
If \[A=\left\{ -2,-1,0,1,2 \right\}\] and \[f:A\to B\] is a surjection defined by\[f\left( x\right)={{x}^{2}}+x+1\], then find B?
Answer
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Hint:in the given question, we have been asked to find the value of B and it is given that \[f:A\to B\] and the values of A are defined as \[A=\left\{ -2,-1,0,1,2 \right\}\]. In order to solve the question, first we have to find the values of the function by putting all the values of A that are given in the question. After getting the values, we have given the condition that\[f:A\to B\], then we will get the values of B. a function \[f:A\to B\] is said to be surjective if every element of A i.e. f(A)=B and the surjective function is also known as onto function.
Complete step by step solution:
We have given that,
\[\Rightarrow A=\left\{ -2,-1,0,1,2 \right\}\]And
\[\Rightarrow f\left( x \right)={{x}^{2}}+x+1\] From \[f:A\to B\]
A function \[f:A\to B\] is said to be surjective if every element of A i.e. \[f\left( A \right)=B\]
We substitute the values of A defined by the set i.e. \[A=\left\{ -2,-1,0,1,2 \right\}\] in the given function,
Now,
\[\Rightarrow f\left( -2 \right)=-{{2}^{2}}+\left( -2 \right)+1=4-2+1=3\]
\[\Rightarrow f\left( -1 \right)=-{{1}^{2}}+\left( -1 \right)+1=1-1+1=1\]
\[\Rightarrow f\left( 0 \right)={{0}^{2}}+\left( 0 \right)+1=1\]
\[\Rightarrow f\left( 1 \right)={{1}^{2}}+\left( 1 \right)+1=1+1+1=3\]
\[\Rightarrow f\left( 2 \right)={{2}^{2}}+\left( 2 \right)+1=4+2+1=7\]
As it is given that,
\[\Rightarrow f:A\to B\]
Therefore, the values of B are 1,3,7.
Thus, \[\Rightarrow B=\left\{ 1,3,7 \right\}\]
Note: While solving these types of questions, we need to read the question carefully and see what we have been asked. Your concepts regarding the function and the surjection should be explicitly cleared. You should always remember all the properties of surjective function. A proper and well defined assumption should be made in order to solve the question. While substitute choose and substitute all the values given in the question very carefully and avoid making any error in calculation. Equate all the relations that are defined in the question and the function also.
Complete step by step solution:
We have given that,
\[\Rightarrow A=\left\{ -2,-1,0,1,2 \right\}\]And
\[\Rightarrow f\left( x \right)={{x}^{2}}+x+1\] From \[f:A\to B\]
A function \[f:A\to B\] is said to be surjective if every element of A i.e. \[f\left( A \right)=B\]
We substitute the values of A defined by the set i.e. \[A=\left\{ -2,-1,0,1,2 \right\}\] in the given function,
Now,
\[\Rightarrow f\left( -2 \right)=-{{2}^{2}}+\left( -2 \right)+1=4-2+1=3\]
\[\Rightarrow f\left( -1 \right)=-{{1}^{2}}+\left( -1 \right)+1=1-1+1=1\]
\[\Rightarrow f\left( 0 \right)={{0}^{2}}+\left( 0 \right)+1=1\]
\[\Rightarrow f\left( 1 \right)={{1}^{2}}+\left( 1 \right)+1=1+1+1=3\]
\[\Rightarrow f\left( 2 \right)={{2}^{2}}+\left( 2 \right)+1=4+2+1=7\]
As it is given that,
\[\Rightarrow f:A\to B\]
Therefore, the values of B are 1,3,7.
Thus, \[\Rightarrow B=\left\{ 1,3,7 \right\}\]
Note: While solving these types of questions, we need to read the question carefully and see what we have been asked. Your concepts regarding the function and the surjection should be explicitly cleared. You should always remember all the properties of surjective function. A proper and well defined assumption should be made in order to solve the question. While substitute choose and substitute all the values given in the question very carefully and avoid making any error in calculation. Equate all the relations that are defined in the question and the function also.
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