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Question: Is \[g=\left\\{ \left( 1,1 \right),\left( 2,3 \right),\left( 3,5 \right),\left( 4,7 \right) \right\\...

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)=αx+βg(x)=\alpha x+\beta then what values should be assigned to α,β\alpha ,\beta ? Find the value of α+β\alpha +\beta

Explanation

Solution

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)=αx+βg(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)=α(1)+βg(1)=\alpha (1)+\beta
g(1)=α+β\Rightarrow g(1)=\alpha +\beta
g(2)=α(2)+βg(2)=\alpha (2)+\beta
g(2)=2α+β\Rightarrow g(2)=2\alpha +\beta
By substituting the output values of (1,1) and (2,3) we get
1=α+β1=\alpha +\beta
3=2α+β3=2\alpha +\beta …. (1)
By subtracting the above equations, we get
31=2α+β(α+β)3-1=2\alpha +\beta -(\alpha +\beta )
2=2α+βαβ2=2\alpha +\beta -\alpha -\beta
α=2\Rightarrow \alpha =2
By substituting the value of α=2\alpha =2 in equation (1) we get
3=2(2)+β3=2(2)+\beta
3=4+β3=4+\beta
β=34\beta =3-4
β=1\Rightarrow \beta =-1
The value of α+β\alpha +\beta = 2 – 1 = 1
Thus, the values of α,β,α+β\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 α,β,α+β\alpha ,\beta ,\alpha +\beta