Inspired by Arvind Kejriwal

Relevant wiki: Even and Odd Functions

Given $F(x) = (3012 - x)\lfloor x^2 \rfloor +\dfrac{1}{\sqrt{1-x^2}}$

Now the part of the function $\dfrac{1}{\sqrt{1-x^2}}$ is defined in the interval $x \in (-1,1)$ in which $x^2 \in (0,1)$ therefore $\lfloor x^2 \rfloor= 0$

So the function clearly becomes $\dfrac{1}{\sqrt{1-x^2}}$ in its domain which is clearly an $\boxed{\text{even function}}$