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Question: Let \(f = \\{ (1,1,),(2,3),(0, - 1),( - 1, - 3)\\} \) be a function\(f(x) = ax + b\) from Z to Z def...

Let f=(1,1,),(2,3),(0,1),(1,3)f = \\{ (1,1,),(2,3),(0, - 1),( - 1, - 3)\\} be a functionf(x)=ax+bf(x) = ax + b from Z to Z defined by for some integers a, b. Determine a, b.

Explanation

Solution

We have been given some ordered pair that lies on function f and the function f(x)f\left( x \right)with some unknown constant. Now using those ordered pairs we’ll create the number of equations equivalent to the number of unknown constants by substituting the value in f(x)f\left( x \right).
Now solving those equations we’ll get the required value of the unknown constant.

Complete step-by-step answer:
Given data: f=(1,1),(2,3),(0,1),(1,3)f = \\{ (1,1),(2,3),(0, - 1),( - 1, - 3)\\}
f(x)=ax+bf(x) = ax + b
Since f is a function containing (1,1,),(2,3),(0,1),(1,3)(1,1,),(2,3),(0, - 1),( - 1, - 3)
We can say that f(1)=1f(1) = 1
And, f(2)=3f(2) = 3
Substituting  x=1\;x = 1 in equation(i)
f(1)=a(1)+b\Rightarrow f(1) = a(1) + b
Substituting the value of f(1)f(1)
1=a+b..........(ii)\Rightarrow 1 = a + b..........(ii)
Now, substituting x=2x = 2in equation(i)
f(2)=a(2)+b\Rightarrow f(2) = a(2) + b
Substituting the value of f(2)f(2)
3=2a+b\Rightarrow 3 = 2a + b
b=32a\Rightarrow b = 3 - 2a
Now, substituting the value of b in equation(ii)
1=a+32a\Rightarrow 1 = a + 3 - 2a
Solving for like terms
a=31\Rightarrow a = 3 - 1
a=2\therefore a = 2
Substituting the value of ‘a’ in equation(ii)
1=2+b\Rightarrow 1 = 2 + b
12=b\Rightarrow 1 - 2 = b
b=1\therefore b = - 1
Therefore the function f(x)=2x1f(x) = 2x - 1
And a=2a = 2,b=1b = - 1

Note: An alternative method for the above solution can be
Since it is given that the f(x)=ax+bf(x) = ax + b, which is an equation of line
And since f is a function containing (1,1,),(2,3),(0,1),(1,3)(1,1,),(2,3),(0, - 1),( - 1, - 3), therefore they are the points lying on the line
Now using the two-point form of a line i.e. if we have two points let say (x1,y1)({x_1},{y_1})and (x2,y2)({x_2},{y_2}) that lie on a line then the equation of the line is given by (yy1)=(y1y2)(x1x2)(xx1)(y - {y_1}) = \dfrac{{({y_1} - {y_2})}}{{({x_1} - {x_2})}}(x - {x_1})
Therefore the equation of line passes through (1,1,)(1,1,) and, (0,1)(0, - 1)will be
(y1)=(1(1))(10)(x1)\Rightarrow (y - 1) = \dfrac{{(1 - ( - 1))}}{{(1 - 0)}}(x - 1), where y=f(x)y = f(x)
(y1)=(1+1)(x1)\Rightarrow (y - 1) = (1 + 1)(x - 1)
(y1)=2(x1)\Rightarrow (y - 1) = 2(x - 1)
Simplifying the brackets
y1=2x2\Rightarrow y - 1 = 2x - 2
y=2x1\therefore y = 2x - 1
f(x)=2x1\therefore f(x) = 2x - 1, which is resulting in the same answer as above