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The domain of the function $$f(x) = \frac{1}{\sqrt{[x]^2 - 3[x] - 10}}$$ is (where $$[x]$$ denotes the greatest integer less than or equal to $$x$$)
Let
$$[x]=n,$$
where $$n$$ is an integer.
Since the square root is in the denominator, we require
$$n^2-3n-10>0.$$
Factorising,
$$(n-5)(n+2)>0.$$
Therefore,
$$n<-2\quad\text{or}\quad n>5.$$
Now,
$$[x]\le-3$$
implies
$$x<-2,$$
and
$$[x]\ge6$$
implies
$$x\ge6.$$
Hence, the domain of the function is
$$(-\infty,-2)\cup[6,\infty).$$
Therefore,
$$\boxed{(-\infty,-2)\cup[6,\infty)}$$
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