Answers:

A = {–2, –1, 0, 1, 2}

f : A → Z

f(x) = x² – 2x – 3

(i)

A – Domain of the function f.

The range is the set of elements f(x) for all x ∈ A.

Substitute, x = –2 in f(x)

f(–2) = (–2)² – 2(–2) – 3

f(–2) = 4 + 4 – 3

f(–2) = 5

Substitute, x = –1 in f(x)

f(–1) = (–1)² – 2(–1) – 3

f(–1) = 1 + 2 – 3

f(–1) = 0

Substitute, x = 0 in f(x)

f(0) = (0)² – 2(0) – 3

f(0) = 0 – 0 – 3

f(0) = – 3

Substitute, x = 1 in f(x)

f(1) = 1² – 2(1) – 3

f(1) = 1 – 2 – 3

f(1) = – 4

Substitute, x = 2 in f(x)

f(2) = 2² – 2(2) – 3

f(2) = 4 – 4 – 3

f(2) = –3

Hence, range of f is {-4, -3, 0, 5}.

(ii)

Consider,

x – pre-image of 6 ⇒ f(x) = 6

x² – 2x – 3 = 6

x² – 2x – 9 = 0

x = [-(-2) ± √ ((-2)² – 4(1) (-9))] / 2(1)

x = [2 ± √ (4+36)] / 2

x = [2 ± √40] / 2

x = 1 ± √10

1 ± √10 ∉ A

So there is no pre-image of 6.

x – 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.

So, 0 and 2 are the pre-images of –3.

x – 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

4 ∉ A but –2 ∈ A

Hence, –2 is the pre-images of 5.

∴ Ø, {0, 2}, -2 are the pre-images of 6, -3, 5