Question 74

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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