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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. - Mathematics

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प्रश्न

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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उत्तर

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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2018-2019 (March) 65/4/3

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