Question

Let \( f(x) = \max\{\cos x, \sin x, 0\} \). If the number of points at which \( f(x) \) is not differentiable in \( (0, 2024\pi) \) is \( 1012k \), then \( k = \)

• A. \( 3/2 \)
• B. \( 6 \)
• C. \( 3 \)
• D. \( 2 \)

Hint:

The function \( \max(f(x), g(x)) \) is not differentiable where \( f(x) = g(x) \) and \( f'(x) \neq g'(x) \). For \( \max(f(x), g(x), h(x)) \), check pairwise equalities and points where the function transitions through the maximum value. Consider the graphs of \( \cos x \), \( \sin x \), and \( 0 \) to visualize the points where the maximum function might not be smooth.

Answer 1

The Correct Option is C

The function \( f(x) = \max\{\cos x, \sin x, 0\} \) is not differentiable at the points where the functions inside the \(\max\) are equal or where the \(\max\) transitions from 0 to a non-zero value or vice versa.
First, consider where \( \cos x = \sin x \).
This occurs when \( \tan x = 1 \), which in the interval \( (0, 2\pi) \) are \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).
Next, consider where \( \cos x = 0 \).
This occurs at \( x = \frac{\pi}{2}, \frac{3\pi}{2} \) in \( (0, 2\pi) \).
Finally, consider where \( \sin x = 0 \).
This occurs at \( x = \pi \) in \( (0, 2\pi) \).
Now, let's analyze the sign of \( \cos x \) and \( \sin x \) in \( (0, 2\pi) \): - In \( (0, \frac{\pi}{2}) \), \( \cos x>0 \) and \( \sin x>0 \).
\( \max\{\cos x, \sin x, 0\} = \max\{\cos x, \sin x\} \), non-differentiable at \( x = \frac{\pi}{4} \).
- In \( (\frac{\pi}{2}, \pi) \), \( \cos x<0 \) and \( \sin x>0 \).
\( \max\{\cos x, \sin x, 0\} = \max\{\sin x, 0\} \), non-differentiable at \( x = \pi \) (transition from 0).
- In \( (\pi, \frac{3\pi}{2}) \), \( \cos x<0 \) and \( \sin x<0 \).
\( \max\{\cos x, \sin x, 0\} = 0 \), differentiable.
- In \( (\frac{3\pi}{2}, 2\pi) \), \( \cos x>0 \) and \( \sin x<0 \).
\( \max\{\cos x, \sin x, 0\} = \max\{\cos x, 0\} \), non-differentiable at \( x = \frac{3\pi}{2} \) (transition to 0).
In the interval \( (0, 2\pi) \), the points of non-differentiability are \( \frac{\pi}{4}, \pi, \frac{3\pi}{2} \).
There are 3 such points.
The interval is \( (0, 2024\pi) = (0, 1012 \times 2\pi) \).
This interval contains 1012 cycles of \( 2\pi \).
In each cycle of \( 2\pi \), there are 3 points of non-differentiability.
So, the total number of points of non-differentiability in \( (0, 2024\pi) \) is \( 1012 \times 3 \).
Given that the number of points is \( 1012k \), we have \( 1012k = 1012 \times 3 \), which implies \( k = 3 \).