(a) (3 pts) Let [tex]f: \{2k \mid k \in \mathbb{Z}\} \to \mathbb{Z}[/tex] be defined by [tex]f(x) = y \in \mathbb{Z}[/tex] such that [tex]2y = x[/tex].

Determine the nature of [tex]f[/tex]:
A. One-to-one only
B. Onto only
C. Bijection
D. Not one-to-one or onto
E. Not a function

(b) (3 pts) Let [tex]R^+ \to \mathbb{R}[/tex] be defined by [tex]g(u) = v \in \mathbb{R}[/tex] such that [tex]v^2 = u[/tex].

Determine the nature of [tex]g[/tex]:
A. One-to-one only
B. Onto only
D. Not one-to-one or onto
E. Not a function

(c) (3 pts) Let [tex]h: \mathbb{R} - \{2\} \to \mathbb{R}[/tex] be defined by [tex]h(t) = 3t - 1[/tex].

Determine the nature of [tex]h[/tex]:
A. One-to-one only
B. Onto only
C. Bijection
D. Not one-to-one or onto
E. Not a function

(d) (3 pts) Let [tex]K: \{\mathbb{Z}, \mathbb{Q}, \mathbb{R} - \mathbb{Q}\} \to \{\mathbb{R}, \mathbb{Q}\}[/tex] be defined by [tex]K(A) = A \cup \mathbb{Q}[/tex].

Determine the nature of [tex]K[/tex]:
A. One-to-one only
B. Onto only
C. Bijection
D. Not one-to-one or onto
E. Not a function

To determine if f is one-to-one, we need to check if different inputs map to different outputs. In this case, for any given input x, there is a unique value y such that 2y = x. This means that no two different inputs can have the same output, satisfying the condition for one-to-one.

To determine if f is onto, we need to check if every element in the codomain (Z) is mapped to by at least one element in the domain ({2k | k ∈ Z}). In this case, for any y in Z, we can find an x such that 2y = x. Therefore, every element in Z has a preimage in the domain, satisfying the condition for onto.

Since f is both one-to-one and onto, it is a bijection.

Learn more about bijections

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