Possible values of z? Let $f:[-2,2]\to \mathbb {R} $ where $$f(x)=x^3+2(\sin x)^5+3(\tan x)^7+\left\lfloor\frac{x^2+1}{z}\right\rfloor $$ is an odd function then what are possible values of $z$?
$\lfloor\cdot\rfloor $ is the floor function.
 A: There are some troubles with the domain ($\pi/2 \in [-2,2]$ but $\tan(\pi/2)$ is not defined). But I will ignore these.
Now, $f$ is odd if and only if for all $x$
$$f(x)+f(-x)=0$$
Now, you have
$$f(x)+f(-x) = \dots = 2 \left[\frac{x^2+1}{z} \right]$$
This can be identically zero when $0 \le \frac{x^2+1}{z} < 1$ for all $x$. From this you have $z >0$. For such $z$, you have that $\frac{x^2+1}{z}$ has a maximum at $x=2$, so you need 
$$\frac{2^2+1}{z}<1$$
i.e. $z>5$.
A: We know that the functions $x^n$ for $n$ odd, $\sin x$ and $\tan x$ are all odd functions. Furthermore, if $h$ and $g$ are odd, then $h(g(-x)) = h(-g(-x))=-h(g(-x))$, so the composition of two odd functions is again odd.
And we can therefore conclude that $x^3+ 2 (\sin x)^5 + 3 ( \tan x)^7$ is an odd function. So for $f$ to be odd, we want $y(x)=[\frac{x^2+1}{z}]$ to be odd. 
Since $x^2+1$ is non negative, we have for positive $z$ that $[\frac{x^2+1}{z}] \ge 0$ for all $x \in [-2,2]$. And for negative $z$ we have $y(x) \le 0$. So for $y$ to be odd, we must find $z$ such that $[\frac{x^2+1}{z}] =0$. 
We have $x^2+1 \in [1,5]$, so we want $z$ such that $[1/z,5/z] \subset [0, 1)$. So if $z >5$ we have $[1/z,5/z] \subset [0,1)$. 
Thus for $z >5 $ we have $f$ is odd.
