# Problem 10 Solution

This page contains the NCERT mathematics class 12 chapter Relations and Functions Exercise 1.2 Problem 10 Solution. Solutions for other problems are available at Exercise 1.2 Solutions
Exercise 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_2A 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 yB in the co-domain should be a image of an element xA 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 yB in the co-domain is an image of an element xA
f is onto.