Given:

A = {–2, –1, 0, 1, 2}

f : A → Z such that f(x) = x^{2} – 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

= 4 + 4 – 3

= 5

Substituting x = –1 in f(x), we get

f(–1) = (–1)^{2} – 2(–1) – 3

= 1 + 2 – 3

= 0

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

f(0) = (0)^{2} – 2(0) – 3

= 0 – 0 – 3

= – 3

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

f(1) = 1^{2} – 2(1) – 3

= 1 – 2 – 3

= – 4

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

f(2) = 2^{2} – 2(2) – 3

= 4 – 4 – 3

= –3

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

**(ii)** pre-images of 6, –3 and 5

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

x^{2} – 2x – 3 = 6

x^{2} – 2x – 9 = 0

x = \( -(-2) ± \sqrt{ (-2)^2– 4(1) (-9)} \)

=\( \dfrac{2 ± \sqrt{ (4+36)}}{2}\)

= \( \dfrac{2 ± \sqrt{ (40)}}{2}\)

= \( 1 ± \sqrt{10}\)

However, \( 1 ± \sqrt{10}\) ∉ A

Thus, there exists no pre-image of 6.

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

x^{2} – 2x – 3 = –3

x^{2} – 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

x^{2} – 2x – 3 = 5

x^{2} – 2x – 8= 0

x^{2} – 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.

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

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