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

Concept: The greatest integer function $[g(x)]$ always outputs a pure integer value (e.g., $0, \pm 1, \pm 2, \dots$) for any real input. We can evaluate trigonometric functions containing integer arguments by using standard properties of periodic functions, such as knowing that the tangent function vanishes at all integer multiples of $\pi$: $$\tan(k\pi) = 0 \quad \forall k \in \mathbb{Z}$$ Step 1: Analyze the numerator function expression.
The term inside the tangent function argument is $\pi \cdot [x - \pi]$. By definition, the floor bracket function $[x - \pi]$ outputs a pure integer value for any real number $x$: $$[x - \pi] = k \quad \text{where } k \in \mathbb{Z}$$ Substitute this integer definition back into the numerator: $$\text{Numerator} = \tan(\pi \cdot k)$$ Since the tangent function is zero at every integer multiple of $\pi$, the numerator is exactly zero for all real numbers: $$\text{Numerator} = 0 \quad \forall x \in \mathbb{R}$$

Step 2: Evaluate the complete function $f(x)$.
Now look at the denominator term, $1 + [x]^2$. Since the square of any real number is non-negative ($[x]^2 \ge 0$), the denominator is always greater than or equal to 1, meaning it can never cause a division-by-zero error: $$\text{Denominator} \ge 1 \quad \forall x \in \mathbb{R}$$ Since the numerator is always zero and the denominator is never zero, the entire function collapses to a simple constant function: $$f(x) = \frac{0}{1 + [x]^2} = 0 \quad \forall x \in \mathbb{R}$$

Step 3: Determine the derivative function $f'(x)$.
Since $f(x) = 0$ is a perfectly flat, constant function across the entire real number line, its derivative is also zero everywhere: $$f'(x) = \frac{d}{dx}(0) = 0 \quad \forall x \in \mathbb{R}$$ Because the derivative is zero at every coordinate point, the derivative $f'(x)$ exists everywhere across the entire domain, matching option (A).