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Question
Let A = R – {3}, B = R – {1}. Let f: A → B be defined by f(x) = `(x - 2)/(x - 3)` ∀ x ∈ A . Then show that f is bijective.
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Solution
Given that,
A = R – {3}, B = R – {1}
f: A → B be defined by f(x) = `(x - 2)/x` – 3 ∀ x ∈ A
∴ f(x) = `(x - 3 + 1)/(x - 3)` = 1 + `1/(x - 3)`
Let f(x1) = f (x2)
⇒ `1 + 1/(x_1 - 3) = 1 + 1/(x_2 - 3)`
⇒ `1/(x_1 - 3) = 1/(x_2 - 3)`
⇒ x1 = x2
So, (x) is an injection function
Now let y = `(x - 2)/(x - 3)`
⇒ x – 2 = xy – 3y
⇒ x(1 – y) = 2 – 3y
⇒ x = `(2 - 3y)/(1 - y)`
⇒ x = `(3y - 2)/(y - 1)`
⇒ y ∈ R – {1} = B
Since y ∈ R – {1} = B for every y ∈ B, there exists an x ∈ A such that f(x) = y.
Therefore, f(x) is surjective (onto).
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