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Question
Let A = {–1, 0, 1, 2}, B = {–4, –2, 0, 2} and f, g : A → B be functions defined by f(x) = x2 – x, x ∈ A and g(x) = `2|x - 1/2| – 1`, x ∈ A. Are f and g equal?
Justify your answer. (Hint: One may note that two functions f : A → B and g : A → B such that f(a) = g(a) ∀ a ∈ A are called equal functions.)
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Solution
It is given that A = {–1, 0, 1, 2}, B = {–4, –2, 0, 2}.
Also, it is given that f, g: A → B are defined by f(x) = x2 – x, x ∈ A and `g(x) = 2|x - 1/2| - 1, x ∈ A`.
It is observed that:
When x = –1,
f(–1) = 12 – (–1)
= 1 + 1
= 2
g(–1) = `2|(-1)-1/2| - 1`
= `2(3/2) - 1`
= 3 – 1
= 2
⇒ f(–1) = g(–1)
When x = 0,
f(0) = (0)2 – 0 = 0
g(0) = `2|0 - 1/2| - 1`
= `2(1/2) - 1`
= 1 – 1
= 0
⇒ f(0) = g(0)
When x = 1,
f(1) = (1)2 – 1
= 1 – 1
= 0
g(1) = `2|1 - 1/2| - 1`
= `2(1/2) - 1`
= 1 – 1
= 0
⇒ f(1) = g(1)
When x = 2,
f(2) = (2)2 – 2
= 4 – 2
= 2
g(2) = `2|2-1/2| - 1`
= `2(3/2)-1`
= 3 – 1
= 2
⇒ f(2) = g(2)
∴ f(a) = g(a) ∀ a ∈ A
Hence, the functions f and g are equal.
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