k:X → Y, X = {t| t ∈ Z, -3 < t ≤ 3};Y = {a| a ∈ N, 1 ≤ a ≤

1.	k:X → Y, X = {t\| t ∈ Z, -3 < t ≤ 3};Y = {a\| a ∈ N, 1 ≤ a ≤ 20}; k(t) = t2 + 2.State the type of function k.
Answer» k:X → Y Domain X={t\| t ∈ Z, -3 ≤ t ≤ 3) = {-3, -2, -1, 0, 1, 2, 3} k (t) = t² + 2 ∴ k(- 3) = (-3)² + 2 = 9 + 2 = 11, k (3) = (3)² + 2=11, k (- 2) = (- 2)² + 2 = 4 + 2 = 6, k (2) = (2)² + 2 = 6, k(- 1) = (- 1)² + 2 = 1 + 2 = 3,k (1) = (1)² + 2 = 3, k (0) = 0 + 2 = 2 Co-domain Y = {a\|a ∈ N ⇒ 1, a ≤ 20} = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, …, 18, 19, 20} Here, for two different elements (-3, 3) of domain, their images in co-domain are k (- 3) = k (3) = 11. Therefore, function k is many one function.

k:X → Y, X = {t| t ∈ Z, -3 < t ≤ 3};Y = {a| a ∈ N, 1 ≤ a ≤ 20}; k(t) = t2 + 2.State the type of function k.

Answer»

k:X → Y

Domain X={t| t ∈ Z, -3 ≤ t ≤ 3)

= {-3, -2, -1, 0, 1, 2, 3}

k (t) = t² + 2

∴ k(- 3) = (-3)² + 2 = 9 + 2 = 11,

k (3) = (3)² + 2=11,

k (- 2) = (- 2)² + 2 = 4 + 2 = 6,

k (2) = (2)² + 2 = 6,

k(- 1) = (- 1)² + 2 = 1 + 2 = 3,k (1) = (1)² + 2 = 3,

k (0) = 0 + 2 = 2

Co-domain Y = {a|a ∈ N ⇒ 1, a ≤ 20}

= {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, …, 18, 19, 20}

Here, for two different elements (-3, 3) of domain, their images in co-domain are k (- 3) = k (3) = 11. Therefore, function k is many one function.