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:
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:
Show Hint
A function is odd if \( f(-x) = -f(x) \) and even if \( f(-x) = f(x) \).
Updated On: Feb 4, 2025
- an odd function
- an even function
- neither odd nor even
- None of these
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The Correct Option is A
Solution and Explanation
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.
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