The function \[f{\text{ }}:{\text{ }}\left[ {0,\infty } \right){\text{ }} \to {\text{ }}\left[ {0,\infty } \right)\] defined by \[f(x) = \dfrac{{2x}}{{1 + 2x}}\] is
(1) one-one and onto
(2) one-one but not onto
(3) not one-one but onto
(4) neither one-one nor onto

Hint: We have to check whether the function f (x) is one - one or onto . We solve this question using the concept of one - one and onto functions . We should also have the knowledge of one - one function and about the onto function . The concept of range of the function is also used to determine the onto function .

Complete step-by-step answer:
Given :
Check for one - one function :
\[f(x) = \dfrac{{2x}}{{1 + 2x}}\]
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
\[\dfrac{{2{x_1}}}{{1 + 2{x_1}}} = \dfrac{{2{x_2}}}{{1 + 2{x_2}}}\]
Cross multiplying the terms , we get
\[\left( 2{{x}_{1}} \right)\left( 1+2{{x}_{2}} \right)=\left( 2{{x}_{2}} \right)\left( 1+2{{x}_{1}} \right)\]
\[\left( {{x_1}} \right) + 2 \times \left( {{x_1}} \right) \times \left( {{x_2}} \right){\text{ }} = {\text{ }}\left( {{x_2}} \right) + 2 \times \left( {{x_1}} \right) \times \left( {{x_2}} \right)\]
Cancelling the terms , we get
\[{x_1} = {x_2}\]
Hence proved that the function is a one - one function .
Check for onto function :
\[f(x) = \dfrac{{2x}}{{1 + 2x}}\]
A function is said to be onto if the domain of the function changed in terms of independence of x is equal to the domain of the original function .
Proof :
Let us consider that \[f\left( x \right){\text{ }} = {\text{ }}y\]
Then , we get
\[f(x) = \dfrac{{2x}}{{1 + 2x}}\]
Now , making the function independent of x
Cross multiplying , we get
\[y + 2xy = 2x\]
Taking terms with x on one side , we get
\[y = 2x - 2xy\]
Taking 2x common , we get
\[y = 2x(1 - y)\]
\[x = \dfrac{y}{{2(1 - y)}}\]
Domain of this function \[f\left( y \right){\text{ }} = {\text{ }}R - \left\{ 1 \right\}\]
Where R is a set of real numbers .
But the domain of the original function is \[\left[ {0,\infty } \right)\] .
As , the domain of \[f(y)\] is not equal to that of function \[f(x)\] . We conclude that the function is not an onto function .
( The pre - image of 1 does not exist )
Thus , we conclude that the function is one - one but not onto .
Hence , the correct option is \[(2)\].
So, the correct answer is “Option 2”.

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