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Is \[g=\left\{ \left( 1,1 \right),\left( 2,3 \right),\left( 3,5 \right),\left( 4,7 \right) \right\}\] a function? If this is described by function \[g(x)=\alpha x+\beta \] then what values should be assigned to \[\alpha ,\beta \] ? Find the value of \[\alpha +\beta \]

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Hint: To solve the question, we have to check whether g is function, which can be checked by using the property of function that states for each input value of domain has only one output value in the range. To calculate the constant values, use the data of the set of g.

Complete step-by-step answer:
We know that a function is a relation for which each input value of a domain has only one output value in the range.
The domain of \[g=\left\{ 1,2,3,4 \right\}\] and the range of \[g=\left\{ 1,3,5,7 \right\}\]
Each input value of the domain of g has only one output value in the range of g. Thus, \[g=\left\{ \left( 1,1 \right),\left( 2,3 \right),\left( 3,5 \right),\left( 4,7 \right) \right\}\] is a function.
Given that the function of g is \[g(x)=\alpha x+\beta \]
The values of set g satisfy the above function. Thus, by substituting the input values of (1,1) and (2,3) we get
\[g(1)=\alpha (1)+\beta \]
\[\Rightarrow g(1)=\alpha +\beta \]
\[g(2)=\alpha (2)+\beta \]
\[\Rightarrow g(2)=2\alpha +\beta \]
By substituting the output values of (1,1) and (2,3) we get
\[1=\alpha +\beta \]
\[3=2\alpha +\beta \] …. (1)
By subtracting the above equations, we get
\[3-1=2\alpha +\beta -(\alpha +\beta )\]
\[2=2\alpha +\beta -\alpha -\beta \]
\[\Rightarrow \alpha =2\]
By substituting the value of \[\alpha =2\] in equation (1) we get
\[3=2(2)+\beta \]
\[3=4+\beta \]
\[\beta =3-4\]
\[\Rightarrow \beta =-1\]
The value of \[\alpha +\beta \] = 2 – 1 = 1
Thus, the values of \[\alpha ,\beta ,\alpha +\beta \] are 2, -1, 1 respectively.
Note: The possibility of mistake can be not able to apply the concept of condition of a relation to be a function. The other possibility of mistake can be not applying the set of numbers of function to calculate the values of \[\alpha ,\beta ,\alpha +\beta \]