Question

If $ f(x) = \dfrac{{[{x^2} - x]}}{{[{x^2} + 2x]}} $ , find the domain of\[f{\text{ }}\left( x \right)\]. Show that f is one - one .

Answer 1

Hint: We have to find the domain of the function f (x)\[f{\text{ }}\left( x \right)\]and show that the function f is one - one . We solve this question using the concepts of functions . We will find for what value of $ x $ the function is defined . We should also have the knowledge of one - one function .

Complete step-by-step answer:
Given :
For domain of function \[f{\text{ }}\left( x \right)\] :
 $ f(x) = \dfrac{{[{x^2} - x]}}{{[{x^2} + 2x]}} $
The domain of function is the value of x for which function \[f{\text{ }}\left( x \right)\] is defined .
For the function to be defined, the denominator should not be equal to zero for any value of $ x $ .
I.e.
 $ [{x^2} + 2x] \ne 0 $
Taking $ x $ common , we get
\[x{\text{ }} \times {\text{ }}\left[ {{\text{ }}x{\text{ }} + {\text{ }}2{\text{ }}} \right]{\text{ }} \ne {\text{ }}0\]
So,
\[x{\text{ }} \ne {\text{ }}0\]or \[\left[ {{\text{ }}x{\text{ }} + {\text{ }}2{\text{ }}} \right]{\text{ }} \ne {\text{ }}0\]
\[x{\text{ }} \ne {\text{ }} - 2\]or \[x{\text{ }} \ne {\text{ }}0\]
The function \[f\left( x \right)\] is defined for every value except for \[x{\text{ }} = {\text{ }}0\]and\[x{\text{ }} = {\text{ }} - {\text{ }}2\] .
Hence , the domain of the function \[f{\text{ }}\left( x \right)\]is\[R{\text{ }} - {\text{ }}\left\{ {{\text{ }}0{\text{ }},{\text{ }} - 2{\text{ }}} \right\}\] .
Where R is the set of real numbers .
To show that function is one - one :
A function is said to be a one - one function if we consider that \[f\left( {{x_1}} \right){\text{ }} = {\text{ }}f\left( {{x_2}} \right)\] and we get the result as \[x_1 = x_2\] .
Proof :
Let us consider that \[f\left( {{x_1}} \right){\text{ }} = {\text{ }}f\left( {{x_2}} \right)\]
Then , we get
$ \dfrac{{[{{({x_1})}^2} - {x_1}]}}{{[{{({x_1})}^2} + 2({x_1})]}} = \dfrac{{[{{({x_2})}^2} - {x_2}]}}{{[{{({x_2})}^2} + 2({x_2})]}} $
Cross multiplying the terms , we get
$ [{({x_1})^2} - {x_1}] \times [{({x_2})^2} + 2({x_2})] = [{({x_2})^2} - {x_2}] \times [{({x_1})^2} + 2({x_1})] $
Expanding the terms , we get
$ {({x_1})^2} \times {({x_2})^2} - {({x_2})^2} \times ({x_1}) + 2({x_2}) \times {({x_1})^2} - 2({x_1}) \times ({x_1}) = {({x_1})^2} \times {({x_2})^2} - {({x_1})^2} \times ({x_2}) + 2({x_1}) \times {({x_2})^2} - 2({x_1}) \times ({x_1}) $
Cancelling the terms , we get
$ - {({x_2})^2} \times ({x_1}) + 2({x_2}) \times {({x_1})^2} = - {({x_1})^2} \times ({x_2}) + 2({x_1}) \times {({x_2})^2}3({x_2}) \times {({x_1})^2} = 3{({x_2})^2} \times ({x_1}) $
Also , the expression can be written as
 $ ({x_2}) \times {({x_1})^2} - {({x_2})^2} \times ({x_1}) = 0 $
Taking \[\left( {{x_1}} \right){\text{ }} \times {\text{ }}\left( {{x_2}} \right)\]common , we get
\[\left( {{x_1}} \right){\text{ }} \times {\text{ }}\left( {{x_2}} \right){\text{ }}\left[ {{\text{ }}\left( {{x_1}} \right){\text{ }} - {\text{ }}\left( {{x_2}} \right){\text{ }}} \right]{\text{ }} = {\text{ }}0\]
So ,
\[\left( {{x_1}} \right){\text{ }} \times {\text{ }}\left( {{x_2}} \right){\text{ }} = {\text{ }}0{\text{ }}or{\text{ }}\left( {{x_1}} \right){\text{ }} - {\text{ }}\left( {{x_2}} \right){\text{ }} = {\text{ }}0\]
Neglecting \[\left( {{x_1}} \right){\text{ }} \times {\text{ }}\left( {{x_2}} \right){\text{ }} = {\text{ }}0\] as if \[\left( {{x_1}} \right){\text{ }} \times {\text{ }}\left( {{x_2}} \right){\text{ }} = {\text{ }}0\] then either \[\left( {{x_1}} \right){\text{ }} = {\text{ }}0\]or\[\left( {{x_2}} \right){\text{ }} = {\text{ }}0\] , which is not possible as \[x{\text{ }} \ne {\text{ }}0\] .
Prove above that the domain of \[f\left( x \right){\text{ }},{\text{ }}x\] cannot be zero .
So ,
\[\left( {{x_1}} \right){\text{ }} - {\text{ }}\left( {{x_2}} \right){\text{ }} = {\text{ }}0\]
Hence ,
\[\left( {{x_1}} \right){\text{ = }}\left( {{x_2}} \right)\]
Hence proved that the function is a one - one function .

Note: A function \[f{\text{ }}:{\text{ }}X{\text{ }} \geqslant {\text{ }}Y\]is defined to be one - one ( or injective ) , if the images of distinct elements of X under f are distinct , I.e. for every \[{x_1}{\text{ }},{\text{ }}{x_2} \in X{\text{ }},{\text{ }}f\left( {{x_1}} \right){\text{ }} = {\text{ }}f\left( {{x_2}} \right)\] implies that \[\left( {{x_1}} \right){\text{ = }}\left( {{x_2}} \right)\] . Otherwise , f is called many - one .
A function \[f{\text{ }}:{\text{ }}X{\text{ }} \geqslant {\text{ }}Y\] is defined to be onto ( or surjective ) , if every images of element of Y is the image of some element of X under f i.e. for every \[y \in Y\] , there exist an element x in X such that \[\;f{\text{ }}\left( x \right){\text{ }} = {\text{ }}y\] .