Question

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

Answer 1

Answer:

Hey There!!!

Here, we have f={(1,1),(2,3),(0,-1),(-1,-3)}

We are given that f is a linear function from Z into Z. This simply means that f(x) is a polynomial of degree 1.

Let f(x) = ax+b

All elements of the set f satisfy this equation.

Let us put (1,1).

That is, when x=1, f(x)=1

So, 1 = a + b -----(1)

Also, let us put (0,-1) there.

We have:-1 = a(0) + b

So, we have b = -1

Put b = -1 in equation (1), we have

1 = a - 1So, a = 2

Finally we have:

f(x)= 2x-1

We can easily see that all elements of set f satisfy the equation. So it is correct.

Hope it helps

Answer 2

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$\Large\fbox{\color{purple}{QUESTION}}$

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

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$\bf\Huge\red{\mid{\overline{\underline{ ANSWER }}}\mid }$

➡️Given, f = {(1, 1), (2, 3), (0, –1), (–1, –3)}

➡️And the function defined as, f(x) = ax + b

➡️For (1, 1) ∈ f

➡️We have, f(1) = 1

➡️So, a × 1 + b = 1

➡️a + b = 1 …. (i)

➡️And for (0, –1) ∈ f

➡️We have f(0) = –1

➡️a × 0 + b = –1

➡️b = –1

➡️On substituting b = –1 in (i), we get

➡️a + (–1) = 1 ⇒ a = 1 + 1 = 2.

➡️Therefore, the values of a and b are 2 and –1 respectively.

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