Let's analyze each function: ### 1. \( f = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 4 & 3 & 1 & 2 \end{pmatrix} \) - **Domain:** {1, 2, 3, 4, 5} - **Codomain:** {1, 2, 3, 4, 5} - **Mapping:** 1→2, 2→4, 3→3, 4→1, 5→2 - **Injective?** No, since 1 and 5 both map to 2. - **Surjective?** No, 5 is not in the range. - **Answer:** **Neither** ### 2. \( f = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 4 & 4 & 2 & 3 & 1 \end{pmatrix} \) - **Mapping:** 1→4, 2→4, 3→2, 4→3, 5→1 - **Injective?** No, since 1 and 2 both map to 4. - **Surjective?** Yes, range is {1, 2, 3, 4}. - **But codomain is {1, 2, 3, 4, 5}.** So 5 is missing, not surjective. - **Correction:** **Neither** ### 3. \( f = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 4 & 3 & 1 & 5 \end{pmatrix} \) - **Mapping:** 1→2, 2→4, 3→3, 4→1, 5→5 - **Injective?** Yes, all outputs are unique. - **Surjective?** Yes, covers all elements in codomain. - **Answer:** **Bijective** ### 4. \( f(x) = \begin{cases} 4-x & \text{if } x<4 \\ 9-x & \text{if } x \geq 4 \end{cases} \) - Let's check for x = 1, 2, 3, 4, 5: - x = 1: 4-1 = 3 - x = 2: 4-2 = 2 - x = 3: 4-3 = 1 - x = 4: 9-4 = 5 - x = 5: 9-5 = 4 - Outputs: {1, 2, 3, 4, 5} - **Injective?** Yes, all outputs are unique. - **Surjective?** Yes, all elements of codomain are hit. - **Answer:** **Bijective** **Summary of correct answers:** 1. Neither 2. Neither (not Surjective) 3. Bijective 4. Bijective **Note:** You have marked 2 as "Surjective", but it should be "Neither".