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Question
Let A = R – {2} and B = R – {1}. If f: A `→` B is a function defined by f(x) = `(x - 1)/(x - 2)` then show that f is a one-one and an onto function.
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Solution
Let A = R – {2}, B = R – {1}
f: A `→` B s.t. f(x) = `(x - 1)/(x - 2)`
For x1, x2 ∈ A
f(x1) = f(x2)
`\implies (x_1 - 1)/(x_1 - 2) = (x_2 - 1)/(x_2 - 2)`
`\implies (x_1 - 1)/(x_1 - 2) - 1 = (x_2 - 1)/(x_2 - 2) - 1`
`\implies 1/(x_1 - 2) = 1/(x_2 - 2)`
`\implies` x1 – 2 = x2 – 2
`\implies` x1 = x2
∴ f(x) is one-one function.
Also if f(x) = y, where y ∈ B.
`\implies (x - 1)/(x - 2)` = y
`\implies` x – 1 = xy – 2y
`\implies` 2y – 1 = xy – x
`\implies` 2y – 1 = x(y – 1)
x = `(2y - 1)/(y - 1) ∈ A`
Clearly every element y ∈ B is associated to x = `(2y - 1)/(y - 1)` of set A.
So Range of f = B `\implies` f is into
Hence f is one-one and onto function.
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