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Let A = {–2, –1, 0, 1, 2} and f: A → Z be a function defined by f(x) = x2 – 2x – 3. Find:
(i) range of f i.e. f (A)
(ii) pre-images of 6, –3 and 5

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Given as

A = {–2, –1, 0, 1, 2}

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

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

By substituting x = –2 in f(x), we get

f(–2) = (–2)2 – 2(–2) – 3

= 4 + 4 – 3

= 5

By substituting x = –1 in f(x), we get

f(–1) = (–1)2 – 2(–1) – 3

= 1 + 2 – 3

= 0

By substituting x = 0 in f(x), we get

f(0) = (0)2 – 2(0) – 3

= 0 – 0 – 3

= – 3

By substituting x = 1 in f(x), we get

f(1) = 12 – 2(1) – 3

= 1 – 2 – 3

= – 4

By substituting x = 2 in f(x), we get

f(2) = 22 – 2(2) – 3

= 4 – 4 – 3

= –3

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

(ii) pre-images of 6, –3 and 5

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

 x2 – 2x – 3 = 6

x2 – 2x – 9 = 0

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

= [2 ±  (4+36)] / 2

= [2 ± 40] / 2

= 1 ± 10

However, 1 ± 10 ∉ A

Hence, there exists no pre-image of 6.

Then, suppose 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

It is clear, both 0 and 2 are elements of A.

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

Then, suppose 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

Hence, –2 is the pre-images of 5.

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

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