a. Since [tex]f:\Bbb Z\to\Bbb z[/tex], the function [tex]f(x)=2x^2+1[/tex] is not onto because not every element in the codomain [tex]\Bbb Z[/tex] is represented by [tex]f(x)[/tex]. For any [tex]x\in Z[/tex], we have [tex]x^2\ge0[/tex], so [tex]x^2+1\ge1[/tex], which means there is no choice of [tex]x[/tex] such that [tex]f(x)\le0[/tex].
b. [tex]f[/tex] is not one-to-one because there are infinitely many choices of [tex]x[/tex] for which [tex]f(-x)=f(x)[/tex]. For example, [tex]f(1)=1^2+1=2[/tex] and [tex]f(-1)=(-1)^2+1=2[/tex].
