Question

The function \[ f(x) = \frac{\cos x}{\left\lfloor \frac{2x}{\pi} \right\rfloor + \frac{1}{2}}, \] where \( x \) is not an integral multiple of \( \pi \) and \( \lfloor \cdot \rfloor \) denotes the greatest integer function, is:

• A. an odd function
• B. an even function
• C. neither odd nor even
• D. None of these

Hint:

A function is odd if \( f(-x) = -f(x) \) and even if \( f(-x) = f(x) \).

Answer 1

The Correct Option is A

Step 1: {Compute \( f(-x) \)} 
\[ f(-x) = \frac{\cos(-x)}{\left\lfloor \frac{2(-x)}{\pi} \right\rfloor + \frac{1}{2}}. \] Using \( \cos(-x) = \cos x \) and property of floor function: \[ \left\lfloor \frac{2(-x)}{\pi} \right\rfloor = - \left\lfloor \frac{2x}{\pi} \right\rfloor - 1. \] Step 2: {Compare \( f(-x) \) with \( -f(x) \)} 
\[ f(-x) = -f(x). \] Step 3: {Conclusion} 
Since \( f(-x) = -f(x) \), the function is odd.

Answer 2

Step 1: Check if the function is odd
We are given the function \( f(x) = \frac{\cos x}{\left\lfloor \frac{2x}{\pi} \right\rfloor + \frac{1}{2}} \), where \( x \) is not an integral multiple of \( \pi \) and \( \lfloor \cdot \rfloor \) denotes the greatest integer function. To check if the function is odd, we need to verify if \( f(-x) = -f(x) \). Let's compute \( f(-x) \): \[ f(-x) = \frac{\cos(-x)}{\left\lfloor \frac{2(-x)}{\pi} \right\rfloor + \frac{1}{2}}. \] Using the identity \( \cos(-x) = \cos(x) \), we have: \[ f(-x) = \frac{\cos x}{\left\lfloor \frac{-2x}{\pi} \right\rfloor + \frac{1}{2}}. \] Now, consider the behavior of the greatest integer function \( \lfloor \cdot \rfloor \). For negative values of \( x \), the floor function \( \left\lfloor \frac{-2x}{\pi} \right\rfloor \) will be equal to \( \left\lfloor -\frac{2x}{\pi} \right\rfloor \), which is the negative of \( \left\lfloor \frac{2x}{\pi} \right\rfloor \). Therefore: \[ \left\lfloor \frac{-2x}{\pi} \right\rfloor = -\left\lfloor \frac{2x}{\pi} \right\rfloor - 1. \] Substitute this back into \( f(-x) \): \[ f(-x) = \frac{\cos x}{-\left\lfloor \frac{2x}{\pi} \right\rfloor - 1 + \frac{1}{2}}. \] Simplify: \[ f(-x) = \frac{\cos x}{-\left\lfloor \frac{2x}{\pi} \right\rfloor - \frac{1}{2}}. \] Now, if you observe carefully, this is the negative of the original expression for \( f(x) \). Thus: \[ f(-x) = -f(x). \] Step 2: Conclusion
Since \( f(-x) = -f(x) \), we conclude that the function is odd. Final Answer:
The function is:

an odd function