Functions - Domain & Range

The domain of $f(x)$ requires that $x-\lfloor x \rfloor > 0$ , which is only satisfied for real $x \notin \mathbb{Z}$ . Thus, $A = x \in \mathbb{R} \backslash \mathbb{Z}$ .

If we now consider the sub-interval $(n,n+1)$ and a positive value $k$ (where $n \in \mathbb{Z}, k \in (0,1)$ ) on the domain of $f(x)$ , then $\ln((n+1-k) - \lfloor n+1-k \rfloor) = \ln((n+1-k) - n) = \ln(1-k)$ . We now find the following limits yield:

$lim_{k \rightarrow 0} \ln(1-k) \rightarrow 0;$

$lim_{k \rightarrow 1} \ln(1-k) \rightarrow -\infty;$

which the range of $f(x)$ is the set $B = f(x) \in (-\infty, 0).$

We now obtain $A \cap B = [\mathbb{R} \backslash \mathbb{Z}] \cap (-\infty,0) = (-\infty,0) \backslash \mathbb{Z}$ . Thus, $[-10,-9,...,0] \notin A \cap B \Rightarrow$ all eleven of these integers are not elements of $A \cap B$ .

11 is the only possible answer. The domain of ln(any variable) is,"any variable" must be greater than zero. X-[X] has only 3 possible answers.

Case 1:

X€I or X=0

Then X-[X]=0.

Case 2

X is not a Integer

Then X-[X]= {X}

{} Represent fractional part of X.

This is always positive

Hence it can be said that integers are not included in the domain of the function.

Therefore you must remove all the integers from the given interval and zero (many mathandticians do not consider zero a Integer) then you will have to the answer.

P.S. If this wasn't helpful, i would appreciate if you emailed me for further explanation.

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$A = (0,\infty)$ and $B = (-\infty,0)$

$A \cap B = \phi$

Hence, no integer is there.

Raj Magesh , please change the question from 'numbers' to 'integers'. It would be more apt.