Question

MATHEMATICS - IX PART -
The set builder form of the relation R= {(-2, -7), (-1, –4), (0, – 1), (1,2), (2,5)
a) R= {(x, y) / y = 2x – 3;X, y e Z}
b) R = {(x, y)/ y = 3x – 1;X, Y EZ}
c) R = {(x, y) / y = 3x + 1;X, y e Z}
d) R = {(x, y)/y = 3x – 1 ; -2 <x<2 and xe Z
Among the followin​

Answer 1

Answer:

R = {(x, y) I y = 3x – 1 ∧ -2 ≤ x ≤ 2 ∧ x ∈ Z }

Step-by-step explanation:

We are given that

R = { (-2 , -7) , (-1 , –4) , (0 , – 1) , ( 1 , 2 ) , ( 2 , 5 ) }

Consider the relation given in the d) part that is

R = {(x, y) I y = 3x – 1 ∧ -2 ≤ x ≤ 2 ∧ x ∈ Z }

Then

y = 3x – 1

Putting x = - 2 we get

y = 3(-2) - 1 = - 6 - 1 = - 7

So

(-2 , -7) ∈ R

Putting x = -1 we get

y = 3(-1) - 1 = - 3 - 1 = - 4

So

(-1 , - 4) ∈ R

Putting x = 0 we get

y = 3(0) - 1 = 0 - 1 = - 1

So

(0 , - 1) ∈ R

Putting x = 1 we get

y = 3(1) - 1 = 3 - 1 = 2

So

(1 , 2) ∈ R

Putting x = 2 we get

y = 3(2) - 1 = 6 - 1 = 5

So

(2 , 5) ∈ R

Hence

Part d is the correct choice for set building notation of relation R  

Answer 2

Answer:

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