Problem 74 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{S}\) be the set of all points in \((-\pi, \pi)\) at which the function \(f(x)=\min \\{\sin x, \cos x\\}\) is not differentiable. Then \(S\) is a subset of which of the following? (a) \(\left\\{-\frac{\pi}{4}, 0, \frac{\pi}{4}\right\\}\) (b) \(\left\\{-\frac{3 \pi}{4},-\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{\pi}{4}\right\\}\) (c) \(\left\\{-\frac{\pi}{2},-\frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{2}\right\\}\) (d) \(\left\\{-\frac{3 \pi}{4},-\frac{\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{4}\right\\}\)

Step-by-Step Solution

Answer

S is a subset of (b) \(\{-\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{3\pi}{4}, \frac{\pi}{4}\}\).

Step 1: Understand the Function

The given function is a piecewise function defined as \( f(x) = \min \{\sin x, \cos x\} \). This means that at any point \( x \), \( f(x) \) will be the lesser of \( \sin x \) or \( \cos x \). The derivatives of \( \sin x \) and \( \cos x \) are \( \cos x \) and \(-\sin x \) respectively.

Step 2: Equate \(\sin x = \cos x\)

To determine where the function switches from \(\sin x\) to \(\cos x\) (or vice versa), we set \( \sin x = \cos x \). This gives \( \tan x = 1 \), which occurs at \( x = \frac{\pi}{4} + n\pi \) for integer \( n \). Within \((-\pi, \pi)\), the relevant values are \(-\frac{3\pi}{4}, \frac{\pi}{4}\).

Step 3: Consider the Interval Edges

The interval given is \((-\pi, \pi)\). Check if close to these bounds, \( \sin x \) and \( \cos x \) continue their cycle pattern. Both repeat in such a way that only points identified by solving \( \sin x = \cos x \) within the interval matter, not \(-\pi\) or \(\pi\).

Step 4: Check Differentiability at \(x=0\)

At \(x=0\), \(\sin x = 0\) and \(\cos x = 1\). Therefore, \( f(x) = \sin x \) near \(-0.5\pi\) to \(\frac{0\pi}{4}\) and remains differentiable at 0. Thus, 0 is not a point where the function is not differentiable.

Step 5: Determine Non-Differentiable Points

Given the points where \( \sin x = \cos x \), the function switches between \( \sin x \) and \( \cos x \) and may be non-differentiable. The points are \(-\frac{3\pi}{4}\) and \(\frac{\pi}{4}\) only, because at these points, the slopes \(\cos x\) and \(-\sin x\) meet at different values.

Step 6: Check Candidate Sets for Containment

Match the identified non-differentiable points \(-\frac{3\pi}{4}, \frac{\pi}{4}\) with the options given to find out the appropriate subset. Option (b) \(\{-\frac{3 \pi}{4},-\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{\pi}{4} \}\) contains both our points.

Key Concepts

Piecewise Functions

A piecewise function is one that is defined by different expressions for different intervals of its domain. For the function in this exercise, we have \( f(x) = \min \{\sin x, \cos x\} \). This means that for any value of \( x \), \( f(x) \) selects the smaller value between \( \sin x \) and \( \cos x \).

Understanding how these functions work is crucial in identifying the points where transitions happen between different sub-functions—when \( \sin x \) takes over from \( \cos x \) or vice versa. These transition points often lead to non-differentiability because the function may have a sharp "corner" at these points where the rate of change is discontinuous.

In the current problem, finding where \( \sin x = \cos x \) helped identify these transition points, as the function switches from one piece to another.

Trigonometric Functions

Trigonometric functions, such as \( \sin x \) and \( \cos x \), exhibit a wave-like pattern that repeats every \( 2\pi \). These functions are vital in many areas of calculus.

- **Sine Function (\( \sin x \))**: It starts at 0 when \( x = 0 \), reaches its maximum of 1 at \( x = \frac{\pi}{2} \), goes back to 0 at \( x = \pi \), reaches -1 at \( x = \frac{3\pi}{2} \), and back to 0 at \( x = 2\pi \).

- **Cosine Function (\( \cos x \))**: It begins at 1 when \( x = 0 \), falls to 0 at \( x = \frac{\pi}{2} \), -1 at \( x = \pi \), back to 0 at \( x = \frac{3\pi}{2} \), and returns to 1 at \( x = 2\pi \).

When solving the exercise, finding where these functions equalize \( (\sin x = \cos x) \) was essential. This equation simplifies to finding \( x \) values where \( \tan x = 1 \), revealing angles at which both functions cross each other, leading to potential non-differentiability.

Calculus Problem Solving

Solving problems in calculus often involves multiple steps. This exercise on differentiability requires logical thinking and the application of several calculus concepts:

- **Understand Problem Context**: Begin by clearly understanding what the function represents. Here, the key was knowing it behaves like a piecewise function. This means you carefully evaluate each interval.

- **Identify Critical Points**: For piecewise functions and potential non-differentiable points, crucial interactions were explored. In this example, solving \( \sin x = \cos x \) to find transition points was the starting point.

- **Use Function Properties**: Calculating derivatives of trigonometric functions—knowing \( \frac{d}{dx} [\sin x] = \cos x \) and \( \frac{d}{dx} [\cos x] = -\sin x \) —helped to check if the function is smooth and continuous at the given points.

- **Validate Solutions with Options**: Once the non-differentiable points were identified, they had to be compared with the given multiple-choice options to find the correct subset. This final step solidifies the solution verification.