Question

For a real number \(y\), consider \([y]\) denotes the greatest integer less than or equal to \(y\). If \[ f(x)=\frac{\tan(\pi[x-\pi])}{1+[x]^{2}}, \] then:

• A. $f^{\prime}(x)$ exists for all x
• B. $f^{\prime}(x)$ does not exist
• C. $f^{\prime}(1)=\frac{\pi}{4}$
• D. $f^{\prime}(1)=-\frac{\pi}{4}$

Hint:

Do not let the discontinuous floor brackets fool you into thinking the function is not differentiable! Because the numerator stays locked at zero across all interval transitions, it completely flattens out any step discontinuities from the brackets, leaving a smooth constant line.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The greatest integer function \( [x] \) always outputs an integer. The expression \( [x-\pi] \) is an integer for any real \( x \). We know that \( \tan(n\pi) = 0 \) for any integer \( n \).
Step 2: Key Formula or Approach:
1. Evaluate the numerator \( \tan(\pi[x-\pi]) \).
2. Determine the value of the function \( f(x) \) for all real \( x \).
Step 3: Detailed Explanation:
Let \( k = [x-\pi] \). Since \( k \) is an integer for any value of \( x \), the numerator of the function is:
\[ \tan(\pi k) = 0 \]
Thus, the function \( f(x) \) is:
\[ f(x) = \frac{0}{1+[x]^2} = 0 \]
This holds for all real \( x \) because the denominator \( 1+[x]^2 \) is always at least 1 and never zero.
Since \( f(x) \) is a constant function \( (f(x) = 0) \), it is continuous and differentiable everywhere.
The derivative \( f'(x) = 0 \) for all \( x \).
Step 4: Final Answer:
The function is identically zero, so its derivative exists for all \( x \).