Problem 48 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f:(-1,1) \rightarrow \mathbf{R}\) be a function defined by \(f(x)=\max\) \(\left\\{-|x|,-\sqrt{1-x^{2}}\right\\}\). If \(\mathrm{K}\) be the set of all points at which \(f\) is not differentiable, then \(K\) has exactly: (a) five elements (b) one element (c) three elements (d) two elements

Step-by-Step Solution

Answer

(d) two elements

Step 1: Understanding the Function

The function is given by \( f(x) = \max\{-|x|, -\sqrt{1-x^2}\} \). This means for each \(x\), we compare \(-|x|\) and \(-\sqrt{1-x^2}\) and take the larger of the two values. The function will be either \(-|x|\) or \(-\sqrt{1-x^2}\) depending on which is larger at that point.

Step 2: Identifying the Critical Points

To determine where \(f(x)\) might not be differentiable, we look for critical points where the expressions \(-|x|\) and \(-\sqrt{1-x^2}\) are equal, since the maximum function can change expression there. Solving \(-|x| = -\sqrt{1-x^2}\), we get \(|x| = \sqrt{1-x^2}\), leading to \(x^2 = 1-x^2\) or \(2x^2 = 1\), giving \(x = \pm\frac{1}{\sqrt{2}}\). Hence, \(x_1 = \frac{1}{\sqrt{2}}\) and \(x_2 = -\frac{1}{\sqrt{2}}\) are critical points.

Step 3: Analyzing Non-Differentiability at Critical Points

At \(x_1 = \frac{1}{\sqrt{2}}\), calculate \(f(x)\) from both sides: when approaching from less than \(x_1\), \(f(x) = -|x|\), and from more than \(x_1\), \(f(x) = -\sqrt{1-x^2}\). Since these derivatives differ, \(f\) is not differentiable here. Similarly, at \(x_2 = -\frac{1}{\sqrt{2}}\), \(f(x)\) switches, causing non-differentiability.

Step 4: Boundary Considerations

The function is defined on \((-1, 1)\). Consider endpoints where \(-|x|\) or \(-\sqrt{1-x^2}\) may sharply change: at \(x = 0\), observe that the max is \(-0 = -0\) based on \(-|x|\) inside \((-1,1)\). Check the slopes' continuity: as \(-|x|\) is already continuous and differentiable across the origin, no issue arises here.

Step 5: Evaluating the Set \(K\)

From the analysis in steps 3 and 4, \(f(x)\) is non-differentiable at \(x_1 = \frac{1}{\sqrt{2}}\) and \(x_2 = -\frac{1}{\sqrt{2}}\). Since no other points on \((-1,1)\) lead to a derivative change (based on cutoffs in sign change), \(K\), the set of non-differentiable points, has \(2\) elements: \(x_1\) and \(x_2\).

Key Concepts

Critical Points

Critical points are places on a graph where a function changes its behavior, such as switching from increasing to decreasing or vice versa. For the function we are exploring, critical points are particularly important because they help us identify where the function might not be differentiable.

In our specific function, which is defined as the maximum of two expressions, critical points occur where these two expressions are equal. This is because at these points, the choice of the maximum might switch from one expression to the other. In mathematical terms, for our function, this means solving \(-|x| = -\sqrt{1-x^2}\). By simplifying, we find \(x = \pm\frac{1}{\sqrt{2}}\).

These are the points at which we will closely look for clues about differentiability, as anything unusual, such as abrupt changes, can cause the function to be not differentiable.

Non-Differentiable Points

Non-differentiable points are locations on the graph of a function where the derivative does not exist. These are often found at sharp corners, cusps, or discontinuities. In the context of our function, we suspect non-differentiability at the critical points identified earlier.

The reason for non-differentiability at these critical points is because the derivative, which measures the slope of the function, changes abruptly. We examined the behavior around the points \(x=\frac{1}{\sqrt{2}}\) and \(x=-\frac{1}{\sqrt{2}}\). At these points, the expression used in \(f(x)\) switches, meaning the derivative is different on each side of these points.
Let's break this down:

• Approaching \(x=\frac{1}{\sqrt{2}}\) from the left, the function behaves as \(-|x|\), giving one derivative.
• Approaching from the right, it changes to \(-\sqrt{1-x^2}\), resulting in a different derivative.

This sharp change means the function is not differentiable at \(x=\frac{1}{\sqrt{2}}\) and \(x=-\frac{1}{\sqrt{2}}\). These are, therefore, the non-differentiable points of our function.

Max Function

The max function is a special operator that takes several inputs and returns the largest one. For our exercise, the function is defined by taking the maximum of two expression values: \(-|x|\) and \(-\sqrt{1-x^2}\). This operator is crucial because it dictates the behavior of the function at each point across its domain.
When dealing with max functions, particularly when expressions intersect, it hints at critical points where differentiability might be compromised. For instance:

• At a point where \(-|x|\) and \(-\sqrt{1-x^2}\) intersect, the max function could "switch" from one expression to the other.
• This transition is where the derivative might not exist, making it non-differentiable.

The use of max functions can thus create subtle "jumps" or changes in slope, which are not always visually evident, requiring careful algebraic analysis to uncover.

Function Analysis

Function analysis involves a thorough examination of a function's behavior across its domain, identifying important properties such as continuity, differentiability, and critical points. For the function in this exercise, this analysis involves several key steps.

First, we recognize that both parts of the function, \(-|x|\) and \(-\sqrt{1-x^2}\), are continuously defined for the domain \((-1, 1)\). We then compare these parts to find the potential non-differentiable points by solving \(-|x| = -\sqrt{1-x^2}\), identifying where these two expressions are equal.
Through analysis:

• We find critical points at \(x=\frac{1}{\sqrt{2}}\) and \(x=-\frac{1}{\sqrt{2}}\).
• By examining both sides of these points, we conclude that the function is non-differentiable there due to changes in the slope or derivative.

Finally, assessing the boundaries confirms that \(x=0\) does not require additional consideration for non-differentiability in this particular range, thus completing the analysis.