Question

let A = {-2,-1,0,1,2} and f : A➡️ z given by f(x)= x^2-2x-3 then pre image of 5 is ​

Answer 1

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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