Question -
Answer -
Given:
A = {–2, –1, 0, 1, 2}
f : A → Z such that f(x) = x2 –2x – 3
(i) Rangeof f i.e. f (A)
A is the domain of the function f. Hence, range is the set of elementsf(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) = 12 – 2(1) – 3
= 1 – 2 – 3
= – 4
Substituting x = 2 in f(x), we get
f(2) = 22 – 2(2) – 3
= 4 – 4 – 3
= –3
Thus, the range of f is {-4, -3, 0, 5}.
(ii) pre-imagesof 6, –3 and 5
Let 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
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.
∴Ø, {0, 2}, -2 are the pre-images of 6, -3, 5