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