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Question
Let A = R − {3} and B = R − {1}. Consider the function f : A → B defined by f(x) = `((x- 2)/(x -3))`. Is f one-one and onto? Justify your answer.
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Solution
A = R − {3}, B = R − {1}
f : A → B is defined as f(x) = `((x -2)/(x -3))`
Let x, y ∈ A such that f(x) = f(y)
⇒ `(x -2)/(x - 3) = (y - 2)/(y - 3)`
⇒ (x − 2) (y − 3) = (y − 2) (x − 3)
⇒ xy − 3x − 2y + 6 = xy − 3y − 2x + 6
⇒ − 3x − 2y = − 3y − 2x
⇒ 3x − 2x = 3y − 2y
⇒ x = y
∴ f is one-one.
Let y ∈ B = R − {1}. Then, y ≠ 1.
The function f is onto if there exists an x ∈ A such that f(x) = y.
f(x) = y
⇒ `(x -2)/(x - 3) = y`
⇒ x − 2 = xy − 3y
⇒ x(1 − y) = −3y + 2
⇒ `x = (2-3y) / (1- y) ∈ A` ...[y ≠ 1]
Thus, for any y ∈ B, there exists `(2 - 3y)/(1 - y) ∈ A` such that:
`f(2 - 3y)/(1- y) = (((2-3y)/(1-y)) -2)/(((2-3y)/(1-y)) - 3)`
= `(2-3y - 2 + 2y)/(2-3y - 3 + 3y)`
= `(-y)/(-1)`
= y
∴ f is onto.
Hence, function f is one-one and onto.
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