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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),` how that f is one-one and onto. Hence, find f−1.
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Solution
A =R − {2}, B = R − {1}
f: A → B is defined as `"f"("x") = ("x"-1)/("x"-2)`.
Let x, y ∈ A such that f (x) = f (y).
⇒ `("x"-1)/("x"-2) = ("y"-1)/("y"-2)`
⇒ ( x -1) (y - 2) = (x - 2) (y - 1)
⇒ xy - 2x - y + 2 = xy - x - 2y + 2
⇒ -2x - y = -x - 2y
⇒ 2x - x = 2y - y
⇒ 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.
Now,
f (x) = y
⇒`("x"-1)/("x"-2) = "y"`
⇒ x - 1 = y (x - 2)
⇒ x (1 - y) = 1 - 2y
⇒ `"x" = (1-2"y")/(1-"y")∈ "A"` .........[y ≠ 1]
Thus, for any y ∈ B, there exists `"x" = (1-2"y")/(1-"y")` ∈ A such that
`"f"((1-2"y")/(1-"y")) =((1-2"y")/(1-"y")-1)/((1-2"y")/(1-"y") - 2) = (1-2"y"-1+"y")/(1-2"y"-2+2"y") = (-"y")/-1 = "y"`
Therefore, f is onto.
Hence, function f is one-one and onto.
`"f"^-1("x") = (1-2"x")/(1-"x")`
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