Let $$f(x)=\log_{[x]}\{x\}$$ where $[.]$ denotes the greatest integer function and $\{.\}$ denotes the fractional part function. Select the correct options.
Options
$1.$ domain of $f(x)$ is $[2,\infty)$
$2.$ range of $f(x)$ is $[\log2,\infty)$
$3.$ period of $f(x)$ is $1$
$4.$ $f(x)$ is injective
My attempt
I quickly saw that option $1$ can't be true as it includes the set of integers as well. Then I saw that option $2$ also can't be true as there's no way $f(x)\to\infty$ as $\{x\}$ always lie between $[0,1)$
Then I also concluded that option $3$ is false as if the period is $1$ then $x=2.4$ and $x=3.4$ must give same values but it is not true. Lastly for option $4$, I don't have any mathematical proof but my intuition says that it's true.
But the answer given says that all four options are correct. How is this possible$?$ I mean I've contradictions for the first three. Any help is greatly appreciated.
For options $1,2,$ and $3,$ you are right. However, $f$ is not injective. We have:
$$1.\mathbb{dom}(f)=(2,\infty)\setminus\mathbb{N}.$$ $$2.\mathbb{range}(f)=(-\infty,0).$$
For $3,$ like you noted, $f(2.4)≠f(3.4),$ so the period cannot be $1.$ In fact, $f(2.4)=-\log_{2}(5)+1,$ and $f(3.4)=\log_3(2)-\log_3(5).$
For $4,$ graphing on desmos shows that any line parallel to the $x-$axis and below it intersects the graph of $f$ infinitely many times. As $\{x\}$ goes from $0$ to $1$ (it doesn't achieve the value $1,$ but you get what I'm saying)$, \log_{a}\left(\{x\}\right),a>1$ takes every value in $(-\infty,0).$ This should help you see why $f$ is not injective.