Question

If \( f(x) = \sin[\lfloor x^2 \rfloor] - \sin[\lfloor -x^2 \rfloor] \), where \( \lfloor x \rfloor \) denotes the greatest integer less than or equal to \( x \), then which of the following is not true?

• A. \( f(\frac{\pi}{2}) = 1 \)
• B. \( f(\frac{\pi}{4}) = 1 + \frac{1}{\sqrt{2}} \)
• C. \( f(\pi) = -1 \)
• D. \( f(0) = 0 \)

Hint:

In problems involving floor functions, remember to carefully compute the greatest integer less than or equal to the value for both positive and negative inputs.

Answer 1

The Correct Option is C

Given the function: \[ f(x) = \sin[\lfloor x^2 \rfloor] - \sin[\lfloor -x^2 \rfloor] \] Let’s evaluate the options: - \( f(\frac{\pi}{2}) \): For \( x = \frac{\pi}{2} \), we have: \[ x^2 = \left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4} \] Since \( \lfloor \frac{\pi^2}{4} \rfloor = 1 \), and \( \lfloor -\frac{\pi^2}{4} \rfloor = -2 \), we get: \[ f\left(\frac{\pi}{2}\right) = \sin(1) - \sin(-2) \] The answer holds true. - \( f(\frac{\pi}{4}) \): For \( x = \frac{\pi}{4} \), we have: \[ x^2 = \left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{16} \] Since \( \lfloor \frac{\pi^2}{16} \rfloor = 0 \), and \( \lfloor -\frac{\pi^2}{16} \rfloor = -1 \), we get: \[ f\left(\frac{\pi}{4}\right) = \sin(0) - \sin(-1) \] The answer holds true. - \( f(\pi) \): For \( x = \pi \), we have: \[ x^2 = \pi^2 \] Since \( \lfloor \pi^2 \rfloor = 9 \), and \( \lfloor -\pi^2 \rfloor = -10 \), we get: \[ f(\pi) = \sin(9) - \sin(-10) \] The value is not equal to \( -1 \), so this statement is false. - \( f(0) \): For \( x = 0 \), we have: \[ x^2 = 0 \] Since \( \lfloor 0 \rfloor = 0 \), and \( \lfloor 0 \rfloor = 0 \), we get: \[ f(0) = \sin(0) - \sin(0) = 0 \] The answer holds true. Thus, option (3) is the correct answer as it is false.