Given A = {–2, –1, 0, 1, 2}

f : A → Z such that f(x) = x² – 2x – 3

i. range of f i.e. f(A)

A is the domain of the function f. Hence, range is the set of elements f(x) for all x ∈ A.

Substituting x = –2 in f(x), we get

f(–2) = (–2)² – 2(–2) – 3

⇒ f(–2) = 4 + 4 – 3

∴ f(–2) = 5

Substituting x = –1 in f(x), we get

f(–1) = (–1)² – 2(–1) – 3

⇒ f(–1) = 1 + 2 – 3

∴ f(–1) = 0

Substituting x = 0 in f(x), we get

f(0) = (0)² – 2(0) – 3

⇒ f(0) = 0 – 0 – 3

∴ f(0) = –3

Substituting x = 1 in f(x), we get

f(1) = 1² – 2(1) – 3

⇒ f(1) = 1 – 2 – 3

∴ f(1) = –4

Substituting x = 2 in f(x), we get

f(2) = 2² – 2(2) – 3

⇒ f(2) = 4 – 4 – 3

∴ f(2) = –3

Thus, the range of f is {5, 0, –3, –4}.

ii. pre-images of 6, –3 and 5

Let x be the pre-image of 6 ⇒ f(x) = 6

⇒ x² – 2x – 3 = 6

⇒ x² – 2x – 9 = 0

However,

Thus, there exists no pre-image of 6.

Now, let x be the pre-image of –3 ⇒ f(x) = –3

⇒ x² – 2x – 3 = –3

⇒ x² – 2x = 0

⇒ x(x – 2) = 0

∴ x = 0 or 2

Clearly, both 0 and 2 are elements of A.

Thus, 0 and 2 are the pre-images of –3.

Now, let x be the pre-image of 5 ⇒ f(x) = 5

⇒ x² – 2x – 3 = 5

⇒ x² – 2x – 8= 0

⇒ x² – 4x + 2x – 8= 0

⇒ x(x – 4) + 2(x – 4) = 0

⇒ (x + 2)(x – 4) = 0

∴ x = –2 or 4

However, 4 ∉ A but –2 ∈ A

Thus, –2 is the pre-images of 5.