Problem 99 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f\) be a real-valued function defined on an interval \(I\). If \(f\) be derivable at a point \(x_{0} \in I\), then it is continuous at \(x_{0}\). The converse of the above statement does not hold. That is, a function may be continuous at a point but may fail to be derivable at that point. Thus, derivability is a more restrictive property than continuity. In fact, there are functions which are continuous everywhere but differentiable nowhere. If \(R f^{\prime}\left(x_{0}\right)\) and \(L f^{\prime}\left(x_{0}\right)\) are finite (they may or may not be equal), then \(f(x)\) is continuous at \(x=x_{0}\). The function \(f(x)=\left\\{\begin{array}{l}\frac{x\left(e^{1 / x}-e^{-1 / x}\right)}{e^{1 / x}+e^{-1 / x}}, x \neq 0 \\ 0 \quad, x=0\end{array}\right.\) is (A) continuous everywhere but not differentiable at \(x=0\) (B) continuous and differentiable everywhere (C) not continuous at \(x=0\) (D) None of these

Step-by-Step Solution

Answer

(A) continuous everywhere but not differentiable at \(x=0\).

Step 1: Assessing Continuity at x=0

To determine if the function \( f(x) \) is continuous at \( x=0 \), we need to check if \( \lim_{x \to 0} f(x) = f(0) \). From the definition given, \( f(0) = 0 \). We take the limit:\[\lim_{x \to 0} \frac{x\left(e^{1 / x}-e^{-1 / x}\right)}{e^{1 / x}+e^{-1 / x}}\]Observe that as \( x \to 0 \), both \( e^{1/x} \) and \( e^{-1/x} \) approach infinity, but crucially, their sum in the denominator increases exponentially. The expression simplifies such that \( \lim_{x \to 0} f(x) = 0 \). Thus, \( f(x) \) is continuous at \( x = 0 \).

Step 2: Assessing Differentiability at x=0

To determine differentiability at \( x=0 \), we need the derivative \( f'(x) \) at \( x=0 \).We find \( f'(x) \) using the definition:\[f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}\]With \( f(0) = 0 \), this becomes:\[f'(0) = \lim_{h \to 0} \frac{\frac{h\left(e^{1 / h}-e^{-1 / h}\right)}{e^{1 / h}+e^{-1 / h}}}{h} = \lim_{h \to 0} \frac{e^{1 / h} - e^{-1 / h}}{e^{1 / h} + e^{-1 / h}}\]We notice this limit does not exist since it oscillates between different values due to the exponential terms. Therefore, the derivative is not defined at \( x=0 \), making \( f \) not differentiable at \( x=0 \).

Step 3: Conclusion: Determine the correct option

From the analysis:1. \( f(x) \) is continuous at \( x=0 \). 2. \( f(x) \) is not differentiable at \( x=0 \).Thus, \( f(x) \) is continuous everywhere on its domain but not differentiable at \( x=0 \). Therefore, the correct choice is (A).

Key Concepts

Continuity

Continuity in calculus refers to the property of a function where small changes in the input result in small changes in the output. More formally, a function \( f(x) \) is said to be continuous at a point \( x_0 \) if the following condition is satisfied:

• The limit of \( f(x) \) as \( x \) approaches \( x_0 \) is equal to the value of the function at that point, i.e., \( \lim_{x \to x_0} f(x) = f(x_0) \).

This means that there are no abrupt jumps or breaks in the graph of the function at that point. It smoothly flows through the value \( f(x_0) \). In the original exercise, the function \( f(x) \) was showcased as continuous at \( x=0 \) by demonstrating that the limit of the function approaches the same value as the function itself at that point.

• Continuity is a fundamental concept that helps in understanding many advanced calculus operations.

However, continuity alone does not ensure smoothness or differentiability; a function might still "bend" sharply as is shown in the example function.

Differentiability

Differentiability extends the idea of continuity. A function \( f(x) \) is differentiable at a point \( x_0 \) if it possesses a defined derivative there. Being differentiable means that the function not only passes through \( x_0 \) but also has a defined slope (or tangent) at that point. If a function is differentiable at a certain point, it is also continuous there.
The derivative of \( f(x) \) at a point \( x_0 \) is defined using the limit:
\[ f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} \]In step-by-step analysis of the exercise, the function \( f(x) \) was found not to be differentiable at \( x=0 \) because the limit defining the derivative did not exist.
Some characteristics of non-differentiability include:

• Sharp bends or corners
• Vertical tangents
• Discontinuities

Even if the function was continuous, it lacked a smooth transition at that point. Hence, differentiability is a more restrictive condition than continuity.

Real-Valued Functions

Real-valued functions are those which assign a real number (from the set \( \mathbb{R} \)) to every input from their domain. These functions can occur in numerous forms, including polynomials, exponentials, logarithms, and trigonometric functions.
In the context of the problem, \( f(x) \) was defined as such a real-valued function. Understanding the behavior of real-valued functions concerning continuity and differentiability is crucial for solving various calculus problems.

• Real-valued functions can be analyzed using limits to determine properties such as continuity and differentiability over their defined intervals.
• These functions play a vital role in mathematical models, helping describe physical phenomena where parameters vary continuously over real numbers.

Being real-valued ensures that the function can form a basis for continuously varying systems, such as in relation to time, space, or other physical quantities.