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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.
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 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) in A` [`y != 1`]
Thus, for any y ∈ B, there exists `(2 - 3y)/(1 - y) in 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)`
Hence, function f is one - one and onto.
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