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**NCERT mathematics class 12 chapter Relations and Functions Exercise 1.2 Problem 10 Solution**. Solutions for other problems are available at Exercise 1.2 SolutionsExercise 1.2 Problem 10 Solution

10. Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by {f(x) = \left(\dfrac{x - 2}{x - 3}\right)}. Is f one-one and onto? Justify your answer.

The function f : A → B is given as

{f(x) = \left(\dfrac{x - 2}{x - 3}\right)}

{⇒ f(x) = \left(\dfrac{x - 3 + 1}{x - 3}\right)}

{⇒ f(x) = \left(\dfrac{x - 3}{x - 3} + \dfrac{1}{x - 3}\right)}

{⇒ f(x) = 1 + \dfrac{1}{x - 3}}

To Check whether f is one-one:

Let x_1, x_2 ∈ A in the domain such that

{f(x_1) = f(x_2)}

{⇒ 1 + \dfrac{1}{x_1 - 3} = 1 + \dfrac{1}{x_2 - 3}}

{⇒ \dfrac{1}{x_1 - 3} = \dfrac{1}{x_2 - 3}}

⇒ {x_1 - 3 = x_2 - 3}

⇒ {x_1 = x_2}

∴ f is one-one.

To Check whether f is onto.

For f to be onto, for every element y ∈ B in the co-domain should be a image of an element x ∈ A in the domain such that

{y = f(x) = 1 + \dfrac{1}{x - 3}}

{⇒ \dfrac{1}{x - 3} = y - 1}

{⇒ x - 3 = \dfrac{1}{y - 1}}

{⇒ x = 3 + \dfrac{1}{y - 1}}

The above expression is valid for all real values of y, except 1 (∵ {y - 1 = 0} when when {y = 0} and division by 0 is not defined)

It is given in the problem that 1 ∉ B,

⇒ Every element y ∈ B in the co-domain is an image of an element x ∈ A

∴ f is onto.