Question

Give an example of a function which is:
(i) One-one but not onto
(ii) One-one and onto
(iii Neither one-one nor onto
(iv) Onto but not one-one

Answer 1

Hint:One-one function means every domain has distinct range i.e. mapping of elements of range and domain are unique. Mathematically, one-one is given for any function $f\left( x \right)$ as if $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$ then ${{x}_{1}}={{x}_{2}}$,where ${{x}_{1}},{{x}_{2}}$ $\in$ domain of $f\left( x \right)$.Onto function defined as every image (elements in range) has some pre-image in domain of the function. i.e. Range should belong to the co-domain of the function.Use the above definition and proceed to give examples for each part.

Complete step-by-step answer:
Here, we need to give an example of a function for all four parts of the problem. So, let us first understand the definition of one-one and onto function:-
One-one function:- One-one basically denotes the mapping of two sets, A function is one to one if every element of the range of that function corresponds to exactly one element of the domain of that function.
In brief, let us consider ‘f’ is a function whose domain is set A and mapped with unique element of set B, then ‘f’ is said to be one-one if for all x and y in set A :-
Whenever, $f\left( x \right)=f\left( y \right)$ then $x=y$ .
Where, $f\left( x \right)$ and $f\left( y \right)$ should lie in set B.
Onto function:- Onto function could be explained by considering two sets, A and set B which consists of elements, where, A is acting as domain and set B is range to which every element of A is mapped and if for every element of B there is at least one or more than one element matching with A, then the function is said to be on to functions.
Now, coming to the question:-

(i) Which is One=one but not on to:-
Let the function $f:N\to N$ , given by $f\left( x \right)=2x$ .
Let us consider two elements ${{x}_{1}}$ and ${{x}_{2}}$, in the domain of ‘f’. So’ we get –
$f\left( {{x}_{1}} \right)=2{{x}_{1}}$ and $f\left( {{x}_{2}} \right)=2{{x}_{2}}$.
Now, we know condition for one-one is given as –
$f\left( {{x}_{1}} \right)=2{{x}_{1}}$ and $f\left( {{x}_{2}} \right)=2{{x}_{2}}$.
Now, putting $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$ , we get

& f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right) \\\ & 2{{x}_{1}}=2{{x}_{2}} \\\ & {{x}_{1}}={{x}_{2}} \\\ \end{aligned}$$ Hence, the given function is One-one. Now Let $f\left( x \right)=y$ , such that $y\in N$ We get – $\begin{aligned} & 2x=y \\\ & x=\dfrac{y}{2} \\\ \end{aligned}$ Put $y=1$ $x=\dfrac{1}{2}=0.5$, which cannot be true as $x\in N$ as supposed in solution. Hence, the given function is not onto. So, $f\left( x \right)=2x$ is an example of One-one but not onto function. (ii) One-one and onto: Suppose $f\left( x \right)=x$ , for $x:R\to R$ Calculate $$f\left( {{x}_{1}} \right)$$ and $$f\left( {{x}_{2}} \right)$$ as $$f\left( {{x}_{1}} \right)={{x}_{1}}$$ , and $$f\left( {{x}_{2}} \right)={{x}_{2}}$$ . Now, equating $$f\left( {{x}_{1}} \right)$$ and $f\left( {{x}_{2}} \right)$ , we get $$\begin{aligned} & f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right) \\\ & {{x}_{1}}={{x}_{2}} \\\ \end{aligned}$$ Hence, $$f\left( x \right)=x$$ for $$x:R\to R$$ is a one-one function. Now, suppose $$f\left( x \right)=y$$ , where $$y\in R$$ , $$x=y$$ . Now, for all $$y\in R,x\in R$$ , as well. Hence, $$f\left( x \right)=x$$ is onto function as well. So, $$f\left( x \right)=x$$ , given as $$x:R\to R$$ is one-one and onto both. (iii) Let function $$f:R\to R$$ , given by $$f\left( x \right)={{x}^{2}}$$ Calculate $$f\left( {{x}_{1}} \right):$$ $$\begin{aligned} & f\left( {{x}_{1}} \right)=x_{1}^{2} \\\ & f\left( {{x}_{2}} \right): \\\ & f\left( {{x}_{2}} \right)=x_{2}^{2} \\\ \end{aligned}$$ Now, $$f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$$ $$\begin{aligned} & x_{1}^{2}=x_{2}^{2} \\\ & x_{1}^{2}-x_{2}^{2}=0 \\\ \end{aligned}$$ We know the algebraic identity of $${{a}^{2}}-{{b}^{2}}$$ can be given as – $${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$$ , So, we get – $$\left( {{x}_{1}}-{{x}_{2}} \right)\left( {{x}_{1}}+{{x}_{2}} \right)=0$$ . $${{x}_{1}}-{{x}_{2}}=0$$ or $${{x}_{1}}+{{x}_{2}}=0$$ or $${{x}_{1}}=-{{x}_{2}}$$ or $${{x}_{1}}={{x}_{2}}$$. Since, x, does not have a unique image as per the above statements, hence, it is not one-one. Now, $f\left( x \right)={{x}^{2}}$ Let $f\left( x \right)=y$ , such that $y\in R$ . $\begin{aligned} & y={{x}^{2}} \\\ & x=\pm \sqrt{y} \\\ \end{aligned}$ Put $y=-5$ , we get – $x=\pm \sqrt{-5}$ Where, x is not a real value. So, $f\left( x \right)={{x}^{2}}$ is not one-one and not onto as well for $x:R\to R$. (iv) Let the function $f:N\to N$ , given by $f\left( 1 \right)=f\left( 2 \right)=1$ Here, $f\left( x \right)=f\left( 2 \right)=1$ and $f\left( x \right)=f\left( 2 \right)=1$ Since, different elements 1 and 2 have the same image ‘1’. Hence, it cannot be a one-one function. Let $f\left( x \right)=y$ , such that $y\in N$ . Here, y is a natural number for every ‘y’, there is a value of x which is a natural number. Hence, f is onto. So, the function $f:N\to N$ , given by $f\left( 1 \right)=f\left( 2 \right)=1$ is not one-one but onto. Note: Another approach to get any function to be one, we can use graphical approach, means draw the curve $y=f\left( x \right)$ as supposed in all the four option, if a line parallel to x cutting the graph at two or more points, then it will not be one-one. So, it can be another way as well.There are infinite examples of the functions asked in the problem. So, don’t confuse with the examples given in the problem, there are other examples as well. We can verify whether any function is one-one by finding derivation of the function on $f\left( x \right)$ as well, If $f\left( x \right)>0$ only or $f'\left( x \right)<0$ only for the given domain then $f\left( x \right)$ is one-one, otherwise $f\left( x \right)$ will not be one-one. Onto function means Range of the function should belong to the co- domain of the function, which is a key point of the onto function.