Given function

$f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}$

where $\lceil x\rceil$ is the least integer greater than or equal to $x$.

Find the domain $A$

Since the square root is in the denominator, we need

$\lceil x\rceil-x>0$

because:

• inside the square root must be positive,
• it cannot be $0$ since it is in the denominator.

Now for any real $x$:

• if $x\notin \mathbb N$? More generally, if $x\in \mathbb Z$, then $\lceil x\rceil=x$, so $\lceil x\rceil-x=0$ and $f(x)$ is not defined.
• if $x\notin \mathbb Z$, then $\lceil x\rceil>x$, hence $0<\lceil x\rceil-x<1.$

Therefore the function is defined for all non-integers and undefined at integers.

Hence $A=\mathbb R\setminus \mathbb Z.$

Find the range $B$

Let $t=\lceil x\rceil-x.$

For $x\notin \mathbb Z$, we have $0<t<1.$

Then $f(x)=\frac{1}{\sqrt t}.$

Since $t\in(0,1)$:

• as $t\to 1^{-}$, $f(x)\to 1^{+}$,
• as $t\to 0^{+}$, $f(x)\to \infty$.

Also every value $t\in(0,1)$ is achievable. For example, if $x=n+\alpha$ with $n\in\mathbb Z$ and $\alpha\in(0,1)$, then

Check statement (S1)

We need $A\cap B=(\mathbb R\setminus \mathbb Z)\cap (1,\infty).$

This is simply all real numbers greater than $1$ that are not integers:

$A\cap B=(1,\infty)\setminus \mathbb Z.$

Since integers greater than $1$ are precisely natural numbers from $2,3,4,\dots$, this is commonly written as

$A\cap B=(1,\infty)-\mathbb N.$

So (S1) is true.

Check statement (S2)

We need $A\cup B=(\mathbb R\setminus \mathbb Z)\cup (1,\infty).$

Now:

• all non-integers are already in $A$,
• integers greater than $1$ are included through $B$,
• but integers $\le 1$ are not in $A$ and not in $B$.

Thus $A\cup B=\mathbb R\setminus \{\ldots,-2,-1,0,1\}.$

This is not equal to $(1,\infty)$, because many numbers less than or equal to $1$ but non-integral (for example $\tfrac12$, $-\pi$) belong to $A\cup B$.

Hence (S2) is false.

Conclusion

• (S1) is true
• (S2) is false

Therefore the correct option is

$\boxed{\text{B}}$

i.e. only (S1) is true.