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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...

If A = {-1, 0 2, 5, 6, 11} and B = {2, -1, 1, 0, 11, 108} and f(x)=x2−x−2f(x)={{x}^{2}}-x-2 then f:A→Bf:A\to B is
A) A function
B) A one-one function
C) An onto function
D) Not a function

Explanation

Solution

Hint: Here, first we have to check whether f(x)=x2−x−2f(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)=x2−x−2.f(x)={{x}^{2}}-x-2.
Now, we have to determine the nature of f:A→Bf:A\to B, that is to determine, whether f(x)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→Yf: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)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)=x2−x−2f(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=−1x=-1, we will get:
f(−1)=(−1)2−(−1)−2 f(−1)=1+1−2 f(−1)=2−2 f(−1)=0 \begin{aligned} & f(-1)={{(-1)}^{2}}-(-1)-2 \\\ & f(-1)=1+1-2 \\\ & f(-1)=2-2 \\\ & f(-1)=0 \\\ \end{aligned}
Here, 0∈B0\in B.
Now, consider x=0x=0, then we have:
f(0)=02−0−2 f(0)=0−2 f(0)=−2 \begin{aligned} & f(0)={{0}^{2}}-0-2 \\\ & f(0)=0-2 \\\ & f(0)=-2 \\\ \end{aligned}
But −2∉B-2\notin B.
Next, let us take x=2x=2 we will obtain:
f(2)=22−2−2 f(2)=4−2−2 f(2)=2−2 f(2)=0 \begin{aligned} & f(2)={{2}^{2}}-2-2 \\\ & f(2)=4-2-2 \\\ & f(2)=2-2 \\\ & f(2)=0 \\\ \end{aligned}
Here, we can say that 0∈B0\in B.
Now, take x=5x=5we will get:
f(5)=52−5−2 f(5)=25−5−2 f(5)=20−2 f(5)=18 \begin{aligned} & f(5)={{5}^{2}}-5-2 \\\ & f(5)=25-5-2 \\\ & f(5)=20-2 \\\ & f(5)=18 \\\ \end{aligned}
Here, we can say that 18∉B18\notin B
Now, consider x=6x=6 we will get:
f(6)=62−6−2 f(6)=36−6−2 f(6)=30−2 f(6)=28 \begin{aligned} & f(6)={{6}^{2}}-6-2 \\\ & f(6)=36-6-2 \\\ & f(6)=30-2 \\\ & f(6)=28 \\\ \end{aligned}
Here, also clearly we can say that 26∉B26\notin B.
Next, take x=11x=11, we will obtain:
f(11)=112−11−2 f(11)=121−11−2 f(11)=100−2 f11)=98 \begin{aligned} & f(11)={{11}^{2}}-11-2 \\\ & f(11)=121-11-2 \\\ & f(11)=100-2 \\\ & f11)=98 \\\ \end{aligned}
Here, again 98∉B98\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)=x2−x−2f(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.