Let the sets $$A$$ and $$B$$ denote the domain and range respectively of the function $$f(x) = \dfrac{1}{\sqrt{[x] - x}}$$, where $$[x]$$ denotes the smallest integer greater than or equal to $$x$$. Then among the statements
(S1): $$A \cap B = (1, \infty) - \mathbb{N}$$ and
(S2): $$A \cup B = (1, \infty)$$

Given: $$f(x) = \frac{1}{\sqrt{\lceil x \rceil - x}}$$ where $$\lceil x \rceil$$ denotes the smallest integer greater than or equal to $$x$$ (ceiling function).

Domain $$A$$: We need $$\lceil x \rceil - x > 0$$, which holds for all non-integer $$x$$.

For integer $$x$$: $$\lceil x \rceil = x$$, so $$\lceil x \rceil - x = 0$$. Not in domain.

$$A = \mathbb{R} \setminus \mathbb{Z}$$

Range $$B$$: For non-integer $$x$$ with $$\lfloor x \rfloor = n$$, the fractional part $$\{x\} \in (0, 1)$$, and $$\lceil x \rceil - x = 1 - \{x\}$$.

As $$\{x\}$$ ranges over $$(0, 1)$$: $$1 - \{x\} \in (0, 1)$$, so $$\frac{1}{\sqrt{1 - \{x\}}} \in (1, \infty)$$.

$$B = (1, \infty)$$

Checking (S1): $$A \cap B = (\mathbb{R} \setminus \mathbb{Z}) \cap (1, \infty) = (1, \infty) \setminus \{2, 3, 4, \ldots\} = (1, \infty) \setminus \mathbb{N}$$.

The statement says $$A \cap B = (1, \infty) - \mathbb{N}$$. This is correct since $$\mathbb{N} \cap (1, \infty) = \{2, 3, 4, \ldots\}$$. (S1) is TRUE.

Checking (S2): $$A \cup B = (\mathbb{R} \setminus \mathbb{Z}) \cup (1, \infty)$$.

This union contains non-integers less than or equal to 1 (like $$0.5$$, $$-0.5$$, etc.), so $$A \cup B \neq (1, \infty)$$.

For example, $$0.5 \in A$$ but $$0.5 \notin (1, \infty)$$. (S2) is FALSE.

Only (S1) is true.

The correct answer is Option B.