(x eq 0\)** and **\(f(x) = 0\) when \(x = 0\)**. The function \(g(x)\) is defined as **\(g(x) = x f(x)\)**.

Step 2: Analysis of function \(f(x)\) at \(x=0\)

The function \(f(x)\) is somewhat of a classic pathological example since \(\sin\left(\frac{1}{x}\right)\) oscillates wildly as \(x\) approaches 0, but \(x \sin\left(\frac{1}{x}\right)\) does approach 0. However, \(f(x)\) is not differentiable at \(x=0\) because it oscillates with an infinite number of crossings through the x-axis as it approaches zero.

Step 3: Express \(g(x)\) in terms of \(f(x)\)

We have \(g(x) = x f(x) = x^2 \sin\left(\frac{1}{x}\right)\) for \(x eq 0\) and \(g(0) = 0 \cdot f(0) = 0\).

Step 4: Evaluate differentiability of \(g(x)\) at \(x=0\)

To check if \(g\) is differentiable at \(x = 0\), we need to compute the derivative \(g'(0)\). The definition of the derivative is \(g'(0) = \lim_{{x \to 0}} \frac{g(x) - g(0)}{x}\), which simplifies to \(\lim_{{x \to 0}} x \sin\left(\frac{1}{x}\right)\). As \(x \to 0\), this expression tends to 0.

Step 5: Evaluate the continuity of \(g'(x)\) around \(x=0\)

The derivative of \(g\) for \(x eq 0\) is \(g'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)\). Computing the limit as \(x\) approaches 0 requires examining \(g'(x)\). Due to the term \(\cos\left(\frac{1}{x}\right)\), this limit does not exist because it continuously oscillates between \(-1\) and \(1\), making \(g'\) discontinuous at zero.

Key Concepts

Piecewise Functions

Piecewise functions are a type of function defined by different expressions based on the input value of the function. This is especially useful in mathematical modeling where a single rule does not apply to the entire domain. Consider the function \(f(x)\) from the original exercise. It is defined in two parts: \(f(x) = x \sin\left(\frac{1}{x}\right)\) for \(x eq 0\) and \(f(x) = 0\) when \(x = 0\).
This means that the behavior of \(f(x)\) changes based on whether \(x\) is zero or not. This type of function is very common in problems involving limits and continuity, as it allows the function to be adaptable to different scenarios. When analyzing piecewise functions, it's crucial to examine the continuity and differentiability at the points where the expression changes. These are known as \"breakpoints\" or \"junctions\". Understanding piecewise functions helps in analyzing real-world systems where different conditions might exist in different contexts, ensuring a comprehensive grasp over the entire function's behavior.

Oscillating Functions

Oscillating functions are ones that move up and down repeatedly, often between two fixed values. A classic example of an oscillating function is \(\sin\left(\frac{1}{x}\right)\), which appears in the function \(f(x)\) in our exercise.
As \(x\) approaches zero, \(\sin\left(\frac{1}{x}\right)\) oscillates increasingly rapidly, creating a challenging scenario to study in terms of limits and analysis.

• The oscillation of these functions is characterized by frequent crossings through the x-axis.
• They do not settle at any specific value as \(x\) tends to zero or any critical point.

Oscillating behavior brings complexity, especially when considering differentiability and continuity, as discussed in the analysis of \(f(x)=x \sin\left(\frac{1}{x}\right)\). Such behaviors often make it impossible for a function to be differentiable, because the infinite number of oscillations over diminutive intervals disrupts any smoothness. Thus, in these types of functions, it's common to encounter issues regarding differentiability at certain points.

Continuity of Derivatives

Understanding the continuity of derivatives involves analyzing whether the derivative function, \(g'(x)\), behaves in a consistent manner across its domain. In the exercise, we focus on the differentiability of \(g(x)\) and whether its derivative \(g'(x)\) remains continuous, especially around \(x=0\).
Like continuity in regular functions, a derivative is continuous at a point if it does not have abrupt changes as \(x\) approaches that point.

• When a function's derivative is continuous, it indicates a smooth transition with no sudden jumps or breaks.
• Discontinuities in the derivative can occur if there's an oscillation or unexpected behavior, like an infinite derivative.

For the function \(g(x)=x^2\sin\left(\frac{1}{x}\right)\), its derivative \(g'(x)\) is analyzed in step 5 of the solution. Despite \(g(x)\) being differentiable (\(g'(0)\) exists as zero), \(g'(x)\) is not continuous at zero because of the oscillating \(\cos\left(\frac{1}{x}\right)\) term. This results in \(g'(x)\) oscillating endlessly between -1 and 1 for values close to zero. Understanding these discontinuities is vital for fully grasping the behavior of derivatives in more advanced mathematical studies.