Interpret the function

The given function is $f(x)=\frac{\cos\left(\sec^{-1}x\right)}{\sqrt{x-[x]}}$ where $[x]$ is the greatest integer function.

Condition from the denominator

Since the denominator is $\sqrt{x-[x]},$ we need:

• $x-[x]\ge 0$ for the square root to exist, which is always true because $x-[x]$ is the fractional part of $x$.
• But since it is in the denominator, we must also have $x-[x] \ne 0.$

Now, $x-[x]=0$ exactly when $x$ is an integer.

Therefore, from this part, $x \notin \mathbb{Z}.$

Condition from $\sec^{-1}x$

For $\sec^{-1}x$ to be defined as a real value, we need $|x|\ge 1.$

That is, $x\in (-\infty,-1]\cup[1,\infty).$

Combine both conditions

We need both:

• $x$ is not an integer
• $|x|\ge 1$

So the domain is $\{x\in \mathbb{R}: |x|\ge 1,\ x\notin \mathbb{Z}\}.$

This means:

• all non-integers less than or equal to $-1$
• all non-integers greater than or equal to $1$

Since $-1$ and $1$ themselves are integers, they are excluded anyway.

Hence the domain can be described as: all non-integers except the interval $[-1,1]$.

Check options

• A: all real except integers
  Incorrect, because $\sec^{-1}x$ is not defined for $|x|<1$.

• B: all non-integers except the interval $[-1,1]$
  Correct.

• C: all integers except $0,-1,1$
  Incorrect, integers are excluded due to $\sqrt{x-[x]}$ in denominator.

• D: all real except the interval $[-1,1]$
  Incorrect, this includes non-integer values outside that interval, but also includes integers like $2,3,-2$, which are not allowed.

Final answer

$\boxed{\text{B}}$