Question

If A = {-1, 0 2, 5, 6, 11} and B = {2, -1, 1, 0, 11, 108} and $f(x)={{x}^{2}}-x-2$ then $f:A\to B$ is
A) A function
B) A one-one function
C) An onto function
D) Not a function

Answer 1

Hint: Here, first we have to check whether $f(x)={{x}^{2}}-x-2$ is a function or not. To be a function, for every x in A there should be a y in B. So, here we have to find the function values of all elements of set A where, A = {-1, 0, 2, 5, 6, 11} and check whether all the values of the function belongs to the set B = B = {2, -1, 1, 0, 11, 108} or not. If it belongs to set B then f is a function otherwise f is not a function.
Complete step-by-step answer:
Here, two sets are given, A = {-1, 0, 2, 5, 6, 11} and B = {2, -1, 1, 0, 11, 108} where $f(x)={{x}^{2}}-x-2.$
Now, we have to determine the nature of $f:A\to B$, that is to determine, whether $f(x)$ is function or not. If it is a function, then check whether it is one-one or onto.
First we will discuss the function.
We know that a relation f from a set A to Set B is said to be a function if every element of set A has one and only one image in set B.
The notation $f:X\to Y$ means that f is a function from X to Y. X is called the domain of f and Y is called the codomain of f. Given an element x in X, there is a unique element y in Y that is related to x.
The set of all values of $f(x)$ taken together is called the range of f or image of X under f.
So here, let us consider the function $f(x)={{x}^{2}}-x-2$.
Next, we have to find the image of f for the values in A = {-1, 0 2, 5, 6, 11} and check whether it belongs to the set B = {2, -1, 1, 0, 11, 108}. If the value belongs to B then f is a function, if any value doesn’t belong to B then f is not a function.
First, consider $x=-1$, we will get:
$\begin{align}
  & f(-1)={{(-1)}^{2}}-(-1)-2 \\
 & f(-1)=1+1-2 \\
 & f(-1)=2-2 \\
 & f(-1)=0 \\
\end{align}$
Here, $0\in B$.
Now, consider $x=0$, then we have:
$\begin{align}
  & f(0)={{0}^{2}}-0-2 \\
 & f(0)=0-2 \\
 & f(0)=-2 \\
\end{align}$
But $-2\notin B$.
Next, let us take $x=2$ we will obtain:
$\begin{align}
  & f(2)={{2}^{2}}-2-2 \\
 & f(2)=4-2-2 \\
 & f(2)=2-2 \\
 & f(2)=0 \\
\end{align}$
Here, we can say that $0\in B$.
Now, take $x=5$we will get:
$\begin{align}
  & f(5)={{5}^{2}}-5-2 \\
 & f(5)=25-5-2 \\
 & f(5)=20-2 \\
 & f(5)=18 \\
\end{align}$
Here, we can say that $18\notin B$
Now, consider $x=6$ we will get:
$\begin{align}
  & f(6)={{6}^{2}}-6-2 \\
 & f(6)=36-6-2 \\
 & f(6)=30-2 \\
 & f(6)=28 \\
\end{align}$
Here, also clearly we can say that $26\notin B$.
Next, take $x=11$, we will obtain:
$\begin{align}
  & f(11)={{11}^{2}}-11-2 \\
 & f(11)=121-11-2 \\
 & f(11)=100-2 \\
 & f11)=98 \\
\end{align}$
Here, again $98\notin B$.
So, here the range values like -2, 18, 28, 98 do not belong to the codomain B.
Therefore, we can conclude that $f(x)={{x}^{2}}-x-2$ is not a function.
Hence, the correct answer for this question is option (D).

Note: Here, you should keep in mind that codomain refers to the set of values that might come out of a function. But, range refers to the actual definitive set of values that might come out of it. Codomain refers to the definition of a function and range refers to the image of a function.