Problem 61 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The set of points where \(f(x)=\frac{x}{1+|x|}\) is differentiable is (a) \((-\infty, 0) \cup(0, \infty)\) (b) \((-\infty,-1) \cup(-1, \infty)\) (c) \((-\infty, \infty)\) (d) \((0, \infty)\)

Step-by-Step Solution

Answer

(c) \((-\infty, \infty)\)

Step 1: Understand the Function

Consider the function \( f(x) = \frac{x}{1 + |x|} \). The function involves the absolute value of \( x \), which can alter its differentiability at zero or other points where the term inside the absolute value equals zero.

Step 2: Analyze the Function for Different Cases

To analyze where the function is differentiable, break it into two cases based on \( x \):1. If \( x \geq 0 \) (non-negative case), then \( |x| = x \), so the function becomes \( f(x) = \frac{x}{1 + x} \).2. If \( x < 0 \) (negative case), then \( |x| = -x \), so the function becomes \( f(x) = \frac{x}{1 - x} \).

Step 3: Check Differentiability for \( x \geq 0 \)

For \( x \geq 0 \), consider the function \( f(x) = \frac{x}{1 + x} \). Find the derivative using the quotient rule:\[ f'(x) = \frac{(1 + x) \cdot 1 - x \cdot 1}{(1 + x)^2} = \frac{1}{(1 + x)^2} \].For \( x > 0 \), \( f'(x) \) is well-defined. At \( x = 0 \), \( f'(x) = 1 \), so the function is differentiable there as well.

Step 4: Check Differentiability for \( x < 0 \)

For \( x < 0 \), consider the function \( f(x) = \frac{x}{1 - x} \). Apply the quotient rule:\[ f'(x) = \frac{(1 - x) \cdot 1 - (-x) \cdot 1}{(1 - x)^2} = \frac{1}{(1 - x)^2} \].The function is differentiable for all \( x < 0 \).

Step 5: Conclusion on Differentiability

Since both \( x \geq 0 \) and \( x < 0 \) provide differentiable functions and \( f(x) \) is differentiable at \( x = 0 \), the function \( f(x) = \frac{x}{1 + |x|} \) is differentiable for all \( x \).

Step 6: Select the Correct Option

Since we've determined that \( f(x) \) is differentiable for all \( x \), option (c) \((-\infty, \infty)\) is the correct answer.

Key Concepts

Absolute Value Function

The absolute value function, denoted as \(|x|\), is one of the simplest yet most important functions in mathematics. It measures the distance of a number from zero on the number line. This means it always outputs non-negative values, regardless of the sign of the input. For any real number \(x\):- If \(x \geq 0\), then \(|x| = x\). - If \(x < 0\), then \(|x| = -x\).

This piecewise definition makes \(|x|\) continuous but can lead to points where it is not differentiable, such as \(x = 0\). When used in a function like \(f(x) = \frac{x}{1+|x|}\), the absolute value affects differentiability because the nature of \(|x|\) changes depending on whether \(x\) is positive or negative. Knowing how \(|x|\) alters the expression is crucial in analyzing the behavior of such functions.

Quotient Rule

The quotient rule is a tool used in calculus to find the derivative of a function that is the quotient of two other functions. It is applied when you need to differentiate a function written as \( \frac{u(x)}{v(x)} \), where both \(u(x)\) and \(v(x)\) are differentiable functions. The formula for the quotient rule is:\[ f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2} \]

In our case, the function \(f(x) = \frac{x}{1 + |x|}\) can be separated into two main components: - a numerator \(u(x) = x\), and - a denominator \(v(x) = 1 + |x|\).
Using the quotient rule, we compute the derivative for both \(x \geq 0\) and \(x < 0\), resulting in a simplified expression that tells us how \(f(x)\) changes. Understanding the quotient rule helps us examine the differentiability across different intervals by evaluating the validity of \(f'(x)\).

Differentiable Functions

A function is considered differentiable at a point \(x = c\) if it has a derivative at that point. This implies that the function must be smooth and continuous without any sharp corners or breaks at \(x = c\). Differentiability extends over an interval if the derivative exists at every point within that interval.

In the exercise, we assessed if \(f(x) = \frac{x}{1 + |x|}\) is differentiable for all real numbers. To do so, we analyzed whether the derivative exists for each piece of the piecewise-defined function formed by the absolute value. Calculating \(f'(x)\) showed us that each segment of \(f(x)\) is smooth and continuous, affirming differentiability.
By checking for boundaries and computed results, we concluded that \(f(x)\) is indeed differentiable for all real numbers \((-\infty, \infty)\), matching option (c). This comprehensive approach ensures that the changes in behavior due to \(|x|\) do not cause any discontinuities or sharp turns.